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+\section{Set of Ideals forms Complete Lattice}
+Tags: Lattice Theory, Ideal Theory
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $K$.
+Then $\struct {\mathbb K, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
+\end{theorem}
+
+\begin{proof}
+Let $\O \subset \mathbb S \subseteq \mathbb K$.
+By [[Intersection of Ring Ideals is Largest Ideal Contained in all Ideals]]:
+:$\bigcap \mathbb S$ is the largest [[Definition:Ideal of Ring|ideal]] of $K$ contained in each of the elements of $\mathbb S$.
+By [[Intersection of Ring Ideals Containing Subset is Smallest]]:
+:The [[Definition:Set Intersection|intersection]] of the set of all [[Definition:Ideal of Ring|ideals]] of $K$ containing $\bigcup \mathbb S$ is the smallest [[Definition:Ideal of Ring|ideal]] of $K$ containing $\bigcup \mathbb S$.
+Thus, not only is $\bigcap \mathbb S$ a [[Definition:Lower Bound of Set|lower bound]] of $\mathbb S$, but also the largest, and therefore an [[Definition:Infimum of Set|infimum]].
+The [[Definition:Supremum of Set|supremum]] of $\mathbb S$ is the [[Definition:Join of Subgroups|join]] of the set of all [[Definition:Ideal of Ring|ideals]] of $\mathbb S$.
+From [[Sum of Ideals is Ideal/General Result|Sum of Ideals is Ideal: General Result]], this [[Definition:Supremum of Set|supremum]] is:
+:$\displaystyle \sum_{S \mathop \in \mathbb S} \struct {S, +, \circ}$
+That is:
+:$\struct {S_1, +, \circ} + \struct {S_2, +, \circ} + \dotsb$
+where addition of ideals is as defined in [[Definition:Subset Product|subset product]].
+Therefore $\struct {\mathbb K, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Product of Abelian Subgroups}
+Tags: Abelian Groups, Subset Products
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be an [[Definition:Abelian Group|abelian group]].
+Let $H_1$ and $H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
+Then $H_1 \circ H_2$ is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From [[Subgroup of Abelian Group is Normal]], $H_1$ and $H_2$ are [[Definition:Normal Subgroup|normal]].
+The result follows from [[Subset Product with Normal Subgroup is Subgroup]].
+{{qed}}
+[[Category:Abelian Groups]]
+[[Category:Subset Products]]
+42ddr5761fjsy89srlm2c6znwzueywi
+\end{proof}<|endoftext|>
+\section{Sum of Ideals is Ideal}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $J_1$ and $J_2$ be [[Definition:Ideal of Ring|ideals]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct{R, +, \circ}$.
+Then:
+: $J = J_1 + J_2$ is an [[Definition:Ideal of Ring|ideal]] of $R$
+where $J_1 + J_2$ is as defined in [[Definition:Subset Product|subset product]].
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Ring (Abstract Algebra)|definition]], $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]].
+So from [[Subset Product of Abelian Subgroups]], we have that:
+:$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$
+is itself a [[Definition:Subgroup|subgroup]] of $R$.
+Now consider $a \circ b$ where $a \in J, b \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \circ b
+      | r = \paren {a_1 + a_2} \circ b
+      | c = by definition of $a$
+}}
+{{eqn | r = \paren {a_1 \circ b} + \paren {a_2 \circ b}
+      | c = as $\circ$ is [[Definition:Distributive Operation|distributive]] over $+$
+}}
+{{eqn | o = \in
+      | r = J_1 + J_2
+      | c = as $\paren {a_i \circ b} \in J_i$ because $J_i$ is an [[Definition:Ideal of Ring|ideal]] of $R$
+}}
+{{end-eqn}}
+Similarly, $b \circ a \in J_1 + J_2$
+So by definition $J_1 + J_2$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principal Ideal from Element in Center of Ring}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $b \in R$ be in the [[Definition:Center of Ring|center]] of $R$.
+Then:
+:$\left({b}\right) = R \circ b = \left\{{x \circ b: x \in R}\right\}$
+where $\left({b}\right)$ is the [[Definition:Principal Ideal of Ring|principal ideal generated by $b$]].
+\end{theorem}
+
+\begin{proof}
+Let $J = R \circ b$
+The [[Definition:Center of Ring|center]] of $R$ is defined as:
+:$Z \left({R}\right) = \left\{{x \in R: \forall s \in R: s \circ x = x \circ s}\right\}$
+Therefore:
+:$R \circ b = b \circ R = \left\{{x \circ b: x \in R}\right\} = \left\{{b \circ x: x \in R}\right\}$
+and so:
+:$x \in J \implies x \circ b \in J \land b \circ x \in J$
+Thus $J$ is an [[Definition:Ideal of Ring|ideal]] of $R$ and so $J = \left({b}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{External Direct Sum of Rings is Ring}
+Tags: Direct Sums of Rings, Ring Theory
+
+\begin{theorem}
+Let $\left({R_1, +_1, \circ_1}\right), \left({R_2, +_2, \circ_2}\right), \ldots, \left({R_n, +_n, \circ_n}\right)$ be [[Definition:Ring (Abstract Algebra)|rings]].
+Then their [[Definition:Ring Direct Product|(external) direct product]]:
+: $\displaystyle \left({R, +, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k, \circ_k}\right)$
+is a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+Consider the structures $\left({R_1, +_1}\right), \left({R_2, +_2}\right), \ldots, \left({R_n, +_n}\right)$.
+By the definition of a [[Definition:Ring (Abstract Algebra)|ring]], these are all [[Definition:Group|groups]].
+From [[External Direct Product of Groups is Group]] we have that the their [[Definition:External Direct Product|external direct product]]:
+: $\displaystyle \left({R, +}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k}\right)$
+is a [[Definition:Group|group]].
+Similarly, consider the structures $\left({R_1, \circ_1}\right), \left({R_2, \circ_2}\right), \ldots, \left({R_n, \circ_n}\right)$.
+By the definition of a [[Definition:Ring (Abstract Algebra)|ring]], these are all [[Definition:Semigroup|semigroups]].
+From [[External Direct Product of Semigroups]] we have that the their [[Definition:External Direct Product|external direct product]]:
+: $\displaystyle \left({R, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, \circ_k}\right)$
+is a [[Definition:Semigroup|semigroup]].
+Finally we note that from [[External Direct Product Distributivity]], $\circ$ as defined here is [[Definition:Distributive Operation|distributive]] over $+$.
+Hence the result, by definition of [[Definition:Ring (Abstract Algebra)|ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{External Direct Product Distributivity}
+Tags: External Direct Products, Distributive Operations
+
+\begin{theorem}
+Let $\left({R_1, +_1, \circ_1}\right), \left({R_2, +_2, \circ_2}\right), \ldots, \left({R_n, +_n, \circ_n}\right)$ be [[Definition:Ringoid (Abstract Algebra)|ringoids]].
+Let their [[Definition:External Direct Product|external direct product]] be $\displaystyle \left({R, +, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k, \circ_k}\right)$.
+Then the operation $\circ$ is [[Definition:Distributive Operation|distributive]] over $+$.
+\end{theorem}
+
+\begin{proof}
+As all the $\left({R_k, +_k, \circ_k}\right)$ are [[Definition:Ringoid (Abstract Algebra)|ringoids]], $\circ_k$ distributes over $+_k$ for all $k$.
+Let $x, y, z \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \circ \left({y + z}\right)
+      | r = \left({x_1, x_2, \ldots, x_n}\right) \circ \left({\left({y_1, y_2, \ldots, y_n}\right) + \left({z_1, z_2, \ldots, z_n}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \left({x_1, x_2, \ldots, x_n}\right) \circ \left({y_1 + z_1, y_2 + z_2, \ldots, y_n + z_n}\right)
+      | c = 
+}}
+{{eqn | r = \left({x_1 \circ \left({y_1 + z_1}\right), x_2 \circ \left({y_2 + z_2}\right), \ldots, x_n \circ \left({y_n + z_n}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \left({\left({x_1 \circ z_1}\right) + \left({y_1 \circ z_1}\right), \left({x_2 \circ z_2}\right) + \left({y_2 \circ z_2}\right), \ldots, \left({x_n \circ z_n}\right) + \left({y_n \circ z_n}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \left({\left({x_1 \circ z_1}\right), \left({x_2 \circ z_2}\right), \ldots, \left({x_n \circ z_n}\right)}\right) + \left({\left({y_1 \circ z_1}\right) + \left({y_2 \circ z_2}\right) + \ldots + \left({y_n \circ z_n}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \left({\left({x_1, x_2, \ldots, x_n}\right) \circ \left({z_1, z_2, \ldots, z_n}\right)}\right) + \left({\left({x_1, x_2, \ldots, x_n}\right) \circ \left({z_1, z_2, \ldots, z_n}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \left({x \circ y}\right) + \left({x \circ z}\right)
+      | c = 
+}}
+{{end-eqn}}
+In the same way:
+:$\left({y + z}\right) \circ x = \left({y \circ x}\right) + \left({z \circ x}\right)$
+{{qed}}
+[[Category:External Direct Products]]
+[[Category:Distributive Operations]]
+0wp7cqiw919zcvcopuv7m87fb8j34dw
+\end{proof}<|endoftext|>
+\section{Ideal of External Direct Sum of Rings}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\left({R_1, +_1, \circ_1}\right), \left({R_2, +_2, \circ_2}\right), \ldots, \left({R_n, +_n, \circ_n}\right)$ be [[Definition:Ring (Abstract Algebra)|rings]].
+Let 
+:$\displaystyle \left({R, +, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k, \circ_k}\right)$ 
+be their [[Definition:Ring Direct Product|direct product]].
+For each $k \in \left[{1 \,.\,.\, n}\right]$, let:
+:$R\,'_k = \left\{{\left({x_1, \ldots, x_n}\right) \in R: \forall j \ne k: x_j = 0}\right\}$
+Then:
+: $\forall k \in \left[{1 \,.\,.\, n}\right]: R\,'_k$ is an ideal of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $y = \left({y_1, y_2, \ldots, y_n}\right) \in R$.
+Let $x = \left({x_1, x_2, \ldots, x_n}\right) \in R_k$
+By definition of [[Definition:Ring Direct Product|direct product]], we have:
+:$x \circ y = \left({x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}\right)$
+:$y \circ x = \left({y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}\right)$
+But we have, $\forall j \ne k: x_j \circ y_j = 0 = y_j \circ x_j$ as $x_j = 0$.
+So it follows that $x \circ y \in R_k$ and $y \circ x \in R_k$
+Hence $R\,'_k$ is an ideal of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Canonical Injection from Ideal of External Direct Sum of Rings}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \dotsc, \struct {R_n, +_n, \circ_n}$ be [[Definition:Ring (Abstract Algebra)|rings]].
+Let $\displaystyle \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their [[Definition:External Direct Product|external direct product]].
+For each $k \in \closedint 1 n$, let:
+:$R'_k = \set {\tuple {x_1, \dotsc, x_n} \in R: \forall j \ne k: x_j = 0}$
+Let $\inj_k: R_k \to R$ be the [[Definition:Canonical Injection (Abstract Algebra)|canonical injection on the $k$th coordinate]] from $R_k$ into $\struct {R, +, \circ}$.
+Let $\pr_k: R \to R'_k$ be the [[Definition:Projection (Mapping Theory)|projection on the $k$th coordinate]] of $\struct {R, +, \circ}$ onto $R'_k$.
+Then:
+:$\inj_k: R_k \to R'_k$ is an [[Definition:Ring Isomorphism|isomorphism]]
+:Its [[Definition:Inverse Mapping|inverse]] is the [[Definition:Restriction of Mapping|restriction]] of $\pr_k$ to $R'_k$.
+\end{theorem}
+
+\begin{proof}
+From [[Ideal of External Direct Sum of Rings]] we have that $R'_k$ is an [[Definition:Ideal of Ring|ideal]] of $R$, and thus a [[Definition:Subring|subring]] of $R$.
+The result follows by application of [[Canonical Injection is Monomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projection onto Ideal of External Direct Sum of Rings}
+Tags: Ideal Theory, Projections
+
+\begin{theorem}
+Let $\left({R_1, +_1, \circ_1}\right), \left({R_2, +_2, \circ_2}\right), \ldots, \left({R_n, +_n, \circ_n}\right)$ be [[Definition:Ring (Abstract Algebra)|rings]].
+Let $\displaystyle \left({R, +, \circ}\right) = \prod_{k \mathop = 1}^n \left({R_k, +_k, \circ_k}\right)$ be their [[Definition:Ring Direct Product|direct product]].
+For each $k \in \left[{1 \,.\,.\, n}\right]$, let:
+:$R'_k = \left\{{\left({x_1, \ldots, x_n}\right) \in R: \forall j \ne k: x_j = 0}\right\}$
+Let $\operatorname{pr}_k: R \to R'_k$ be the [[Definition:Projection (Mapping Theory)|projection on the $k$th coordinate]] of $\left({R, +, \circ}\right)$ onto $R'_k$.
+Then $\operatorname{pr}_k$ is an [[Definition:Ring Epimorphism|epimorphism]].
+\end{theorem}
+
+\begin{proof}
+From [[Ideal of External Direct Sum of Rings]] we already have that $R'_k$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+The result follows by application of [[Projection is Epimorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conditions for Internal Ring Direct Sum}
+Tags: Subrings, Ideal Theory, Direct Sums of Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\sequence {\struct {S_k, +, \circ} }$ be a [[Definition:Sequence|sequence]] of [[Definition:Subring|subrings]] of $R$.
+Then $R$ is the [[Definition:Ring Direct Sum|ring direct sum]] of $\sequence {S_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}:
+:$(1): \quad R = S_1 + S_2 + \cdots + S_n$
+:$(2): \quad \sequence {\struct {S_k, +} }_{1 \mathop \le k \mathop \le n}$ is a [[Definition:Sequence|sequence]] of [[Definition:Independent Subgroups|independent subgroups]] of $\struct {R, +}$
+:$(3): \quad \forall k \in \closedint 1 n: S_k$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $S$ be the [[Definition:Cartesian Product|cartesian product]] of $\sequence {\struct {S_k, +} }$
+First note that $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]] from $\struct {S, +}$ to $\struct {R, +}$, as:
+:$\displaystyle \sum_{j \mathop = 1}^n \paren {x_j + y_j} = \sum_{j \mathop = 1}^n x_j + \sum_{j \mathop = 1}^n y_j$ from [[Associativity on Indexing Set]].
+So $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]] {{iff}}:
+:$\displaystyle \paren {\sum_{j \mathop = 1}^n x_j} \circ \paren {\sum_{j \mathop = 1}^n y_j} = \sum_{j \mathop = 1}^n \paren {x_j \circ y_j}$
+Now, let:
+:$\struct {S, +, \circ}$ be the [[Definition:Cartesian Product|cartesian product]] of $\sequence {\struct {S_k, +, \circ} }$
+:$\phi: S \to R$ be the [[Definition:Mapping|mapping]] defined as:
+::$\map \phi {x_1, x_2, \ldots, x_n} = x_1 + x_2 + \cdots x_n$
+Clearly $\phi$ is a [[Definition:Surjection|surjection]] {{iff}} $(1)$ holds.
+By [[Internal Direct Product Generated by Subgroups]], $\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]] {{iff}} $(1)$ and $(2)$ hold.
+Let $\phi: S \to R$ be a [[Definition:Ring Isomorphism|ring isomorphism]].
+By [[Canonical Injection from Ideal of External Direct Sum of Rings]], $\map {\inj_k} {S_k}$ is an [[Definition:Ideal of Ring|ideal]] of $S$.
+So $\map \phi {\map {\inj_k} {S_k} }$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+But $\phi$ and $\pr_k$ coincide on $\map {\inj_k} {S_k}$.
+So:
+:$\map \phi {\map {\inj_k} {S_k} } = \map {\pr_k} {\map {\inj_k} {S_k} } = S_k$
+and so $(3)$ holds.
+Now suppose $(3)$ holds, and the $S_k$ are all [[Definition:Ideal of Ring|ideals]] of $R$.
+By $(2)$ and by definition of [[Definition:Independent Subgroups/Definition 2|independent subgroups]]:
+:$i \ne j \implies S_i \cap S_j = \set {0_R}$
+So for all $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in S$:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x_1, x_2, \ldots, x_n} \circ \map \phi {y_1, y_2, \ldots, y_n}
+      | r = \paren {\sum_{i \mathop = 1}^n x_i} \circ \paren {\sum_{j \mathop = 1}^n y_j}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n x_i \circ y_j}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n x_i \circ y_i
+      | c = 
+}}
+{{eqn | r = \map \phi {x_1, x_2, \ldots, x_n} \circ \map \phi {y_1, y_2, \ldots, y_n}
+      | c = 
+}}
+{{end-eqn}}
+because as $S_i, S_j$ are [[Definition:Ideal of Ring|ideals]], we have:
+:$x_i \circ j_j = S_i \cap S_j = \set {0_R}$
+Hence the three conditions are sufficient for $\phi$ to be a [[Definition:Ring Isomorphism|ring isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Structure Induced by Ring Operations is Ring}
+Tags: Rings of Mappings, Examples of Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $S$ be a [[Definition:Set|set]].
+Then $\struct {R^S, +', \circ'}$ is a [[Definition:Ring (Abstract Algebra)|ring]], where $+'$ and $\circ'$ are the [[Definition:Pointwise Operation|pointwise operations induced]] on $R^S$ by $+$ and $\circ$.
+\end{theorem}
+
+\begin{proof}
+As $R$ is a [[Definition:Ring (Abstract Algebra)|ring]], both $+$ and $\circ$ are [[Definition:Closed Operation|closed]] on $R$ by definition.
+From [[Closure of Pointwise Operation on Algebraic Structure]], it follows that both $+'$ and $\circ'$ are [[Definition:Closed Operation|closed]] on $R^S$:
+:$\forall f, g \in R^S: f +' g \in R^S$
+:$\forall f, g \in R^S: f \circ' g \in R^S$
+By [[Structure Induced by Abelian Group Operation is Abelian Group]], $\struct {R^S, +'}$ is an [[Definition:Abelian Group|abelian group]].
+By [[Structure Induced by Associative Operation is Associative]], $\struct {R^S, \circ'}$ is a [[Definition:Semigroup|semigroup]].
+From [[Pointwise Operation on Distributive Structure is Distributive]], $\circ'$ is [[Definition:Distributive Operation|distributive]] over $+'$.
+The result follows by definition of [[Definition:Ring (Abstract Algebra)|ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Subring Test}
+Tags: Subrings
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]], and let $L$ be a [[Definition:Subset|subset]] of $K$.
+Then $\struct {L, +, \circ}$ is a [[Definition:Division Subring|division subring]] of $\struct {K, +, \circ}$ {{iff}} these all hold:
+:$(1) \quad L^* \ne \O$
+:$(2) \quad \forall x, y \in L: x + \paren {-y} \in L$
+:$(3) \quad \forall x, y \in L: x \circ y \in L$
+:$(4) \quad x \in L^* \implies x^{-1} \in L^*$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $\struct {L, +, \circ}$ is a [[Definition:Division Subring|division subring]] of $\struct {K, +, \circ}$.
+The conditions $(1)$ to $(3)$ hold by virtue of the [[Subring Test]].
+Then $(4)$ also holds by the definition of a [[Definition:Division Ring|division ring]]:
+:$\forall x \in L^*: \exists ! x^{-1} \in L^*: x^{-1} \circ x = x \circ x^{-1} = 1_L$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose the conditions $(1)$ to $(4)$ hold.
+By $(1)$ to $(3)$, it follows from [[Subring Test]] that $\struct {L, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {K, +, \circ}$.
+By $(4)$, every element of $L^*$ has a [[Definition:Ring Product Inverse|product inverse]].
+Thus, from the [[Two-Step Subgroup Test]], $\struct {L^*, \circ}$ is a [[Definition:Group|group]].
+Therefore, $\struct {L, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]] in which every [[Definition:Element|element]] has a [[Definition:Ring Product Inverse|product inverse]], which makes $\struct {L, +, \circ}$ a [[Definition:Division Ring|division ring]].
+{{qed}}
+[[Category:Subrings]]
+1o60mugxbp4aeldykaapiwlbxt3ytex
+\end{proof}<|endoftext|>
+\section{Set of Division Subrings forms Complete Lattice}
+Tags: Lattice Theory, Division Subrings
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $K$.
+Then $\struct {\mathbb K, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
+\end{theorem}
+
+\begin{proof}
+Let $\O \subset \mathbb S \subseteq \mathbb K$.
+By [[Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings]]:
+:$\bigcap \mathbb S$ is the largest [[Definition:Division Subring|division subring]] of $K$ contained in each of the elements of $\mathbb S$.
+By [[Intersection of Division Subrings Containing Subset is Smallest]]:
+:The [[Definition:Set Intersection|intersection]] of the set of all [[Definition:Division Subring|division subrings]] of $K$ containing $\bigcup \mathbb S$ is the smallest [[Definition:Division Subring|division subring]] of $K$ containing $\bigcup \mathbb S$.
+Thus:
+:Not only is $\bigcap \mathbb S$ a [[Definition:Lower Bound of Set|lower bound]] of $\mathbb S$, but also the largest, and therefore an [[Definition:Infimum of Set|infimum]].
+:The [[Definition:Supremum of Set|supremum]] of $\mathbb S$ is the [[Definition:Set Intersection|intersection]] of the set of all [[Definition:Division Subring|division subrings]] of $K$.
+Therefore $\struct {\mathbb K, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Subfields forms Complete Lattice}
+Tags: Lattice Theory, Subfields
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathbb F$ be the [[Definition:Set|set]] of all [[Definition:Subfield|subfields]] of $F$.
+Then $\struct {\mathbb F, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
+\end{theorem}
+
+\begin{proof}
+Let $\O \subset \mathbb S \subseteq \mathbb F$.
+By [[Intersection of Subfields is Largest Subfield Contained in all Subfields]]:
+:$\bigcap \mathbb S$ is the largest [[Definition:Subfield|subfield]] of $F$ contained in each of the elements of $\mathbb S$.
+By [[Intersection of Subfields Containing Subset is Smallest]]:
+:The [[Definition:Set Intersection|intersection]] of the set of all [[Definition:Subfield|subfields]] of $F$ containing $\bigcup \mathbb S$ is the smallest [[Definition:Subfield|subfield]] of $F$ containing $\bigcup \mathbb S$.
+Thus:
+:Not only is $\bigcap \mathbb S$ a [[Definition:Lower Bound of Set|lower bound]] of $\mathbb S$, but also the largest, and therefore an [[Definition:Infimum of Set|infimum]].
+:The [[Definition:Supremum of Set|supremum]] of $\mathbb S$ is the [[Definition:Set Intersection|intersection]] of the set of all [[Definition:Subfield|subfields]] of $R$.
+Therefore $\struct {\mathbb F, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphism from Integers into Ring with Unity}
+Tags: Ideal Theory, Group Homomorphisms, Integers, Rings with Unity
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let the [[Definition:Characteristic of Ring|characteristic]] of $R$ be $p$.
+For any $a \in R$, we define the [[Definition:Mapping|mapping]] $g_a: \Z \to R$ from the [[Definition:Integer|integers]] into $R$ as:
+:$\forall n \in \Z: \map {g_a} n = n \cdot a$
+Then $g_a$ is a [[Definition:Group Homomorphism|group homomorphism]] from $\struct {\Z, +}$ to $\struct {R, +}$.
+Also:
+:$\ideal p \subseteq \map \ker {g_a}$
+where:
+:$\map \ker {g_a}$ is the [[Definition:Kernel of Group Homomorphism|kernel]] of $g_a$;
+:$\ideal p$ is the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $p$.
+Also:
+:$p \divides n \implies n \cdot a = 0$
+where $p \divides n$ denotes that $p$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
+\end{theorem}
+
+\begin{proof}
+The fact that $g_a$ is a [[Definition:Group Homomorphism|group homomorphism]] follows directly from [[Index Laws for Monoids]]:
+{{begin-eqn}}
+{{eqn | l = \map {g_a} m + \map {g_a} n
+      | r = m \cdot a + n \cdot a
+}}
+{{eqn | r = \paren {m + n} \cdot a
+}}
+{{eqn | r = \map {g_a} {m + n}
+}}
+{{end-eqn}}
+By [[Multiple of Ring Product]], we have that:
+:$\forall n \in \Z_{>0}: \paren {n \cdot x} \circ y = n \cdot \paren {x \circ y} = x \circ \paren {n \cdot y}$
+So:
+:$\forall n \in \Z_{>0}: n \cdot a = \paren {n \cdot a} \circ 1_R = a \circ \paren {n \cdot 1_R}$
+so when $n \cdot 1_R = 0$ we have $n \cdot a = 0$.
+For $n \in \Z_{<0}$, we have $-n \in \Z_{>0}$.
+So:
+:$n \cdot a = -\paren {-n} \cdot a = -\paren {\paren {-n \cdot a} \circ 1_R} = -\paren {a \circ \paren {-n \cdot 1_R}}$
+so when $-n \cdot 1_R = 0$ we have $n \cdot a = 0$.
+For $n = 0$, we trivially have $n \cdot a = 0$.
+By definition of [[Definition:Characteristic of Ring|characteristic]], we have:
+:$p \divides n \iff n \cdot 1_R = 0$
+by the above, this implies $n \cdot a = 0$.
+Therefore:
+:$p \divides n \implies n \cdot a = 0$
+We also have:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \in
+      | r = \ideal p
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = p
+      | o = \divides
+      | r = n
+      | c = {{Defof|Ideal of Ring}}
+}}
+{{eqn | ll = \leadsto
+      | l = n \cdot a
+      | r = 0
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = n
+      | o = \in
+      | r = \map \ker {g_a}
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{end-eqn}}
+By definition of [[Definition:Subset|subset]]:
+:$\ideal p \subseteq \map \ker {g_a}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Prime Subfield}
+Tags: Subfields
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $K$ be the [[Definition:Prime Subfield|prime subfield]] of $F$.
+Then $K$ is [[Definition:Field Isomorphism|isomorphic]] to either:
+:$\Q$, the [[Definition:Field of Rational Numbers|field of rational numbers]], or
+:$\Z_p$, the [[Ring of Integers Modulo Prime is Field|Ring of Integers Modulo $p$]], where $p$ is [[Definition:Prime Number|prime]].
+\end{theorem}
+
+\begin{proof}
+From [[Field of Characteristic Zero has Unique Prime Subfield]], if $\Char F = 0$, then its [[Definition:Prime Subfield|prime subfield]] is [[Definition:Field Isomorphism|isomorphic]] to $\Q$, the [[Definition:Field of Rational Numbers|field of rational numbers]].
+From [[Field of Prime Characteristic has Unique Prime Subfield]], if $\Char F = p$, then its [[Definition:Prime Subfield|prime subfield]] is [[Definition:Field Isomorphism|isomorphic]] to $\Z_p$, the [[Ring of Integers Modulo Prime is Field|Ring of Integers Modulo $p$]].
+From [[Characteristic of Field is Zero or Prime]], $p$ is [[Definition:Prime Number|prime]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Additive Group Modulo m is Ideal of Ring}
+Tags: Subgroups, Ideal Theory, Additive Groups of Integers Modulo m, Modulo Addition
+
+\begin{theorem}
+Let $m \in \Z: m > 1$.
+Let $\struct {\Z_m, +_m}$ be the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
+Then every [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +_m}$ is an [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]] $\struct {\Z_m, +_m, \times_m}$.
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +_m}$
+Suppose:
+: $(1): \quad h + \ideal m \in H$, where $\ideal m$ is a [[Definition:Principal Ideal of Ring|principal ideal]] of $\struct {\Z_m, +_m, \times_m}$
+and 
+: $(2): \quad n \in \N_{>0}$.
+Then by definition of [[Definition:Integer Multiplication|multiplication on integers]] and [[Homomorphism of Powers/Integers|Homomorphism of Powers as applied to integers]]:
+{{begin-eqn}}
+{{eqn | l = \paren {n + \ideal m} \times \paren {h + \ideal m}
+      | r = \map {q_m} n \times \map {q_m} h
+      | c = where $q_m$ is the [[Definition:Quotient Mapping|quotient mapping]]
+}}
+{{eqn | r = \map {q_m} {n \times h}
+      | c = 
+}}
+{{eqn | r = \map {q_m} {n \cdot h}
+      | c = 
+}}
+{{eqn | r = n \cdot \map {q_m} h
+      | c = 
+}}
+{{end-eqn}}
+But:
+:$n \cdot \map {q_m} h \in \gen {\map {q_m} h}$
+where $\gen {\map {q_m} h}$ is the [[Definition:Generator of Cyclic Group|group generated by $\map {q_m} h$]].
+Hence by [[Epimorphism from Integers to Cyclic Group]], $n \cdot \map {q_m} h \in H$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bijection from Divisors to Subgroups of Cyclic Group}
+Tags: Subgroups, Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$ [[Definition:Generator of Cyclic Group|generated by $a$]].
+Let $S = \left\{{m \in \Z_{>0}: m \divides n}\right\}$ be the [[Definition:Set|set]] of all [[Definition:Divisor of Integer|divisors]] of $n$.
+Let $T$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G$
+Let $\phi: S \to T$ be the [[Definition:Mapping|mapping]] defined as:
+:$\phi: m \to \left\langle{a^{n/m}}\right\rangle$
+where  $\left\langle{a^{n/m}}\right\rangle$ is the [[Definition:Generator of Subgroup|subgroup generated by $a^{n/m}$]].
+Then $\phi$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+From [[Subgroup of Finite Cyclic Group is Determined by Order]], there exists exactly one [[Definition:Subgroup|subgroup]] $\left \langle {a^{n/m}} \right \rangle$ of $G$ with $a$ elements.
+So the mapping as defined is indeed a [[Definition:Bijection|bijection]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generator of Additive Group Modulo m iff Unit of Ring}
+Tags: Additive Groups of Integers Modulo m, Ring Theory
+
+\begin{theorem}
+Let $m \in \Z: m > 1$.
+Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
+Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
+Let $a \in \Z_m$.
+Then:
+:$a$ is a [[Definition:Generator of Cyclic Group|generator]] of $\struct {\Z_m, +_m}$ 
+{{iff}}
+:$a$ is a [[Definition:Unit of Ring|unit]] of $\struct {\Z_m, +_m, \times_m}$
+\end{theorem}
+
+\begin{proof}
+From [[Integers under Addition form Infinite Cyclic Group]], the [[Definition:Identity Element|identity element]] $1_\Z$ of the [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$ is a [[Definition:Generator of Cyclic Group|generator of the group]] $\struct {\Z, +}$.
+Thus from [[Quotient Group of Cyclic Group]], the [[Definition:Identity Element|identity element]] $1_{\Z_m}$ of the [[Definition:Ring of Integers Modulo m|ring]] $\struct {\Z_m, +_m, \times_m}$ is a [[Definition:Generator of Cyclic Group|generator of the group]] $\struct {\Z_m, +_m}$.
+Let $a \in \Z_m$.
+Suppose $1_{\Z_m} \in \gen a$, where $\gen a$ signifies the [[Definition:Generator of Cyclic Group|group generated]] by $a$.
+Then the smallest [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +_m}$ containing $1_{\Z_m}$, that is $\struct {\Z_m, +_m}$ itself, is contained in $\gen a$.
+Thus $\gen a = \struct {\Z_m, +_m}$ iff $1_{\Z_m} \in \gen a$.
+However, from [[Subgroup of Additive Group Modulo m is Ideal of Ring]], $\gen a$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {\Z_m, +_m, \times_m}$, and hence is the [[Definition:Principal Ideal of Ring|principal ideal]] $\ideal a$ generated by $a$.
+But from [[Principal Ideal from Element in Center of Ring]], $1_{\Z_m} \in \gen a$ {{iff}} $a$ is a [[Definition:Unit of Ring|unit of the ring]] $\struct {\Z_m, +_m, \times_m}$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Real Differentiable Functions}
+Tags: Vector Subspaces, Differential Calculus
+
+\begin{theorem}
+Let $\mathbb J$ be an [[Definition:Open Real Interval|open interval]] of the [[Definition:Real Number Line|real number line]] $\R$.
+Let $\map \DD {\mathbb J}$ be the set of all [[Definition:Differentiable on Interval|differentiable real functions]] on $\mathbb J$.
+Then $\struct {\map \DD {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of the [[Definition:Vector Space|$\R$-vector space]] $\struct {\R^{\mathbb J}, +, \times}_\R$.
+\end{theorem}
+
+\begin{proof}
+Note that by definition, $\map \DD {\mathbb J} \subseteq \R^{\mathbb J}$. 
+Let $f, g \in \map \DD {\mathbb J}$. 
+Let $\lambda \in \R$.
+From [[Linear Combination of Derivatives]], we have that: 
+:$f + \lambda g$ is [[Definition:Differentiable Function|differentiable]] on $\mathbb J$. 
+That is: 
+:$f + \lambda g \in \map \DD {\mathbb J}$
+So, by [[One-Step Vector Subspace Test]]:
+:$\struct {\map \DD {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of $\R^{\mathbb J}$. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Real Functions of Differentiability Class}
+Tags: Vector Subspaces, Analysis
+
+\begin{theorem}
+Let $\mathbb J = \set {x \in \R: a < x < b}$ be an [[Definition:Open Real Interval|open interval]] of the [[Definition:Real Number Line|real number line]] $\R$.
+Let $\map {\CC^{\paren m} } {\mathbb J}$ be the set of all [[Definition:Continuous Real Function|continuous real functions]] on $\mathbb J$ in [[Definition:Differentiability Class|differentiability class]] $m$.
+Then $\struct {\map {\CC^{\paren m} } {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of the [[Definition:Vector Space|$\R$-vector space]] $\struct {\R^{\mathbb J}, +, \times}_\R$.
+\end{theorem}
+
+\begin{proof}
+Note that by definition, $\map {\CC^{\paren m} } {\mathbb J} \subseteq \R^{\mathbb J}$.
+Let $f, g \in \map {\CC^{\paren m} } {\mathbb J}$.
+Let $\lambda \in \R$.
+Applying [[Linear Combination of Derivatives]] $m$ times we have: 
+:$f + \lambda g$ is $m$-times [[Definition:Differentiable Function|differentiable]] on $\mathbb J$ with [[Definition:Nth Derivative|$m$th derivative]] $f^{\paren m} + \lambda g^{\paren m}$.
+Since both $f$ and $g$ are of [[Definition:Differentiability Class|differentiability class]] $m$:
+:$f^{\paren m}$ and $g^{\paren m}$ are [[Definition:Continuous Function|continuous]] on $\mathbb J$. 
+From [[Combination Theorem for Continuous Functions/Combined Sum Rule|Combination Theorem for Continuous Functions: Combined Sum Rule]]:
+:$f^{\paren m} + \lambda g^{\paren m} = \paren {f + \lambda g}^{\paren m}$ is [[Definition:Continuous Function|continuous]] on $\mathbb J$.
+So:
+:$f + \lambda g \in \map {\CC^{\paren m} } {\mathbb J}$
+Therefore, by [[One-Step Vector Subspace Test]]:
+:$\struct {\map {\CC^{\paren m} } {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of $\struct {\R^{\mathbb J}, +, \times}_\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Riemann Integrable Functions}
+Tags: Vector Subspaces, Analysis
+
+\begin{theorem}
+Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a [[Definition:Closed Real Interval|closed interval]] of the [[Definition:Real Number Line|real number line]] $\R$.
+Let $\map \RR {\mathbb J}$ be the set of all [[Definition:Riemann Integrable Function|Riemann integrable functions]] on $\mathbb J$.
+Then $\struct {\map \RR {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of the [[Definition:Vector Space|$\R$-vector space]] $\struct {\R^{\mathbb J}, +, \times}_\R$.
+\end{theorem}
+
+\begin{proof}
+Note that by definition, $\map {\mathcal D} {\mathbb J} \subseteq \R^{\mathbb J}$. 
+Let $f, g \in \map \RR {\mathbb J}$.
+Let $\lambda \in \R$.
+By [[Linear Combination of Definite Integrals]]:
+:$f + \lambda g$ is [[Definition:Riemann Integrable Function|Riemann integrable]] on $\mathbb J$.
+That is: 
+:$f + \lambda g \in \map \RR {\mathbb J}$
+So by [[One-Step Vector Subspace Test]]:
+:$\struct {\map \RR {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of $\struct {\R^{\mathbb J}, +, \times}_\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis of Vector Space of Polynomial Functions}
+Tags: Bases of Vector Spaces, Polynomial Theory
+
+\begin{theorem}
+Let $B$ be the [[Definition:Set|set]] of all the [[Definition:Identity Mapping|identity functions]] $I^n$ on $\R^n$ where $n \in \N^*$.
+Then $B$ is a [[Definition:Basis of Vector Space|basis]] of the [[Definition:Vector Space|$\R$-vector space]] $\map P \R$ of all [[Definition:Real Polynomial Function|polynomial functions]] on $\R$.
+\end{theorem}
+
+\begin{proof}
+By definition, every [[Definition:Real Polynomial Function|real polynomial function]] is a [[Definition:Linear Combination|linear combination]] of $B$.
+Suppose:
+:$\displaystyle \sum_{k \mathop = 0}^m \alpha_k I^k = 0, \alpha_m \ne 0$
+Then by [[Definition:Differentiation|differentiating]] $m$ times, we obtain from [[Nth Derivative of Nth Power]]:
+:$m! \alpha_m = 0$
+whence $\alpha_m = 0$ which is a contradiction.
+Hence $B$ is [[Definition:Linearly Independent Set|linearly independent]] and therefore is a [[Definition:Basis of Vector Space|basis]] for $\map P \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordered Basis for Coordinate Plane}
+Tags: Coordinate Systems
+
+\begin{theorem}
+Let $a_1, a_2 \in \R^2$ such that $\left\{{a_1, a_2}\right\}$ forms a [[Definition:Linearly Independent Set|linearly independent set]].
+Then $\left({a_1, a_2}\right)$ is an [[Definition:Ordered Basis|ordered basis]] for the [[Definition:Vector Space|$\R$-vector space]] $\R^2$.
+Hence the points on the [[Definition:Plane|plane]] can be uniquely identified by means of [[Definition:Linear Combination|linear combinations]] of $a_1$ and $a_2$.
+\end{theorem}
+
+\begin{proof}
+:[[File:OrderedBasisForPlane.png|500px|right|thumb]]
+Let $P$ be any [[Definition:Point|point]] in the [[Definition:Plane|plane]] for which we want to provide a linear combination of $a_1$ and $a_2$.
+Let the [[Definition:Length (Linear Measure)|distance]] from $O$ to the point determined by $a_1$ be defined as being $1$ unit of length on the [[Definition:Line|line]] $L_1$.
+Let the [[Definition:Length (Linear Measure)|distance]] from $O$ to the point determined by $a_2$ be defined as being $1$ unit of length on the [[Definition:Line|line]] $L_2$.
+Draw [[Definition:Parallel Lines|lines parallel]] to $L_1$ and $L_2$ through $P$.
+Then the [[Definition:Coordinate (Coordinate System)|coordinates]] $\lambda_1$ and $\lambda_2$ of $P$ are given by:
+:$P = \lambda_1 a_1 + \lambda_2 a_2$
+by the [[Parallelogram Law]].
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Differentiation on Polynomials is Linear Operator}
+Tags: Linear Algebra, Differential Calculus
+
+\begin{theorem}
+Let $P \left({\R}\right)$ be the [[Definition:Vector Space|vector space]] of all [[Definition:Real Polynomial Function|polynomial functions]] on the [[Definition:Real Number Line|real number line]] $\R$.
+Then the [[Definition:Differentiation|differentiation]] operator $D$ on $P \left({\R}\right)$ is a [[Definition:Linear Operator|linear operator]].
+\end{theorem}
+
+\begin{proof}
+Proved in [[Linear Combination of Derivatives]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integration on Polynomials is Linear Operator}
+Tags: Linear Algebra, Integral Calculus
+
+\begin{theorem}
+Let $P \left({\R}\right)$ be the [[Definition:Vector Space|vector space]] of all [[Definition:Real Polynomial Function|polynomial functions]] on the [[Definition:Real Number Line|real number line]] $\R$.
+Let $S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\displaystyle \forall p \in P \left({\R}\right): \forall x \in \R: S \left({p \left({x}\right)}\right) = \int_0^x p \left({t}\right) \mathrm d t$
+Then $S$ is a [[Definition:Linear Operator|linear operator]] on $P \left({\R}\right)$.
+\end{theorem}
+
+\begin{proof}
+Proved in [[Linear Combination of Integrals]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Noether's Theorem (Calculus of Variations)}
+Tags: Partial Differential Equations, Calculus of Variations
+
+\begin{theorem}
+Let $y_i$, $F$, $\Psi_i$, $\Phi$ be real functions.
+Let $x, \epsilon \in \R$.
+Let $\mathbf y = \sequence {y_i}_{1 \mathop \le i \mathop \le n}$ and $\mathbf \Psi = \sequence{\Psi_i}_{1 \mathop \le i \mathop \le n}$ be [[Definition:Vector (Linear Algebra)|vectors]].
+Let 
+:$\Phi = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}, \quad \Psi_i = \map {\Psi_i} {x, \mathbf y, \mathbf y'; \epsilon}$ 
+such that:
+:$\map \Phi {x, \mathbf y, \mathbf y'; 0} = x, \quad \map {\Psi_i} {x, \mathbf y, \mathbf y'; 0} = y_i$
+where $x, \mathbf y, \mathbf y', \epsilon$ are [[Definition:Variable|variables]].
+Let:
+:$\ds J \sqbrk {\mathbf y} = \int_{x_0}^{x_1} \map F {x, \mathbf y, \mathbf y'} \rd x$
+be a [[Definition:Real Functional|functional]].
+Let:
+:$X = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}, \quad \mathbf Y = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}$
+Suppose, $J \sqbrk {\mathbf y}$ is [[Definition:Invariant Functional under Transformation|invariant under transformations]] $x \rightarrow X$ and $\mathbf y \rightarrow \mathbf Y$ for arbitrary $x_0$ and $x_1$.
+Then:
+:$\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F} \phi = C$
+where $C$ is [[Definition:Constant|a constant]] and:
+:$\map {\boldsymbol \psi} {x, \mathbf y, \mathbf y'} = \dfrac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0}$
+:$\map \phi {x, \mathbf y, \mathbf y'} = \dfrac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0}$
+\end{theorem}
+
+\begin{proof}
+Apply [[Taylor's Theorem]] to the transformations $X$, $\mathbf Y$ at the point $\epsilon = 0$:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}
+}}
+{{eqn | r = \map \Phi {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0} + \map \OO \epsilon
+}}
+{{eqn | r = x + \epsilon \phi + \map \OO \epsilon
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \mathbf Y
+      | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}
+}}
+{{eqn | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg \rvert_{\epsilon \mathop = 0} + \map \OO \epsilon
+}}
+{{eqn | r = y + \epsilon \boldsymbol \psi + \map \OO \epsilon
+}}
+{{end-eqn}}
+Use [[General Variation of Integral Functional/Dependent on N Functions|the general variation formula]], and suppose that the curve $\mathbf y = \map {\mathbf y} x$ is an extremal of $J \sqbrk {\mathbf y}$:
+:$\delta x = \epsilon \phi, \quad \delta y = \epsilon\psi$
+:$\delta J = \epsilon \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_0} {x \mathop = x_1}$
+Since $J \sqbrk {\mathbf y}$ is [[Definition:Invariant Functional under Transformation|invariant under the transformation]], the [[Definition:Differential of Functional|variation]] vanishes.
+Then for any $\epsilon \ne 0$ we have:
+$\bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y'\cdot \nabla_{\mathbf y'} F} \phi} {x \mathop = x_0} {} = \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_1} {}$
+This has to hold for arbitrary $x_0$ and $x_1$.
+Since only a [[Definition:Constant Mapping|constant mapping]] produces the same result for any input, the term in brackets has to be a [[Definition:Constant|constant]].
+{{qed}}
+{{Namedfor|Emmy Noether|cat = Noether E}}
+\end{proof}<|endoftext|>
+\section{Transpose of Linear Transformation is a Linear Transformation}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $G$ and $H$ be [[Definition:Module|$R$-modules]].
+Let $G^*$ and $H^*$ be the [[Definition:Algebraic Dual|algebraic duals]] of $G$ and $H$ respectively.
+Let $\map {\LL_R} {G, H}$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
+Let $u \in \map {\LL_R} {G, H}$.
+Let $u^t: H^* \to G^*$ be the [[Definition:Transpose of Linear Transformation|transpose]] of $u$.
+Then $u^t: H^* \to G^*$ is itself a [[Definition:Linear Transformation|linear transformation]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Evaluation Linear Transformation|evaluation linear transformation]]:
+:$\forall x \in G: y' \in H^*: \sequence {x, \map {u^t} {y'} } = \sequence {\map u x, y'}$
+Since we have:
+{{begin-eqn}}
+{{eqn | l = \sequence {x, \map {u^t} {y' + z'} }
+      | r = \sequence {\map u x, y' + z'}
+      | c = 
+}}
+{{eqn | r = \sequence {\map u x, y'} + \sequence {\map u x, z'}
+      | c = 
+}}
+{{eqn | r = \sequence {x, \map {u^t} {y'} } + \sequence {x, \map {u^t} {z'} }
+      | c = 
+}}
+{{eqn | r = \sequence {x, \map {u^t} {y'} + \map {u^t} {z'} }
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \sequence {x, \map {u^t} {\lambda y'} }
+      | r = \sequence {\map u x, \lambda y'}
+      | c = 
+}}
+{{eqn | r = \lambda \sequence {\map u x, y'}
+      | c = 
+}}
+{{eqn | r = \lambda \sequence {x, \map {u^t} {y'} }
+      | c = 
+}}
+{{eqn | r = \sequence {x, \lambda \map {u^t} {y'} }
+      | c = 
+}}
+{{end-eqn}}
+it follows that $u^t: H^* \to G^*$ is a [[Definition:Linear Transformation|linear transformation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Supremum of Real-Valued Function}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $f: S \to \R$ be a [[Definition:Real Function|real function]] on $S$.
+{{TFAE|def = Supremum of Real-Valued Function}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $K \in \R$ be a [[Definition:Supremum of Mapping/Real-Valued Function/Definition 1|supremum of $f$ by definition 1]].
+Then from the definition:
+: $\text{(a)}: \quad K$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] of $f \left({x}\right)$ in $\R$.
+: $\text{(b)}: \quad K \le M$ for all [[Definition:Upper Bound of Real-Valued Function|upper bounds]] $M$ of $f \left({S}\right)$ in $\R$.
+As $K$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] it follows that:
+:$(1): \quad \forall x \in S: f \left({x}\right) \le K$
+Now let $\epsilon \in \R_{>0}$.
+{{AimForCont}} $(2)$ were false:
+:$\forall x \in S: f \left({x}\right) \le K - \epsilon$
+Then by definition, $K - \epsilon$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] of $f$.
+But by definition that means $K \le K + \epsilon$.
+So by [[Real Plus Epsilon]]:
+: $K < K$
+From this [[Proof by Contradiction|contradiction]] we conclude that:
+: $(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < f \left({x}\right)$
+Thus $K$ is a [[Definition:Supremum of Mapping/Real-Valued Function/Definition 2|supremum of $f$ by definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $K$ be a [[Definition:Supremum of Mapping/Real-Valued Function/Definition 2|supremum of $f$ by definition 2]]:
+: $(1) \quad \forall x \in S: f \left({x}\right) \le K$
+: $(2) \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < f \left({x}\right)$
+From $(1)$ we have that $K$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] of $f$.
+{{AimForCont}} that $K$ is ''not'' a [[Definition:Supremum of Mapping/Real-Valued Function/Definition 1|supremum of $f$ by definition 1]].
+Then:
+: $\exists M \in \R, M < K: \forall x \in S: f \left({x}\right) \le M$
+Then:
+:$\exists \epsilon \in \R_{>0}: M = K - \epsilon$
+Hence:
+:$\exists \epsilon \in \R_{>0}: \forall x \in S: f \left({x}\right) \le K - \epsilon$
+This [[Proof by Contradiction|contradicts]] $(2)$.
+Thus $K$ is a [[Definition:Supremum of Mapping/Real-Valued Function/Definition 1|supremum of $f$ by definition 1]].
+{{qed}}
+[[Category:Real Analysis]]
+dhw6ddzg7b8ps4uhwpmz8l3ed78h72m
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Infimum of Real-Valued Function}
+Tags: Infima
+
+\begin{theorem}
+Let $S \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $f: S \to \R$ be a [[Definition:Real Function|real function]] on $S$.
+{{TFAE|def = Infimum of Real-Valued Function}}
+\end{theorem}
+
+\begin{proof}
+{{wtd|Use [[Characterizing Property of Infimum of Subset of Real Numbers]]}}
+=== Definition 1 implies Definition 2 ===
+Let $k \in \R$ be an [[Definition:Infimum of Mapping/Real-Valued Function/Definition 1|infimum of $f$ by definition 1]].
+Then from the definition:
+:$\text{(a)}: \quad k$ is a [[Definition:Lower Bound of Real-Valued Function|lower bound]] of $\map f x$ in $\R$.
+:$\text{(b)}: \quad k \ge m$ for all [[Definition:Lower Bound of Real-Valued Function|lower bounds]] $m$ of $f \sqbrk S$ in $\R$.
+As $k$ is a [[Definition:Lower Bound of Real-Valued Function|lower bound]] it follows that:
+:$(1): \quad \forall x \in S: k \le \map f x$
+Now let $\epsilon \in \R_{>0}$.
+{{AimForCont}} $(2)$ were false:
+:$\forall x \in S: k + \epsilon \le \map f x$
+Then by definition, $k + \epsilon$ is a [[Definition:Lower Bound of Real-Valued Function|lower bound]] of $f$.
+But by definition that means $k \ge k + \epsilon$.
+So by [[Real Plus Epsilon]]:
+:$k > k$
+From this [[Proof by Contradiction|contradiction]] we conclude that:
+:$(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: \map f x < k + \epsilon$
+Thus $k$ is an [[Definition:Infimum of Mapping/Real-Valued Function/Definition 2|infimum of $f$ by definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $k$ be an [[Definition:Infimum of Mapping/Real-Valued Function/Definition 2|infimum of $f$ by definition 2]]:
+:$(1): \quad \forall x \in S: k \le \map f x$
+:$(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: \map f x < k + \epsilon$
+From $(1)$ we have that $k$ is a [[Definition:Lower Bound of Real-Valued Function|lower bound]] of $f$.
+{{AimForCont}} that $k$ is ''not'' an [[Definition:Infimum of Mapping/Real-Valued Function/Definition 1|infimum of $f$ by definition 1]].
+Then:
+:$\exists m \in \R, m > k: \forall x \in S: \map f x \ge m$
+Then:
+:$\exists \epsilon \in \R_{>0}: m = k + \epsilon$
+Hence:
+:$\exists \epsilon \in \R_{>0}: \forall x \in S: k + \epsilon \le \map f x$
+This [[Proof by Contradiction|contradicts]] $(2)$.
+Thus $k$ is an [[Definition:Infimum of Mapping/Real-Valued Function/Definition 1|infimum of $f$ by definition 1]].
+{{qed}}
+[[Category:Infima]]
+k5t6anxwz58e3xa9n3e2rss9dstbjlm
+\end{proof}<|endoftext|>
+\section{Supremum does not Precede Infimum}
+Tags: Suprema, Infima
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$ admit both a [[Definition:Supremum of Set|supremum]] $M$ and an [[Definition:Infimum of Set|infimum]] $m$.
+Then $m \preceq M$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\forall a \in T: a \preceq M$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\forall a \in T: m \preceq a$
+The result follows from [[Definition:Transitive Relation|transitivity]] of [[Definition:Ordering|ordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiples form Commutative Ring}
+Tags: Integers, Commutative Rings, Integer Multiples form Commutative Ring
+
+\begin{theorem}
+Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
+Then $\struct {n \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
+Unless $n = 1$, $\struct {n \Z, +, \times}$ is not a [[Definition:Ring with Unity|ring with unity]].
+\end{theorem}
+
+\begin{proof}
+From [[Integer Multiples under Addition form Infinite Cyclic Group]], $\struct {n \Z, +}$ is a [[Definition:Cyclic Group|cyclic group]]
+From [[Cyclic Group is Abelian]], $\struct {n \Z, +}$ is [[Definition:Abelian Group|abelian]].
+From [[Integer Multiples Closed under Multiplication]] and [[Integer Multiplication is Associative]], we have that $\struct {n \Z, \times}$ is a [[Definition:Semigroup|semigroup]].
+From [[Integer Multiplication Distributes over Addition]] it follows that $\struct {n \Z, +, \times}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+From [[Integer Multiplication is Commutative]] we have that $\struct {n \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
+So $\struct {n \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
+{{qed|lemma}}
+The [[Definition:Unity of Ring|unity]] of $\struct {\Z, +, \times}$ is $1$.
+But unless $n = 1$, we have that $1 \notin n \Z$.
+As $\struct {\Z, +, \times}$ is an [[Integers form Integral Domain|integral domain]], all its elements are [[Definition:Cancellable Element|cancellable]], by the [[Cancellation Law of Ring Product of Integral Domain]].
+We have that [[Subrings of Integers are Sets of Integer Multiples]].
+From [[Identity of Cancellable Monoid is Identity of Submonoid]] it follows that $\struct {n \Z, +, \times}$ has no [[Definition:Unity of Ring|unity]], because if it did, that unity would be $1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quaternions Defined by Matrices}
+Tags: Matrix Algebra, Quaternion Group
+
+\begin{theorem}
+Let $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ denote the following four [[Definition:Element|elements]] of the [[Definition:Matrix Space|matrix space]] $\map {\mathcal M_\C} 2$:
+:$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+\qquad
+\mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+\qquad
+\mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+\qquad
+\mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+$
+where $\C$ is the set of [[Definition:Complex Number|complex numbers]].
+Then $\mathbf 1, \mathbf i, \mathbf j, \mathbf k$ are related to each other in the following way:
+{{begin-eqn}}
+{{eqn | l = \mathbf i \mathbf j = - \mathbf j \mathbf i
+      | r = \mathbf k
+}}
+{{eqn | l = \mathbf j \mathbf k = - \mathbf k \mathbf j
+      | r = \mathbf i
+}}
+{{eqn | l = \mathbf k \mathbf i = - \mathbf i \mathbf k
+      | r = \mathbf j
+}}
+{{eqn | l = \mathbf i^2 = \mathbf j^2 = \mathbf k^2 = \mathbf i \mathbf j \mathbf k
+      | r = -\mathbf 1
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+This is demonstrated by straightforward application of [[Definition:Matrix Product (Conventional)|conventional matrix multiplication]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf i \mathbf j
+      | r = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} i \cdot 0 + 0 \cdot -1 & i \cdot 1 + 0 \cdot 0 \\ 0 \cdot 0 + -i \cdot -1 & 0 \cdot 1 + -i \cdot 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf k
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = -\mathbf j \mathbf i
+      | r = -\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 0 \cdot i + 1 \cdot 0 & 0 \cdot 0 + 1 \cdot -i \\ -1 \cdot i + 0 \cdot 0 & -1 \cdot 0 + 0 \cdot -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf k
+      | c = 
+}}
+{{end-eqn}}
+----
+{{begin-eqn}}
+{{eqn | l = \mathbf j \mathbf k
+      | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \cdot 0 + 1 \cdot i & 0 \cdot i + 1 \cdot 0 \\ -1 \cdot 0 + 0 \cdot i & -1 \cdot i + 0 \cdot 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf i
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = -\mathbf k \mathbf j
+      | r = -\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 0 \cdot 0 + i \cdot -1 & 0 \cdot 1 + i \cdot 0 \\ i \cdot 0 + 0 \cdot -1 & i \cdot 1 + 0 \cdot 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf i
+      | c = 
+}}
+{{end-eqn}}
+----
+{{begin-eqn}}
+{{eqn | l = \mathbf k \mathbf i
+      | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \cdot i + i \cdot 0 & 0 \cdot 0 + i \cdot -i \\ i \cdot i + 0 \cdot 0 & i \cdot 0 + 0 \cdot -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf j
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = -\mathbf i \mathbf k
+      | r = -\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} i \cdot 0 + 0 \cdot i & i \cdot i + 0 \cdot 0 \\ 0 \cdot 0 + -i \cdot i & 0 \cdot i + -i \cdot 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf j
+      | c = 
+}}
+{{end-eqn}}
+----
+{{begin-eqn}}
+{{eqn | l = \mathbf i^2
+      | r = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} i \cdot i + 0 \cdot 0 & i \cdot 0 + 0 \cdot -i \\ 0 \cdot i + -i \cdot 0 & -i \cdot 0 + -i \cdot -i \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\mathbf 1
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \mathbf j^2
+      | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \cdot 0 + 1 \cdot -1 & 0 \cdot 1 + 1 \cdot 0 \\ -1 \cdot 0 + 0 \cdot -1 & -1 \cdot 1 + 0 \cdot 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\mathbf 1
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \mathbf k^2
+      | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \cdot 0 + i \cdot i & 0 \cdot i + i \cdot 0 \\ i \cdot 0 + 0 \cdot i & i \cdot i + 0 \cdot 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = -\mathbf 1
+      | c = 
+}}
+{{end-eqn}}
+----
+{{begin-eqn}}
+{{eqn | l = \mathbf i \mathbf j \mathbf k
+      | r = \paren {\mathbf i \mathbf j} \mathbf k
+      | c = [[Matrix Multiplication is Associative]]
+}}
+{{eqn | r = \mathbf k \mathbf k
+      | c = from above: $\mathbf i \mathbf j = \mathbf k$
+}}
+{{eqn | r = -\mathbf 1
+      | c = from above: $\mathbf k^2 = - \mathbf 1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Form of Quaternion}
+Tags: Quaternions, Matrix Algebra
+
+\begin{theorem}
+Let $\mathbf x$ be a [[Definition:Quaternion|quaternion]] such that:
+: $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$
+When the [[Definition:Quaternion|quaternion]] basis is [[Quaternions Defined by Matrices|expressed in the form of matrices]]:
+:$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+\qquad
+\mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}
+\qquad
+\mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+\qquad
+\mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
+the general [[Definition:Quaternion|quaternion]] $\mathbf x$ has the form:
+:$\mathbf x = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix}$
+or:
+:$\mathbf x = \begin{bmatrix} w & z \\ -\overline z & \overline w \end{bmatrix}$
+where :
+:$w$ and $z$ are [[Definition:Complex Number|complex numbers]]
+:$\overline z$ is the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf x
+      | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
+      | c = 
+}}
+{{eqn | r = a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} + c \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} b i & 0 \\ 0 & - b i \end{bmatrix} + \begin{bmatrix} 0 & c \\ -c & 0 \end{bmatrix} + \begin{bmatrix} 0 & d i \\ d i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end{bmatrix}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quaternion Multiplication}
+Tags: Quaternions
+
+\begin{theorem}
+Let $\mathbf x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$ and $\mathbf x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$ be [[Definition:Quaternion|quaternions]].
+Then their '''product''' is given by:
+{{begin-eqn}}
+{{eqn | ll=\mathbf x_1 \mathbf x_2
+      | l==
+      | o=
+      | r=\left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) \mathbf 1
+      | c=
+}}
+{{eqn | o=+
+      | r=\left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) \mathbf i
+      | c=
+}}
+{{eqn | o=+
+      | r=\left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) \mathbf j
+      | c=
+}}
+{{eqn | o=+
+      | r=\left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) \mathbf k
+      | c=
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Matrix Form of Quaternion]] we have that:
+:$\mathbf x_1 \mathbf x_2 = \begin{bmatrix} a_1 + b_1 i & c_1 + d_1 i \\ -c_1 + d_1 i & a_1 - b_1 i \end{bmatrix} \begin{bmatrix} a_2 + b_2 i & c_2 + d_2 i \\ -c_2 + d_2 i & a_2 - b_2 i \end{bmatrix}$
+Let $\mathbf x_1 \mathbf x_2 = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end{bmatrix}$.
+Throughout we use the definition of [[Definition:Matrix Product (Conventional)|conventional matrix product]].
+So:
+{{begin-eqn}}
+{{eqn | l=p_{11}
+      | r=\left({a_1 + b_1 i}\right) \left({a_2 + b_2 i}\right) + \left({c_1 + d_1 i}\right) \left({-c_2 + d_2 i}\right)
+      | c=
+}}
+{{eqn | r=a_1 a_2 + b_1 a_2 i + a_1 b_2 i + b_1 b_2 i^2 - c_1 c_2 - d_1 c_2 i + c_1 d_2 i + d_1 d_2 i^2
+      | c=
+}}
+{{eqn | r=\left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) + \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i
+      | c=
+}}
+{{eqn | r=a + b i
+      | c=
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l=p_{12}
+      | r=\left({a_1 + b_1 i}\right) \left({c_2 + d_2 i}\right) + \left({c_1 + d_1 i}\right) \left({a_2 - b_2 i}\right)
+      | c=
+}}
+{{eqn | r=a_1 c_2 + b_1 c_2 i + a_1 d_2 i + b_1 d_2 i^2 + c_1 a_2 + d_1 a_2 i - c_1 b_2 i - d_1 b_2 i^2
+      | c=
+}}
+{{eqn | r=\left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) + \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) i
+      | c=
+}}
+{{eqn | r=c + d i
+      | c=
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l=p_{21}
+      | r=\left({-c_1 + d_1 i}\right) \left({a_2 + b_2 i}\right) + \left({a_1 - b_1 i}\right) \left({-c_2 + d_2 i}\right)
+      | c=
+}}
+{{eqn | r=-c_1 a_2 + d_1 a_2 i - c_1 b_2 i + d_1 b_2 i^2 - a_1 c_2 + b_1 c_2 i + a_1 d_2 i - b_1 d_2 i^2
+      | c=
+}}
+{{eqn | r=-\left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) + \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) i
+      | c=
+}}
+{{eqn | r=-c + d i
+      | c=
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l=p_{22}
+      | r=\left({-c_1 + d_1 i}\right) \left({c_2 + d_2 i}\right) + \left({a_1 - b_1 i}\right) \left({a_2 - b_2 i}\right)
+      | c=
+}}
+{{eqn | r=-c_1 c_2 + d_1 c_2 i - c_1 d_2 i + d_1 d_2 i^2 + a_1 a_2 - b_1 a_2 i - a_1 b_2 i + b_1 b_2 i^2
+      | c=
+}}
+{{eqn | r=\left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) - \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i
+      | c=
+}}
+{{eqn | r=a - b i
+      | c=
+}}
+{{end-eqn}}
+Hence the result from [[Matrix Form of Quaternion]].
+{{qed}}
+[[Category:Quaternions]]
+5ppc4f8h3le29g9xdsoky7nc4e6u7im
+\end{proof}<|endoftext|>
+\section{Quaternion Addition forms Abelian Group}
+Tags: Quaternions, Abelian Groups
+
+\begin{theorem}
+Let $\mathbb H$ be the [[Definition:Set|set]] of [[Definition:Quaternion|quaternions]].
+Then $\left({\mathbb H, +}\right)$, where $+$ denotes [[Definition:Quaternion Addition|quaternion addition]], is an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Abelian Group Axioms|abelian group axioms]] in turn:
+=== G0: Closure ===
+Let:
+: $\mathbf x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$
+: $\mathbf x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$
+be [[Definition:Quaternion|quaternions]].
+By definition of [[Definition:Quaternion Addition|quaternion addition]]:
+:$\mathbf x_1 + \mathbf x_2 = \left({a_1 + a_2}\right) \mathbf 1 + \left({b_1 + b_2}\right) \mathbf i + \left({c_1 + c_2}\right) \mathbf j + \left({d_1 + d_2}\right) \mathbf k$
+So as $a_1, a_2, b_1, b_2$ etc. are all elements of $\R$, then so are $a_1 + a_2, b_1 + b_2$ etc.
+So $\left({a_1 + a_2}\right) \mathbf 1 + \left({b_1 + b_2}\right) \mathbf i + \left({c_1 + c_2}\right) \mathbf j + \left({d_1 + d_2}\right) \mathbf k$ is a [[Definition:Quaternion|quaternion]].
+Hence $\left({\mathbb H, +}\right)$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== G1: Associativity ===
+From [[Matrix Form of Quaternion]], we can express a [[Definition:Quaternion|quaternion]] $\mathbf x$ in the form of a [[Definition:Matrix|matrix]]:
+:$\mathbf x = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix}$
+We have that [[Matrix Entrywise Addition is Associative]].
+It follows that [[Definition:Quaternion Addition|quaternion addition]] is also [[Definition:Associative|associative]]
+{{qed|lemma}}
+=== G2: Identity ===
+The [[Definition:Identity Element|identity element]] of $\left({\mathbb H, +}\right)$ is:
+:$\mathbf 0 = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$
+as can be seen:
+{{begin-eqn}}
+{{eqn | l = \mathbf 0 + \mathbf x
+      | r = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k + a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
+      | c = 
+}}
+{{eqn | r = \left({0 + a}\right) \mathbf 1 + \left({0 + b}\right) \mathbf i + \left({0 + c}\right) \mathbf j + \left({0 + d}\right) \mathbf k
+      | c = 
+}}
+{{eqn | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
+      | c = 
+}}
+{{eqn | r = \mathbf x
+      | c = 
+}}
+{{eqn | r = \left({a + 0}\right) \mathbf 1 + \left({b+ 0}\right) \mathbf i + \left({c + 0}\right) \mathbf j + \left({d + 0}\right) \mathbf k
+      | c = 
+}}
+{{eqn | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k + 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k
+      | c = 
+}}
+{{eqn | r = \mathbf x + \mathbf 0
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== G3: Inverses ===
+Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a [[Definition:Quaternion|quaternion]].
+The [[Definition:Inverse Element|inverse]] of $\left({\mathbb H, +}\right)$ is:
+:$- \mathbf x = -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k$
+as can be seen:
+{{begin-eqn}}
+{{eqn | l = \mathbf -x + \mathbf x
+      | r = -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k + a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
+      | c = 
+}}
+{{eqn | r = \left({-a + a}\right) \mathbf 1 + \left({-b + b}\right) \mathbf i + \left({-c + c}\right) \mathbf j + \left({-d + d}\right) \mathbf k
+      | c = 
+}}
+{{eqn | r = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k
+      | c = 
+}}
+{{eqn | r = \mathbf 0
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\mathbf x + \mathbf -x$.
+{{qed|lemma}}
+=== Commutativity ===
+[[Definition:Commutative Operation|Commutativity]] follows from [[Real Addition is Commutative]].
+{{qed|lemma}}
+Thus all the [[Definition:Abelian Group Axioms|abelian group axioms]] are seen to be fulfilled.
+{{qed}}
+[[Category:Quaternions]]
+[[Category:Abelian Groups]]
+f9okedoso5y46sqvj3orbt8k7gacxw9
+\end{proof}<|endoftext|>
+\section{Ring of Quaternions is Ring}
+Tags: Quaternions, Examples of Rings
+
+\begin{theorem}
+The [[Definition:Set|set]] $\mathbb H$ of [[Definition:Quaternion|quaternions]] forms a [[Definition:Ring (Abstract Algebra)|ring]] under the [[Definition:Binary Operation|operations]] of [[Definition:Quaternion Addition|addition]] and [[Definition:Quaternion Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+From [[Quaternion Addition forms Abelian Group]], $\mathbb H$ forms an [[Definition:Abelian Group|abelian group]] under [[Definition:Quaternion Addition|quaternion addition]].
+From the definition it is clear that [[Definition:Quaternion Multiplication|quaternion multiplication]] is [[Definition:Closure (Abstract Algebra)|closed]].
+We have from [[Matrix Form of Quaternion]] that [[Definition:Quaternion|quaternions]] can be expressed in [[Definition:Matrix|matrix form]].
+From [[Quaternion Multiplication]] we have that [[Definition:Quaternion Multiplication|quaternion multiplication]] can be expressed in terms of [[Definition:Matrix Product (Conventional)|matrix multiplication]], which [[Matrix Multiplication is Associative|is associative]].
+So $\mathbb H$ forms a [[Definition:Semigroup|semigroup]] under [[Definition:Quaternion Multiplication|quaternion multiplication]].
+Finally, we have that [[Matrix Multiplication Distributes over Matrix Addition]].
+So all the conditions are fulfilled for $\mathbb H$ to be a [[Definition:Ring (Abstract Algebra)|ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quaternions Subring of Complex Matrix Space}
+Tags: Quaternions, Matrix Algebra
+
+\begin{theorem}
+The [[Ring of Quaternions]] is a [[Definition:Subring|subring]] of the [[Definition:Matrix Space|matrix space]] $\mathcal M_\C \left({2}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Matrix Form of Quaternion]] it is clear that the [[Definition:Quaternion|quaternions]] $\mathbb H$ can be expressed in [[Definition:Matrix|matrix]] form, as elements of $\mathcal M_\C \left({2}\right)$.
+Thus $\mathbb H \subseteq \mathcal M_\C \left({2}\right)$.
+As [[Ring of Quaternions|the quaternions form a ring]], the result follows by definition of [[Definition:Subring|subring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Quaternion with Conjugate}
+Tags: Quaternions
+
+\begin{theorem}
+Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a [[Definition:Quaternion|quaternion]].
+Let $\overline {\mathbf x}$ be the [[Definition:Conjugate Quaternion|conjugate]] of $\mathbf x$.
+Then their [[Definition:Quaternion Multiplication|product]] is given by:
+:$\mathbf x \overline {\mathbf x} = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1 = \overline {\mathbf x} \mathbf x$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Quaternion Multiplication|quaternion multiplication]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf x \overline {\mathbf x}
+      | r = \paren {a^2 - b \paren {-b} - c \paren {-c} - d \paren {-d} } \mathbf 1
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {a \paren {-b} + b a + c \paren {-d} - d \paren {-c} } \mathbf i
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {a \paren {-c} - b \paren {-d} + c a + d \paren {-b} } \mathbf j
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {a \paren {-d} + b \paren {-c} - c \paren {-b} + d a} \mathbf k
+      | c = 
+}}
+{{eqn | r = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {-a b + b a - c d + d c} \mathbf i
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {-a c + b d + c a - d b} \mathbf j
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {-a d - b c + c b + d a} \mathbf k
+      | c = 
+}}
+{{end-eqn}}
+We have that $a, b, c, d \in \R$.
+So from [[Real Multiplication is Commutative]], their products [[Definition:Commute|commute]].
+So the terms in $\mathbf i, \mathbf j, \mathbf k$ vanish.
+The proof that $\overline {\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$ is similar.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Inverse of Quaternion}
+Tags: Quaternions
+
+\begin{theorem}
+Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a [[Definition:Quaternion|quaternion]] such that $\mathbf x \ne \mathbf 0$.
+Then $\mathbf x$ has an [[Definition:Inverse Element|inverse]] $\mathbf x^{-1}$ under the operation of [[Definition:Quaternion Multiplication|quaternion multiplication]]:
+:$\mathbf x^{-1} = \lambda \overline {\mathbf x}$
+where:
+:$\lambda = \dfrac 1 {a^2 + b^2 + c^2 + d^2} \mathbf 1$
+\end{theorem}
+
+\begin{proof}
+From [[Multiplicative Identity for Quaternions]], we need to show that $\lambda \overline{\mathbf x} \mathbf x = \mathbf 1 = \mathbf x \lambda \overline{\mathbf x}$.
+From [[Product of Quaternion with Conjugate]] we have that:
+:$\overline{\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$
+Using the definition of $\lambda$ from above:
+{{begin-eqn}}
+{{eqn | l = \lambda \overline{\mathbf x} \mathbf x
+      | r = \paren {\frac 1 {a^2 + b^2 + c^2 + d^2} } \mathbf 1 \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1
+      | c = 
+}}
+{{eqn | r = \paren {\frac {a^2 + b^2 + c^2 + d^2} {a^2 + b^2 + c^2 + d^2} } \mathbf 1
+      | c = {{Defof|Quaternion Multiplication}}
+}}
+{{eqn | r = \mathbf 1
+      | c = 
+}}
+{{end-eqn}}
+Unless all of $a, b, c, d = 0$ we have that $a^2 + b^2 + c^2 + d^2 \ne 0$.
+Therefore $\lambda$ is defined for all [[Definition:Quaternion|quaternions]] not equal to $\mathbf 0$.
+It follows from [[Left Inverse for All is Right Inverse]] that $\mathbf x \lambda \overline{\mathbf x} = \mathbf 1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Identity for Quaternions}
+Tags: Quaternions
+
+\begin{theorem}
+In the [[Definition:Set|set]] of [[Definition:Quaternion|quaternions]] $\mathbb H$, the [[Definition:Element|element]]:
+:$\mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$
+serves as the [[Definition:Identity Element|identity element]] for [[Definition:Quaternion Multiplication|quaternion multiplication]].
+This [[Definition:Element|element]] is written $\mathbf 1$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$.
+From the definition of [[Definition:Quaternion Multiplication|quaternion multiplication]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf x \mathbf 1
+      | r = \paren {a \cdot 1 - b \cdot 0 - c \cdot 0 - d \cdot 0} \mathbf 1
+      | c = 
+}}
+{{eqn | o = +
+      | r = \paren {a \cdot 0 + b \cdot 1 + c \cdot 0 - d  \cdot 0} \mathbf i
+      | c = 
+}}
+{{eqn | o = +
+      | r = \paren {a \cdot 0 - b \cdot 0 + c \cdot 1 + d \cdot 0} \mathbf j
+      | c = 
+}}
+{{eqn | o = +
+      | r = \paren {a \cdot 0 + b \cdot 0 - c \cdot 0 + d \cdot 1} \mathbf k
+      | c = 
+}}
+{{eqn | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
+      | c = 
+}}
+{{eqn | r = \mathbf x
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\mathbf 1 \mathbf x$.
+{{qed}}
+[[Category:Quaternions]]
+bj6kz4l5i0oylvv53hoxsmzktikilvu
+\end{proof}<|endoftext|>
+\section{Quaternions form Skew Field}
+Tags: Quaternions, Examples of Skew Fields
+
+\begin{theorem}
+The [[Definition:Set|set]] $\H$ of [[Definition:Quaternion|quaternions]] forms a [[Definition:Skew Field|skew field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Quaternion Addition|addition]] and [[Definition:Quaternion Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Quaternions is Ring]] we have that $\H$ forms a [[Definition:Ring (Abstract Algebra)|ring]].
+From [[Multiplicative Identity for Quaternions]] we have that $\mathbf 1$ is the [[Definition:Identity Element|identity]] for [[Definition:Quaternion Multiplication|quaternion multiplication]].
+From [[Multiplicative Inverse of Quaternion]] we have that every element of $\H$ except $\mathbf 0$ has an [[Definition:Inverse Element|inverse]] under [[Definition:Quaternion Multiplication|quaternion multiplication]].
+So $\H \setminus \set 0 = \H^*$ is a [[Definition:Group|group]].
+Hence $\H$ forms a [[Definition:Division Ring|division ring]].
+But [[Definition:Quaternion Multiplication|quaternion multiplication]] is specifically not [[Definition:Commutative Operation|commutative]], for example:
+:$\mathbf i \mathbf j = \mathbf k, \ \mathbf j \mathbf i = -\mathbf k$
+So $\H$ forms a [[Definition:Skew Field|skew field]] under [[Definition:Quaternion Addition|addition]] and [[Definition:Quaternion Multiplication|multiplication]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cancellation Law of Ring Product of Integral Domain}
+Tags: Integral Domains, Named Theorems
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+Let $D^*$ denote $D \setminus \set {0_D}$, that is, $D$ without its [[Definition:Ring Zero|zero]].
+Let $a \in D^*$.
+Then:
+:$\forall x, y \in D: a \circ x = a \circ y \implies x = y$
+That is, all elements of $D^*$ are [[Definition:Cancellable Element|cancellable]] for the [[Definition:Ring Product|ring product]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Integral Domain|integral domain]], no elements of $D^*$ are [[Definition:Zero Divisor of Ring|zero divisors]].
+From [[Ring Element is Zero Divisor iff not Cancellable]], it follows that all elements of $D^*$ are [[Definition:Cancellable Element|cancellable]] for the [[Definition:Ring Product|ring product]] $\circ$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gaussian Integers form Integral Domain}
+Tags: Integral Domains, Gaussian Integers
+
+\begin{theorem}
+The [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]] $\struct {\Z \sqbrk i, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
+\end{theorem}
+
+\begin{proof}
+The [[Complex Numbers form Field|set of complex numbers $\C$ forms a field]], which is [[Definition:Field (Abstract Algebra)|by definition]] a [[Definition:Division Ring|division ring]].
+We have that $\Z \sqbrk i \subset \C$.
+So from [[Cancellable Element is Cancellable in Subset]], all non-zero elements of $\Z \sqbrk i$ are [[Definition:Cancellable Element|cancellable]] for [[Definition:Complex Multiplication|complex multiplication]].
+The [[Definition:Identity Element|identity element]] for [[Definition:Complex Multiplication|complex multiplication]] is $1 + 0 i$ which is in $\Z \sqbrk i$.
+We also have that [[Complex Multiplication is Commutative]].
+From [[Identity of Cancellable Monoid is Identity of Submonoid]], the [[Definition:Identity Element|identity element]] of $\struct {\Z \sqbrk i^*, \times}$ is the same as for $\struct {\C^*, \times}$.
+So we have that:
+:$\struct {\Z \sqbrk i, +, \times}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]
+:All non-zero elements of $\struct {\Z \sqbrk i, +, \times}$ are [[Definition:Cancellable Element|cancellable]].
+Hence the result from definition of [[Definition:Integral Domain|integral domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set Intersection Not Cancellable}
+Tags: Intersection, Set Intersection, Set Intersection
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] and let $\mathcal P \left({S}\right)$ be the [[Definition:Power Set|power set]] of $S$.
+Let $S_1, S_2, T \in \mathcal P \left({S}\right)$.
+Suppose that $S_1 \cap T = S_2 \cap T$.
+Then it is not necessarily the case that $S_1 = S_2$.
+\end{theorem}
+
+\begin{proof}
+Proof by counterexample:
+Let $S = \left\{{1, 2, 3}\right\}$.
+Let $T = \left\{{3}\right\}$.
+Let $S_1 = \left\{{1, 3}\right\}, S_2 = \left\{{2, 3}\right\}$
+Then $S_1 \cap T = S_2 \cap T = \left\{{3}\right\}$ but $S_1 \ne S_2$.
+{{qed}}
+[[Category:Set Intersection]]
+b1v80uc46x69qbuvlgkeqgah7bdxw6a
+\end{proof}<|endoftext|>
+\section{Sum of Triangular Matrices}
+Tags: Triangular Matrices, Matrix Entrywise Addition
+
+\begin{theorem}
+Let $\mathbf A = \left[{a}\right]_{n}, \mathbf B = \left[{b}\right]_{n}$ be [[Definition:Square Matrix|square matrices]] of order $n$.
+Let $\mathbf C = \mathbf A + \mathbf B$ be the [[Definition:Matrix Entrywise Addition|matrix entrywise sum]] of $\mathbf A$ and $\mathbf B$.
+If $\mathbf A$ and $\mathbf B$ are [[Definition:Upper Triangular Matrix|upper triangular matrices]], then so is $\mathbf C$.
+If $\mathbf A$ and $\mathbf B$ are [[Definition:Lower Triangular Matrix|lower triangular matrices]], then so is $\mathbf C$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Matrix Entrywise Addition|matrix addition]], we have:
+:$\forall i, j \in \left[{1 .. n}\right]: c_{ij} = a_{ij} + b_{ij}$
+If $\mathbf A$ and $\mathbf B$ are [[Definition:Upper Triangular Matrix|upper triangular matrices]], we have:
+:$\forall i > j: a_{ij} = b_{ij} = 0$
+Hence:
+:$\forall i > j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
+and so $\mathbf C$ is itself [[Definition:Upper Triangular Matrix|upper triangular]].
+Similarly, if $\mathbf A$ and $\mathbf B$ are [[Definition:Lower Triangular Matrix|lower triangular matrices]], we have:
+:$\forall i < j: a_{ij} = b_{ij} = 0$
+Hence:
+:$\forall i < j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
+and so $\mathbf C$ is itself [[Definition:Lower Triangular Matrix|lower triangular]].
+{{Qed}}
+[[Category:Triangular Matrices]]
+[[Category:Matrix Entrywise Addition]]
+h6dwze7055xyqeo5fwlnt4ak9rikwy0
+\end{proof}<|endoftext|>
+\section{Negative of Triangular Matrix}
+Tags: Negative Matrices, Triangular Matrices
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $n$.
+Let $-\mathbf A$ be the [[Definition:Negative Matrix|negative]] of $\mathbf A$.
+If $\mathbf A$ is an [[Definition:Upper Triangular Matrix|upper triangular matrix]], then so is $-\mathbf A$.
+If $\mathbf A$ is a [[Definition:Lower Triangular Matrix|lower triangular matrix]], then so is $-\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Negative Matrix|negative matrix]], we have:
+:$\forall i, j \in \closedint 1 n: \sqbrk {-a}_{i j} = -a_{i j}$
+If $\mathbf A$ is an [[Definition:Upper Triangular Matrix|upper triangular matrix]], we have:
+:$\forall i > j: a_{i j} = 0$
+Hence:
+:$\forall i > j: \sqbrk {-a}_{i j} = -a_{i j} = 0$
+and so $-\mathbf A$ is itself [[Definition:Upper Triangular Matrix|upper triangular]].
+Similarly, if $\mathbf A$ is a [[Definition:Lower Triangular Matrix|lower triangular matrix]], we have:
+:$\forall i < j: a_{i j} = 0$
+Hence:
+:$\forall i < j: \sqbrk {-a}_{i j} = -a_{i j} = 0$
+and so $-\mathbf A$ is itself [[Definition:Lower Triangular Matrix|lower triangular]].
+{{Qed}}
+[[Category:Negative Matrices]]
+[[Category:Triangular Matrices]]
+ivy08olfqqzixab53xua25egndy4xer
+\end{proof}<|endoftext|>
+\section{Triangular Matrices forms Subring of Square Matrices}
+Tags: Triangular Matrices, Rings of Square Matrices, Subrings
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\map {\mathcal M_R} n$ be the [[Definition:Matrix Space|order $n$ square matrix space]] over a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
+Let $\struct {\map {\mathcal M_R} n, +, \times}$ denote the [[Definition:Ring of Square Matrices|ring of square matrices of order $n$ over $R$]].
+Let $\map {U_R} n$ be the [[Definition:Set|set]] of [[Definition:Upper Triangular Matrix|upper triangular matrices]] of [[Definition:Order of Square Matrix|order]] $n$ over $R$.
+Then $\map {U_R} n$ forms a [[Definition:Subring|subring]] of $\struct {\map {\mathcal M_R} n, +, \times}$.
+Similarly, let $\map {L_R} n$ be the [[Definition:Set|set]] of [[Definition:Lower Triangular Matrix|lower triangular matrices]] of [[Definition:Order of Square Matrix|order]] $n$ over $R$.
+Then $\map {L_R} n$ forms a [[Definition:Subring|subring]] of $\struct {\map {\mathcal M_R} n, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Negative of Triangular Matrix]], if $\mathbf B \in \map {U_R} n$ then $-\mathbf B \in \map {U_R} n$.
+Then from [[Sum of Triangular Matrices]], if $\mathbf A, -\mathbf B \in \map {U_R} n$ then $\mathbf A + \paren {-\mathbf B} \in \map {U_R} n$.
+From [[Product of Triangular Matrices]], if $\mathbf A, \mathbf B \in \map {U_R} n$ then $\mathbf A \mathbf B \in \map {U_R} n$.
+The result follows from the [[Subring Test]].
+The same argument can be applied to matrices in $\map {L_R} n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers form Subsemiring of Integers}
+Tags: Natural Numbers, Integers
+
+\begin{theorem}
+The [[Definition:Semiring of Natural Numbers|semiring of natural numbers]] $\struct {\N, +, \times}$ forms a [[Definition:Subsemiring|subsemiring]] of the [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+We have that [[Natural Numbers form Commutative Semiring]].
+From [[Natural Numbers are Non-Negative Integers]] we have that $\N$ is a [[Definition:Subset|subset]] of $\Z$.
+Hence the result, from the definition of [[Definition:Subsemiring|subsemiring]].
+{{qed}}
+[[Category:Natural Numbers]]
+[[Category:Integers]]
+0io25lnk5ngnlb90vw3fp3kag33fsht
+\end{proof}<|endoftext|>
+\section{Integers form Subdomain of Rationals}
+Tags: Integers, Rational Numbers
+
+\begin{theorem}
+The [[Integers form Integral Domain|integral domain of integers]] $\left({\Z, +, \times}\right)$ forms a [[Definition:Subdomain|subdomain]] of the [[Definition:Field of Rational Numbers|field of rational numbers]].
+\end{theorem}
+
+\begin{proof}
+The rational numbers are defined as the [[Definition:Rational Number|quotient field of the integers]].
+From its method of construction, it follows that the [[Definition:Integer|integers]] $\Z$ are a [[Definition:Subset|subset]] of the [[Definition:Rational Number|rational numbers]] $\Q$.
+Hence the result, from the definition of [[Definition:Subdomain|subdomain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers form Subfield of Complex Numbers}
+Tags: Subfields, Real Numbers, Complex Numbers
+
+\begin{theorem}
+The [[Definition:Field of Real Numbers|field of real numbers]] $\struct {\R, +, \times}$ forms a [[Definition:Subfield|subfield]] of the [[Definition:Field of Complex Numbers|field of complex numbers]] $\struct {\C, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Reals is Subgroup of Complex]], $\struct {\R, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, +}$.
+From [[Multiplicative Group of Reals is Subgroup of Complex]], $\struct {\R, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, \times}$.
+The result follows from the [[Subfield Test]] via the [[One-Step Subgroup Test]] and [[Two-Step Subgroup Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of All Ring Products is Additive Subgroup}
+Tags: Ring Theory, Subset Products
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\left({S, +}\right)$ and $\left({T, +}\right)$ be [[Definition:Additive Subgroup|additive subgroups]] of $\left({R, +, \circ}\right)$.
+Let $S + T$ be defined as [[Definition:Subset Product|subset product]].
+Let $S T$ be defined as:
+:$\displaystyle S T = \left\{{\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1 \,.\,.\ n}\right]}\right\}$
+Then both $S + T$ and $S T$ are [[Definition:Additive Subgroup|additive subgroups]] of $\left({R, +, \circ}\right)$.
+\end{theorem}
+
+\begin{proof}
+As $\left({R, +}\right)$ is [[Definition:Abelian Group|abelian]] (from the definition of a [[Definition:Ring (Abstract Algebra)|ring]]), we have:
+:$S + T = T + S$
+from [[Subset Product of Commutative is Commutative]].
+So from [[Subset Product of Subgroups]] it follows that $S + T$ is an [[Definition:Additive Subgroup|additive subgroup]] of $\left({R, +, \circ}\right)$.
+Let $x, y \in S T$.
+We have that [[Sum of All Ring Products is Closed under Addition|$\left({S T, +}\right)$ is closed]].
+So $x + y \in S T$.
+So, if $\displaystyle y = \sum s_i \circ t_i \in S T$, it follows that:
+: $\displaystyle -y = \sum \left({-s_i}\right) \circ t_i \in S T$
+By the [[Two-Step Subgroup Test]], we have that $S T$ is an [[Definition:Additive Subgroup|additive subgroup]] of $\left({R, +, \circ}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of All Ring Products is Closed under Addition}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\left({S, +}\right)$ and $\left({T, +}\right)$ be [[Definition:Additive Subgroup|additive subgroups]] of $\left({R, +, \circ}\right)$.
+Let $S T$ be defined as:
+:$\displaystyle S T = \left\{{\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1 \,.\,.\ n}\right]}\right\}$
+Then $\left({S T, +}\right)$ is a [[Definition:Closed Algebraic Structure|closed]] subset of $\left({R, +}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $x_1, x_2 \in S T$.
+Then:
+: $\displaystyle x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$
+for some $s_i, t_i, j, k$, etc.
+By renaming the indices, we can express $x_2$ as:
+: $\displaystyle x_2 = \sum_{i \mathop = j+1}^{j+k} s_i \circ t_i$
+and hence:
+:$\displaystyle x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j+1}^{j+k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$
+So $x_1 + x_2 \in S T$ and $\left({S T, +}\right)$ is shown to be [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of All Ring Products is Associative}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\left({S, +}\right), \left({T, +}\right), \left({U, +}\right)$ be [[Definition:Additive Subgroup|additive subgroups]] of $\left({R, +, \circ}\right)$.
+Let $S T$ be defined as:
+:$\displaystyle S T = \left\{{\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1 \,.\,.\,n}\right]}\right\}$
+Then:
+:$\left({S T}\right) U = S \left({T U}\right)$
+\end{theorem}
+
+\begin{proof}
+We have by definition that $S T$ is made up of all finite sums of elements of the form $s \circ t$ where $s \in S, t \in T$.
+From [[Sum of All Ring Products is Closed under Addition]], this set is [[Definition:Closure (Abstract Algebra)|closed]] under [[Definition:Ring Addition|ring addition]].
+Therefore, so are $\left({S T}\right) U$ and $S \left({T U}\right)$.
+Let $z \in \left({S T}\right) U$.
+Then $z$ is a finite sum of elements in the form $x \circ u$ where $x \in ST$ and $u \in U$.
+So $x$ is a finite sum of elements in the form $s \circ t$ where $s \in S, t \in T$.
+Therefore $z$ is a finite sum of elements in the form $\left({s \circ t}\right) \circ u$ where $s \in S, t \in T, u \in U$.
+As $\left({R, +, \circ}\right)$ os a [[Definition:Ring (Abstract Algebra)|ring]], $\circ$ is [[Definition:Associative|associative]].
+So $z$ is a finite sum of elements in the form $s \circ \left({t \circ u}\right)$ where $s \in S, t \in T, u \in U$.
+So these elements all belong to $S \left({T U}\right)$.
+Since $S \left({T U}\right)$ is [[Sum of All Ring Products is Closed under Addition|closed under addition]], $z \in S \left({T U}\right)$.
+So:
+:$\left({S T}\right) U \subseteq S \left({T U}\right)$
+By a similar argument in the other direction, 
+$S \left({T U}\right) \subseteq \left({S T}\right) U $
+and so by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\left({S T}\right) U = S \left({T U}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Ring Products is Subring of Commutative Ring}
+Tags: Commutative Rings, Subrings
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\left({S, +, \circ}\right)$ and $\left({T, +, \circ}\right)$ be [[Definition:Subring|subrings]] of $\left({R, +, \circ}\right)$.
+Let $S T$ be defined as:
+:$\displaystyle S T = \left\{{\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \left[{1 \,.\,.\ n}\right]}\right\}$
+Then $S T$ is a [[Definition:Subring|subring]] of $\left({R, +, \circ}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Sum of All Ring Products is Additive Subgroup]] we have that $\left({S T, +}\right)$ is an [[Definition:Additive Subgroup|additive subgroup]] of $R$.
+Let $x_1, x_2 \in S T$.
+Then:
+: $\displaystyle x_1 = \sum_{i \mathop = 1}^m s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^n s_j \circ t_j$
+for some $s_i, t_i, s_j, t_j, m, n$, etc.
+Then:
+{{begin-eqn}}
+{{eqn | l = x_1 \circ x_2
+      | r = \left({\sum_{i \mathop = 1}^m s_i \circ t_i}\right) \circ \left({\sum_{j \mathop = 1}^n s'_j \circ t'_j}\right)
+      | c = 
+}}
+{{eqn | r = \sum_{ {1 \mathop  \le i \mathop  \le m} \atop {1 \mathop  \le j \mathop  \le n} } \left({s_i \circ s'_j}\right) \circ \left({t_i \circ t'_j}\right)
+      | c = as $\circ$ is [[Definition:Commutative Operation|commutative]]
+}}
+{{end-eqn}}
+So $x_1 \circ x_2 \in S T$ and the result follows from the [[Subring Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers form Subfield of Quaternions}
+Tags: Complex Numbers, Quaternions, Subfields
+
+\begin{theorem}
+The [[Definition:Field of Complex Numbers|field of complex numbers]] $\left({\C, +, \times}\right)$ is [[Definition:Field Isomorphism|isomorphic]] to the [[Definition:Subfield|subfields]] of the [[Definition:Quaternion|quaternions]] $\left({\mathbb H, +, \times}\right)$ whose [[Definition:Underlying Set|underlying subsets]] are:
+:$(1): \quad \mathbb H_\mathbf i = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: c = d = 0}\right\}$
+:$(2): \quad \mathbb H_\mathbf j = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = d = 0}\right\}$
+:$(3): \quad \mathbb H_\mathbf k = \left\{{a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = c = 0}\right\}$
+That is:
+:$(1): \quad \mathbb H_\mathbf i = \left\{{a \mathbf 1 + b \mathbf i \in \mathbb H}\right\}$
+:$(2): \quad \mathbb H_\mathbf j = \left\{{a \mathbf 1 + c \mathbf j \in \mathbb H}\right\}$
+:$(3): \quad \mathbb H_\mathbf k = \left\{{a \mathbf 1 + d \mathbf k \in \mathbb H}\right\}$
+\end{theorem}
+
+\begin{proof}
+Let $\phi_i: \mathbb H_\mathbf i \to \C$  be defined as:
+:$\forall \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = a + b i$
+where in this context $i$ is the [[Definition:Imaginary Unit|imaginary unit]].
+Similarly we can define $\phi_j$ and $\phi_k$:
+:$\forall \mathbf x = \mathbf a \mathbf 1 + c \mathbf j \in \mathbb H_\mathbf j: \phi_j \left({\mathbf x}\right) = a + c i$
+:$\forall \mathbf x = \mathbf a \mathbf 1 + d \mathbf k \in \mathbb H_\mathbf k: \phi_k \left({\mathbf x}\right) = a + d i$
+=== Proof of Bijectivity ===
+First note that each of $\phi_i, \phi_j, \phi_k$ are [[Definition:Bijection|bijections]], as follows:
+* [[Definition:Injection|Injections]]:
+Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$. Then:
+:$\phi_i \left({x_1}\right) = \phi_i \left({x_2}\right) \implies a_1 + b_1 i = a_2 + b_2 i \implies a_1 = a_2, b_1 = b_2 \implies \mathbf x_1 = \mathbf x_2$
+and similarly with $\phi_j$ and $\phi_k$.
+* [[Definition:Surjection|Surjections]]:
+Let $x = a + b i \in \C$.
+Then:
+: $\exists \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = x$
+and similarly with $\phi_j$ and $\phi_k$.
+Thus it is established that $\phi_i, \phi_j, \phi_k$ are [[Definition:Bijection|bijections]].
+{{qed|lemma}}
+=== Proof of Morphism Property ===
+Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$.
+First we show that $+$ has the [[Definition:Morphism Property|morphism property]] under $\phi_i$:
+{{begin-eqn}}
+{{eqn | l=\phi_i \left({x_1}\right) + \phi_i \left({x_2}\right)
+      | r=\phi_i \left({a_1 \mathbf 1 + b_1 \mathbf i}\right) + \phi_i \left({a_2 \mathbf 1 + b_2 \mathbf i}\right)
+      | c=
+}}
+{{eqn | r=\left({a_1 + b_1 i}\right) + \left({a_2 + b_2 i}\right)
+      | c=
+}}
+{{eqn | r=\left({a_1 + a_2}\right) + \left({b_1 + b_2}\right) i
+      | c=
+}}
+{{eqn | r=\phi_i \left({x_1 + x_2}\right)
+      | c=
+}}
+{{end-eqn}}
+and similarly for $\phi_j$ and $\phi_k$.
+Next we show that $\times$ has the [[Definition:Morphism Property|morphism property]] under $\phi_i$:
+{{begin-eqn}}
+{{eqn | l=\phi_i \left({x_1 \times x_2}\right)
+      | r=\phi_i \left({a_1 a_2 - b_1 b_2 \mathbf i}\right) + \phi_i \left({a_1 b_2 + b_1 a_2 \mathbf i}\right)
+      | c=by [[Definition:Quaternion Multiplication|quaternion multiplication]] - most terms are zero
+}}
+{{eqn | r=\left({a_1 a_2 - b_1 b_2}\right) + \left({a_1 b_2 + b_1 a_2}\right) i
+      | c=
+}}
+{{eqn | r=\left({a_1 + b_1 i}\right) \times \left({a_2 + b_2 i}\right)
+      | c=by [[Definition:Complex Multiplication|complex multiplication]]
+}}
+{{eqn | r=\phi_i \left({x_1}\right) \times \phi_i \left({x_2}\right)
+      | c=
+}}
+{{end-eqn}}
+Similarly for  $\phi_j$ and $\phi_k$.
+So we have shown that $\phi_i, \phi_j, \phi_k$ are [[Definition:Ring Homomorphism|ring homomorphisms]].
+{{qed|lemma}}
+So all of these mappings are [[Definition:Bijection|bijective]] [[Definition:Ring Homomorphism|homomorphisms]], that is, [[Definition:Ring Isomorphism|isomorphisms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Homomorphism whose Kernel contains Ideal}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\nu: R \to R / J$ be the [[Definition:Quotient Ring Epimorphism|quotient epimorphism]].
+Let $\phi: R \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]] such that:
+:$J \subseteq \map \ker \phi$
+where $\map \ker \phi$ is the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$.
+Then there exists a unique [[Definition:Ring Homomorphism|ring homomorphism]] $\psi: R / J \to S$ such that:
+:$\phi = \psi \circ \nu$
+where $\circ$ denotes [[Definition:Composition of Mappings|composition of mappings]].
+:[[File:CommDiagRingHomKerIdeal.png|300px]]
+Also:
+:$\map \ker \psi = \map \ker \phi / J$
+\end{theorem}
+
+\begin{proof}
+Suppose $\phi = \psi \circ \nu$.
+Let $J + x$ be an arbitrary element of $R / J$.
+Then:
+:$(1) \quad \map \psi {J + x} = \psi \circ \map \nu x = \map \phi x$
+So there is only one possible way to define $\psi$.
+Now suppose $J + x = J + x'$.
+Then $x + \paren {-x'} \in J$.
+So $x + \paren {-x'} \in \map \ker \phi$ as $J \subseteq \map \ker \phi$.
+That is, $\map \phi {x + \paren {-x'} } = 0_S$.
+So $\map \phi x = \map \phi {x'}$ and so $\psi$ as defined in $(1)$ is [[Definition:Well-Defined Mapping|well-defined]].
+Now suppose $J + x, J + y \in R / J$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \psi {\paren {J + x} + \paren {J + y} }
+      | r = \map \psi {J + \paren {x + y} }
+      | c = {{Defof|Quotient Ring}}
+}}
+{{eqn | r = \map \phi {x + y}
+      | c = Definition of $\psi$ in $(1)$ above
+}}
+{{eqn | r = \map \phi x + \map \phi y
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | r = \map \psi {J + x} + \map \psi {J + y}
+      | c = 
+}}
+{{end-eqn}}
+So $\psi$ preserves [[Definition:Ring Addition|ring addition]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \psi {\paren {J + x} \paren {J + y} }
+      | r = \map \psi {J + \paren {x y} }
+      | c = {{Defof|Quotient Ring}}
+}}
+{{eqn | r = \map \phi {x y}
+      | c = Definition of $\psi$ in $(1)$ above
+}}
+{{eqn | r = \map \phi x \map \phi y
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | r = \map \psi {J + x} \map \psi {J + y}
+      | c = 
+}}
+{{end-eqn}}
+So $\psi$ preserves [[Definition:Ring Product|ring product]], and so $\psi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+Finally:
+{{begin-eqn}}
+{{eqn | l = \map \psi {J + x}
+      | r = 0_S
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \phi x
+      | r = 0_S
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \map \ker \phi
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\map \ker \psi = \map \ker \phi / J$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Decay Equation}
+Tags: Examples of Linear First Order ODEs, Decay Equation, Examples of Separation of Variables
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$\dfrac {\d y} {\d x} = k \paren {y_a - y}$
+where $k \in \R: k > 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = y_a + C e^{-k x}$
+where $C$ is an [[Definition:Arbitrary Constant|arbitrary constant]].
+If $y = y_0$ at $x = 0$, then:
+:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
+This differential equation is known as the '''decay equation'''.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {\d y} {\d x}
+      | r = -k \paren {y - y_a}
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d y} {y - y_a}
+      | r = -\int k \rd x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \ln {y - y_a}
+      | r = -k x + C_1
+      | c = [[Primitive of Reciprocal]] and [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]
+}}
+{{eqn | ll= \leadsto
+      | l = y - y_a
+      | r = e^{-k x + C_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = y_a + C e^{-k x}
+      | c = where we put $C = e^{C_1}$
+}}
+{{end-eqn}}
+This is our [[Definition:General Solution|general solution]].
+{{qed|lemma}}
+Suppose we have the [[Definition:Initial Condition|initial condition]]:
+:$y = y_0$ when $x = 0$
+Then:
+:$y_0 = y_a + C e^{-k \cdot 0} = y_a + C$
+and so:
+:$C = y_0 - y_a$
+Hence the solution:
+:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation on Empty Set is Equivalence}
+Tags: Relation Theory, Equivalence Relations, Empty Set
+
+\begin{theorem}
+Let $S = \varnothing$, that is, the [[Definition:Empty Set|empty set]].
+Let $\mathcal R \subseteq S \times S$ be a [[Definition:Relation|relation]] on $S$.
+Then $\mathcal R$ is the [[Definition:Null Relation|null relation]] and is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+As $S = \varnothing$, we have from [[Cartesian Product is Empty iff Factor is Empty]] that $S \times S = \varnothing$.
+Then it follows that $\mathcal R \subseteq S \times S = \varnothing$.
+=== Reflexivity ===
+From the definition:
+:$\mathcal R = \varnothing \implies \forall x \in S: \left({x, x}\right) \notin \mathcal R$
+But as $\neg \, \exists x \in S$ it follows [[Definition:Vacuous Truth|vacuously]] that $\mathcal R$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+It follows [[Definition:Vacuous Truth|vacuously]] that:
+:$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R$
+and so $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+It follows [[Definition:Vacuous Truth|vacuously]] that:
+:$\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \in \mathcal R$
+and so $\mathcal R$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+It follows from the definition that $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+[[Category:Relation Theory]]
+[[Category:Equivalence Relations]]
+[[Category:Empty Set]]
+ayuky9nnbh7tsmmmcvwaj4wgkcpliug
+\end{proof}<|endoftext|>
+\section{Fourth Isomorphism Theorem}
+Tags: Quotient Rings, Named Theorems, Isomorphisms, Isomorphism Theorems
+
+\begin{theorem}
+Let $\phi: R \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Let $K = \map \ker \phi$ be the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$.
+Let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Subring|subrings]] of $R$ which contain $K$ as a [[Definition:Subset|subset]].
+Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Subring|subrings]] of $\Img \phi$.
+Let $\phi^\to: \powerset R \to \powerset S$ be the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $\phi$.
+Then its [[Definition:Restriction of Mapping|restriction]] $\phi^\to: \mathbb K \to \mathbb S$ is a [[Definition:Bijection|bijection]].
+Also:
+:$(1): \quad \phi^\to$ and its [[Definition:Inverse Mapping|inverse]] both preserve [[Definition:Subset|subsets]].
+:$(2): \quad \phi^\to$ and its [[Definition:Inverse Mapping|inverse]] both preserve [[Definition:Ideal of Ring|ideals]]:
+:::: If $J$ is an [[Definition:Ideal of Ring|ideal]] of $R$, then $\map {\phi^\to} J$ is an [[Definition:Ideal of Ring|ideal]] of $S$
+:::: If $J'$ is an [[Definition:Ideal of Ring|ideal]] of $S$, then $\map {\paren {\phi^\to}^{-1} } {J'}$ is an [[Definition:Ideal of Ring|ideal]] of $R$
+\end{theorem}
+
+\begin{proof}
+=== Proof of Preservation of Subsets ===
+From [[Subset Maps to Subset]], we have:
+:$(a) \quad \forall X, X' \in \mathbb K: X \subseteq X' \implies \map {\phi^\to} X \subseteq \map {\phi^\to} {X'}$
+:$(b) \quad \forall Y, Y' \in \mathbb S: Y \subseteq Y' \implies \map {\paren {\phi^\to}^{-1} } Y \subseteq \map {\paren {\phi^\to}^{-1} } {Y'}$
+So $\phi^\to$ and its [[Definition:Inverse Mapping|inverse]] both preserve [[Definition:Subset|subsets]], and $(1)$ has been demonstrated to hold.
+{{qed|lemma}}
+=== Proof that Inverse Image is a Subring ===
+Let $U \in \mathbb S$, that is, let $U$ be a [[Definition:Subring|subring]] of $\Img \phi$.
+From [[Preimage of Subring under Ring Homomorphism is Subring]], we have that $\map {\paren {\phi^\to}^{-1} } U$ is a [[Definition:Subring|subring]] of $R$ such that $K \subseteq R$ and so:
+:$\map {\paren {\phi^\to}^{-1} } U \in \mathbb K$
+{{qed|lemma}}
+=== Proof that Image is a Subring ===
+Let $V \in \mathbb K$, that is, a [[Definition:Subring|subring]] of $R$ containing $K$.
+From [[Ring Homomorphism Preserves Subrings]], we have that $\map {\phi^\to} V$ is a [[Definition:Subring|subring]] of $S$ and so:
+:$\map {\phi^\to} V \in \mathbb S$
+{{qed|lemma}}
+=== Proof that $f$ is a Bijection ===
+By [[Subset of Codomain is Superset of Image of Preimage]], we have that:
+:$U = \map {\phi^\to} {\map {\paren {\phi^\to}^{-1} } U}$
+{{questionable|It doesn't, actually, it shows it's the superset. IIRC it needs to be a surjection for that. Check this to see whether $\phi^\to$ is indeed a surjection.}}
+We also have from [[Subset of Domain is Subset of Preimage of Image]] that:
+:$V \subseteq \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$
+Now let $r \in \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$.
+Thus $\map \phi r \in \map {\phi^\to} V$ and so:
+:$\exists v \in V: \map \phi r = \map \phi v$
+So $\map \phi {r + \paren {-v} } = 0_S$ and so $r - v \in K$ by definition of [[Definition:Kernel of Ring Homomorphism|kernel]].
+So:
+:$\exists k \in K: r + k = v$
+But by assumption, $K \subseteq V$ as $V \in \mathbb K$.
+So $k \in V$ and so it follows that $r \in V$ as well, by the fact that $V$ is [[Definition:Subring|subring]] and so [[Definition:Closure (Abstract Algebra)|closed]] for $+$.
+So:
+:$\map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V} \subseteq V$
+Putting that together with $V \subseteq \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$ and we see:
+:$V = \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$
+So we have:
+:$U = \map {\phi^\to} {\map {\paren {\phi^\to}^{-1} } U}$
+:$V = \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$
+That is:
+:$\phi^\to \circ \paren {\phi^\to}^{-1} = I_\mathbb S$
+:$\paren {\phi^\to}^{-1} \circ \phi^\to = I_\mathbb K$
+where $I_\mathbb S$ and $I_\mathbb K$ are the [[Definition:Identity Mapping|identity mappings]] of $\mathbb S$ and $\mathbb K$ respectively.
+By [[Composite of Bijection with Inverse is Identity Mapping]], it follows that $\phi^\to: \mathbb K \to \mathbb S$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+=== Proof of Preservation of Ideals ===
+Let $V \in \mathbb K$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $U = \map {\phi^\to} V$.
+Then from [[Ring Epimorphism Preserves Ideals]], $U$ is an [[Definition:Ideal of Ring|ideal]] of $S$.
+Similarly, let $U = \map {\phi^\to} V$ be an [[Definition:Ideal of Ring|ideal]] of $S$.
+Then by [[Preimage of Ideal under Ring Homomorphism is Ideal]], $V = \map {\paren {\phi^\to}^{-1} } U$ is an [[Definition:Ideal of Ring|ideal]] of $R$ such that $K \subseteq V$.
+Hence [[Definition:Ideal of Ring|ideals]] are preserved in both directions.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Increasing Union of Subrings is Subring}
+Tags: Set Union, Subrings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$ be [[Definition:Subring|subrings]] of $R$.
+Then the [[Definition:Increasing Union|increasing union]] $S$:
+:$\displaystyle S = \bigcup_{i \mathop \in \N} S_i$
+is a [[Definition:Subring|subring]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle S = \bigcup_{i \mathop \in \N} S_i$.
+Clearly $0_R \in S$.
+Let $a, b \in S$.
+Then $\exists i, j \in \N: a \in S_i, b \in S_j$.
+From the construction, we have that either of $S_i$ and $S_j$ contains the other.
+Let $\displaystyle l = \max \set {i, j}$ so $a, b \in S_l$.
+Then $a + \paren {-b} \in S_l$ and $a b \in S_l$, as $S_l$ is a [[Definition:Subring|subring]].
+Thus $a + \paren {-b} \in S$ and $a b \in S$.
+By the [[Subring Test]], $S$ is a [[Definition:Subring|subring]] of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subrings of Integers are Sets of Integer Multiples}
+Tags: Subrings, Integers, Subrings of Integers are Sets of Integer Multiples
+
+\begin{theorem}
+Let $\struct {\Z, +, \times}$ be the [[Integers form Integral Domain|integral domain of integers]].
+The [[Definition:Subring|subrings]] of $\struct {\Z, +, \times}$ are the [[Integer Multiples form Commutative Ring|rings of integer multiples]]:
+:$\struct {n \Z, +, \times}$
+where $n \in \Z: n \ge 0$.
+There are no other [[Definition:Subring|subrings]] of $\struct {\Z, +, \times}$ but these.
+\end{theorem}
+
+\begin{proof}
+From [[Integer Multiples form Commutative Ring]], it is clear that $\struct {n \Z, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {\Z, +, \times}$ when $n \ge 1$.
+We also note that when $n = 0$, we have:
+:$\struct {n \Z, +, \times} = \struct {0, +, \times}$
+which is the [[Definition:Null Ring|null ring]].
+When $n = 1$, we have:
+:$\struct {n \Z, +, \times} = \struct {\Z, +, \times}$
+From [[Null Ring and Ring Itself Subrings]], these extreme cases are also [[Definition:Subring|subrings]] of $\struct {\Z, +, \times}$.
+From [[Subgroups of Additive Group of Integers]], the only [[Definition:Additive Subgroup|additive subgroups]] of $\struct {\Z, +, \times}$ are $\struct {n \Z, +}$.
+So there can be no [[Definition:Subring|subrings]] of $\struct {\Z, +, \times}$ which do not have $\struct {n \Z, +}$ as their [[Definition:Additive Group of Ring|additive group]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subring of Integers is Ideal}
+Tags: Subrings, Integers, Ideal Theory
+
+\begin{theorem}
+Every [[Definition:Subring|subring]] of $\struct {\Z, +, \times}$ is an [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring (Abstract Algebra)|ring]] $\struct {\Z, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Subrings of Integers are Sets of Integer Multiples]]
+and:
+:[[Subgroup of Integers is Ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial Ring of Sequences is Ring}
+Tags: Polynomial Theory, Ring Theory
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $P \left[{R}\right]$ be the [[Definition:Polynomial Ring over Sequence|polynomial ring over sequences in $R$]].
+Then $P \left[{R}\right]$ is itself a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+We have by definition of [[Definition:Polynomial Ring over Sequence|polynomial ring over sequences in $R$]] that:
+:$P \left[{R}\right] = \left\{{\left \langle {r_0, r_1, r_2, \ldots}\right \rangle}\right\}$
+where each $r_i \in R$, and all but a [[Definition:Finite|finite number]] of [[Definition:Term of Sequence|terms]] is [[Definition:Ring Zero|zero]].
+=== Proof that Operations are Closed ===
+We need to ensure that the operations as defined are [[Definition:Closed Operation|closed]].
+Let $r = \left \langle {r_0, r_1, r_2, \ldots}\right \rangle, s = \left \langle {s_0, s_1, s_2, \ldots}\right \rangle \in P \left[{R}\right]$.
+As all but a finite number of terms of $r$ and $s$ are [[Definition:Ring Zero|zero]], there exist $m, n \ge 0$ such that:
+:$\forall i > m: r_i = 0$
+:$\forall j > n: s_j = 0$
+Let $l = \max \left\{{m, n}\right\}$.
+We can express the operations on $P \left[{R}\right]$ as:
+{{begin-eqn}}
+{{eqn | n=1
+      | l=\left \langle {r_0, r_1, \ldots, r_m}\right \rangle + \left \langle {s_0, s_1, \ldots, s_n}\right \rangle
+      | r=\left \langle {r_0 + s_0, r_1 + s_1, \ldots, r_l + s_l}\right \rangle
+      | c=
+}}
+{{eqn | n=2
+      | l=-\left \langle {r_0, r_1, \ldots, r_m}\right \rangle
+      | r=\left \langle {-r_0, -r_1, \ldots, -r_m}\right \rangle
+      | c=
+}}
+{{eqn | n=3
+      | l=\left \langle {r_0, r_1, \ldots, r_m}\right \rangle \left \langle {s_0, s_1, \ldots, s_n}\right \rangle
+      | r=\left \langle {t_0, t_1, \ldots, t_{m+n} }\right \rangle
+      | c=where $\displaystyle t_k = \sum_{i \mathop + j \mathop = k} r_i s_j$
+}}
+{{end-eqn}}
+We have that:
+: $\forall i > l: r_i + s_i = 0$
+and so 
+: $r + s \in P \left[{R}\right]$
+Equally clearly:
+: $-r \in P \left[{R}\right]$
+Now consider:
+: $\displaystyle \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k$
+Let $i > m + n$.
+Then in any $r_j s_k$ such that $j + k = i$, either $j > m$ or k > n.
+In the first case $r_j = 0$ and in the second $s_j = 0$.
+In either case $r_j s_k = 0$.
+So:
+: $\displaystyle \forall i > m + n: \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k = 0$
+So:
+: $r s \in P \left[{R}\right]$
+{{qed|lemma}}
+=== Proof of Additive Group ===
+The addition operation $r + s$ is clearly [[Definition:Commutative Operation|commutative]] and [[Definition:Associative|associative]], and:
+:$\left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$
+Equally clearly:
+:$\forall r \in P \left[{R}\right]: r + \left({-r}\right) = \left \langle {r_i + \left({-r_i}\right)}\right \rangle = \left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$
+and so $-r$ is the [[Definition:Inverse Element|inverse]] of $r$ for [[Definition:Ring Addition|addition]].
+So $\left({P \left[{R}\right], +}\right)$ is an [[Definition:Abelian Group|abelian group]], as it needs to be for $P \left[{R}\right]$ to be a [[Definition:Ring (Abstract Algebra)|ring]].
+{{qed|lemma}}
+=== Proof of Ring Product ===
+We need to establish that the [[Definition:Ring Product|ring product]] is [[Definition:Associative|associative]].
+Let $r = \left \langle {r_0, r_1, \ldots}\right \rangle, s = \left \langle {s_0, s_1, \ldots}\right \rangle, t = \left \langle {t_0, t_1, \ldots}\right \rangle \in P \left[{R}\right]$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left({\left({r s}\right) t}\right)_n
+      | r = \sum_{i \mathop + j \mathop = n} \left({r s}\right)_i t_j
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop + j \mathop = n} \left({\sum_{k \mathop + l \mathop = i} r_k s_l}\right) t_j
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop + l \mathop + j \mathop = n} r_k s_l t_j
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\left({r \left({s t}\right)}\right)_n$.
+So the [[Definition:Ring Product|ring product]] is [[Definition:Associative|associative]], and so forms a [[Definition:Semigroup|semigroup]].
+{{qed|lemma}}
+=== Proof of Distributivity ===
+Finally we need to show that the [[Definition:Ring Product|ring product]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Ring Addition|ring addition]].
+{{begin-eqn}}
+{{eqn | l = \left({\left({r + s}\right) t}\right)_n
+      | r = \sum_{i \mathop + j \mathop = n} \left({r_i + s_i} \right) t_j
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop + j \mathop = n} r_i t_j + \sum_{i \mathop + j \mathop = n} s_i t_j
+      | c = 
+}}
+{{eqn | r = \left({r t}\right)_n + \left({s t}\right)_n
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\left({t \left({r + s}\right)}\right)_n$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition of Polynomial from Polynomial Ring over Sequence}
+Tags: Ring Theory, Polynomial Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $\struct {P \sqbrk R, \oplus, \odot}$ be the [[Definition:Polynomial Ring over Sequence|polynomial ring]] over the [[Definition:Set|set]] of all [[Definition:Sequence|sequences in $R$]]:
+:$P \sqbrk R = \set {\sequence {r_0, r_1, r_2, \ldots} }$
+where the [[Definition:Operations on Polynomial Ring of Sequences|operations $\oplus$ and $\odot$ on $P \sqbrk R$]] be defined as:
+{{:Definition:Operations on Polynomial Ring of Sequences}}
+Let $\struct {R \sqbrk X, +, \circ}$ be the [[Definition:Ring of Polynomials|ring of polynomials over $R$ in $X$]].
+{{explain|Strictly speaking the definition in terms of [[Definition:Polynomial Form]] is needed here, with $X$ specifically being an [[Definition:Indeterminate (Polynomial Theory)]] or [[Definition:Transcendental over Integral Domain]].}}
+Then $\struct {R \sqbrk X, +, \circ}$ and $\struct {P \sqbrk R, \oplus, \odot}$ are [[Definition:Ring Isomorphism|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+Let $P \sqbrk R$ be the [[Definition:Polynomial Ring over Sequence|polynomial ring over $R$]].
+Consider the [[Definition:Injection|injection]] $\phi: R \to P \sqbrk R$ defined as:
+:$\forall r \in R: \map \phi r = \sequence {r, 0, 0, \ldots}$
+It is easily checked that $\phi$ is a [[Definition:Ring Monomorphism|ring monomorphism]].
+So the set $\set {\sequence {r, 0, 0, \ldots}: r \in R}$ is a [[Definition:Subring|subring]] of $P \sqbrk R$ which is [[Definition:Ring Isomorphism|isomorphic]] to $R$.
+So we ''identify'' $r \in R$ with the [[Definition:Sequence|sequence]] $\sequence {r, 0, 0, \ldots}$.
+Next we note that $P \sqbrk R$ contains the element $\sequence {0, 1, 0, \ldots}$ which we can call $x$.
+From the definition of [[Definition:Ring Product|ring product]] on the [[Definition:Polynomial Ring over Sequence|polynomial ring over $R$]], we have that:
+:$x^2 = \sequence {0, 1, 0, \ldots} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, 0, 1, 0, 0, \ldots}$
+:$x^3 = \sequence {0, 0, 1, 0, 0, \ldots} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, 0, 0, 1, 0, \ldots}$
+and in general:
+:$x^n = \sequence {0, 1, 0, \ldots}^{n - 1} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}$
+for all $n \ge 1$.
+Hence we see that:
+{{begin-eqn}}
+{{eqn | l = \sequence {r_0, r_1, \ldots, r_n, 0, \ldots \ldots}
+      | r = \sequence {r_0, 0, 0, \ldots \ldots} \odot \sequence {1, 0, 0, \ldots}
+      | c = 
+}}
+{{eqn | ro= \oplus
+      | o = 
+      | r = \sequence {r_1, 0, 0, \ldots \ldots} \odot \sequence {0, 1, 0, \ldots}
+      | c = 
+}}
+{{eqn | ro= \oplus
+      | o = 
+      | r = \cdots
+      | c = 
+}}
+{{eqn | ro= \oplus
+      | o = 
+      | r = \sequence {r_n, 0, 0, \ldots \ldots} \odot \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}
+      | c = 
+}}
+{{eqn | r = r_0 \oplus r_1 \circ x \oplus r_2 \circ x^2 \oplus \ldots \oplus r_n \circ x^n
+      | c = 
+}}
+{{end-eqn}}
+So by construction, $R \sqbrk X$ is seen to be equivalent to $P \sqbrk R$.
+{{qed}}
+{{handwaving|The entirety of the above argument needs to be made properly rigorous.}}
+It can also be shown that this proof works for the general [[Definition:Ring (Abstract Algebra)|ring]] whether it be a [[Definition:Ring with Unity|ring with unity]] or not.
+\end{proof}<|endoftext|>
+\section{Integers under Multiplication form Countably Infinite Semigroup}
+Tags: Integer Multiplication, Examples of Semigroups
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+From [[Integers under Multiplication form Semigroup]], $\struct {\Z, \times}$ is a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Semigroup|semigroup]].
+Then we have that the [[Integers are Countably Infinite]].
+The criteria for $\struct {\Z, \times}$ to be a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Semigroup|semigroup]] are seen to be satisfied. 
+{{Qed}}
+[[Category:Integer Multiplication]]
+[[Category:Examples of Semigroups]]
+ofhk6z2f8lxht1xp6vezhh6swr7te1j
+\end{proof}<|endoftext|>
+\section{Integer Addition is Closed}
+Tags: Integer Addition, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Addition|addition]]:
+:$\forall a, b \in \Z: a + b \in \Z$
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
+$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by:
+:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+[[Definition:Integer Addition|Integer addition]] is defined as:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$
+We have that:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \in \Z, \eqclass {c, d} {} \in \Z$
+Also:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$
+But:
+:$a + c \in \N, b + d \in \N$
+So:
+:$\forall a, b, c, d \in \N: \eqclass {a + c, b + d} {} \in \Z$
+Therefore [[Definition:Integer Addition|integer addition]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Addition Identity is Zero}
+Tags: Integer Addition
+
+\begin{theorem}
+The [[Definition:Identity Element|identity]] of [[Definition:Integer Addition|integer addition]] is $0$:
+:$\exists 0 \in \Z: \forall a \in \Z: a + 0 = a = 0 + a$
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxtimes$.
+$\boxtimes$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by:
+:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+From [[Construction of Inverse Completion|the method of construction]], $\eqclass {c, c} {}$, where $c$ is any element of the [[Definition:Natural Numbers|natural numbers]] $\N$, is [[Definition:Isomorphic Copy|the isomorphic copy]] of $0 \in \N$.
+So, we need to show that:
+:$\forall a, b, c \in \N: \eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {} + \eqclass {c, c} {}
+      | r = \eqclass {a + c, b + c} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {a, b} {}
+      | c = [[Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes|Construction of Inverse Completion: Members of Equivalence Classes]]
+}}
+{{end-eqn}}
+So:
+:$\eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {}$
+The identity $\eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$ is demonstrated similarly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverses for Integer Addition}
+Tags: Integer Addition, Inverse Elements
+
+\begin{theorem}
+Each element $x$ of the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ has an [[Definition:Inverse Element|inverse element]] $-x$ under the operation of [[Definition:Integer Addition|integer addition]]:
+:$\forall x \in \Z: \exists -x \in \Z: x + \paren {-x} = 0 = \paren {-x} + x$
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxtimes$.
+$\boxtimes$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by:
+:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+From [[Construction of Inverse Completion/Invertible Elements in Quotient Structure|the method of construction]], the element $\eqclass {a, a + x} {}$ has an [[Definition:Inverse Element|inverse]] $\eqclass {a + x, a} {}$ where $a$ and $x$ are elements of the [[Definition:Natural Numbers|natural numbers]] $\N$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, a + x} {} + \eqclass {a + x, a} {}
+      | r = \eqclass {a + a + x, a + x + a} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {a, a} {}
+      | c = [[Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes|Construction of Inverse Completion: Members of Equivalence Classes]]
+}}
+{{eqn | r = \eqclass {a + x + a , a + a + x} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {a + x, a} {} + \eqclass {a, a + x} {}
+      | c = 
+}}
+{{end-eqn}}
+So $\eqclass {a, a + x} {}$ has the inverse $\eqclass {a + x, a} {}$.
+{{explain|Needs more detail explaining the connection to the natural numbers. The Inverse Completion page is okay, but the link to what's going on here is hard to follow.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Ring}
+Tags: Examples of Rings, Rational Numbers
+
+\begin{theorem}
+The [[Definition:Rational Number|set of rational numbers]] $\Q$ forms a [[Definition:Ring (Abstract Algebra)|ring]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]: $\struct {\Q, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Recall that [[Rational Numbers form Field|$\struct {\Q, +, \times}$ is a field]].
+As a [[Definition:Field (Abstract Algebra)|field]] is also by definition a [[Definition:Division Ring|division ring]], which is an example of a [[Definition:Ring (Abstract Algebra)|ring]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers form Ring}
+Tags: Examples of Rings, Real Numbers
+
+\begin{theorem}
+The [[Definition:Real Number|set of real numbers]] $\R$ forms a [[Definition:Ring (Abstract Algebra)|ring]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Numbers under Addition form Abelian Group]], $\struct {\R, +}$ is an [[Definition:Abelian Group|abelian group]].
+We also have that:
+:[[Real Multiplication is Closed]]:
+::$\forall x, y \in \R: x \times y \in \R$
+:[[Real Multiplication is Associative]]:
+::$\forall x, y, z \in \R: x \times \paren {y \times z} = \paren {x \times y} \times z$
+Thus $\struct {\R, +}$ is a [[Definition:Semigroup|semigroup]].
+Finally we have that [[Real Multiplication Distributes over Addition]]:
+{{begin-eqn}}
+{{eqn | lo= \forall x, y, z \in \R:
+      | l = x \times \paren {y + z}
+      | r = x \times y + x \times z
+}}
+{{eqn | l = \paren {y + z} \times x
+      | r = y \times x + z \times x
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Ring (Abstract Algebra)|ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers form Ring}
+Tags: Examples of Rings, Complex Numbers
+
+\begin{theorem}
+The [[Definition:Complex Number|set of complex numbers]] $\C$ forms a [[Definition:Ring (Abstract Algebra)|ring]] under [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]]: $\struct {\C, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Complex Numbers under Addition form Abelian Group]], $\struct {\C, +}$ is an [[Definition:Abelian Group|abelian group]].
+We also have that:
+:[[Complex Multiplication is Closed]]:
+::$\forall x, y \in \C: x \times y \in \C$
+:[[Complex Multiplication is Associative]]:
+::$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$
+Thus $\struct{\C, +}$ is a [[Definition:Semigroup|semigroup]].
+Finally we have that [[Complex Multiplication Distributes over Addition]]:
+{{begin-eqn}}
+{{eqn | lo= \forall x, y, z \in \C:
+      | l = x \times \paren {y + z}
+      | r = x \times y + x \times z
+}}
+{{eqn | l = \paren {y + z} \times x
+      | r = y \times x + z \times x
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Ring (Abstract Algebra)|ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Addition Identity is Zero}
+Tags: Complex Addition
+
+\begin{theorem}
+Let $\C$ be the set of [[Definition:Complex Number|complex numbers]].
+The [[Definition:Identity Element|identity element]] of $\struct {\C, +}$ is the [[Definition:Complex Number|complex number]] $0 + 0 i$.
+\end{theorem}
+
+\begin{proof}
+We have:
+:$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$
+:$\paren {0 + 0 i} + \paren {x + i y} = \paren {0 + x} + i \paren {0 + y} = x + i y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Addition Identity is Zero}
+Tags: Rational Addition
+
+\begin{theorem}
+The [[Definition:Identity Element|identity]] of [[Definition:Rational Addition|rational number addition]] is $0$:
+:$\exists 0 \in \Q: \forall a \in \Q: a + 0 = a = 0 + a$
+\end{theorem}
+
+\begin{proof}
+From the definition, the [[Definition:Field (Abstract Algebra)|field]] $\struct {\Q, +, \times}$ of [[Definition:Rational Number|rational numbers]] is the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+From [[Zero of Inverse Completion of Integral Domain]], for any $k \in \Z^*$, the element $\dfrac {0_D} k$ of $\Q$ serves as the [[Definition:Field Zero|zero]] of $\struct {\Q, +, \times}$.
+Hence $\dfrac 0 k$ is the [[Definition:Identity Element|identity]] for $\left({\Q, +}\right)$:
+{{begin-eqn}}
+{{eqn | l = \frac a b + \frac 0 k
+      | r = \frac {a \times k + b \times 0} {b \times k}
+      | c = 
+}}
+{{eqn | r = \frac {a \times k} {b \times k}
+      | c = 
+}}
+{{eqn | r = \frac a b
+      | c = [[Equal Elements of Field of Quotients]]
+}}
+{{end-eqn}}
+Similarly for $\dfrac 0 k + \dfrac a b$.
+Next we note that it ''is'' a [[Definition:Field Zero|zero]]:
+{{begin-eqn}}
+{{eqn | l = \frac a b \times \frac 0 k
+      | r = \frac {a \times 0} {b \times k}
+      | c = 
+}}
+{{eqn | r = \frac 0 {b \times k}
+      | c = 
+}}
+{{eqn | r = \frac {0 \times b} {k \times b}
+      | c = 
+}}
+{{eqn | r = \frac 0 k
+      | c = [[Equal Elements of Field of Quotients]]
+}}
+{{end-eqn}}
+Hence we ''define'' the [[Definition:Field Zero|zero]] of $\struct {\Q, +, \times}$ as $0$ and identify it with the set of all elements of $\Q$ of the form $\dfrac 0 k$ where $k \in \Z^*$.
+{{qed}}
+[[Category:Rational Addition]]
+7xbsf9voemuqibzxl3t8s0tzo1ramkn
+\end{proof}<|endoftext|>
+\section{Inverse for Complex Addition}
+Tags: Complex Addition
+
+\begin{theorem}
+Let $z = x + i y \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $-z = -x - i y \in \C$ be the [[Definition:Negative of Complex Number|negative]] of $z$.
+Then $-z$ is the [[Definition:Inverse Element|inverse element]] of $z$ under the operation of [[Definition:Complex Addition|complex addition]]:
+:$\forall z \in \C: \exists -z \in \C: z + \paren {-z} = 0 = \paren {-z} + z$
+\end{theorem}
+
+\begin{proof}
+From [[Complex Addition Identity is Zero]], the [[Definition:Identity Element|identity element]] for $\struct {\C, +}$ is $0 + 0 i$.
+Then:
+{{begin-eqn}}
+{{eqn | o =  
+      | r = \paren {x + i y} + \paren {-x - i y}
+      | c = 
+}}
+{{eqn | r = \paren {x - x} + i \paren {y - y}
+      | c = 
+}}
+{{eqn | r = 0 + 0 i
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\paren {-x - i y} + \paren {x + i y}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equal Elements of Field of Quotients}
+Tags: Integral Domains, Inverse Completions, Fields of Quotients
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+Let $\struct {K, +, \circ}$ be the [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$.
+Let $x = \dfrac p q \in K$.
+Then:
+:$\forall k \in D^*: x = \dfrac {p \circ k} {q \circ k}$
+where:
+:$D^* := D \setminus \set {0_D}$
+that is, $D$ with its [[Definition:Ring Zero|zero]] removed.
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Field of Quotients|field of quotients]] $\struct {K, +, \circ}$ of an [[Definition:Integral Domain|integral domain]] is its [[Inverse Completion of Integral Domain Exists|inverse completion]].
+Thus we have:
+:$\forall x_1, x_2 \in D, y_1, y_2 \in D^*: \dfrac {x_1} {y_1} = \dfrac {x_2} {y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
+So:
+{{begin-eqn}}
+{{eqn | l = \paren {p \circ k} \circ q
+      | r = p \circ \paren {q \circ k}
+      | c = as $D$ is an [[Definition:Integral Domain|integral domain]]: $\circ$ is [[Definition:Commutative Operation|commutative]] and [[Definition:Associative|associative]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac p q
+      | r = \frac {p \circ k} {q \circ k}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+Note that in order for $\dfrac {p \circ k} {q \circ k}$ to be defined, $q \circ k \ne 0_D$, that is, $k \ne 0_D$.
+{{qed}}
+[[Category:Integral Domains]]
+[[Category:Inverse Completions]]
+[[Category:Fields of Quotients]]
+2vqbfdk1ql793ab95ow031s3pvw1stq
+\end{proof}<|endoftext|>
+\section{Rational Addition is Closed}
+Tags: Rational Addition, Algebraic Closure
+
+\begin{theorem}
+The operation of [[Definition:Rational Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ is [[Definition:Well-Defined Operation|well-defined]] and [[Definition:Closed Algebraic Structure|closed]]:
+:$\forall x, y \in \Q: x + y \in \Q$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the [[Definition:Rational Number|definition of rational numbers]] as the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+So $\struct {\Q, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]], and therefore [[Definition:A Priori|a priori]] $+$ is [[Definition:Well-Defined Operation|well-defined]] and [[Definition:Closed Algebraic Structure|closed]] on $\Q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverses for Rational Addition}
+Tags: Rational Addition, Inverse Elements
+
+\begin{theorem}
+Each element $x$ of the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ has an [[Definition:Inverse Element|inverse element]] $-x$ under the operation of [[Definition:Rational Addition|rational number addition]]:
+:$\forall x \in \Q: \exists -x \in \Q: x + \paren {-x} = 0 = \paren {-x} + x$
+\end{theorem}
+
+\begin{proof}
+Let $x = \dfrac a b$ where $b \ne 0$.
+We take the [[Definition:Rational Number|definition of rational numbers]] as the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+From [[Existence of Field of Quotients]], we have that the [[Definition:Inverse Element|inverse]] of $\dfrac a b$ for $+$ is $\dfrac {-a} b$:
+{{begin-eqn}}
+{{eqn | l = \frac a b + \frac {-a} b
+      | r = \frac {a \times b + b \times \paren {-a} } {b \times b}
+      | c = 
+}}
+{{eqn | r = \frac {b \times \paren {a + \paren {-a} } } {b \times b}
+      | c = 
+}}
+{{eqn | r = \frac {b \times 0} {b \times b}
+      | c = 
+}}
+{{eqn | r = \frac 0 {b \times b}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = [[Rational Addition Identity is Zero]]
+}}
+{{end-eqn}}
+From [[Negative of Division Product]], we have that:
+:$\displaystyle -\frac a b = \frac {-a} b = \frac a {-b}$
+So $\dfrac a b$ has a unique and unambiguous [[Definition:Inverse Element|inverse]] for $+$.
+{{qed}}
+[[Category:Rational Addition]]
+[[Category:Inverse Elements]]
+beycyloc3bpkkhd2qgufhmeb4pfzacf
+\end{proof}<|endoftext|>
+\section{Rational Multiplication is Closed}
+Tags: Rational Multiplication, Algebraic Closure, Rational Multiplication is Closed
+
+\begin{theorem}
+The operation of [[Definition:Rational Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ is [[Definition:Well-Defined Operation|well-defined]] and [[Definition:Closed Algebraic Structure|closed]]:
+:$\forall x, y \in \Q: x \times y \in \Q$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the [[Definition:Rational Number|definition of rational numbers]] as the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+So $\struct {\Q, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]], and therefore [[Definition:A Priori|a priori]] $\times$ is [[Definition:Well-Defined Operation|well-defined]] and [[Definition:Closed Algebraic Structure|closed]] on $\Q$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the definition of [[Definition:Rational Number|rational numbers]], there exists four [[Definition:Integer|integers]] $p$, $q$, $r$, $s$, where:
+:$\displaystyle q \ne 0$
+:$\displaystyle s \ne 0$
+:$\displaystyle \frac p q = x$
+:$\displaystyle \frac r s = y$
+We have that:
+:$\displaystyle p \times r \in \Z$
+:$\displaystyle q \times s \in \Z$
+Since $q \ne 0$ and $s \ne 0$, we have that $q \times s \ne 0$.
+Therefore, by the definition of [[Definition:Rational Number|rational numbers]]:
+:$\displaystyle x \times y = \frac {p \times r} {q \times s} \in \Q$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Multiplication Identity is One}
+Tags: Rational Multiplication
+
+\begin{theorem}
+The [[Definition:Identity Element|identity]] of [[Definition:Rational Multiplication|rational number multiplication]] is $1$:
+:$\exists 1 \in \Q: \forall a \in \Q: a \times 1 = a = 1 \times a$
+\end{theorem}
+
+\begin{proof}
+From the definition, the [[Definition:Field (Abstract Algebra)|field]] $\struct {\Q, +, \times}$ of [[Definition:Rational Number|rational numbers]] is the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+From [[Construction of Inverse Completion/Identity of Quotient Structure|the properties of the quotient structure]], elements of $\Q$ of the form $\dfrac p p$ where $p \ne 0$ act as the [[Definition:Identity Element|identity]] for [[Definition:Rational Multiplication|multiplication]].
+From [[Equal Elements of Field of Quotients]], we have that:
+:$\displaystyle \frac p p = \frac {1 \times p} {1 \times p} = \frac 1 1$
+Hence $\dfrac p p$ is the [[Definition:Identity Element|identity]] for $\struct {\Q, \times}$:
+{{begin-eqn}}
+{{eqn | l = \frac a b \times \frac p p
+      | r = \frac {a \times p} {b \times p}
+      | c = 
+}}
+{{eqn | r = \frac a b
+      | c = [[Equal Elements of Field of Quotients]]
+}}
+{{end-eqn}}
+Similarly for $\dfrac p p \times \dfrac a b$.
+Hence we ''define'' the [[Definition:Unity of Field|unity]] of $\struct {\Q, +, \times}$ as $1$ and identify it with the set of all elements of $\Q$ of the form $\dfrac p p$ where $ \in \Z^*$.
+{{qed}}
+[[Category:Rational Multiplication]]
+biysaxcjqkesm7cn7xak6e3e1dz6nqb
+\end{proof}<|endoftext|>
+\section{Inverses for Rational Multiplication}
+Tags: Rational Multiplication, Inverse Elements
+
+\begin{theorem}
+Each element $x$ of the [[Definition:Set|set]] of [[Definition:Rational Number|non-zero rational numbers]] $\Q^*$ has an [[Definition:Inverse Element|inverse element]] $\dfrac 1 x$ under the operation of [[Definition:Rational Multiplication|rational number multiplication]]:
+:$\displaystyle \forall x \in \Q^*: \exists \frac 1 x \in \Q^*: x \times \frac 1 x = 1 = \frac 1 x \times x$
+\end{theorem}
+
+\begin{proof}
+From the definition, the [[Definition:Field (Abstract Algebra)|field]] $\struct {\Q, +, \times}$ of [[Definition:Rational Number|rational numbers]] is the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+From [[Rational Multiplication Identity is One]], the [[Definition:Identity Element|identity]] for $\struct {\Q, \times}$ is $1 = \dfrac 1 1 = \dfrac p p$ where $p \in \Z$ and $p \ne 0$.
+From the definition of [[Existence of Field of Quotients#Product Inverses in K|product inverses in field of quotients]], we have that:
+:$\paren {\dfrac p q}^{-1} = \dfrac q p$
+which can be demonstrated directly:
+{{begin-eqn}}
+{{eqn | l = \frac p q \times \frac q p
+      | r = \frac {p \times q} {q \times p}
+      | c = 
+}}
+{{eqn | r = \frac {p \times q} {p \times q}
+      | c = 
+}}
+{{eqn | r = \frac 1 1
+      | c = [[Equal Elements of Field of Quotients]]
+}}
+{{end-eqn}}
+Note that this is defined only when $p \ne 0$ and $q \ne 0$.
+Now let $x = \dfrac p q \in \Q$.
+We define $\dfrac 1 x$ as $1 \times \paren {\dfrac p q}^{-1}$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \frac 1 x
+      | r = \frac 1 1 \times \paren {\frac p q}^{-1}
+      | c = 
+}}
+{{eqn | r = \frac 1 1 \times \frac q p
+      | c = 
+}}
+{{eqn | r = \frac {1 \times q} {1 \times p}
+      | c = 
+}}
+{{eqn | r = \frac q p
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Rational Multiplication]]
+[[Category:Inverse Elements]]
+7hiyuzjr3bjoa27wg3xduwqvvmxanph
+\end{proof}<|endoftext|>
+\section{Real Addition Identity is Zero}
+Tags: Additive Group of Real Numbers
+
+\begin{theorem}
+The [[Definition:Identity Element|identity]] of [[Definition:Real Addition|real number addition]] is $0$:
+:$\exists 0 \in \R: \forall x \in \R: x + 0 = x = 0 + x$
+\end{theorem}
+
+\begin{proof}
+From the definition, the [[Definition:Real Number|real numbers]] are the set of all [[Definition:Equivalence Class|equivalence classes]] $\eqclass {\sequence {x_n} } {}$ of [[Definition:Cauchy Sequence|Cauchy sequences]] of [[Definition:Rational Number|rational numbers]].
+Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such [[Definition:Equivalence Class|equivalence classes]].
+From the definition of [[Definition:Real Addition|real addition]], $x + y$ is defined as $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$.
+Let $\sequence {0_n}$ be such that $\forall i: 0_n = 0$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {\sequence {0_n} } {} + \eqclass {\sequence {x_n} } {}
+      | r = \eqclass {\sequence {0_n + x_n} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\sequence {0 + x_n} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\sequence {x_n} } {}
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {0_n} } {}$.
+Thus the [[Definition:Identity Element|identity element]] of $\struct {\R, +}$ is the [[Definition:Real Number|real number]] $0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverses for Real Addition}
+Tags: Real Addition, Inverse Elements
+
+\begin{theorem}
+Each element $x$ of the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ has an [[Definition:Inverse Element|inverse element]] $-x$ under the operation of [[Definition:Real Addition|real number addition]]:
+:$\forall x \in \R: \exists -x \in \R: x + \paren {-x} = 0 = \paren {-x} + x$
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {\sequence {x_n} } {} + \paren {-\eqclass {\sequence {x_n} } {} }
+      | r = \eqclass {\sequence {x_n - x_n} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\sequence {0_n} } {}
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\paren {-\eqclass {\sequence {x_n} } {} } + \eqclass {\sequence {x_n} } {}$.
+Thus the inverse of $x \in \struct {\R, +}$ is $-x$.
+{{qed}}
+[[Category:Real Addition]]
+[[Category:Inverse Elements]]
+ksgam98thmvevfig81r8xqlyjtt6xt7
+\end{proof}<|endoftext|>
+\section{Real Multiplication Identity is One}
+Tags: Multiplicative Group of Real Numbers
+
+\begin{theorem}
+The [[Definition:Identity Element|identity element]] of [[Definition:Real Multiplication|real number multiplication]] is the real number $1$:
+:$\exists 1 \in \R: \forall a \in \R_{\ne 0}: a \times 1 = a = 1 \times a$
+\end{theorem}
+
+\begin{proof}
+From the definition, the [[Definition:Real Number|real numbers]] are the set of all [[Definition:Equivalence Class|equivalence classes]] $\eqclass {\sequence {x_n} } {}$ of [[Definition:Cauchy Sequence|Cauchy sequences]] of [[Definition:Rational Number|rational numbers]].
+Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such [[Definition:Equivalence Class|equivalence classes]].
+From the definition of [[Definition:Real Multiplication|real multiplication]], $x \times y$ is defined as:
+:$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$
+Let $\sequence {1_n}$ be such that $\forall i: 1_n = 1$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {\sequence {1_n} } {} \times \eqclass {\sequence {x_n} } {}
+      | r = \eqclass {\sequence {1_n \times x_n} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\sequence {1 \times x_n} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\sequence {x_n} } {}
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {1_n} } {}$.
+So the [[Definition:Identity Element|identity element]] of $\struct {\R_{\ne 0}, \times}$ is the [[Definition:Real Number|real number]] $1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverses for Real Multiplication}
+Tags: Real Multiplication, Inverse Elements
+
+\begin{theorem}
+Each element $x$ of the [[Definition:Set|set]] of [[Definition:Real Number|non-zero real numbers]] $\R_{\ne 0}$ has an [[Definition:Inverse Element|inverse element]] $\dfrac 1 x$ under the operation of [[Definition:Real Multiplication|real number multiplication]]:
+:$\forall x \in \R_{\ne 0}: \exists \dfrac 1 x \in \R_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Real Number|real number]]:
+:$\forall \epsilon > 0 : \exists t \in \N : \forall i > t: \size {x_i - x} < \epsilon$
+{{explain|In order for this line to make sense, it needs the definition (or at least the relevant part) extracted and posted here, so the context of the convergence condition is clear.}}
+Let $\epsilon = \size x$. This is possible because $x \ne 0$, since it is required that $\epsilon > 0$.
+{{begin-eqn}}
+{{eqn | ll= \exists t \in \N : \forall i > t :
+      | l = \size {x_i - x}
+      | o = <
+      | r = \size x
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | ll= \exists t \in \N : \forall i > t :
+      | l = \size {x_i - x}
+      | o = <
+      | r = \size {x_i - x} + \size {x_i}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | lll=\leadsto
+      | ll= \exists t \in \N : \forall i > t :
+      | l = \size {x_i}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Construct a sequence $\sequence {y_n}$ as follows:
+:$y_n = \begin {cases} \dfrac 1 {x_n} & n > t \\ 0 & n \le t \end {cases}$
+Let $u_n$ denote the [[Definition:Heaviside Step Function|Heaviside step function]] with parameter $t$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}
+      | r = \eqclass {\sequence {x_n y_n} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\sequence {u_n} } {}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}$.
+So the [[Definition:Inverse Element|inverse]] of $x \in \struct {\R_{\ne 0}, \times}$ is $x^{-1} = \dfrac 1 x = y$.
+{{qed}}
+[[Category:Real Multiplication]]
+[[Category:Inverse Elements]]
+5yxnk8q1yhk8t3wadof1vjkb4mirq68
+\end{proof}<|endoftext|>
+\section{Complex Multiplication Identity is One}
+Tags: Complex Multiplication
+
+\begin{theorem}
+Let $\C_{\ne 0}$ be the set of [[Definition:Complex Number|complex numbers]] without [[Definition:Zero (Number)|zero]].
+The [[Definition:Identity Element|identity element]] of $\struct {\C_{\ne 0}, \times}$ is the [[Definition:Complex Number|complex number]] $1 + 0 i$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {x + i y} \paren {1 + 0 i}
+      | r = \paren {x \cdot 1 - y \cdot 0} + i \paren {x \cdot 0 + y \cdot 1}
+      | c = 
+}}
+{{eqn | r = \paren {x + i y}
+      | c = 
+}}
+{{end-eqn}}
+and similarly:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + 0 i} \paren {x + i y}
+      | r = \paren {1 \cdot x - 0 \cdot y} + i \paren {0 \cdot x + 1 \cdot y}
+      | c = 
+}}
+{{eqn | r = \paren {x + i y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse for Complex Multiplication}
+Tags: Complex Multiplication
+
+\begin{theorem}
+Each element $z = x + i y$ of the [[Definition:Set|set]] of [[Definition:Complex Number|non-zero complex numbers]] $\C_{\ne 0}$ has an [[Definition:Inverse Element|inverse element]] $\dfrac 1 z$ under the operation of [[Definition:Complex Multiplication|complex multiplication]]:
+:$\forall z \in \C_{\ne 0}: \exists \dfrac 1 z \in \C_{\ne 0}: z \times \dfrac 1 z = 1 + 0 i = \dfrac 1 z \times z$
+This inverse can be expressed as:
+:$\dfrac 1 z = \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$
+where $\overline z$ is the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {x + i y} \frac {x - i y} {x^2 + y^2}
+      | r = \frac {\paren {x \cdot x - y \cdot \paren {-y} } + i \paren {x \cdot \paren {-y} + x \cdot y} } {x^2 + y^2}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {x^2 + y^2} + 0 i} {x^2 + y^2}
+      | c = 
+}}
+{{eqn | r = 1 + 0 i
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\dfrac {x - i y} {x^2 + y^2} \paren {x + i y}$.
+So the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is $\dfrac {x - i y} {x^2 + y^2}$.
+As $x^2 + y^2 > 0 \iff x, y \ne 0$ the inverse is defined for all $z \in \C: z \ne 0 + 0 i$.
+{{qed|lemma}}
+From the definition, the [[Definition:Complex Conjugate|complex conjugate]] $\overline z$ of $z = x + i y$ is $x - i y$.
+From the definition of the [[Definition:Modulus of Complex Number|modulus of a complex number]], we have:
+:$\cmod z = \sqrt {a^2 + b^2}$
+From [[Modulus in Terms of Conjugate]], we have that:
+:$\cmod z^2 = z \overline z$
+Hence the result:
+:$\dfrac 1 z = \dfrac {\overline z} {z \overline z}$
+{{qed}}
+{{expand|Add the polar and exponential forms of this}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Integers is Normal Subgroup of Complex}
+Tags: Additive Group of Integers, Additive Group of Complex Numbers, Examples of Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
+Then $\struct {\Z, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\C, +}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Integers is Normal Subgroup of Reals]], $\struct {\Z, +} \lhd \struct {\R, +}$.
+From [[Additive Group of Reals is Subgroup of Complex]], $\struct {\R, +} \lhd \struct {\C, +}$.
+Thus $\struct {\Z, +} \le \struct {\C, +}$.
+From [[Complex Numbers under Addition form Abelian Group]], $\struct {\C, +}$ is [[Definition:Abelian Group|abelian]].
+From [[Subgroup of Abelian Group is Normal]], it follows that $\struct {\Z, +} \lhd \struct {\C, +}$.
+{{qed}}
+[[Category:Additive Group of Integers]]
+[[Category:Additive Group of Complex Numbers]]
+[[Category:Examples of Normal Subgroups]]
+ikeecuhel3tpbsr0yhe0vfstw909fqw
+\end{proof}<|endoftext|>
+\section{Additive Group of Rationals is Subgroup of Complex}
+Tags: Additive Group of Rational Numbers, Additive Group of Complex Numbers, Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\Q, +}$ be the [[Definition:Additive Group of Rational Numbers|additive group of rational numbers]].
+Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
+Then $\struct {\Q, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\C, +}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Rationals is Subgroup of Reals]], $\struct {\Q, +} \lhd \struct {\R, +}$.
+From [[Additive Group of Reals is Subgroup of Complex]], $\struct {\R, +} \lhd \struct {\C, +}$.
+Thus $\struct {\Q, +} \le \struct {\C, +}$.
+From [[Complex Numbers under Addition form Abelian Group]], $\struct {\C, +}$ is [[Definition:Abelian Group|abelian]].
+From [[Subgroup of Abelian Group is Normal]], it follows that $\struct {\Q, +} \lhd \struct {\C, +}$.
+{{qed}}
+[[Category:Additive Group of Rational Numbers]]
+[[Category:Additive Group of Complex Numbers]]
+[[Category:Normal Subgroups]]
+4boyza8yljy6ipg9wn24xyz3fvfxgft
+\end{proof}<|endoftext|>
+\section{Multiplicative Group of Rationals is Subgroup of Complex}
+Tags: Multiplicative Group of Rational Numbers, Multiplicative Group of Complex Numbers, Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\Q, \times}$ be the [[Definition:Multiplicative Group of Rational Numbers|multiplicative group of rational numbers]].
+Let $\struct {\C, \times}$ be the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]].
+Then $\struct {\Q, \times}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\C, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Multiplicative Group of Rationals is Subgroup of Reals]], $\struct {\Q, \times} \lhd \struct {\R, \times}$.
+From [[Multiplicative Group of Reals is Subgroup of Complex]], $\struct {\R, \times} \lhd \left({\C, \times}\right)$.
+Thus $\struct {\Q, \times} \le \struct {\C, \times}$.
+From [[Non-Zero Complex Numbers under Multiplication form Abelian Group]], $\struct {\C, \times}$ is [[Definition:Abelian Group|abelian]].
+From [[Subgroup of Abelian Group is Normal]] it follows that $\struct {\Q, \times} \lhd \struct {\C, \times}$.
+{{qed}}
+[[Category:Multiplicative Group of Rational Numbers]]
+[[Category:Multiplicative Group of Complex Numbers]]
+[[Category:Normal Subgroups]]
+r2dngx9z4talkqh8mt674djbe9q9wyu
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Subfield of Complex Numbers}
+Tags: Rational Numbers, Complex Numbers, Subfields
+
+\begin{theorem}
+Let $\struct {\Q, +, \times}$ denote the [[Definition:Field of Rational Numbers|field of rational numbers]].
+Let $\struct {\C, +, \times}$ denote the [[Definition:Field of Complex Numbers|field of complex numbers]].
+$\struct {\Q, +, \times}$ is a [[Definition:Subfield|subfield]] of $\struct {\C, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Rational Numbers form Subfield of Real Numbers]], $\struct {\Q, +, \times}$ is a [[Definition:Subfield|subfield]] of $\struct {\R, + \times}$.
+From [[Real Numbers form Subfield of Complex Numbers]], $\struct {\R, +, \times}$ is a [[Definition:Subfield|subfield]] of $\struct {\C, + \times}$.
+Thus from [[Subfield of Subfield is Subfield]] $\struct {\Q, +, \times}$ is a [[Definition:Subfield|subfield]] of $\struct {\C, + \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subfield of Subfield is Subfield}
+Tags: Subfields
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $K_1, K_2$ be [[Definition:Field (Abstract Algebra)|fields]], such that:
+: $K_1$ is a [[Definition:Subfield|subfield]] of $R$
+: $K_2$ is a [[Definition:Subfield|subfield]] of $K_1$
+Then $K_2$ is a [[Definition:Subfield|subfield]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $K_1$ be a [[Definition:Subfield|subfield]] of $R$ and $K_2$ be a [[Definition:Subfield|subfield]] of $K_1$.
+Then by definition:
+:$K_1 \subseteq R$
+:$K_2 \subseteq K_1$
+From [[Subset Relation is Transitive]] it follows that $K_2 \subseteq R$
+So by definition $K_2$ is a [[Definition:Subfield|subfield]] of $R$.
+{{qed}}
+[[Category:Subfields]]
+g8qj1gxkqsgvlury7jpgaahfpiooiz9
+\end{proof}<|endoftext|>
+\section{Characterisation of Jacobson Radical}
+Tags: Commutative Algebra
+
+\begin{theorem}
+Let $A$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $A^\times$ be the [[Definition:Group of Units of Ring|group of units]] of $A$.
+Let $\map {\operatorname {Jac} } A$ be the [[Definition:Jacobson Radical|Jacobson radical]] of $A$.
+Then:
+:$\map {\operatorname {Jac} } A = \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$
+where $1_A$ is the [[Definition:Unity of Ring|unity]] of $A$.
+\end{theorem}
+
+\begin{proof}
+Recall the definition of [[Definition:Jacobson Radical|Jacobson radical]] of $A$:
+:$\map {\operatorname {Jac} } A$ is the [[Definition:Set Intersection|intersection]] of all [[Definition:Maximal Ideal of Ring|maximal ideals]] of $R$.
+=== $\subset$ Direction ===
+{{AimForCont}} that $x \in \map {\operatorname {Jac} } A$ such that there exists $y \in A$ such that $1_A - x y \notin A^\times$. 
+By definition $x$ is contained in all [[Definition:Maximal Ideal of Ring|maximal ideals]] of $A$.
+Let $\mathfrak m \subseteq A$ be one particular such [[Definition:Maximal Ideal of Ring|maximal ideal]] of $A$.
+Then $x \in \map {\operatorname {Jac} } A \subseteq \mathfrak m$ implies that $xy \in \mathfrak m$ and therefore $1_A \in \mathfrak m$.
+But if $1_A \in \mathfrak m$ then from [[Ideal of Unit is Whole Ring/Corollary|Ideal of Unit is Whole Ring: Corollary]]:
+:$\mathfrak m = R$
+This [[Definition:Contradiction|contradicts]] the definition of $\mathfrak m$ as a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $A$.
+Hence by [[Proof by Contradiction]]:
+:$\map {\operatorname {Jac} } A \subseteq \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$
+{{qed|lemma}}
+=== $\supset$ Direction ===
+Now suppose that $x \notin \map {\operatorname {Jac} } A$.
+Then $x \notin \mathfrak m$ for some [[Definition:Maximal Ideal of Ring|maximal ideal]] $\mathfrak m$ of $A$.
+Because $\mathfrak m$ is [[Definition:Maximal Ideal of Ring|maximal]], $x$ and $\mathfrak m$ [[Definition:Generator of Ring|generate]] $A$.
+Therefore there exist $w \in \mathfrak m$ and $y \in A$ such that $w + x y = 1_A$.
+Thus $1_A - x y \in \mathfrak m$, and $1_A - x y \notin A^\times$.
+Hence:
+:$\map {\operatorname {Jac} } A \supseteq \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$
+{{qed|lemma}}
+The result follows by definition of [[Definition:Set Equality|set equality]].
+{{qed}}
+[[Category:Commutative Algebra]]
+sr2jx31q78xi1evsr3ym5z0zl65mp6a
+\end{proof}<|endoftext|>
+\section{Prime Ideal iff Quotient Ring is Integral Domain}
+Tags: Quotient Rings, Prime Ideals of Rings, Integral Domains
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Then $J$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $R$ {{iff}} the [[Definition:Quotient Ring|quotient ring]] $R / J$ is an [[Definition:Integral Domain|integral domain]].
+\end{theorem}
+
+\begin{proof}
+Since $J \subset R$, it follows from:
+:[[Quotient Ring of Commutative Ring is Commutative]]
+and:
+:[[Quotient Ring of Ring with Unity is Ring with Unity]]
+that $R / J$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]]. 
+Let $0_{R / J}$ be the [[Definition:Ring Zero|zero]] of $R / J$.
+=== Sufficient Condition ===
+Let $J$ be a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]].
+We need to show that if:
+:$x + J, \ y + J \in \struct {R / J, +, \circ}$
+such that:
+:$\paren {x + J} \circ \paren {y + J} = \paren {x \circ y} + J = 0_{R / J}$
+then:
+:$x + J = 0_{R/J}$ or $y + J = 0_{R/J}$.
+The [[Definition:Ring Zero|zero]] of $R / J$ is $0_{R / J} = J$.
+Therefore:
+:$\paren {x \circ y} + J = 0_{R / J} \implies x \circ y \in J$
+Because $J$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]], it follows that either $x \in J$ or $y \in J$.
+{{WLOG}} we assume that $x \in J$.
+But then:
+:$x + J = J = 0_{R / J}$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $R / J$ be an [[Definition:Integral Domain|integral domain]].
+Let $x, y \in R$ be such that $x \circ y \in J$.
+Then:
+:$0_{R / J} = J = x \circ y + J = \paren {x + J} \circ \paren {y + J}$
+Because $R / J$ is an [[Definition:Integral Domain|integral domain]] it follows that:
+:$x + J = 0_{R / J}$
+or:
+:$y + J = 0_{R / J}$
+If $x + J = 0_{R / J}$, then $x \in J$.
+If $y + J = 0_{R / J}$, then $y \in J$.
+Thus either $x \in J$ or $y \in J$, and so by definition $J$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximal Ideal of Commutative and Unitary Ring is Prime Ideal}
+Tags: Maximal Ideals of Rings, Prime Ideals of Rings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
+Let $M$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $R$.
+Then $M$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $R$.
+\end{theorem}
+
+\begin{proof}
+From [[Maximal Ideal iff Quotient Ring is Field]]:
+:the [[Definition:Quotient Ring|quotient ring]] $R / M$ is a [[Definition:Field (Abstract Algebra)|field]].
+It follows from [[Field is Integral Domain]] that $R / M$ is an [[Definition:Integral Domain|integral domain]].
+Finally it follows from [[Prime Ideal iff Quotient Ring is Integral Domain]] that $M$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Radical Ideal iff Quotient Ring is Reduced}
+Tags: Quotient Rings
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Then $J$ is a [[Definition:Radical Ideal of Ring|radical ideal]] {{iff}} the [[Definition:Quotient Ring|quotient ring]] $R / J$ is a [[Definition:Reduced Ring|reduced ring]].
+\end{theorem}
+
+\begin{proof}
+Since $J \subset R$, it follows from:
+:[[Quotient Ring of Commutative Ring is Commutative]]
+and:
+: [[Quotient Ring of Ring with Unity is Ring with Unity]]
+that $R / J$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $0_{R / J}$ be the [[Definition:Ring Zero|zero]] of $R / J$.
+=== Sufficient Condition ===
+Let $J$ be [[Definition:Radical Ideal|radical]].
+We need to show that:
+:if $x + J \in R / J$ such that $\left({x + J}\right)^n = 0_{R / J}$ for some [[Definition:Positive Integer|positive integer]] $n$
+:then $x + J = 0_{R / J}$.
+Let $\left({x + J}\right)^n = 0_{R / J}$.
+Then:
+:$x^n + J = 0_{R / J}$
+and therefore:
+:$x^n \in J$
+Because $J$ is [[Definition:Radical Ideal|radical]] it follows that:
+:$x \in J$
+That is:
+:$x + J = 0_{R / J}$
+Therefore $R / J$ is [[Definition:Reduced Ring|reduced]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $R / J$ be [[Definition:Reduced Ring|reduced]].
+Let $x \in R$ be such that $x^n \in J$.
+Then:
+:$0_{R / J} = J = x^n + J = \left({x + J}\right)^n$
+Because $A / J$ is [[Definition:Reduced Ring|reduced]], this implies that:
+:$x + J = 0_{R / J}$
+That is:
+:$x \in J$
+This shows that $J$ is [[Definition:Radical Ideal|radical]].
+{{qed}}
+[[Category:Quotient Rings]]
+c8zunqlot5mf4vpehyl9wn5288fmfge
+\end{proof}<|endoftext|>
+\section{Properties of Degree}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $R \sqbrk X$ denote the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over $R$ in the indeterminate $X$.
+For $f \in R \sqbrk X$ let $\map \deg f$ denote the [[Definition:Degree (Polynomial)|degree]] of $f$.
+Then the following hold:
+\end{theorem}<|endoftext|>
+\section{Universal Property of Polynomial Ring/Free Monoid on Set}
+Tags: Polynomial Theory, Universal Properties
+
+\begin{theorem}
+Let $R, S$ be [[Definition:Commutative and Unitary Ring|commutative and unitary rings]].
+Let $\left\langle{s_j}\right\rangle_{j \mathop \in J}$ be an [[Definition:Indexed Family|indexed family]] of elements of $S$.
+Let $\psi: R \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Let $R \left[{\left\{{X_j: j \in J}\right\}}\right]$ be a [[Definition:Polynomial Ring|polynomial ring]].
+Then there exists a unique [[Definition:Polynomial Evaluation Homomorphism|evaluation homomorphism]] $\phi: R \left[{\left\{{X_j: j \in J}\right\}}\right] \to S$ at $\langle s_j\rangle_{j \in J}$ [[Definition:Extension of Mapping|extending]] $\psi$.
+\end{theorem}
+
+\begin{proof}
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $J$.
+Let $k_j$ be the $j$th component of a [[Definition:Multiindex|multiindex]] $k$.
+Let $\displaystyle f = \sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j}$ be a [[Definition:Polynomial (Abstract Algebra)|polynomial]] over $R$.
+Define:
+:$\displaystyle \phi \left({f}\right) = \sum_{k \mathop \in Z} \psi \left({a_k}\right) \prod_{j \mathop \in J}s_j^{k_j}$
+It is clear that $\phi$ extends $\psi$.
+If $\displaystyle g = \sum_{k \mathop \in Z} b_k \prod_{j \mathop \in J} X_j^{k_j}$, then:
+{{begin-eqn}}
+{{eqn | l = \phi \left({f + g}\right)
+      | r = \phi \left({\sum_{k \mathop \in Z} \left({a_k + b_k}\right) \prod_{j \mathop \in J} X_j^{k_j} }\right)
+      | c = {{Defof|Addition of Polynomial Forms}}
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \left({\psi \left({a_k + b_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \left({\psi \left({a_k}\right) + \psi \left({b_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j}
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | r =\sum_{k \mathop \in Z} \psi \left({a_k}\right) \prod_{j \mathop \in J} s_j^{k_j} + \sum_{k \mathop \in Z} \psi \left({b_k}\right) \prod_{j \mathop \in J} s_j^{k_j}
+      | c = [[Definition:Ring Axioms|Ring Axioms of $S$]]
+}} 
+{{eqn | r =\phi \left({f}\right) + \phi \left({g}\right)
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+Therefore $\phi$ preserves addition.
+Also:
+{{begin-eqn}}
+{{eqn | l = \phi \left({f g}\right)
+      | r = \phi \left({\sum_{k \mathop \in Z} \left({\sum_{p \mathop + q \mathop = k} a_p b_q}\right) \prod_{j \mathop \in J} X_j^{k_j} }\right)
+      | c = {{Defof|Multiplication of Polynomial Forms}}
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \psi \left({\sum_{p \mathop + q \mathop = k} a_p b_q}\right) \prod_{j \mathop \in J} s_j^{k_j}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \left({\sum_{p \mathop + q \mathop = k} \psi \left({a_p}\right) \psi \left({b_q}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j}
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | r = \left({\sum_{k \mathop \in Z} \left({\psi \left({a_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j} }\right) \left({\sum_{k \mathop \in Z} \left({\psi \left({b_k}\right)}\right) \prod_{j \mathop \in J} s_j^{k_j} }\right)
+      | c = [[Definition:Ring Axioms|Ring Axioms of $S$]]
+}}
+{{eqn | r =\phi \left({f}\right) \phi \left({g}\right)
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+This shows that $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]].
+Now suppose that $\phi'$ is another such [[Definition:Ring Homomorphism|homomorphism]].
+For each $j \in J$, $\phi'$ must satisfy $\phi'(X_j)=s_j$ and $\phi' \left({r}\right) = \psi \left({r}\right)$ for all $r \in R$.
+In addition $\phi'$ must be a [[Definition:Ring Homomorphism|homomorphism]], so we compute:
+{{begin-eqn}}
+{{eqn | l = \phi' \left({\sum_{k \mathop \in Z} a_k \prod_{j \mathop \in J} X_j^{k_j} }\right)
+      | r = \sum_{k \mathop \in Z} \phi' \left({a_k \prod_{j \mathop \in J} X_j^{k_j} }\right)
+      | c = $\phi'$ preserves [[Definition:Ring Addition|Ring Addition]]
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \phi' \left({a_k}\right) \prod_{j \mathop \in J} \phi' \left({X_j}\right)^{k_j} 
+      | c = $\phi'$ preserves [[Definition:Ring Product|Ring Product]]
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \psi \left({a_k}\right) \prod_{j \mathop \in J} s_j^{k_j}
+      | c = as $\phi' \left({X_j}\right) = s_j$ and $\phi' \left({r}\right) = \psi \left({r}\right)$ for all $r \in R$
+}}
+{{end-eqn}}
+and therefore $\phi' = \phi$.
+This concludes the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Monomials is Closed Under Multiplication}
+Tags: Monomials
+
+\begin{theorem}
+Let $M$ be the set of all [[Definition:Monomial|monomials]] on the set $\set {X_j: j \in J}$, with multiplication $\circ$ defined by:
+:$\ds \paren {\prod_{j \mathop \in J} X_j^{k_j} } \circ \paren {\prod_{j \mathop \in J} X_j^{k_j'} } = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
+Then $M$ is closed under $\circ$.
+\end{theorem}
+
+\begin{proof}
+Let $\ds m_1 = \prod_{j \mathop \in J} X_j^{k_j}, m_2 = \prod_{j \mathop \in J} X_j^{k_j'}$ be two [[Definition:Monomial|monomials]].
+Their product is:
+:$\ds m_1 \circ m_2 = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
+If $k_j + k_j' \ne 0$ then either $k_j \ne 0$ or $k_j' \ne 0$ (or both are nonzero).
+Therefore if $k_j + k_j' \ne 0$ for infinitely many $j$, then either $m_1$ or $m_2$ is not a [[Definition:Monomial|monomial]].
+{{qed}}
+[[Category:Monomials]]
+h2lb3nxhe53agesh51ic1k8xmovh20p
+\end{proof}<|endoftext|>
+\section{Free Commutative Monoid is Commutative Monoid}
+Tags: Examples of Monoids, Polynomial Theory
+
+\begin{theorem}
+The [[Definition:Free Commutative Monoid|free commutative monoid]] on a set $\family {X_j: j \in J}$ is a [[Definition:Commutative Monoid|commutative monoid]].
+\end{theorem}
+
+\begin{proof}
+Let $M$ be the set of all [[Definition:Monomial|monomials]] on the indexed set $\family {X_j: j \in J}$.
+We are required to show that the following properties hold:
+{{begin-axiom}}
+{{axiom | q = \forall m_1, m_2 \in M
+        | m = m_1 \circ m_2 \in M
+        | t = [[Definition:Closed Algebraic Structure|Closure]]
+}}
+{{axiom | q = \forall m_1, m_2, m)3 \in M
+        | m = \paren {m_1 \circ m_2} \circ m_3 = m_1 \circ \paren {m_2 \circ m_3}
+        | t = [[Definition:Associative|Associativity]]
+}}
+{{axiom | q = \forall m_1, m_2 \in M
+        | m = m_1 \circ m_2 = m_2 \circ m_1
+        | t = [[Definition:Commutative Operation|Commutativity]]
+}}
+{{axiom | q = \exists e_m \in M: \forall m_1 \in M
+        | m = m_1 \circ e_M = m_1
+        | t = [[Definition:Identity Element|Identity]]
+}}
+{{end-axiom}}
+First note that using the [[Definition:Multiindex|multiindex]] notation described in the [[Definition:Monomial|definition of monomials]], for $r \in \N$, $m_i = \mathbf X^{k^i} \in M$, $i = 1, \ldots, r$, the product of the $m_i$ is given by:
+:$m_1 \circ \cdots \circ m_r = \mathbf X^{k^1 + \cdots + k^r}$
+Here the superscripts enumerate the [[Definition:Multiindex|multiindices]], and do not indicate raising to a power.
+So to show the closure, associativity and commutativity of monomials under $\circ$, it is sufficient to show the corresponding properties for multiindices under addition defined by:
+:$\family {k^1 + k^2}_j := k^1_j + k^2_j$
+In the following $k^1, k^2, k^3$ are multiindices, that is, families of non-negative integers indexed by $J$ such that only finitely many entries are non-zero.
+=== Proof of Closure ===
+Let $\family {k^1 + k^2}_j = k^1_j + k^2_j \ne 0$.
+By definition of [[Definition:Multiindex|multiindex]], at least one of $k^1_j$ and $k^2_j$ must be non-zero, and this can only be true for a finite number of entries.
+Furthermore, since $k^1_j,\ k^2_j \ge 0$, we have $k^1_j + k^2_j \ge 0$.
+Therefore $k^1 + k^2$ has finitely many non-zero entries, and these are all positive, and multiindices are [[Definition:Closed Algebraic Structure|closed]] under addition.
+=== Proof of Associativity ===
+Using associativity of integer addition, we have: 
+:$\family {\paren {k^1 + k^2} + k^3}_j = \paren {k^1_j + k^2_j} + k^3_j = k^1_j + \paren {k^2_j + k^3_j} = \family {k^1 + \paren {k^2 + k^3} }_j$
+So addition of multiindices is [[Definition:Associative|associative]].
+=== Proof of Commutativity ===
+Using commutativity of integer addition, we have 
+:$\family {k^1 + k^2}_j = k^1_j + k^2_j = k^2_j + k^1_j = \family {k^2 + k^1}_j$
+So addition of multiindices is [[Definition:Commutative Operation|commutative]].
+=== Proof of Existence of Identity ===
+Let $e_M$ be the multiindex such that $\family {e_M}_j = 0$ for all $j \in J$.
+Then:
+:$\family {e_M + k^1}_j = \family {e_M}_j + k^1_j = \family {k^1}_j$
+so $e_M$ is an [[Definition:Identity Element|identity]] for the set of monomials.
+{{qed}}
+[[Category:Examples of Monoids]]
+[[Category:Polynomial Theory]]
+igkuc9y7o4iylu6pr4yx1afrbjflqrh
+\end{proof}<|endoftext|>
+\section{Polynomial Addition is Associative}
+Tags: Polynomial Theory
+
+\begin{theorem}
+[[Definition:Addition of Polynomial Forms|Addition of polynomials]] is an [[Definition:Associative|associative]] operation.
+\end{theorem}
+
+\begin{proof}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $\left\{{X_j: j \in J}\right\}$ be a set of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
+Let:
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
+:$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
+:$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$ 
+be arbitrary [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j \in J}\right\}$ over $R$.
+Then:
+{{begin-eqn}}
+{{eqn|l = \left({ f + g }\right) + h
+     |r = \sum_{k \mathop \in Z} \left({ \left({ a_k + b_k }\right) + c_k }\right) \mathbf X^k
+     |c = applying the definition of [[Definition:Addition of Polynomial Forms|polynomial addition]] twice
+}}
+{{eqn|r = \sum_{k \mathop \in Z} \left({ a_k + \left({ b_k + c_k }\right) }\right) \mathbf X^k
+     |c = because $+$ in $R$ is [[Definition:Associative|associative]]
+}} 
+{{eqn|r = f + \left({ g + h }\right)
+     |c = applying the definition of [[Definition:Addition of Polynomial Forms|polynomial addition]] twice
+}} 
+{{end-eqn}}
+{{qed}}
+[[Category:Polynomial Theory]]
+khbinhw56oq156ek6gvmmv21h544jhg
+\end{proof}<|endoftext|>
+\section{Null Polynomial is Additive Identity}
+Tags: Polynomial Theory
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Polynomial Form|polynomial forms]] has an [[Definition:Identity Element|additive identity]].
+{{explain|Context}}
+\end{theorem}
+
+\begin{proof}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] with [[Definition:Ring Zero|zero]] $0_R$.
+Let $\left\{{X_j: j  \in J}\right\}$  be a [[Definition:Set|set]] of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the [[Definition:Set|set]] of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
+Let:
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
+be an arbitrary [[Definition:Polynomial Form|polynomial form]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j \in J}\right\}$ over $R$.
+Let:
+:$\displaystyle N = \sum_{k \mathop \in Z} 0_R \mathbf X^k$ 
+be the [[Definition:Null Polynomial/Polynomial Form|null polynomial]].
+Then:
+{{begin-eqn}}
+{{eqn | l = f + N
+      | r = \sum_{k \mathop \in Z} \left({ a_k + 0_R }\right) \mathbf X^k
+      | c = Definition of [[Definition:Polynomial Addition|Polynomial Addition]]
+}}
+{{eqn | r = \sum_{k \mathop \in Z} a_k \mathbf X^k
+      | c = 
+}} 
+{{eqn | r = f
+      | c = Definition of [[Definition:Polynomial Addition|Polynomial Addition]]
+}} 
+{{end-eqn}}
+Therefore, $N + f = f$ for all [[Definition:Polynomial Form|polynomial forms]] $f$.
+Therefore, $N$ is an [[Definition:Identity Element|additive identity]] for the set of [[Definition:Polynomial Form|polynomial forms]].
+{{qed}}
+[[Category:Polynomial Theory]]
+owfhks8vopppzlob0sxveyfurzp2jir
+\end{proof}<|endoftext|>
+\section{Polynomial Addition is Commutative}
+Tags: Polynomial Theory
+
+\begin{theorem}
+[[Definition:Addition of Polynomial Forms|Addition of polynomials]] is [[Definition:Commutative Operation|commutative]].
+\end{theorem}
+
+\begin{proof}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] with [[Definition:Ring Zero|zero]] $0_R$.
+Let $\left\{{X_j: j  \in J}\right\}$ be a set of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
+Let:
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
+:$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
+be arbitrary [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j \in J}\right\}$ over $R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = f + g
+      | r = \sum_{k \mathop \in Z} \left({ a_k + b_k }\right) \mathbf X^k
+      | c = by definition of [[Definition:Addition of Polynomial Forms|polynomial addition]]
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \left({ b_k + a_k }\right) \mathbf X^k
+      | c = because addition in $R$ is [[Definition:Commutative Operation|commutative]]
+}} 
+{{eqn | r = g + f
+      | c = by definition of [[Definition:Addition of Polynomial Forms|polynomial addition]]
+}} 
+{{end-eqn}}
+So $f + g = g + f$ for all [[Definition:Polynomial|polynomials]] $f$ and $g$.
+Therefore [[Definition:Addition of Polynomial Forms|polynomial addition]] is [[Definition:Commutative Operation|commutative]].
+{{qed}}
+[[Category:Polynomial Theory]]
+9es01ie7qkr8hwqtnlzft4e1kqkmze8
+\end{proof}<|endoftext|>
+\section{Multiplication of Polynomials is Associative}
+Tags: Polynomial Theory
+
+\begin{theorem}
+[[Definition:Multiplication of Polynomial Forms|Multiplication of polynomials]] is [[Definition:Associative|associative]].
+\end{theorem}
+
+\begin{proof}
+Let $(R, +, \circ)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] with [[Definition:Ring Zero|zero]] $0_R$.
+To improve readability of the expressions used, we will write the [[Definition:Ring Product|ring product]] $\circ$ in [[Definition:Multiplicative Notation|multiplicative notation]].
+Let $\left\{{X_j: j \in J}\right\}$ be a set of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
+Let:
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
+:$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
+:$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$ 
+be arbitrary [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j  \in J}\right\}$ over $R$.
+Then it follows from [[Polynomials Closed under Ring Product]] that:
+:$\displaystyle f \circ \left({ g \circ h }\right) = \sum_{k \mathop \in Z} m_k \mathbf X^k$
+for some $m_k \in R$, and:
+:$\displaystyle \left({ f \circ g }\right) \circ h = \sum_{k \mathop \in Z} n_k \mathbf X^k$
+for some $n_k \in R$.
+To establish [[Definition:Associative|associativity]] of $\circ$ we compute $m_k$ and $n_k$, and check that they are equal. 
+We have:
+{{begin-eqn}}
+{{eqn|l = m_k
+     |r = \sum_{r + s \mathop = k} a_r \sum_{p + q \mathop = s} b_p c_q
+     |c = by definition of [[Definition:Multiplication of Polynomial Forms|polynomial multiplication]]
+}}
+{{eqn|r = \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = s} a_r b_p c_q
+     |c = by the properties of finite sums
+}} 
+{{eqn|r = \sum_{p + q + r \mathop = k} a_r b_p c_q
+     |c = by the properties of finite sums
+}} 
+{{end-eqn}}
+{{explain|what properties?}}
+Similarly we compute:
+{{begin-eqn}}
+{{eqn|l = n_k
+     |r = \sum_{r + s \mathop = k} \left({ \sum_{p + q \mathop = r} a_p b_q }\right) c_s
+     |c = by definition of [[Definition:Multiplication of Polynomial Forms|polynomial multiplication]]
+}}
+{{eqn|r = \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = r} a_p b_q c_s
+     |c = by the properties of finite sums
+}} 
+{{eqn|r = \sum_{p + q + s \mathop = k} a_p b_q c_s
+     |c = by the properties of finite sums
+}} 
+{{end-eqn}}
+{{explain|again, what properties?}}
+Since $p$, $q$, $r$ and $s$ are all dummy variables, it follows that $m_k = n_k$ for all $k \in Z$.
+Therefore, $f \circ \left({ g \circ h }\right) = \left({ f \circ g }\right) \circ h$ for all [[Definition:Polynomial Form|polynomials]] $f$, $g$ and $h$.
+Hence [[Definition:Multiplication of Polynomial Forms|multiplication of polynomials]] is [[Definition:Associative|associative]].
+{{qed}}
+[[Category:Polynomial Theory]]
+hx02dcqwubyh5v5javvpj1i7mldfrmq
+\end{proof}<|endoftext|>
+\section{Polynomials Contain Multiplicative Identity}
+Tags: Polynomial Theory
+
+\begin{theorem}
+The set of [[Definition:Polynomial|polynomials]] has a [[Definition:Identity Element|multiplicative identity]].
+\end{theorem}
+
+\begin{proof}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] with multiplicative identity $1_R$ and additive identity $0_R$.
+Let $\left\{{X_j: j \in J}\right\}$ be a set of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
+Let:
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
+be an arbitrary [[Definition:Polynomial|polynomial]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j \in J}\right\}$ over $R$.
+Let:
+:$\displaystyle N = 1_R \mathbf X^0 = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
+where $b_k = 0_R$ if $k \ne 0$ and $b_0 = 1_R$.
+Then:
+{{begin-eqn}}
+{{eqn|l = f \circ N
+     |r = \sum_{k \mathop \in Z} \left({\sum_{p \mathop + q \mathop = k} a_p b_q}\right) \mathbf X^k
+     |c = applying the definition of polynomial multiplication
+}}
+{{eqn|r = \sum_{k \mathop \in Z} \left({\sum_{p \mathop + 0 \mathop = k} a_p 1_R}\right) \mathbf X^k
+     |c = by the definition of $b_k$
+}} 
+{{eqn|r = \sum_{k \mathop \in Z} a_p \mathbf X^k
+     |c = because $1_R$ is a multiplicative identity
+}} 
+{{eqn|r = f
+     |c = 
+}} 
+{{end-eqn}}
+Therefore, $f \circ N = f$ for all polynomials $f$.
+Therefore, $N$ is a multiplicative identity for the set of polynomials.
+{{qed}}
+[[Category:Polynomial Theory]]
+74q1ml9so06jni8k2c3xa3zhy3b7g4k
+\end{proof}<|endoftext|>
+\section{Multiplication of Polynomials is Commutative}
+Tags: Polynomial Theory
+
+\begin{theorem}
+[[Definition:Polynomial Multiplication|Multiplication of polynomials]] is [[Definition:Commutative Operation|commutative]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $\set {X_j: j \in J}$ be a set of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\set {X_j: j \in J}$.
+Let:
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
+:$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
+be arbitrary [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set {X_j: j \in J}$ over $R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = f \circ g
+      | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k
+      | c = {{Defof|Polynomial Multiplication}}
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_q b_p} \mathbf X^k
+      | c = Addition of [[Definition:Multiindex|multiindices]] is [[Definition:Commutative Operation|commutative]], and $p$ and $q$ are dummy variables
+}} 
+{{eqn | r = g \circ f
+      | c = {{Defof|Polynomial Multiplication}}
+}} 
+{{end-eqn}}
+Therefore, $f \circ g = g \circ f$ for all [[Definition:Polynomial|polynomials]] $f$ and $g$.
+Therefore, [[Definition:Polynomial Multiplication|polynomial multiplication]] is [[Definition:Commutative Operation|commutative]].
+{{qed}}
+[[Category:Polynomial Theory]]
+19n11usqlh1gwbqe8bbn35tlbqa9asg
+\end{proof}<|endoftext|>
+\section{Multiplication of Polynomials Distributes over Addition}
+Tags: Polynomial Theory, Distributive Operations
+
+\begin{theorem}
+Multiplication of [[Definition:Polynomial|polynomials]] is [[Definition:Left Distributive Operation|left]]- and [[Definition:Right Distributive Operation|right]]- [[Definition:Distributive Operation|distributive]] over addition.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $\set {X_j: j \in J}$ be a [[Definition:Set|set]] of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
+Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\set {X_j: j \in J}$.
+Let
+:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
+:$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
+:$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$ 
+be  arbitrary [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\set{X_j: j \in J}$ over $R$.
+By [[Multiplication of Polynomials is Commutative]], it is sufficient to prove that $\circ$ is [[Definition:Left Distributive Operation|left distributive]] over addition only.
+Then
+{{begin-eqn}}
+{{eqn | l = f \circ \paren {g + h}
+      | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} a_p \paren {b_q + c_q} \mathbf X^k
+      | c = by the definitions of polynomial multiplication and addition
+}}
+{{eqn | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} \paren {a_p b_q + a_p c_q} \mathbf X^k
+      | c = by the properties of finite sums
+}} 
+{{eqn | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} a_p b_q \mathbf  X^k + \sum_{k \mathop \in Z} \sum_{p + q = k} a_p c_q \mathbf X^k
+      | c = by the properties of finite sums
+}} 
+{{eqn | r = f \circ g + f \circ h
+      | c = by the definitions of polynomial multiplication and addition
+}} 
+{{end-eqn}}
+Therefore, $f \circ \paren {g + h} = f \circ g + f \circ h$ for all polynomials $f, g, h$.
+Therefore, polynomial multiplication is distributive over addition.
+{{qed}}
+[[Category:Polynomial Theory]]
+[[Category:Distributive Operations]]
+tqn5xtor14xks6rbwjlg4k61mt8a8uh
+\end{proof}<|endoftext|>
+\section{Distributivity is Preserved in Induced Structure}
+Tags: Pointwise Operations, Distributive Operations
+
+\begin{theorem}
+Let $\struct {T, \oplus, \otimes}$ be an [[Definition:Algebraic Structure|algebraic structure]], and let $S$ be a [[Definition:Set|set]].
+Let $T^S$ denote the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$.
+Let $\struct {T^S, \oplus}$ be the structure on $T^S$ [[Definition:Induced Structure|induced]] by $\oplus$.
+Let $\struct {T^S, \otimes}$ be the structure on $T^S$ [[Definition:Induced Structure|induced]] by $\otimes$.
+If $\otimes$ is [[Definition:Distributive Operation|distributive]] over $\oplus$, then the [[Definition:Pointwise Operation|pointwise operation induced]] on $T^S$ by $\otimes$ is [[Definition:Distributive Operation|distributive]] over the operation induced by $\oplus$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {T, \oplus, \otimes}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\otimes$ distributes over $\oplus$.
+Let $f, g, h \in T^S$. 
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {f \otimes \paren {g \oplus h} } } x
+      | r = \map f x \otimes \paren {\map g x \oplus \map h x}
+      | c = {{Defof|Pointwise Operation}}
+}}
+{{eqn | r = \paren {\map f x \otimes \map g x} \oplus \paren {\map f x \otimes \map h x}
+      | c = because $\otimes$ distributes over $\oplus$ in $T$
+}}
+{{eqn | r = \map {\paren {\paren {f \otimes g} \oplus \paren {f \otimes h} } } x
+      | c = {{Defof|Pointwise Operation}}
+}}
+{{end-eqn}}
+This shows [[Definition:Left Distributive Operation|left distributivity]].
+The proof for [[Definition:Right Distributive Operation|right distributivity]] is identical.
+{{qed}}
+[[Category:Pointwise Operations]]
+[[Category:Distributive Operations]]
+c2unl9bb8a9qwficzc2xs3madjanz72
+\end{proof}<|endoftext|>
+\section{Dirichlet Series Absolute Convergence Lemma}
+Tags: Dirichlet Series
+
+\begin{theorem}
+Let $\displaystyle \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
+Suppose that $f$ [[Definition:Absolutely Convergent Series|converges absolutely]] at $s_0 = \sigma_0 + i t_0 \in \C$.
+Then $f$ [[Definition:Absolutely Convergent Series|converges absolutely]] at all points $s = \sigma + i t \in \C$ with $\sigma \ge \sigma_0$.
+\end{theorem}
+
+\begin{proof}
+Suppose that $f$ [[Definition:Absolutely Convergent Series|converges absolutely]] at $\sigma_0 + i t_0$.
+If $\sigma \ge \sigma_0$, then:
+{{begin-eqn}}
+{{eqn | l = \size {\frac {a_n} {n^s} }
+      | r = \frac {\size {a_n} } {n^\sigma}
+}}
+{{eqn | o = \le
+      | r = \frac {\size {a_n} } {n^{\sigma_0} }
+}}
+{{eqn | r = \size {\frac {a_n} { n^{s_0} } }
+}}
+{{end-eqn}}
+Therefore [[Definition:Absolutely Convergent Series|absolute convergence]] of $\map f {s_0}$ directly implies [[Definition:Absolutely Convergent Series|absolute convergence]] of $\map f s$ for all $s = \sigma + i t$ with $\sigma > \sigma_0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Abscissa of Absolute Convergence}
+Tags: Dirichlet Series
+
+\begin{theorem}
+Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
+Let the series $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert { a_n n^{-s} } \right\vert$ not converge for all $s \in \C$, or diverge for all $s \in \C$.
+Then there exists a real number $\sigma_a$ such that $f \left({s}\right)$ [[Definition:Absolutely Convergent Series|converges absolutely]] for all $s = \sigma + it$ with $\sigma > \sigma_a$, and does not converge absolutely for all $s$ with $\sigma < \sigma_a$.
+We call $\sigma_a$ the '''abscissa of absolute convergence''' of the Dirichlet series.
+\end{theorem}
+
+\begin{proof}
+Let $S$ be the set of all complex numbers $s$ such that $f \left({s}\right)$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $f \left({s_0}\right)$ converges absolutely, so $S$ is not [[Definition:Empty Set|empty]].
+Moreover, $S$ is bounded below, for otherwise it follows from [[Dirichlet Series Absolute Convergence Lemma]] that $f \left({s}\right)$ converges absolutely for all $s \in \C$, a contradiction of our assumptions.
+Therefore the [[Definition:Infimum of Set|infimum]]:
+:$\sigma_a = \inf \left\{{\sigma: s = \sigma + i t \in S}\right\} \in \R$
+is well defined.
+Now if $s = \sigma + it$ with $\sigma > \sigma_a$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $f \left({s'}\right)$ is absolutely convergent.
+Then it follows from [[Dirichlet Series Absolute Convergence Lemma]] that $f \left({s}\right)$ is absolutely convergent.
+If $s = \sigma + it$ with $\sigma < \sigma_a$, and $f \left({s}\right)$ is absolutely convergent then $s$ contradicts the definition of $\sigma_a$.
+Therefore, $\sigma_a$ has the claimed properties.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dirichlet Series Convergence Lemma}
+Tags: Dirichlet Series
+
+\begin{theorem}
+Let $\displaystyle f \left({s}\right) = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
+Let $f \left({s}\right)$ [[Definition:Convergent Series|converge]] at $s_0 = \sigma_0 + i t_0$.
+Then $f \left({s}\right)$ [[Definition:Convergent Series|converge]] for all $s = \sigma + i t$ where $\sigma > \sigma_0$.
+\end{theorem}
+
+\begin{proof}
+We begin with a lemma.
+=== [[Dirichlet Series Convergence Lemma/Lemma|Lemma]] ===
+{{:Dirichlet Series Convergence Lemma/Lemma}}
+=== Proof of Theorem ===
+Suppose that $f \left({s}\right)$ [[Definition:Convergent Series|converges]] at $s_0 = \sigma_0 + it_0$.
+Then by [[Convergent Sequence in Metric Space is Bounded]], $\left\vert \displaystyle \sum_{k\mathop=1}^n \frac {a_k} {k^{s_0}} \right\vert$ is bounded. Thus the results of the lemma hold.
+Choose any $s = \sigma + it$ with $\sigma > \sigma_0$.
+By [[Cauchy's Convergence Criterion]], it suffices to show that for all $\epsilon>0$, there exists an $N>0$ such that for all $m>n>N$
+:$\displaystyle \left\vert \sum_{k \mathop = 1}^m \frac {a_k} {k^s} - \sum_{k \mathop = 1}^n \frac {a_k} {k^s} \right\vert = \left\vert \sum_{k \mathop = n+1}^m \frac {a_k} {k^s} \right\vert < \epsilon$
+The lemma shows that for a given $s$ there exists a constant $C$ independent of $N$ such that:
+:$\displaystyle \left\vert{\sum_{k \mathop = n+1}^n a_n n^{-s} }\right\vert \le C \left({n+1}\right)^{\sigma_0 - \sigma}$
+Since $\sigma_0 - \sigma <0$, the right hand side tends to zero as $n \to \infty$.
+Thus we may choose $N$ large enough so that for $n>N$
+:$\displaystyle \left({n+1}\right)^{\sigma_0 - \sigma} < \dfrac \epsilon C$
+Which gives us
+$\displaystyle \left\vert{\sum_{k \mathop = n+1}^n a_n n^{-s} }\right\vert < \epsilon$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Abscissa of Convergence/General}
+Tags: General Dirichlet Series
+
+\begin{theorem}
+Let $s = \sigma + i t$
+Let $\displaystyle \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s}$ be a [[Definition:General Dirichlet Series|general Dirichlet series]].
+Then there exists a [[Definition:Extended Real Number Line|extended real number]], $\sigma_0$, such that
+:$(1): \quad$ For $\sigma < \sigma_0$, $\map f s$ diverges
+:$(2): \quad$ For $\sigma > \sigma_0$, $\map f s$ converges.
+\end{theorem}
+
+\begin{proof}
+If there does not exist an $s_0$ such that $\map f {s_0}$ converges, then $\sigma_0 = \infty$ and the theorem is [[Definition:Vacuous Truth|vacuously true]]
+If there does exist such an $s_0$, let $\sigma_0$ be the [[Definition:Infimum of Subset of Real Numbers|infimum]] of the real part of all such $s_0$, where $\sigma_0= -\infty$ if the set is not bounded from below.
+*It is clear that $\map f s$ diverges if $\sigma < \sigma_0$ by construction of $\sigma_0$ as the infimum of all convergent $s$
+*As $\sigma_0$ is the infimum of the the real part of the set all s such that $\map f s$ converges, for any $\epsilon > 0$, there exists an $s$ such that $\map f s$ converges and $\sigma_0 \le \sigma < \sigma_0+ \epsilon$ as otherwise $\sigma_0 + \epsilon$ would be the infimum of the set. 
+:Therefore, for any $s$ such that $\sigma > \sigma_0$, we can pick $\epsilon = \sigma - \sigma_0$ and find a $q$ such that $\map f q$ converges and $\map \Re q < \sigma$.
+:By [[Dirichlet Series Convergence Lemma]], We therefore have that $\map f s$ converges.
+{{qed}}
+[[Category:General Dirichlet Series]]
+1gzj420mgixxuwfb1mzgc4zjamfldt3
+\end{proof}<|endoftext|>
+\section{Difference of Abscissae of Convergence}
+Tags: Dirichlet Series
+
+\begin{theorem}
+Let $\displaystyle \map f s = \sum_{n \mathop = 1}^\infty a_n n^{-s}$ be a [[Definition:Dirichlet Series|Dirichlet series]].
+Suppose that $\map f s$ has finite [[Abscissa of Convergence]] $\sigma_c$.
+Then the [[Abscissa of Absolute Convergence]] $\sigma_a$ is finite, and:
+:$0 \le \sigma_a - \sigma_c \le 1$
+\end{theorem}
+
+\begin{proof}
+It is trivial that $\sigma_a \ge \sigma_c$.
+Suppose $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ [[Definition:Convergent Series|converges]].
+It is sufficient to show that $\map f s$ [[Definition:Absolutely Convergent Series|converges absolutely]] for all $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$.
+Pick an upper bound $M$ for the real numbers $\size {a_n n^{-s_0} }$.
+Then for $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$:
+:$\size {\dfrac {a_n} {n^s} } = \size {\dfrac {a_n} {n^{s_0} n^{s - s_0} } } = \dfrac M {n^{\sigma - \sigma_0} }$
+Then by the [[Comparison Test]], $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ converges.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completion Theorem (Metric Space)}
+Tags: Isometries, Metric Spaces, Complete Metric Spaces, Named Theorems, Completion Theorem
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then there exists a [[Definition:Completion (Metric Space)|completion]] $\tilde M = \struct {\tilde A, \tilde d}$ of $\struct {A, d}$.
+Moreover, this [[Definition:Completion (Metric Space)|completion]] is unique up to [[Definition:Isometry (Metric Spaces)|isometry]].
+That is, if $\struct {\hat A, \hat d}$ is another [[Definition:Completion (Metric Space)|completion]] of $\struct {A, d}$, then there is a [[Definition:Bijection|bijection]] $\tau: \tilde A \leftrightarrow \hat A$ such that:
+:$(1): \quad \tau$ [[Definition:Restriction of Mapping|restricts]] to the [[Definition:Identity Mapping|identity]] on $x$:
+:::: $\forall x \in A: \map \tau x = x$
+:$(2): \quad \tau$ preserves [[Definition:Metric|metrics]]:
+:::: $\forall x_1, x_2 \in A : \map {\hat d} {\map \tau {x_1}, \map \tau {x_2} } = \map {\tilde d} {x_1, x_2}$
+\end{theorem}
+
+\begin{proof}
+We construct the [[Definition:Completion (Metric Space)|completion of a metric space]] as [[Definition:Equivalence Class|equivalence classes]] of the [[Definition:Set|set]] of [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequences]] in the [[Definition:Metric Space|space]] under a suitable [[Definition:Equivalence Relation|equivalence relation]].
+Let $\struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\CC \sqbrk A$ denote the [[Definition:Set|set]] of all [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequences]] in $A$.
+Define a [[Definition:Relation|relation]] $\sim$ on $\CC \sqbrk A$ by:
+:$\displaystyle \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
+By [[Equivalence Relation on Cauchy Sequences]], $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]] on $\CC \sqbrk A$.
+Denote the [[Definition:Equivalence Class|equivalence class]] of $\sequence {x_n} \in \CC \sqbrk A$ by $\sqbrk {x_n}$.
+Denote the set of [[Definition:Equivalence Class|equivalence classes]] under $\sim$ by $\tilde A$.
+By [[Relation Partitions Set iff Equivalence]] this is a [[Definition:Partition (Set Theory)|partition]] of $\CC \sqbrk A$.
+That is, each $\sequence {x_n} \in \CC \sqbrk A$ lies in one and only one [[Definition:Equivalence Class|equivalence class]] under $\sim$.
+Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:
+:$\displaystyle \map {\tilde d} {\sqbrk {x_n}, \sqbrk {y_n} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
+=== [[Completion Theorem (Metric Space)/Lemma 1|Lemma 1: $\tilde d$ is Well-Defined]] ===
+{{:Completion Theorem (Metric Space)/Lemma 1}}
+We claim that $\struct {\tilde A, \tilde d}$ is a [[Definition:Completion (Metric Space)|completion]] of $\struct {A, d}$.
+Therefore we must show that:
+:$\tilde d$ is a [[Definition:Metric|metric]] on $\tilde A$
+:There exists an [[Definition:Everywhere Dense|everywhere dense]] [[Definition:Inclusion Mapping|inclusion]] $\struct {A, d} \to \struct {\tilde A, \tilde d}$ preserving $d$.
+In addition the theorem claims that $\struct {\tilde A, \tilde d}$ is unique up to [[Definition:Isometry (Metric Spaces)|isometry]].
+=== [[Completion Theorem (Metric Space)/Lemma 2|Lemma 2: $\tilde d$ is a Metric]] ===
+{{:Completion Theorem (Metric Space)/Lemma 2}}
+=== [[Completion Theorem (Metric Space)/Lemma 3|Lemma 3: $\tilde A$ is a Completion of $A$]] ===
+{{:Completion Theorem (Metric Space)/Lemma 3}}
+=== [[Completion Theorem (Metric Space)/Lemma 4|Lemma 4: Uniqueness of $\tilde A$]] ===
+{{:Completion Theorem (Metric Space)/Lemma 4}}{{qed}}
+{{ACC|Completion Theorem (Metric Space)/Lemma 3}}
+\end{proof}<|endoftext|>
+\section{Completeness Criterion (Metric Spaces)}
+Tags: Metric Spaces, Complete Metric Spaces, Completeness Criterion (Metric Spaces)
+
+\begin{theorem}
+Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
+Let $A \subseteq S$ be a [[Definition:Everywhere Dense|dense]] [[Definition:Subset|subset]].
+Suppose that every [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$ [[Definition:Convergent Sequence (Metric Space)|converges]] in $M$.
+Then $M$ is [[Definition:Complete Metric Space|complete]].
+\end{theorem}
+
+\begin{proof}
+Let $\left\langle{x_n}\right\rangle_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $S$.
+For each $n$ pick a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] $\left\langle{y_{n, m}}\right\rangle_{m \mathop \in \N}$ in $A$ converging to $x_n$ like so:
+:[[File:CompletenessCriterionProof.png|400px|Caption]]
+Let $N \in \N$ be such that $d \left({x_{n_1}, x_{n_2}}\right) < \epsilon / 3$ for all $n_1, n_2 > N$.
+Let $M \in \N$ be such that $d \left({y_{n_i, m}, x_{n_i}}\right) < \epsilon / 3$ for all $m > M$ and all $n_1, n_2 > N$. 
+Let $m > M$.
+Let $n_1, n_2 > N$.
+We have:
+{{begin-eqn}}
+{{eqn | l = d \left({y_{n_1, m}, y_{n_2, m} }\right)
+      | o = \le
+      | r = d \left({y_{n_1, m}, x_{n_1} }\right) + d \left({x_{n_1}, x_{n_2} }\right) + d \left({x_{n_2}, y_{n_2, m} }\right)
+      | c = two applications of the [[Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+Therefore $\left\langle{y_{m, n}}\right\rangle_{n \in \N}$ is [[Definition:Cauchy Sequence (Metric Space)|Cauchy]] in $A$ for $m > M$.
+So $\left\langle{y_{m, n}}\right\rangle_{n \mathop \in \N}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to some [[Definition:Limit of Sequence (Metric Space)|limit]] $y_n \in S$.
+{{finish|Proof to complete}}
+\end{proof}
+
+\begin{proof}
+Let $\left\langle{x_n}\right\rangle$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $M$.
+Since $A$ is [[Definition:Everywhere Dense|dense]], we can invoke the [[Axiom:Axiom of Countable Choice|Axiom of Countable Choice]] to choose for each $n$ some $y_n \in A$ which is within $1 / n$ of $x_n$.
+We will show that $\left\langle{y_n}\right\rangle$ is [[Definition:Cauchy Sequence (Metric Space)|Cauchy]].
+Let $\epsilon > 0$.
+Since $\left\langle{x_n}\right\rangle$ is [[Definition:Cauchy Sequence (Metric Space)|Cauchy]], there exists some $N'$ such that:
+:$\forall n, m \ge N': d \left({x_n, x_m}\right) < \frac 1 3 \epsilon$
+Let $N \in \N$ be greater than $N'$ and $3 \epsilon^{-1}$.
+Note that $n, m \ge N$ implies that both:
+: $(1): \quad n, m \ge N'$
+: $(2): \quad \dfrac 1 n, \dfrac 1 m \le \dfrac 1 N < \dfrac 1 3 \epsilon$
+Thus for $n, m \ge N$, by the [[Triangle Inequality]] and the above:
+{{begin-eqn}}
+{{eqn | l = d \left({y_n, y_m}\right)
+      | o = \le
+      | r = d \left({y_n, x_n}\right) + d \left({x_n, x_m}\right) + d \left({x_m, y_m}\right)
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 n + \frac 1 3 \epsilon + \frac 1 m 
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 3 \epsilon + \frac 1 3 \epsilon + \frac 1 3 \epsilon
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+Hence $\left\langle{y_n}\right\rangle$ is a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$.
+By assumption, it [[Definition:Convergent Sequence (Metric Space)|converges]] to some [[Definition:Limit of Sequence (Metric Space)|limit]] $y \in S$.
+Now, we verify that $\left\langle{x_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $y$ as well.
+Let $\epsilon > 0$.
+Since $\left\langle{y_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]], there exists some $N'$ such that:
+:$\forall n \ge N': d \left({y_n, y}\right) < \dfrac 1 2 \epsilon$
+Let $N \in \N$ be greater than $N'$ and $2 \epsilon^{-1}$.
+Note that $n \ge N$ implies that both:
+: $(1): \quad n \ge N'$
+: $(2): \quad \dfrac 1 n \le \dfrac 1 N < \dfrac 1 2 \epsilon$
+Thus, for $n \ge N$, we have by the [[Triangle Inequality]] and above that:
+{{begin-eqn}}
+{{eqn | l = d \left({x_n, y}\right)
+      | o = \le
+      | r = d \left({x_n, y_n}\right) + d \left({y_n, y}\right)
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 n + \frac 1 2 \epsilon
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 2 \epsilon + \frac 1 2 \epsilon
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+Hence $\left\langle{x_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Number Addition is Closed}
+Tags: Natural Numbers, Natural Number Addition
+
+\begin{theorem}
+The operation of [[Definition:Natural Number Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ is [[Definition:Closed Algebraic Structure|closed]]:
+:$\forall x, y \in \N: x + y \in \N$
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Natural Numbers under Addition form Commutative Monoid]].
+A [[Definition:Monoid|monoid]] by definition is a [[Definition:Semigroup|semigroup]].
+Again by definition, the [[Definition:Binary Operation|operation]] in a [[Definition:Semigroup|semigroup]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Ordered Field}
+Tags: Ordered Fields
+
+\begin{theorem}
+Let $\struct {F, +, \cdot}$ be an [[Definition:Ordered Field|ordered field]] with [[Definition:Unity of Field|unity]] $1$, [[Definition:Field Zero|zero]] $0$.
+Denote the [[Definition:Strict Ordering|strict order]] by $<$ and the [[Definition:Weak Ordering|weak order]] by $\le$.
+Let $\Char F$ denote the [[Definition:Characteristic of Ring|characteristic]] of $F$.
+Then the following hold for all $x, y, z \in k$:
+:$(1): \quad x < 0 \iff -x > 0$
+:$(2): \quad x > y \iff x-y > 0$
+:$(3): \quad x < y \iff -x > -y$
+:$(4): \quad \paren {z < 0} \land \paren {x < y} \implies x z > y z$
+:$(5): \quad x \ne 0 \implies x^2 > 0$
+:$(6): \quad 1 > 0$
+:$(7): \quad \Char k = 0$
+:$(8): \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Ordering|ordering]], the relation $\le$ is:
+:[[Definition:Reflexive Relation|reflexive]]
+:[[Definition:Transitive Relation|transitive]]
+:[[Definition:Antisymmetric Relation|antisymmetric]]
+and furthermore, every pair of elements is [[Definition:Comparable|comparable]].
+The [[Definition:Ordering Compatible with Ring Structure|order is compatible with $F$]] in the sense that, for all $x, y, z, c \in F$:
+:$x < y \implies x + z < y + z$
+:$c > 0,\ x < y \implies c x < c y$
+The proof is by repeated deduction from these properties.
+* $(1): \quad x < 0 \iff -x > 0$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = x - x
+      | o = <
+      | r = 0 - x
+}}
+{{eqn | ll= \leadsto
+      | l = -x
+      | o = >
+      | r = 0
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = x - x
+      | o = >
+      | r = 0 - x
+}}
+{{eqn | ll= \leadsto
+      | l = -x
+      | o = <
+      | r = 0
+}}
+{{end-eqn}}
+* $(2): \quad x > y \iff x - y > 0$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = x - y
+      | o = >
+      | r = y - y = 0
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = x - y > 0
+      | o = >
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = x - y + y
+      | o = >
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = >
+      | r = y
+}}
+{{end-eqn}}
+* $(3): \quad x < y \iff -x > -y$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = x - x - y
+      | o = <
+      | r = y - x - y
+}}
+{{eqn | ll= \leadsto
+      | l = -y
+      | o = <
+      | r = -x
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = -y
+      | o = <
+      | r = -x
+}}
+{{eqn | ll= \leadsto
+      | l = -y + y + x
+      | o = <
+      | r = -x + y + x
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = <
+      | r = y
+}}
+{{end-eqn}}
+* $(4): \quad (z < 0) \land (x < y) \implies x z > y z$:
+By parts 1 and 3 above, if $z < 0$, $x < y$ then $-z > 0$ and $-x > -y$.
+Then:
+:$x z = \paren {-x} \paren {-z} > \paren {-y} \paren {-z} = y z$
+* $(5): \quad x \ne 0 \implies x^2 > 0$:
+If $x > 0$, then:
+:$x^2 = x \cdot x > x \cdot 0 = 0$
+If $x < 0$, then by 1, $-x > 0$, so:
+:$x^2 = \paren {-x} \cdot \paren {-x} > \paren {-x} \cdot 0 = 0$
+* $(6): \quad 1 > 0$:
+This is immediate from $(5)$, noting that $1 = 1^2$ is a square.
+* $(7): \quad \Char k = 0$:
+By $(6)$, we have:
+:$0 < 1 < 1 + 1 < 1 + 1 + 1 < \cdots$
+so $n \cdot 1 \ne 0$ for all $n \in \N$.
+* $(8): \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$:
+First let $x > 0$, and suppose that $x^{-1} < 0$.
+Then by $(4)$:
+:$0 = 0 \cdot x^{-1} > x \cdot x^{-1} = 1$
+which contradicts $(6)$, so $x^{-1} > 0$.
+Now let $x > y > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x^{-1} y^{-1} x
+      | o = >
+      | r = x^{-1} y^{-1} y > 0
+}}
+{{eqn | ll= \leadsto
+      | l = y^{-1}
+      | o = >
+      | r = x^{-1} > 0
+}}
+{{end-eqn}}
+The converse follows upon interchanging $x^{-1} \leftrightarrow x$ and $y^{-1} \leftrightarrow y$.
+{{qed}}
+[[Category:Ordered Fields]]
+jfttok3h9u1ed8f6bf74k928uc01v1b
+\end{proof}<|endoftext|>
+\section{Characterisation of Ordered Fields}
+Tags: Field Theory, Ordered Fields
+
+\begin{theorem}
+Let $\struct {k, +, \cdot}$ be a [[Definition:Field (Abstract Algebra)|field]] with [[Definition:Unity of Field|unity]] $1$ and [[Definition:Field Zero|zero]] $0$.
+Then the following are equivalent:
+:$(1): \quad$ There exists a [[Definition:Total Ordering|total ordering]] $\le$ on $k$ such that $\struct {k, \le}$ is an [[Definition:Ordered Field|ordered field]]
+:$(2): \quad$ $-1$ cannot be written as a sum of squares of elements of $k$
+:$(3): \quad$ $0$ cannot be written as a non-empty sum of squares of non-zero elements of $k$
+\end{theorem}
+
+\begin{proof}
+=== $(2)$ iff $(3)$ ===
+Suppose there exist $\set {x_i: i \in I}$ such that
+:$\displaystyle -1 = \sum_{i \mathop \in I} x_i^2$
+Then
+:$\displaystyle 0 = 1^2 + \sum_{i \mathop \in I} x_i^2$
+a non-empty sum of non-zero squared of $k$.
+Conversely, suppose that there is a set $\set {x_i: i \in I} \ne \O$ with $x_i \ne 0$ for all $i \in I$ such that:
+:$\displaystyle 0 = \sum_{i \mathop \in I} x_i^2$
+Then for any $j \in I$:
+:$\displaystyle -x_j^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } x_i^2$
+Dividing through by $x_j^2$ we find that:
+:$\displaystyle -1 = -\paren {\frac {x_j} {x_j} }^2 = \sum_{\substack {i \mathop \in I \\ i \mathop \ne j} } \paren {\frac {x_i} {x_j} }^2$
+=== $(1)$ implies $(2)$ ===
+By [[Properties of Ordered Field]], in an ordered field we have, $-1 < 0$, and squares are non-negative.
+Therefore for any subset $\set {x_i : i \mathop \in I} \subseteq k$,
+:$\displaystyle -1 < 0 \le \sum_{i \mathop \in I} x_i ^2$
+=== $(2)$ implies $(1)$ ===
+Suppose that $-1$ is not a sum of squares in $k$. 
+Let $S$ be the set of non-empty sums of squares of non-zero elements of $k$.
+Then by supposition and  $(2) \iff (1)$:
+:$0, 1 \notin S$
+{{explain|but $(2) \iff (1)$ has not yet been established.}}
+Trivially, $S$ is [[Definition:Closure (Abstract Algebra)|closed]] under addition.
+Also, for any subsets $\set {x_i: i \in I}, \set {y_j: j \in J} \subseteq k$:
+:$\displaystyle \paren {\sum_{i \mathop \in I} x_i^2} \cdot \paren {\sum_{j \mathop \in J} y_j^2} = \sum_{i, j} \paren {x_i y_j}^2 \in S$
+so $S$ is [[Definition:Closure (Abstract Algebra)|closed]] under multiplication. 
+It follows that $S$ is a [[Definition:Multiplicative Notation|multiplicative]] [[Definition:Subgroup|subgroup]] of the [[Definition:Set Difference|set difference]] $k \setminus \set 0$.
+Now let $\Gamma$ be the collection of all subsets $M$ of $k$ such that:
+:$S \subseteq M$
+:$M$ is closed under addition
+:$M$ is a multiplicative subgroup of $k \setminus \set 0$
+Then every chain $\mathscr C = \set {M_i : i \in \N}$ has an upper bound:
+:$\displaystyle \bigcup_{i \mathop \in \N} M_i \in \Gamma$
+So by [[Zorn's Lemma]] there is a maximal element $M$ with these properties.
+Clearly $0 \notin M$.
+{{explain|why "clearly"?}}
+Let:
+:$\paren {-M} := \set {x \in k : -x \in M}$
+Suppose $x, -x \in M$.
+Then:
+:$x - x = 0 \in M$
+and this [[Definition:Contradiction|contradicts]] $0 \notin M$.
+Hence $M$, $\set 0$ and $-M$ are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+At this point the reader should think of $M$ as a [[Definition:Partition (Set Theory)|partition]] of $k$ into positive elements, $\set 0$ and negative elements.
+The following two claims justify this statement.
+==== Lemma 1 ====
+$k = M \cup \set 0 \cup \paren {-M}$
+==== Proof of Lemma 1 ====
+Let $a \in k$, $a \ne 0$, $-a \notin M$, and:
+:$M' = \set {x + a y: x, y \in M \cup \set 0, \paren {x = 0 \lor y = 0} }$
+We have:
+:$S \subseteq M \subseteq M'$
+and it is trivial to check that $M'$ is closed under addition.
+{{explain|Trivial though it be, it needs to be demonstrated.}}
+Let $x + a y$, $z + a w \in M'$.
+Then:
+:$\paren {x + a y} \paren {z + a w} = \paren {x z + a^2 y w} + a \paren {y z + x w}$
+Since:
+:$a^2 \in S \subseteq M'$
+and:
+:$M$ is closed under multiplication and addition
+it follows that:
+:$M'$ is also closed under multiplication.
+Since $a$ and at least one of $x, y \ne 0$, it follows that:
+:$0 \notin M'$
+Since $1 \in S \subseteq M'$:
+:$1 \in M'$
+Let $t = x + a y \in M'$.
+Then:
+{{begin-eqn}}
+{{eqn | l = t \paren {\frac x {t^2} + a \frac y {t^2} }
+      | r = \frac x t + a \frac y t
+}}
+{{eqn | r = \frac {x + a y} {x + a y}
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+Therefore every $t \in M'$ has a multiplicative inverse in $M'$
+So $M'$ is a multiplicative subgroup of $k \setminus \set 0$.
+Therefore $M'$ satisfies all the conditions subject to which we chose $M$.
+This and the fact that $M \subseteq M'$ imply that:
+:$M = M'$
+Therefore:
+:$a = 1 + 1 a \in M$
+Let $-a \in M$.
+Then:
+:$a \in -M$
+so every $a \in k$ lies in exactly one of $M$, $\set 0$ and $-M$.
+Thus $k = M \cup \set 0 \cup \paren {-M}$.
+{{qed|lemma}}
+Now define a binary relation on $k$ by $x < y \iff y - x \in M$.
+Equivalently:
+:$x \le y \iff \paren {x < y} \lor \paren {x = y}$
+==== Lemma 2 ====
+$\struct {k, \le}$ is an ordered field.
+==== Proof of Lemma 2 ====
+The proof is an elementary check of the relevant axioms.
+First we check that $\le$ is a [[Definition:Total_Ordering|total order]] on $k$.
+[[Definition:Reflexive Relation|Reflexivity]]:
+Trivial from the definition.
+[[Definition:Antisymmetric Relation|Antisymmetry]]:
+Suppose that $x \le y$ and $y \le x$.
+We cannot have $x - y \in M$ and $y - x \in M$, because since $M$ is closed under addition, this would imply $0 \in M$.
+Therefore, $x - y = y - x = 0$ and $y = x$.
+[[Definition:Transitive Relation|Transitivity]]:
+Suppose $x \le y$ and $y \le z$.
+Then $y - x \in M$ and $z - y \in M$.
+$M$ is closed under addition, so $z - x = \paren {y - x} + \paren {z - y} \in M$.
+Therefore $x \le z$.
+[[Definition:Non-comparable|Comparability]]:
+The comparability of each pair of elements of $\struct {k, \le}$ is immediate from the above mentioned fact that $M$, $\set 0$ and $\paren {-M}$ [[Definition:Partition (Set Theory)|partition]] $k$:
+Suppose $x \nleq y$.
+Then:
+:$y - x \notin M$
+and:
+:$y - x \ne 0$
+Then:
+:$y - x \in \paren {-M}$
+so:
+:$x - y \in M$
+and:
+:$y \le x$
+Thus $x$ and $y$ are [[Definition:Comparable|comparable]].
+It remains to show that the total order $\le$ is [[Definition:Ordering Compatible with Ring Structure|compatible]] with the field structure.
+Let $x, y, z \in k$ such that $x < y$.
+Then:
+:$\paren {y + z} - \paren {x + z} = y - x \in M$
+So:
+:$x + z \le y + z$.
+Now suppose further that $z > 0$.
+We have that:
+:$z \in M$ and $y - x \in M$ by hypothesis
+and:
+:$M$ is closed under multiplication.
+Therefore:
+:$y z - x z = z \paren {y - x} \in M$
+This establishes that $k$ is an ordered field.
+{{qed|lemma}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hahn-Banach Theorem}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $E$ be a [[Definition:Vector Space|vector space]] over $\R$.
+Let $p: E \to \R$ be a [[Definition:Minkowski Functional|Minkowski functional]].
+Let $G \subseteq E$ be a [[Definition:Vector Subspace|linear subspace]] of $E$.
+Let $f: G \to \R$ be a [[Definition:Linear Functional|linear functional]] such that:
+:$\forall x \in G: \map f x \le \map p x$
+Then there exists a [[Definition:Linear Functional|linear functional]] $\tilde f$ defined on the whole [[Definition:Vector Space|space]] $E$ which [[Definition:Extension of Mapping|extends]] $f$.
+That is:
+:$\forall x \in G: \map {\tilde f} x = \map f x$
+such that:
+:$\forall x \in E: \map {\tilde f} x \le \map p x$
+\end{theorem}
+
+\begin{proof}
+Let a [[Definition:Linear Functional|linear functional]] $g$ be called '''admissible''' {{iff}}
+:$\forall x \in \Dom g: \map g x \le \map p x$
+A [[Definition:Linear Functional|linear functional]] $h_1$ [[Definition:Extension of Mapping|extends]] a [[Definition:Linear Functional|linear functional]] $h_2$ {{iff}}:
+:$\Dom {h_2} \subseteq \Dom {h_1}$
+and:
+:$\forall x \in \Dom {h_2}: \map {h_2} x = \map {h_1} x$
+The proof consists of two steps:
+First, the set of '''admissible''' [[Definition:Linear Functional|linear functionals]] that [[Definition:Extension of Mapping|extend]] $f$ is [[Definition:Inductive Set as Subset of Real Numbers|inductive]].
+Using [[Zorn's Lemma]] the existence of a [[Definition:Maximal Element|maximal element]] is derived.
+Second, it is proved by contradiction that this [[Definition:Linear Functional|functional]] is defined on the whole [[Definition:Vector Space|space]] $E$.
+{{finish}}
+{{Namedfor|Hans Hahn|name2 = Stefan Banach|cat = Hahn|cat2 = Banach}}
+\end{proof}<|endoftext|>
+\section{Completion of Valued Field}
+Tags: Valued Fields
+
+\begin{theorem}
+Let $\struct {k, \norm {\,\cdot\,} }$ be a [[Definition:Valued Field|valued field]].
+Then there exists a [[Definition:Completion (Normed Division Ring)|completion]] $\struct {k', \norm {\,\cdot\,}'}$ of $\struct {k, \norm {\,\cdot\,} }$ such that $\struct {k', \norm {\,\cdot\,}'}$ is a [[Definition:Valued Field|valued field]].
+Furthermore, every [[Definition:Completion (Normed Division Ring)|completion]] of $\struct{k, \norm {\,\cdot\,} }$ is [[Definition:Isometric Isomorphism|isometrically isomorphic]] to $\struct {k', \norm {\,\cdot\,}'}$.
+\end{theorem}
+
+\begin{proof}
+By [[Completion of Normed Division Ring]] then $\struct {k, \norm {\, \cdot \,} }$ has a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] $\struct {k', \norm {\, \cdot \,}'}$
+By [[Normed Division Ring is Field iff Completion is Field]] then $\struct {k', \norm {\, \cdot \,}'}$ is a [[Definition:Field (Abstract Algebra)|field]].
+By [[Normed Division Ring Completions are Isometric and Isomorphic]] then every [[Definition:Completion (Normed Division Ring)|completion]] of $\struct {k, \norm {\,\cdot\,} }$ is [[Definition:Isometric Isomorphism|isometrically isomorphic]] to $\struct {k', \norm {\,\cdot\,}'}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hardy-Littlewood Circle Method}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\AA$ be a [[Definition:Subset|subset]] of the non-negative [[Definition:Integer|integers]].
+Let:
+:$\displaystyle \map T s = \sum_{a \mathop \in \AA} s^a$
+be the [[Definition:Generating Function|generating function]] for $\AA$.
+For $N \in \N$, let $\map {r_{\AA, \ell} } N$ be the number of solutions $\tuple {x_1, \ldots, x_\ell} \in \AA^\ell$ to the equation:
+:$x_1 + \cdots + x_\ell = N$
+Then:
+:$\displaystyle \forall \rho \in \openint 0 1: \map {r_{\AA, \ell} } N = \oint_{\size s \mathop = \rho} \frac {\map T s^\ell} {s^{N + 1} } \rd s$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$\displaystyle \map T s^\ell = \sum_{N \mathop = 0}^\infty \map {r_{\AA, \ell} } N s^N$
+and:
+:$\map {\dfrac {\d^N} {\d s^N} } {\map T s^\ell} = N! \cdot \map {r_{\AA, \ell} } N + \map \OO s$
+so:
+:$\map {r_{\AA, \ell} } N = \dfrac 1 {N!} \dfrac {\d^N} {\d s^N} \bigintlimits {\map T s^\ell} {s \mathop = 0}{}$
+Now recall [[Cauchy's Integral Formula for Derivatives]] for a [[Definition:Complex Function|complex function]] $f$ [[Definition:Holomorphic Function|holomorphic]] on a [[Definition:Domain of Mapping|domain]] $D$, and a [[Definition:Path (Complex Plane)|path]] $\gamma \subseteq D$ [[Definition:Winding Number|winding]] once around $a$:
+:$\displaystyle \valueat {\frac {\d^N} {\d s^N} \map f s} {s \mathop = a} = \frac {N!} {2 \pi i} \oint_\gamma \frac {\map f s} {\paren {s - a}^{N + 1} } \rd s$
+Since $\map T s$ is defined by a [[Definition:Generating Function|generating function]], $\map T s^\ell$ has a [[Definition:Taylor Series|Taylor series]] about $s = 0$ which converges for all $\size s < 1$.
+Applying Cauchy's formula:
+{{begin-eqn}}
+{{eqn | l = \map {r_{\AA, \ell} } N
+      | r = \frac 1 {N!} \frac {\d^N} {\d s^N} \bigintlimits {\map T s^\ell} {s \mathop = 0} {}
+}}
+{{eqn | r = \frac 1 {2 \pi i} \oint_\gamma \frac {\map T s^\ell} {s^{N + 1} } \rd s
+}}
+{{end-eqn}}
+where $\gamma$ a [[Definition:Circle|circle]] about zero of radius $\rho$ for any $\rho < 1$.
+{{qed}}
+{{Namedfor|Godfrey Harold Hardy|name2 = John Edensor Littlewood|cat = Hardy|cat2 = Littlewood}}
+{{explain|Plenty of links to be included and explanations to be added.}}
+[[Category:Analytic Number Theory]]
+3m6riycwzi0hkml8ggvt8kgztxgf2bn
+\end{proof}<|endoftext|>
+\section{Vinogradov Circle Method}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+$\newcommand{\A}{\mathcal A}$Let $\A$ be a subset of the non-negative integers.
+For $\alpha \in \R$, let:
+:$e \left({\alpha}\right) := \exp \left({2 \pi i \alpha} \right)$
+Let:
+:$\displaystyle T_N \left({s}\right) = \sum_{\substack {a \mathop \in \A \\ a \mathop \le N}} s^a$
+be the truncated [[Definition:Generating Function|generating function]] for $\A$.
+{{explain|"truncated" [[Definition:Generating Function|generating function]]}}
+Let:
+:$V_N \left({\alpha}\right) := T_N \left({e \left({\alpha}\right)}\right)$
+Let $r_{\A, \ell} \left({N}\right)$ be the number of solutions $\left({x_1, \dotsc, x_\ell}\right) \in \A^\ell$ to the equation:
+:$(1): \quad \displaystyle x_1 + \cdots + x_l = N$
+Then:
+:$\displaystyle r_{\A, \ell} \left({N}\right) = \int_0^1 V_N \left({\alpha}\right)^\ell e \left({-N \alpha}\right) \, \mathrm d \alpha$
+\end{theorem}
+
+\begin{proof}
+For $m \in \N$ let $r_{\A, \ell} \left({m; N}\right)$ be the number of solutions to $(1)$ with no $x_i$ exceeding $N$.
+Then:
+:$\forall m \le N: r_{\A, \ell} \left({m; N}\right) = r_{\A, \ell} \left({m}\right)$
+and:
+:$\forall m > \ell N: r_{\A, \ell} \left({m; N}\right) = 0$
+Then we compute:
+:$\displaystyle T_N \left({s}\right)^\ell = \sum_{m \mathop = 0}^{\ell N} r_{\A, \ell} \left({m; N}\right) s^m$
+and:
+:$(2): \quad \displaystyle V_N \left({\alpha}\right)^\ell =  \sum_{m \mathop = 0}^{\ell N} r_{\A, \ell} \left({m; N}\right) e \left({\alpha m}\right)$
+Now it follows from [[Exponentials Form Orthonormal Basis for L^2|Exponentials Form Orthonormal Basis for $\mathcal L^2$]] that:
+:$\displaystyle \int_0^1 e \left({\alpha m}\right) e \left({-\alpha n}\right) \, \mathrm d \alpha = \delta_{m n}$
+where $\delta_{m n}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Therefore, we multiply $(2)$ by $e \left({-N \alpha}\right)$ and integrate:
+:$\displaystyle r_{\A, \ell} \left({N}\right) = r_{\A, \ell} \left({N; N}\right) = \int_0^1 \displaystyle V_N \left({\alpha}\right)^\ell e \left({-N \alpha}\right) \, \mathrm d \alpha $
+{{qed}}
+{{Namedfor|Ivan Matveevich Vinogradov|cat = Vinogradov I M}}
+[[Category:Analytic Number Theory]]
+mo7rlqmgqxyozsgkmoi5hrxm6bywtv6
+\end{proof}<|endoftext|>
+\section{Euler Product}
+Tags: Analytic Number Theory, Dirichlet Series
+
+\begin{theorem}
+Let $a_n : \N \to \C$ be an [[Definition:Arithmetic Function|arithmetic function]].
+Let $\displaystyle f \left({s}\right) = \sum_{n \mathop \in \N} a_n n^{-s}$ be its [[Definition:Dirichlet Series|Dirichlet series]].
+Let $\sigma_a$ be its [[Definition:Abscissa of Absolute Convergence|abscissa of absolute convergence]].
+Then for $\Re \left({s}\right) > \sigma_a$:
+:$\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-s} = \prod_p \frac 1 {1 - a_p p^{-s} }$
+where $p$ ranges over the [[Definition:Prime Number|primes]].
+This representation for $f$ is called an '''Euler product''' for the [[Definition:Dirichlet Series|Dirichlet series]].
+{{stub|Completely multiplicative hypothesis not mentioned. Needs also the statement: $\displaystyle f \left({s}\right) {{=}} \prod_p \left\{{\sum_{k \mathop \ge 1} a_{p^k} p^{-k s}}\right\}$ or however it goes for multiplicative functions which are not completely multiplicative}}
+\end{theorem}
+
+\begin{proof}
+This is immediate from [[Product Form of Sum on Completely Multiplicative Function]].
+{{namedfor|Leonhard Paul Euler|cat = Euler}}
+[[Category:Analytic Number Theory]]
+[[Category:Dirichlet Series]]
+p2ub82xp2f0y4q8r5tnwmokhanfez7x
+\end{proof}
+
+\begin{proof}
+From [[Euler Product]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$
+{{iff}} $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]].
+For all $n \in \Z_{\ge 1}$, let $a_n = 1$.
+From [[Convergence of P-Series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]]
+{{iff}}:
+:$\cmod z \ge 1$
+It then follows that:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sum of Geometric Sequence]]:
+:$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$
+From [[Convergence of P-Series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]]
+{{iff}}:
+:$\cmod z \ge 1$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_p \dfrac 1 {1 - p^{-z} }
+      | r = \sum_p \paren {1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | r = \paren {1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = \paren {1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = \paren {1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = \cdots
+      | c = 
+}}
+{{eqn | r = 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \cdots
+      | c = 
+}}
+{{end-eqn}}
+The result follows from the [[Fundamental Theorem of Arithmetic]].
+{{qed}}
+{{handwaving}}
+\end{proof}<|endoftext|>
+\section{Correspondence between Linear Group Actions and Linear Representations}
+Tags: Group Actions, Representation Theory
+
+\begin{theorem}
+Let $\left({k, +, \cdot}\right)$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $V$ be a [[Definition:Vector Space|vector space]] over $k$ of [[Definition:Finite Dimensional Module|finite]] [[Definition:Dimension (Linear Algebra)|dimension]].
+Let $\left({G, *}\right)$ be a [[Definition:Finite Group|finite group]].
+There is a one-to-one correspondence between [[Definition:Linear Group Action|linear group actions]] of $G$ on $V$ and [[Definition:Linear Representation|linear representations]] of $G$ in $V$, as follows:
+Let $\phi : G \times V \to V$ be a [[Definition:Group Action|group action]].
+Let $\rho : G \to \operatorname{Sym}(V)$ be a [[Definition:Permutation Representation|permutation representation]] of $G$ on $V$.
+The following are [[Definition:Logically Equivalent|equivalent]]:
+:$(1): \quad$ $\rho$ is the [[Definition:Permutation Representation/Group Action|permutation representation associated to]] $\phi$
+:$(2): \quad$ $\phi$ is the [[Definition:Group Action/Permutation Representation|group action associated to]] $\rho$
+If this is the case, the following are [[Definition:Logically Equivalent|equivalent]]:
+:$(1): \quad$ $\rho$ is a [[Definition:Linear Representation|linear representation]]
+:$(2): \quad$ $\phi$ is a [[Definition:Linear Group Action|linear group action]]
+\end{theorem}
+
+\begin{proof}
+The first equivalence follows from [[Correspondence Between Group Actions and Permutation Representations]].
+=== 1 implies 2 ===
+Let $\rho : G \to \operatorname{GL} \left({V}\right)$ be a [[Definition:Linear Representation|linear representation]] of $G$ on $V$.
+Then for all $g \in G$, $v_1, v_2 \in V$:
+{{begin-eqn}}
+{{eqn | l = \phi \left({g, v_1 + v_2}\right)
+      | r = \rho \left({g}\right) \left({v_1 + v_2}\right)
+      | c = {{Defof|Group Action Associated to Permutation Representation}}
+}}
+{{eqn | r = \rho \left({g}\right) \left({v_1}\right) + \rho \left({g}\right) \left({v_2}\right)
+      | c = $\rho \left({g}\right)$ is linear
+}}
+{{eqn | r =\phi \left({g, v_1}\right) + \phi \left({g, v_2}\right)
+      | c = {{Defof|Group Action Associated to Permutation Representation}}
+}}
+{{end-eqn}}
+and for all $g \in G$, $v \in V$, $\lambda \in k$:
+{{begin-eqn}}
+{{eqn | l = \phi \left({g, \lambda \cdot v }\right)
+      | r = \rho \left({g}\right) \left({\lambda \cdot v}\right)
+      | c = {{Defof|Group Action Associated to Permutation Representation}}
+}}
+{{eqn | r = \lambda \cdot \rho\left({g}\right) \left({v}\right)
+      | c = $\rho \left({g}\right)$ is linear
+}}
+{{eqn | r = \lambda \cdot \phi \left({g, v}\right)
+      | c = {{Defof|Group Action Associated to Permutation Representation}}
+}}
+{{end-eqn}}
+Therefore $\phi$ is a [[Definition:Linear Group Action|linear group action]] of $G$ on $V$.
+{{qed|lemma}}
+=== 2 implies 1 ===
+Let $\phi: G \times V \to V$ be a [[Definition:Linear Group Action|linear action]] of $G$ on $V$.
+Then for all $g \in G$, $v_1, v_2 \in V$:
+{{begin-eqn}}
+{{eqn | l = \rho \left({g}\right) \left({v_1 + v_2}\right)
+      | r = \phi \left({g, v_1 + v_2}\right)
+      | c = {{Defof|Permutation Representation Associated to Group Action}}
+}}
+{{eqn | r = \phi \left({g, v_1}\right) + \phi \left({g, v_2}\right)
+      | c = {{Defof|Linear Group Action}}
+}}
+{{eqn | r =  \rho \left({g}\right) \left({v_1}\right) + \rho \left({g}\right) \left({v_2}\right)
+      | c = {{Defof|Permutation Representation Associated to Group Action}}
+}}
+{{end-eqn}}
+and for all $g \in G$, $v \in V$, $\lambda \in k$:
+{{begin-eqn}}
+{{eqn | l = \rho \left({g}\right) \left({\lambda \cdot v}\right)
+      | r = \phi \left({g, \lambda \cdot v }\right)
+      | c = {{Defof|Permutation Representation Associated to Group Action}}
+}}
+{{eqn | r = \lambda \cdot \phi \left({g, v}\right)
+      | c = {{Defof|Linear Group Action}}
+}}
+{{eqn | r = \lambda \cdot \rho\left({g}\right) \left({v}\right)
+      | c = {{Defof|Permutation Representation Associated to Group Action}}
+}}
+{{end-eqn}}
+Therefore $\rho$ is a [[Definition:Linear Representation|linear representation]] of $G$ on $V$.
+{{qed|lemma}}
+{{qed}}
+{{refactor|the below can safely be merged into [[Group Action defines Permutation Representation]]}}
+Now let $g_1, g_2 \in G$.
+We have for all $v \in V$:
+{{begin-eqn}}
+{{eqn | l = \rho \left({g_1 * g_2}\right) \left({v}\right)
+      | r = \phi \left({g_1 * g_2, v}\right) 
+      | c =
+}}
+{{eqn | r = \phi \left({g_1, \phi\left({g_2, v}\right) }\right)
+      | c = $\phi$ is an action
+}}
+{{eqn | r = \rho \left({g_2}\right) \left({\rho \left({g_1}\right) \left({v}\right) }\right)
+      | c = 
+}}
+{{eqn | r = \left({ \rho \left({g_2}\right) \circ \rho \left({g_1}\right) }\right) \left({v}\right) 
+      | c = where $\circ$ is the [[Definition:Composition of Mappings|composition of mappings]]
+}}
+{{end-eqn}}
+Thus $\rho$ satisfies the [[Definition:Morphism Property|homomorphism property]].
+{{refactor|the below is better merged into the theorem statement}}
+Therefore:
+:$\hat{} : ($linear representations$) \to ($linear actions$)$
+:$\tilde{} : ($linear actions$) \to ($linear representations$)$
+give a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set Difference with Disjoint Set}
+Tags: Set Difference, Disjoint Sets
+
+\begin{theorem}
+Let $S, T$ be [[Definition:Set|sets]].
+Then:
+:$S \cap T = \O \iff S \setminus T = S$
+where:
+:$S \cap T$ denotes [[Definition:Set Intersection|set intersection]]
+:$\O$ denotes the [[Definition:Empty Set|empty set]]
+:$S \setminus T$ denotes [[Definition:Set Difference|set difference]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = S \cap T
+      | r = \O
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = S
+      | o = \subseteq
+      | r = \map \complement T
+      | c = [[Intersection with Complement is Empty iff Subset]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = S \cap \map \complement T
+      | r = S
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = S \setminus T
+      | r = S
+      | c = [[Set Difference as Intersection with Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Subtraction is Closed}
+Tags: Integer Subtraction
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Subtraction|subtraction]]:
+:$\forall a, b \in \Z: a - b \in \Z$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Integer Subtraction|subtraction]]:
+:$a - b := a + \paren {-b}$
+where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Integer Addition|integer addition]].
+From [[Integers under Addition form Abelian Group]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z, +}$ is a [[Definition: Group|group]].
+Thus:
+:$\forall a, b \in \Z: a + \paren {-b} \in \Z$
+Therefore [[Definition:Integer Subtraction|integer subtraction]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Subtraction is Closed}
+Tags: Rational Subtraction, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Rational Subtraction|subtraction]]:
+:$\forall a, b \in \Q: a - b \in \Q$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Rational Subtraction|subtraction]]:
+:$a - b := a + \paren {-b}$
+where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Rational Addition|rational number addition]].
+From [[Rational Numbers under Addition form Abelian Group]], $\struct {\Q, +}$ forms a [[Definition:Group|group]].
+Thus:
+:$\forall a, b \in \Q: a + \paren {-b} \in \Q$
+Therefore [[Definition:Rational Subtraction|rational number subtraction]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Subtraction is Closed}
+Tags: Real Subtraction, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Subtraction|subtraction]]:
+:$\forall a, b \in \R: a - b \in \R$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Real Subtraction|real subtraction]]:
+:$a - b := a + \paren {-b}$
+where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Real Addition|real number addition]].
+From [[Real Numbers under Addition form Abelian Group]]:
+:$\forall a, b \in \R: a + \paren {-b} \in \R$
+Therefore [[Definition:Real Subtraction|real number subtraction]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Subtraction is Closed}
+Tags: Complex Subtraction
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Complex Subtraction|subtraction]]:
+:$\forall a, b \in \C: a - b \in \C$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Complex Subtraction|complex subtraction]]:
+:$a - b := a + \paren {-b}$
+where $-b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Complex Addition|complex number addition]].
+From [[Complex Numbers under Addition form Abelian Group]], it follows that:
+:$\forall a, b \in \C: a + \paren {-b} \in \C$
+Therefore [[Definition:Complex Subtraction|complex number subtraction]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Positive Real Numbers Closed under Division}
+Tags: Real Division, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] $\R_{>0}$ of [[Definition:Strictly Positive Real Number|strictly positive real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Division|division]]:
+:$\forall a, b \in \R_{>0}: a \div b \in \R_{>0}$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Division|division]]:
+:$a \div b := a \times \paren {\dfrac 1 b}$
+where $\dfrac 1 b$ is the [[Definition:Inverse Element|inverse]] for [[Definition:Real Multiplication|real number multiplication]].
+From [[Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\R_{>0}, \times}$ forms a [[Definition:Group|group]].
+Thus it follows that:
+:$\forall a, b \in \R_{>0}: a \times \paren {\dfrac 1 b} \in \R$
+Therefore [[Definition:Division|real number division]] is [[Definition:Closed Algebraic Structure|closed]] in $\R_{>0}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Integral Domain}
+Tags: Integral Domains, Rational Numbers
+
+\begin{theorem}
+The [[Definition:Rational Number|set of rational numbers]] $\Q$ forms an [[Definition:Integral Domain|integral domain]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]: $\struct {\Q, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Recall that [[Rational Numbers form Field]].
+The result then follows directly from [[Field is Integral Domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers form Integral Domain}
+Tags: Integral Domains, Real Numbers
+
+\begin{theorem}
+The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Integral Domain|integral domain]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Recall that [[Real Numbers form Field]].
+The result then follows directly from [[Field is Integral Domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Content}
+Tags: Content of Polynomial
+
+\begin{theorem}
+Let $f, g \in \Q \sqbrk X$ and $q \in \Q$.
+The [[Definition:Content of Polynomial|content]] of a [[Definition:Polynomial (Abstract Algebra)|polynomial]] satisfies the following:
+\end{theorem}<|endoftext|>
+\section{Complex Numbers form Integral Domain}
+Tags: Integral Domains, Complex Numbers
+
+\begin{theorem}
+The [[Definition:Complex Number|set of complex numbers]] $\C$ forms an [[Definition:Integral Domain|integral domain]] under [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]]: $\struct {\C, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Recall that [[Complex Numbers form Field]].
+The result then follows directly from [[Field is Integral Domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Root of Polynomial iff Divisible by Minimal Polynomial}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $L / K$ be a [[Definition:Field Extension|field extension]] of $K$.
+Let $\alpha \in L$ be [[Definition:Algebraic over Field|algebraic over $K$]].
+Then there is a unique [[Definition:Monic Polynomial|monic polynomial]] $\mu_\alpha \in K \left[{X}\right]$ of least [[Definition:Degree (Polynomial)|degree]] such that $\mu_\alpha \left({\alpha}\right) = 0$.
+Moreover $f \in K \left[{X}\right]$ is such that $f \left({\alpha}\right) = 0$ {{iff}} $\mu_\alpha$ [[Definition:Divisor of Polynomial|divides]] $f$.
+{{explain|Link to an appropriate page defining divisibility among polynomials.}}
+\end{theorem}
+
+\begin{proof}
+By definition, there is some polynomial $f \in K \left[{X}\right]$ such that $f \left({\alpha}\right) = 0$.
+Let $\mu$ be such a polynomial of minimal degree, and define:
+:$\mu_\alpha = \dfrac 1 a \mu $
+where $a$ is the [[Definition:Leading Coefficient (Polynomial)|leading coefficient]] of $\mu$.
+Clearly this polynomial has the required properties, so we show uniqueness.
+Let $\mu'$ be another such polynomial. Then
+:$\mu_\alpha \left({\alpha}\right) - \mu' \left({\alpha}\right) = 0 - 0 = 0$
+Since both polynomials are monic, $\mu_\alpha - \mu'$ has degree strictly less than $\mu_\alpha$.
+So we must have $\mu_\alpha - \mu' = 0$.
+Let $\mu_\alpha \divides f$.
+Then:
+:$f = g \mu_\alpha$ for some $g \in K \left[{X}\right]$
+and:
+:$f \left({\alpha}\right) = 0 \cdot g \left({\alpha}\right) = 0$
+Conversely, let $f \in K \left[{X}\right]$ be any polynomial such that $f \left({\alpha}\right) = 0$.
+By the [[Division Theorem for Polynomial Forms over Field]], there exists $q, r \in K \left[{X}\right]$ such that:
+:$f = q \mu_\alpha + r$
+and:
+:$\deg r < \deg \mu_\alpha$.
+Evaluating this expression at $\alpha$ we find that:
+:$f \left({\alpha}\right) = q \left({\alpha}\right) \mu_\alpha \left({\alpha}\right) + r \left({\alpha}\right) \implies r \left({\alpha}\right) = 0$
+since $\mu_\alpha \left({\alpha}\right) = f \left({\alpha}\right) = 0$.
+But $\mu_\alpha$ has minimal degree among the non-zero polynomials that are zero at $\alpha$.
+Therefore as $\deg r < \deg \mu_\alpha$ we must have $r = 0$.
+Therefore:
+:$f = q \mu_\alpha$
+That is, $\mu_\alpha$ divides $f$.
+{{qed}}
+[[Category:Field Extensions]]
+mg9uxte3mllfsldwer50b4uqk1lqt16
+\end{proof}<|endoftext|>
+\section{Gauss's Lemma on Unique Factorization Domains}
+Tags: Factorization, Unique Factorization Domains
+
+\begin{theorem}
+Let $R$ be a [[Definition:Unique Factorization Domain|unique factorization domain]].
+Then the [[Definition:Ring of Polynomials|ring of polynomials]] $R \sqbrk X$ is also a [[Definition:Unique Factorization Domain|unique factorization domain]].
+\end{theorem}
+
+\begin{proof}
+Since a [[UFD is Noetherian]], and a [[Noetherian Domain is UFD if every irreducible element is prime]], it is sufficient to prove that every [[Definition:Irreducible Polynomial|irreducible]] element of $R \sqbrk X$ is [[Definition:Prime Number|prime]].
+etc,
+{{ProofWanted|this possibly works}}
+{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
+[[Category:Factorization]]
+[[Category:Unique Factorization Domains]]
+bljlmlki1jz1fkjcafneorlm61c8zzm
+\end{proof}<|endoftext|>
+\section{Product with Inverse equals Identity iff Equality}
+Tags: Group Theory
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$.
+Then:
+:$\forall a, b \in G: a \circ b^{-1} = e \iff a = b$
+\end{theorem}
+
+\begin{proof}
+Using various properties of [[Definition:Group|groups]]:
+{{begin-eqn}}
+{{eqn | l = a \circ b^{-1}
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {a \circ b^{-1} } \circ b
+      | r = e \circ b
+      | c = [[Cancellation Laws]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a \circ \paren {b^{-1} \circ b}
+      | r = e \circ b
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a
+      | r = b
+      | c = {{GroupAxiom|2}} and {{GroupAxiom|3}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Group Theory]]
+hn59z6bi84g0y3d4d3ijnrryeortqfn
+\end{proof}<|endoftext|>
+\section{Gröbner Basis}
+Tags: 
+
+\begin{theorem}
+Let $F$ be a finite set of polynomials.
+{{explain|Explain what the underlying field of the polynomials is. Real numbers, abstract rings, general fields? Link to the appropriate definition for the case.}}
+Let $SP \left({f_1, f_2}\right)$ be the [[Definition:S-Polynomial|$S$-polynomial]] of $f_1$ and $f_2$.
+Let $g$ be a polynomial.
+Let $RF \left({g}\right)$ be the [[Definition:Reduced Form of Polynomial|reduced form]] of $g$.
+Then $F$ is a [[Definition:Gröbner Basis|Gröbner basis]] {{iff}}:
+:$\forall f_1, f_2 \in F \left({RF \left({F, SP \left({f_1, f_2}\right)}\right) = 0}\right)$
+{{explain|The meaning of the above needs to be made clear. Presumably it should be interpreted something like: $\forall f_1, f_2 \in F: RF \left({F, SP \left({f_1, f_2}\right)}\right) {{=}} 0$. But further to this: $RF$ as defined takes a single argument, a polynomial (still to be defined upon what domain). As this statement stands, $RF$ seems to take two parameters: $F$ and $SP$. Clarification is needed.}}
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $F$ be a [[Definition:Gröbner Basis|Gröbner basis]].
+Let $f_1, f_2 \in F$.
+Then:
+:$SP \left({f_1, f_2}\right) \in Ideal \left({F}\right)$
+That is:
+:$SP \left({f_1, f_2}\right) \equiv_F 0$
+{{explain|Link to a result explaining the above.}}
+By the relation between reduction and congruence this implies that:
+:$SP \left({f_1, f_2}\right) \leftrightarrow_F^* 0$
+{{explain|Define via links to the appropriate pages the meaning of the above.}}
+Hence:
+:$RF \left({F, SP \left({f_1, f_2}\right)}\right) = RF \left({F, 0}\right) = 0$ 
+because $F$ is a [[Definition:Gröbner Basis|Gröbner basis]] (and because of the equivalence
+between the [[Definition:Church-Rosser Property|Church-Rosser property]] and the [[Definition:Normal Form Church-Rosser Property|normal form Church-Rosser property]]).
+{{explain|Establish the above equivalence and the conclusion drawn from it in a separate page.}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let:
+:$\forall f_1, f_2 \in F \left({RF \left({F, SP \left({f_1, f_2}\right)}\right) = 0}\right)$
+By the [[Definition:Generalized Newman Lemma|generalized Newman lemma]] and:
+: $\gets_F \mathop \subseteq \succ$
+it suffices to prove [[Definition:Local Connectibility|local connectibility]].
+That is, it suffices to prove that under the assumption:
+:$g_1 \to_F h \gets_F g_2$
+we have:
+:$g_1 \stackrel{\prec h} {\longleftrightarrow^*_F} g_2$
+{{explain|All the above notation needs to be explained.}}
+Let $LPP \left({f}\right)$ be the [[Definition:Leading Power Product|leading power product]] of $f$.
+Then by the assumption, there exist $f_1, f_2 \in F$ and $t_1, t_2 \in S \left({h}\right)$ with:
+: $LPP \left({f_1}\right) | t_1$
+and:
+: $LPP \left({f_2}\right) | t_2$
+such that:
+: $h \rightarrow_{f_1, t_1} g_1$
+and:
+: $h \rightarrow_{f_2, t_2} g_2$
+Now we have three cases.
+''Case $t_1 \succ t_2$'':
+In this case:
+{{begin-eqn}}
+{{eqn | l = g_1
+      | r = H \left({h, t_1}\right) + 0 \cdot t_1 + B \left({h, t_1, t_2}\right) + C \left({h, t_2}\right) \cdot t_2
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = L \left({h, t_2}\right) - C \left({h, t_1}\right) \cdot u_1 \cdot R \left({f_1}\right)
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = g_2
+      | r = H \left({h, t_1}\right) + C \left({h, t_1}\right) \cdot t_1 + B \left({h, t_1, t_2}\right) + 0 \cdot t_2
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = L \left({h, t_2}\right) - C \left({h, t_2}\right) \cdot u_2 \cdot R \left({f_2}\right)
+      | c = 
+}}
+{{end-eqn}}
+where:
+: $u_1 := \dfrac {t_1} {LPP \left({f_1}\right)}$
+and:
+: $u_2 := \dfrac {t_2} {LPP \left({f_1}\right)}$
+Furthermore:
+{{begin-eqn}}
+{{eqn | l = g_2 \to_{f_1} g_{1, 2}
+      | r = H \left({h, t_1}\right) + 0 \cdot t_1 + B \left({h, t_1, t_2}\right) + 0 \cdot t_2 + L \left({h, t_2}\right)
+      | c = 
+}}
+{{eqn | o =
+      | ro= -
+      | r = C \left({h, t_1}\right) \cdot u_1 \cdot R \left({f_1}\right) - C \left({h, t_2}\right) \cdot u_2 \cdot R \left({f_2}\right)
+      | c = 
+}}
+{{end-eqn}}
+Now:
+: $g_1 = h - C \left({h, t_1}\right) \cdot u_1 \cdot f_1$
+and:
+: $g_{1, 2} = g_2 - C \left({h, t_1}\right) \cdot u_1 \cdot f_1$
+and, by assumption:
+: $h \to_F g_2$
+Hence, by [[Definition:Sum Semi-Compatibility|sum semi-compatibility]]:
+:$g_1 \downarrow_{\stackrel{*} {F} } g_{1, 2}$
+and hence:
+:$g_1 \stackrel{\prec h}{\longleftrightarrow^{*}_F} g_2$
+(Note that, in general, $g_1 \to_{f_1} g_{1, 2}$ need not be the case. Why not?)
+{{explain|Not {{ProofWiki}} style. Needs to be linked to a page demonstrating this fact.}}
+''Case $t_1 \prec t_2$'':
+Analogous.
+''Case $t := t_1 = t_2$'':
+In this case:
+:$g_1 = H \left({h, t}\right) + 0 \cdot t + L \left({h, t}\right) - C \left({h, t}\right) \cdot u_1 \cdot R \left({f_1}\right)$
+and:
+:$g_2 = H \left({h, t}\right) + 0 \cdot t + L \left({h, t}\right) - C \left({h, t}\right) \cdot u_2 \cdot R \left({f_2}\right)$
+Hence:
+{{begin-eqn}}
+{{eqn | l = g_1 - g_2
+      | r = - C \left({h, t}\right) \cdot \left({u_1 \cdot R \left({f_1}\right) - u_2 \cdot R \left({f_2}\right)}\right)
+      | c = 
+}}
+{{eqn | r = - C \left({h, t}\right) \cdot \left({u_1 \cdot f_1 - u_2 \cdot f_2}\right)
+      | c = 
+}}
+{{eqn | r = - C \left({h, t}\right) \cdot v \cdot SP \left({f_1, f_2}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{explain|Explain why the above steps hold, by links to explanatory pages if necessary}}
+where:
+:$v := t / LCM \left({LPP \left({f_1}\right), LPP \left({f_2}\right)}\right)$
+{{explain|Presumably here LCM means lowest common multiple?}}
+                                     
+We have assumed that:
+: $RF \left({F, SP \left({f_1, f_2}\right)}\right) = 0$
+That is:
+: $SP \left({f_1, f_2}\right) \to_{\stackrel{\ast}{F}} 0$
+Hence,by [[Definition:Product Compatibility|product compatibility]]:
+:$g_1 - g_2 = - C \left({h, t}\right) \cdot v \cdot SP \left({f_1, f_2}\right) \to_{\stackrel{*}{F}} 0$ 
+This means that there exists a [[Definition:Sequence|sequence]] $p \in P^{\ast}$ such that:
+:
+{{begin-eqn}}
+{{eqn | l = p_1
+      | r = g_1 - g_2
+      | c = 
+}}
+{{eqn | n = 1
+      | l = \forall 1 \le i < \left\vert{p}\right\vert:
+      | o = 
+      | r = \left({p_i \to_F p_{i + 1} }\right)
+      | c = 
+}}
+{{end-eqn}}
+{{explain|What is $P^*$?}}
+and:
+:$p_{\left\vert{p}\right\vert} = 0 $
+Furthermore note that, because of $\to_F \subseteq \succ$:
+:$\forall 1 \le i < \left\vert{p}\right\vert: p_i\preceq g_1 -g_2 \prec h$
+Thus, by [[Definition:Sum Semi-Compatibility|sum semi-compatibility]] applied to $(1)$:
+:$g_1 = p_1 + g_2$
+:$\forall 1 \le i < \left\vert{p}\right\vert: p_i + g_2 \downarrow_{\stackrel{*}{F} } p_{i+1} + g_2$
+:$g_2 = p_{\left\vert{p}\right\vert} + g + 2$
+Also, we have:
+:$\forall 1 \le i < \left\vert{p}\right\vert: p_i + g_2 \prec h$
+because:
+:$\forall 1 \le i < \left\vert{p}\right\vert: H \left({p_i + g_2, t}\right) = H \left({h, t}\right) \wedge C \left({p_i + g_2, t}\right) = 0$
+Thus, summarizing:
+: $g_1 \stackrel{\prec h} {\longleftrightarrow^{*}_F} g_2$ also in this case.
+\end{proof}<|endoftext|>
+\section{Numbers of Type Integer a plus b root 2 form Subdomain of Reals}
+Tags: Integral Domains, Numbers of Type Integer a plus b root 2 form Subdomain of Reals
+
+\begin{theorem}
+Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
+:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
+That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
+Then the [[Definition:Algebraic Structure|algebraic structure]]:
+:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
+where $+$ and $\times$ are conventional [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] on [[Definition:Real Number|real numbers]], form an [[Definition:Subdomain|integral subdomain]] of the [[Definition:Real Number|real numbers]] $\R$.
+\end{theorem}
+
+\begin{proof}
+$\Z \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of [[Definition:Real Number|real numbers]], so we immediately have that [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] are [[Definition:Well-Defined Operation|well-defined]].
+=== Closure ===
+Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a_1 + b_1 \sqrt 2} + \paren {a_2 + b_2 \sqrt 2}
+      | r = \paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \Z \sqbrk {\sqrt 2}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {a_1 + b_1 \sqrt 2} \times \paren {a_2 + b_2 \sqrt 2}
+      | r = \paren {a_1 \times a_2 + 2 \times b_1 \times b_2} + \paren {a_1 \times b_2 + b_1 \times a_2} \sqrt 2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \Z \sqbrk {\sqrt 2}
+      | c = 
+}}
+{{end-eqn}}
+So both the operations $+$ and $\times$ are [[Definition:Closed Algebraic Structure|closed on $\Z \sqbrk {\sqrt 2}$]].
+{{qed|lemma}}
+=== Associativity ===
+We have that [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] are [[Definition:Associative|associative]] on $\R$.
+Therefore it follows from [[Restriction of Associative Operation is Associative]] that they are also [[Definition:Associative|associative]] on $\Z \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Commutativity ===
+We have that [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] are [[Definition:Commutative Operation|commutative]] on $\R$.
+Therefore it follows from [[Restriction of Commutative Operation is Commutative]] that they are also [[Definition:Commutative Operation|commutative]] on $\Z \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Identities ===
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b \sqrt 2} + \paren {0 + 0 \sqrt 2}
+      | r = \paren {a + 0} + \paren {b + 0} \sqrt 2
+      | c = 
+}}
+{{eqn | r = a + b \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$.
+So $\paren {0 + 0 \sqrt 2}$ is the [[Definition:Identity Element|identity]] for $+$ on $\Z \sqbrk {\sqrt 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}
+      | r = \paren {a \times 1 + 2 \times b \times 0} + \paren {b \times 1 + a \times 0} \sqrt 2
+      | c = 
+}}
+{{eqn | r = a + b \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}$.
+So $\paren {1 + 0 \sqrt 2}$ is the [[Definition:Identity Element|identity]] for $\times$ on $\Z \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Inverses ===
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b \sqrt 2} + \paren {-a + \paren {-b} \sqrt 2}
+      | r = \paren {a - a} + \paren {b - b} \sqrt 2
+      | c = 
+}}
+{{eqn | r = 0 + 0 \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {-a + \paren {-b} \sqrt 2} + \paren {a + b \sqrt 2}$.
+So $\paren {-a + \paren {-b} \sqrt 2}$ is the [[Definition:Inverse Element|inverse]] of $\paren {a + b \sqrt 2}$ for $+$ on $\Z \sqbrk {\sqrt 2}$.
+We have no need to investigate inverses for $\times$ (which is convenient as $\times$ happens not to be closed for inverses on $\Z \sqbrk {\sqrt 2}$).
+{{qed|lemma}}
+=== Distributivity ===
+We have that [[Real Multiplication Distributes over Addition]], so by [[Restriction of Operation Distributivity]], $\times$ is [[Definition:Distributive Operation|distributive]] over $+$ on $\Z \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Divisors of Zero ===
+By [[Real Numbers form Field]], $\R$ is a [[Definition:Field (Abstract Algebra)|field]] and so by [[Field is Integral Domain]] is itself an [[Definition:Integral Domain|integral domain]].
+Hence it has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+{{qed|lemma}}
+The result follows by putting all the pieces together.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Integers form Subdomain of Reals]], $\struct {\Z, +, \times}$ is an [[Definition:Subdomain|integral subdomain]] of the [[Definition:Real Number|real numbers]] $\R$.
+We have that $\sqrt 2 \in \R$.
+Every expression of the form:
+:$a_0 + a_1 \sqrt 2 + a_2 \paren {\sqrt 2}^2 + \cdots + a_n \paren {\sqrt 2}^n$
+can be simplified to a number of the form $a + b \sqrt 2$, where $a, b \in \Z$.
+The result follows from [[Set of Polynomials over Integral Domain is Subring]].
+\end{proof}<|endoftext|>
+\section{Sufficient Conditions for Uncountability}
+Tags: Uncountable Sets
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+The following are equivalent:
+:$(1): \quad X$ contains an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]]
+:$(2): \quad X$ is [[Definition:Uncountable Set|uncountable]]
+:$(3): \quad $ Every [[Definition:Sequence of Distinct Terms|sequence of distinct points]] $\sequence {x_n}_{n \mathop \in \N}$ in $X$ omits at least one $x \in X$
+:$(4): \quad $ There is no [[Definition:Surjection|surjection]] $\N \twoheadrightarrow X$
+:$(5): \quad X$ is [[Definition:Infinite Set|infinite]] and there is no [[Definition:Bijection|bijection]] $X \leftrightarrow \N$
+Assuming the [[Continuum Hypothesis]] holds, we also have the equivalent [[Definition:Uncountable Set|uncountability]] condition:
+:$(6): \quad $There exist [[Definition:Extended Real Number Line|extended real numbers]] $a < b$ and a [[Definition:Surjection|surjection]] $X \to \closedint a b$
+\end{theorem}
+
+\begin{proof}
+Recall that $X$ is [[Definition:Uncountable Set|uncountable]] if there is no [[Definition:Injection|injection]] $X \hookrightarrow \N$.
+=== $(1)$ implies $(2)$ ===
+Suppose there exists an [[Definition:Injection|injection]] $f: X \hookrightarrow \N$.
+Let $Y \subseteq X$ be [[Definition:Uncountable Set|uncountable]].
+Then $\hat f: Y \to \N$ defined as:
+:$\hat f = \set {\map f x: x \in Y}$
+is an [[Definition:Injection|injection]] $Y \hookrightarrow N$, which is a [[Definition:Contradiction|contradiction]].
+{{qed|lemma}}
+=== $(2)$ implies $(3)$ ===
+Suppose $\sequence {x_n}_{n \mathop \in \N}$ is a [[Definition:Sequence of Distinct Terms|sequence of distinct points]] such that every $x \in X$ equals $x_n$ for some $n$.
+Since the $x_n$ are [[Definition:Distinct|distinct]], this $n$ is unique.
+Let $f: X \to \N$ be defined by $\map f x = n$, where $n$ is the unique [[Definition:Natural Number|natural number]] such that $x_n = x$.
+Then $f$ is [[Definition:Injection|injective]] because $n = m$ implies that $x_n = x_m$.
+So there is an [[Definition:Injection|injection]] $X \hookrightarrow \N$, a contradiction.
+{{qed|lemma}}
+=== $(3)$ implies $(4)$ ===
+Suppose there exists a [[Definition:Surjection|surjection]] $f: \N \to X$.
+Therefore every $x \in X$ is $\map f n$ for some $n \in \N$.
+Define $g: X \to \N$ by $\map g x = \inf \set {n \in \N : \map f n = x}$.
+Then if $\map g {x_1} = \map g {x_2} = n$, we have $\map f n = x_1$ and $\map f n = x_2$.
+So by the definition of a [[Definition:Mapping|mapping]], $x_1 = x_2$.
+Therefore $g$ is an [[Definition:Injection|injection]] $X \to \N$, a contradiction.
+{{qed|lemma}}
+=== $(4)$ implies $(5)$ ===
+Since there exists no [[Definition:Surjection|surjection]] $\N \to X$, and a [[Definition:Bijection|bijection]] $\N \to X$ must be [[Definition:Surjection|surjective]], there is no such [[Definition:Mapping|mapping]].
+{{qed|lemma}}
+=== $(5)$ implies $(1)$ ===
+Suppose $f: X \hookrightarrow \N$ is an [[Definition:Injection|injection]].
+We essentially relabel the [[Definition:Image of Mapping|image]] of $f$ to obtain a [[Definition:Bijection|bijection]], and thus a contradiction.
+Construct a map $g: X \to \N$ as follows.
+Let $S_1 = \Img f$ and:
+:$x_1 = \map {f^{-1} } {\inf \set {n \in \N : n \in S_1} }$
+and define $\map f x = 1$.
+Note that $x$ is a [[Definition:Singleton|singleton]] because $f$ is [[Definition:Injection|injective]].
+Given $S_n \subseteq \Img f$, let:
+:$x_n = \map {f^{-1} } {\inf \set {n \in \N : n \in S_n} }$
+and set $S_{n + 1} = S_n \setminus x_n$.
+The [[Definition:Mapping|mapping]]:
+:$n \stackrel {f^{-1} } \longrightarrow x_k \stackrel g \longrightarrow k$
+defines a [[Definition:Bijection|bijection]] from $\Img f$ to $\N$, so $g$ is [[Definition:Injection|injective]] because $f$ is.
+Furthermore $g$ assigns each $n \in \N$ to some $x \in X$ (because $X$ is [[Definition:Infinite Set|infinite]]) so $g$ is [[Definition:Surjection|surjective]].
+Thus $g$ is a [[Definition:Bijection|bijection]], a contradiction.
+{{qed|lemma}}
+=== $(6)$ implies $(1)$ ===
+{{proof wanted|to do}}
+{{qed}}
+[[Category:Uncountable Sets]]
+951p86012533eucwpxfwrtezutjl1c8
+\end{proof}<|endoftext|>
+\section{Direct Product iff Nontrivial Idempotent}
+Tags: Commutative Algebra
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Then $R$ is the direct product of two [[Definition:Non-Trivial Ring|non-trivial rings]] {{iff}} $R$ contains an [[Definition:Idempotent Element|idempotent element]] not equal to $0_R$ or $1_R$.
+\end{theorem}
+
+\begin{proof}
+If $R$ has a nontrivial decomposition $R = R_1 \times R_2$ then $\tuple {1_R, 0_R}$ is a non-trivial idempotent element of $R$.
+{{explain|non-trivial decomposition, non-trivial element}}
+Conversely suppose there is $0_R, 1_R \ne e \in R$ with $e^2 = e$.
+Let $R_1 = \ideal e$, the [[Definition:Ideal of Ring|ideal]] [[Definition:Generator|generated]] by $e$, and $R_2 = R / \ideal e$.
+Since $e \paren {e - 1_R} = 0_R$, it follows by definition that $e$ is a [[Definition:Zero Divisor of Ring|zero divisor]].
+So by [[Unit Not Zero Divisor]] it is not a [[Definition:Unit of Ring|unit]].
+Therefore, $1 \notin R_1$ and $\ideal e \subsetneqq R$.
+Also $R_1 \cap R_2 = \set {0_R}$ so the product is direct (that is, the [[Universal Property for Direct Products]] is satisfied).
+Finally we define the "gluing [[Definition:Ring Homomorphism|homomorphism]]" $\phi : R \to R_1 \times R_2$ by
+:$\phi: a \mapsto \tuple {a e, a + \ideal e}$
+which is easily shown to be an [[Definition:Ring Isomorphism|isomorphism]].
+{{qed}}
+[[Category:Commutative Algebra]]
+fh03qu2of6klj77b3msijjski733boh
+\end{proof}<|endoftext|>
+\section{Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+Then:
+:$\forall x \in D: x \ne 0_D \iff \map P {x \times x}$
+where $\map P {x \times x}$ denotes that $x \times x$ has the [[Definition:Strict Positivity Property|(strict) positivity property]].
+That is, the [[Definition:Square Element|square]] of any [[Definition:Element|element]] of an [[Definition:Ordered Integral Domain|ordered integral domain]] is [[Definition:Strict Positivity Property|(strictly) positive]] {{iff}} that [[Definition:Element|element]] is non-[[Definition:Ring Zero|zero]].
+\end{theorem}
+
+\begin{proof}
+Let $x = 0_D$.
+Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the [[Definition:Ring Zero|ring zero]].
+Thus by definition of [[Definition:Strict Positivity Property|strict positivity property]]:
+:$\neg \map P {0_D \times 0_D}$
+{{qed|lemma}}
+Now suppose $x \ne 0_D$.
+One of two cases applies:
+:$\map P x$
+:$\neg \map P x$
+Let $\map P x$.
+Then by definition of [[Definition:Strict Positivity Property|(strict) positivity]]:
+:$\map P {x \times x}$
+Now suppose $\neg \map P x$.
+Then by the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]]:
+:$\map P {-x}$
+Then again by definition:
+:$\map P {\paren {-x} \times \paren {-x} }$.
+But by [[Product of Ring Negatives]]:
+:$\paren {-x} \times \paren {-x} = x \times x$
+So again:
+:$\map P {x \times x}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simultaneous Equation With Two Unknowns}
+Tags: Linear Algebra
+
+\begin{theorem}
+A pair of [[Definition:Simultaneous Equations|simultaneous linear equations]] of the form:
+{{begin-eqn}}
+{{eqn | l=a x + b y
+      | r=c
+}}
+{{eqn | l=d x + e y
+      | r=f
+}}
+{{end-eqn}}
+where $ ae \ne b d$, has as its only [[Definition:Simultaneous Equations#Solution|solution]]:
+{{begin-eqn}}
+{{eqn | l=x
+      | r= \frac {c e - b f} {a e - b d}
+}}
+{{eqn | l=y
+      | r=\frac {a f - c d} {a e - b d}
+}}
+{{end-eqn}}
+\end{theorem}<|endoftext|>
+\section{Open Sets in Real Number Line}
+Tags: Open Sets, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Every non-empty [[Definition:Open Set (Real Analysis)|open set]] $I \subseteq \R$ can be expressed as a [[Definition:Countable|countable]] [[Definition:Set Union|union]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Open Real Interval|open intervals]].
+If:
+:$\displaystyle I = \bigcup_{n \mathop \in \N} J_n$
+:$\displaystyle I = \bigcup_{n \mathop \in \N} K_n$
+are two such expressions, then there exists a [[Definition:Permutation|permutation]] $\sigma$ of $\N$ such that:
+:$\forall n \in \N: J_n = K_{\map \sigma n}$
+\end{theorem}
+
+\begin{proof}
+We know that [[Real Number Line is Complete Metric Space|$\R$ is a complete metric space]].
+Let $x \in I$.
+By the [[Definition:Open Set (Metric Space)|definition of open sets in a metric space]], there is an [[Definition:Open Real Interval|open interval]] $I_x$, [[Definition:Subset|included]] in $I$, that contains $x$.
+Define the following:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map a x
+      | r = \inf \set {z: \openint z x \subseteq I}
+}}
+{{eqn | n = 2
+      | l = \map b x
+      | r = \sup \set {y: \openint x y \subseteq I}
+}}
+{{end-eqn}}
+The existence of $I_x$ assures us that we have $\map a x < x < \map b x$.
+In this way we associate a non-empty [[Definition:Open Real Interval|open interval]]:
+:$\map J x = \openint {\map a x} {\map b x}$
+to $x \in I$.
+Let $\epsilon > 0$ be arbitrary.
+As $\map a x$ is an [[Definition:Infimum of Set|infimum]], there exists $z < \map  a x + \epsilon$ such that:
+:$\openint z x \subseteq I$
+Thus certainly:
+:$\openint {\map a x + \epsilon} x \subseteq I$
+It follows that we have:
+:$\displaystyle \openint {\map a x} x = \bigcup_{n \mathop = 1}^\infty \openint {\map a x + \frac 1 n} x  \subseteq I$
+By a similar argument:
+:$\openint x {\map b x} \subseteq I$
+Therefore, as we have:
+:$\map J x = \openint {\map a x} x \cup \set x \cup \openint x {\map b x}$
+we conclude that:
+:$\map J x \subseteq I$
+Suppose now that we have $\map a x = -\infty$ or $\map b x = +\infty$.
+If both are the case, it must be that $I = \R$, and hence $I$ is open.
+It can then also be written uniquely as the [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Set Union|union]] of the single open interval $\R$.
+In other cases, we observe that intervals of the type:
+:$\map {K_-} a = \openint {-\infty} a$
+or:
+:$\map {K_+} a = \openint a {+\infty}$
+are open.
+Assume $I$ contains such an interval $\map {K_\pm} a$ with $a \notin I$.
+Then we have:
+:$\displaystyle I = \paren {I \cap \map {K_-} a} \cup \paren {I \cap \map {K_+} a}$
+Thus $I$ is open {{iff}} $I \setminus \map {K_\pm} a$ is open, since the $\map {K_\pm} a$ are disjoint, open sets.
+It is also clear that a unique decomposition of $I \setminus \map {K_\pm} a$ into disjoint open intervals will give rise to such a decomposition for $I$.
+Therefore, without loss of generality, we assume $I$ contains no such interval.
+Define a [[Definition:Relation|relation]] $\sim$ on $I$ by $x \sim y$ {{iff}} $\map J x = \map J y$ (the [[Definition:Equivalence Relation Induced by Mapping|induced equivalence]] of $J$).
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]] on $I$ by [[Relation Induced by Mapping is Equivalence Relation]].
+Therefore, by the [[Fundamental Theorem on Equivalence Relations]], $\sim$ [[Definition:Partition (Set Theory)|partitions]] $I$.
+In fact, the equivalence classes are [[Definition:Open Real Interval|open intervals]]. 
+To prove this, let $x < y$ with $x, y\in I$. We see that when $x \sim y$, we have $x \in \map J x = \map J y \ni y$. 
+Also, when $x \in \map J y$, it follows that $\openint x y \subseteq I$. 
+Therefore, we must have:
+:$\map a x = \map a y$
+and:
+:$\map b x = \map b y$
+hence:
+:$\map J x = \map J y$
+This implies that the equivalence class of $x$ is precisely $\map J x$.
+Finally, as:
+:$\openint {\map a x} {\map b x} \ne \O$
+by [[Rationals are Everywhere Dense in Topological Space of Reals]] there exists:
+:$q \in \openint {\map a x} {\map b x} \cap \Q$
+Therefore each set in the partition of $I$ can be labelled with a [[Definition:Rational Number|rational number]].
+Since the [[Rational Numbers are Countably Infinite]], the partition is [[Definition:Countable|countable]].
+Then [[Definition:Enumeration|enumerating]] the disjoint intervals of $I$, we have an expression:
+:$\displaystyle I = \bigcup_{n \mathop \in \N} J_n $
+Now let:
+:$\displaystyle I = \bigcup_{n \mathop \in \N} J_n,\qquad I = \bigcup_{n \mathop \in \N} K_n$
+be two such expression for $I$.
+Suppose that there exists $m_0 \in \N$ such that for all $n \in \N$, $K_{m_0} \ne J_n$.
+Write $K_{m_0} = \openint {a_0} {b_0}$.
+Since $I$ contains $K_{m_0}$ there exists $J_{m_0'} = \openint {a_0'} {b_0'}$ such that $J_{m_0'} \cap K_{m_0} \ne \O$.
+This means that either $a_0' \in \openint {a_0} {b_0}$ or $b_0' \in \openint {a_0} {b_0}$.
+Therefore either:
+:$J_{m_0'} \setminus K_{m_0} = \hointr {b_0'} {b_0}$
+or:
+:$J_{m_0'} \setminus K_{m_0} = \hointl {a_0} {a_0'}$
+Then we have one of the following two expressions:
+:$\displaystyle I \setminus K_{m_0} = \bigcup_{n \mathop \ne m_0} \paren {J_n \setminus K_{m_0} } \cup \hointr {b_0'} {b_0}$
+:$\displaystyle I \setminus K_{m_0} = \bigcup_{n \mathop \ne m_0} \paren {J_n \setminus K_{m_0} } \cup \hointl {a_0} {a_0'}$
+Now by [[Half-Open Real Interval is neither Open nor Closed]], and since this union is disjoint, we have that $I\setminus K_{m_0}$ is not open.
+On the other hand, we have:
+:$\displaystyle I \setminus K_{m_0} = \bigcup_{n \mathop \ne m_0} K_n$
+which is a union of open sets, hence open.
+This contradiction shows that for every $m \in \N$ there exists $\map \sigma m \in \N$ such that $K_m = J_{\map \sigma m}$.
+Moreover since the $J_n$, $n \in \N$ are disjoint, there can be only one such $\map \sigma m$, so $\sigma$ is a permutation.
+{{qed}}
+[[Category:Open Sets]]
+[[Category:Real Number Line with Euclidean Metric]]
+biss2fk2311lbh0sxdynpd1iji8vold
+\end{proof}<|endoftext|>
+\section{Unity of Ordered Integral Domain is Strictly Positive}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Then:
+:$\map P {1_D}$
+where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
+\end{theorem}
+
+\begin{proof}
+We have by definition of the [[Definition:Unity of Ring|unity]] that:
+:$\forall a \in D: 1_D \times a = a = a \times 1_D$
+This particularly applies to $1_D$ itself:
+:$1_D = 1_D \times 1_D$
+But then by [[Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive]]:
+:$\map P {1_D \times 1_D} \implies \map P {1_D}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers form Ordered Integral Domain}
+Tags: Integers, Ordered Integral Domains
+
+\begin{theorem}
+The [[Definition:Integer|integers]] $\Z$ form an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Integral Domain]] we have that $\struct {\Z, +, \times}$ forms an [[Definition:Integral Domain|integral domain]].
+From [[Natural Numbers are Non-Negative Integers]] we have that the set $\N$ can be considered as a [[Definition:Subset|subset]] of the [[Definition:Integer|integers]].
+Then we have that [[Non-Zero Natural Numbers form Commutative Semiring|$\struct {\N_{> 0}, +, \times}$ is a (commutative) semiring]].
+So it follows by definition of [[Definition:Semiring (Abstract Algebra)|semiring]], in particular the fact that $+$ and $\times$ on $\N_{> 0}$ are [[Definition:Closed Algebraic Structure|closed]], that:
+:$\forall a, b \in \N_{> 0}: a + b \in \N_{> 0}$
+:$\forall a, b \in \N_{> 0}: a \times b \in \N_{> 0}$
+It follows that we can define a property $P$ on $\Z$ such that:
+:$\forall x \in \Z: \map P x \iff x \in \N_{> 0}$
+Checking that $P$ fulfils the conditions for it to be the [[Definition:Strict Positivity Property|(strict) positivity property]]:
+: $(1): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a + b}$
+Follows directly from the fact that $P \left({x}\right) \iff x \in \N_{> 0}$.
+: $(2): \quad \forall a, b \in \Z: \map P a \land \map P b \implies \map P {a \times b}$
+Follows directly from the fact that $P \left({x}\right) \iff x \in \N_{> 0}$.
+: $(3): \quad \forall a \in \Z: \map P a \lor \map P {-a} \lor a = 0$
+This follows from the definition of the [[Definition:Integer|integers]] as the [[Inverse Completion of Natural Numbers|inverse completion of the natural numbers]].
+If $a \in \Z - \N$ then $\exists b \in \N: a + b = 0$ and so $a = -b$.
+Hence the conditions are fulfilled and $\struct {\Z, +, \times}$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Ordered Integral Domain}
+Tags: Rational Numbers, Ordered Integral Domains, Rational Numbers form Ordered Integral Domain
+
+\begin{theorem}
+The [[Definition:Rational Number|rational numbers]] $\Q$ form an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from [[Rational Numbers form Ordered Field]]:
+The [[Definition:Rational Number|set of rational numbers]] $\Q$ forms an [[Definition:Ordered Field|ordered field]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]: $\struct {\Q, +, \times, \le}$.
+An [[Definition:Ordered Field|ordered field]] is also an [[Definition:Ordered Integral Domain|ordered integral domain]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers form Ordered Integral Domain}
+Tags: Ordered Integral Domains, Real Numbers, Real Numbers form Ordered Integral Domain
+
+\begin{theorem}
+The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$.
+\end{theorem}
+
+\begin{proof}
+This follows directly from [[Real Numbers form Ordered Field]].
+The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Ordered Field|ordered field]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$.
+An [[Definition:Ordered Field|ordered field]] is also an [[Definition:Ordered Integral Domain|ordered integral domain]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+We have that the [[Real Numbers form Integral Domain|real numbers $\struct {\R, +, \times}$ form an integral domain]].
+It remains to specify a property $P$ on $\R$ such that:
+:$(1): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a + b}$
+:$(2): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a \times b}$
+:$(3): \quad \forall a \in \R: \map P a \lor \map P {-a} \lor a = 0$
+We have that the [[Rational Numbers form Ordered Integral Domain|integers $\struct {\Q, +, \times}$ form an ordered integral domain]], where $\Q$ denotes the [[Definition:Rational Number|set of rational numbers]].
+Let $P'$ be the [[Definition:Strict Positivity Property|(strict) positivity property]] on $\struct {\Q, +, \times}$.
+Let us define the property $P$ on $\R$ as:
+:$\forall a \in \R: \map P a \iff a = \eqclass {\sequence {a_n} } {}: \forall n \in \N: \map {P'} {a_n}$
+That is, an element $a = \eqclass {\sequence {a_n} } {}$ has $P$ {{iff}} $a_n$ has [[Definition:Strict Positivity Property|(strict) positivity property]] in $\Q$ for all $n \in \N$.
+Now let $a = \eqclass {\sequence {a_n} } {}$ and $b = \eqclass {\sequence {b_n} } {}$ such that $\map P a$ and $\map P b$.
+Then by definition of [[Definition:Real Addition|real addition]]:
+:$\eqclass {\sequence {a_n} } {} + \eqclass {\sequence {b_n} } {} = \eqclass {\sequence {a_n + b_n} } {}$
+and [[Definition:Real Multiplication|real multiplication]]:
+:$\eqclass {\sequence {a_n} } {} \times \eqclass {\sequence {b_n} } {} = \eqclass {\sequence {a_n \times b_n} } {}$
+It can be seen from the definition of [[Definition:Strict Positivity Property|(strict) positivity]] $P'$ on $\Q$ that $\map P {a + b}$ and $\map P {a \times b}$.
+It can be seen that if $\map P a$ then $\neg \map P {-a}$ and vice versa.
+Also we note that $\neg \map P 0$ and of course $\neg \map P {-0}$.
+So the property $P$ we defined fulfils the criteria for the [[Definition:Strict Positivity Property|(strict) positivity property]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Numbers of Type Integer a plus b root 2 Form Ordered Integral Domain}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
+:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
+that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
+Then the [[Definition:Algebraic Structure|algebraic structure]]:
+:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
+where $+$ and $\times$ are conventional [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] on [[Definition:Real Number|real numbers]], is an [[Definition:Ordered Integral Domain|ordered integral domain]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Numbers of Type Integer a plus b root 2 form Subdomain of Reals]].
+We also have that such numbers are [[Definition:Real Number|real]].
+The result follows from [[Real Numbers form Ordered Integral Domain]].
+{{explain|So is a subdomain of an ordered integral domain always an ordered integral domain? Does this need to be proved?}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{General Positivity Rule in Ordered Integral Domain}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]], whose [[Definition:Strict Positivity Property|(strict) positivity property]] is denoted $P$.
+Let $S \subset D$ be a [[Definition:Subset|subset]] of $D$ such that:
+:$\forall s \in S: \map P x$
+Then the following are true:
+:$\displaystyle \forall n \in \N_{>0}: \forall s_i \in S: \map P {\sum_{i \mathop = 1}^n s_i}$
+:$\displaystyle \forall n \in \N_{>0}: \forall s_i \in S: \map P {\prod_{i \mathop = 1}^n s_i}$
+where:
+:$\displaystyle \sum_{i \mathop = 1}^n s_i = s_1 + s_2 + \cdots + s_n$
+:$\displaystyle \prod_{i \mathop = 1}^n s_i = s_1 \times s_2 \times \cdots \times s_n$
+That is, the [[Definition:Ring Addition|ring sum]] and [[Definition:Ring Product|ring product]] of any number of elements in $D$ which have the [[Definition:Strict Positivity Property|(strict) positivity property]] also have the [[Definition:Strict Positivity Property|(strict) positivity property]].
+\end{theorem}
+
+\begin{proof}
+Trivially true for $n = 1$, and true by definition for $n = 2$.
+Suppose it is true for all $n$ up to $n = k$ for some $k \in \N$.
+Then:
+:$\displaystyle \forall s_i \in S: \map P {\sum_{i \mathop = 1}^k s_i}$
+:$\displaystyle \forall s_i \in S: \map P {\prod_{i \mathop = 1}^k s_i}$
+Take any $\displaystyle \sum_{i \mathop = 1}^{k + 1} s_i = \sum_{i \mathop = 1}^k s_i + s_{k + 1}$.
+Then:
+:$\displaystyle \map P {\sum_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$
+so:
+:$\displaystyle \map P {\sum_{i \mathop = 1}^{k + 1} s_i}$
+because the [[Definition:Strict Positivity Property|(strict) positivity property]] holds for all $s_i \in S$.
+Similarly:
+:$\displaystyle \map P {\prod_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$
+so:
+:$\displaystyle \map P {\prod_{i \mathop = 1}^{k + 1} s_i}$
+The result follows by the [[Principle of Complete Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Integral Domain cannot be Ordered}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$ and whose [[Definition:Order of Structure|order]] is [[Definition:Finite Order (Structure)|finite]].
+Then $D$ cannot be an [[Definition:Ordered Integral Domain|ordered integral domain]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Order of Structure|order]] is $n$.
+{{AimForCont}} there exists a [[Definition:Strict Positivity Property|(strict) positivity property]] $P$ on $\struct {D, +, \times}$.
+From [[Unity of Ordered Integral Domain is Strictly Positive]], we have that $\map P {1_D}$.
+Let $\struct {D, +}$ be the [[Definition:Additive Group of Ring|additive group]] of $\struct {D, +, \times}$.
+From [[Order of Element Divides Order of Finite Group]], the [[Definition:Order of Group Element|order of $1_D$]] is also [[Definition:Finite Order Element|finite]].
+So:
+:$\exists m \in \N: m \cdot 1_D = 0_D$
+where $0_D$ is the [[Definition:Ring Zero|zero]] of $D$.
+But then from the [[General Positivity Rule in Ordered Integral Domain]] that would mean $\map P {0_D}$ which contradicts the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of the positivity property]].
+Further, note that from [[Inverse Element is Power of Order Less 1]] we have that $-1_D = \paren {m - 1} \cdot 1_D$.
+But from the [[General Positivity Rule in Ordered Integral Domain]] $\map P {\paren {m - 1} \cdot 1_D}$ and so $\map P {-1_D}$ which is an even more dramatic way of showing that the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of the positivity property]] is contradicted.
+Hence an [[Definition:Integral Domain|integral domain]] whose [[Definition:Order of Structure|order]] is [[Definition:Finite Order (Structure)|finite]] can not be [[Definition:Ordered Integral Domain|ordered]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Integers Modulo m cannot be Ordered Integral Domain}
+Tags: Ring of Integers Modulo m, Ordered Integral Domains
+
+\begin{theorem}
+Let $m \in \Z: m \ge 2$.
+Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
+Then $\struct {\Z_m, +, \times}$ cannot be an [[Definition:Ordered Integral Domain|ordered integral domain]].
+\end{theorem}
+
+\begin{proof}
+First note that from [[Ring of Integers Modulo Prime is Integral Domain]], $\struct {\Z_m, +, \times}$ is an [[Definition:Integral Domain|integral domain]] only when $m$ is [[Definition:Prime Number|prime]].
+So for $m$ [[Definition:Composite Number|composite]] the result holds.
+If $m$ is [[Definition:Prime Number|prime]], and $\struct {\Z_m, +, \times}$ is therefore an [[Definition:Integral Domain|integral domain]], its [[Definition:Order of Structure|order]] is [[Definition:Finite Structure|finite]]. 
+The result follows from [[Finite Integral Domain cannot be Ordered]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Positivity Property induces Total Ordering}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+Let $D$ be endowed with a [[Definition:Strict Positivity Property|(strict) positivity property]] $P: D \to \set {\mathrm T, \mathrm F}$.
+Then there exists a [[Definition:Total Ordering|total ordering]] $\le$ on $\struct {D, +, \times}$ [[Definition:Total Ordering Defined by Strict Positivity Property|induced by $P$]] which is [[Definition:Ordering Compatible with Ring Structure|compatible with the ring structure]] of $\struct {D, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Strict Positivity Property|strict positivity property]]:
+{{:Definition:Strict Positivity Property}}
+Let us define a [[Definition:Relation|relation]] $<$ on $D$ as:
+:$\forall a, b \in D: a < b \iff \map P {-a + b}$
+Setting $a = 0$:
+:$\forall b \in D: 0 < b \iff \map P b$
+demonstrating that [[Definition:Strict Positivity Property|(strictly) positive]] elements of $D$ are those which are greater than [[Definition:Ring Zero|zero]].
+From [[Relation Induced by Strict Positivity Property is Compatible with Addition]] we have that $<$ is [[Definition:Relation Compatible with Operation|compatible with $+$]].
+From [[Relation Induced by Strict Positivity Property is Transitive]] we have that $<$ is [[Definition:Transitive Relation|transitive]].
+From [[Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive]] we have that $<$ is [[Definition:Asymmetric Relation|asymmetric]] and [[Definition:Antireflexive Relation|antireflexive]].
+Thus by definition, $<$ is a [[Definition:Strict Ordering|strict ordering]].
+Let the [[Definition:Relation|relation]] $\le$ be defined as the [[Definition:Reflexive Closure|reflexive closure]] of $<$.
+From [[Reflexive Closure of Strict Ordering is Ordering]] we have that $\le$ is an [[Definition:Ordering|ordering]] on $D$.
+From [[Relation Induced by Strict Positivity Property is Trichotomy]], and from the [[Trichotomy Law (Ordering)]], we have that $\le$ is a [[Definition:Total Ordering|total ordering]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Induced by Strict Positivity Property is Transitive}
+Tags: Ordered Integral Domains, Transitive Relations
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
+Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
+:$\forall a, b \in D: a < b \iff \map P {-a + b}$
+Then $<$ is a [[Definition:Transitive Relation|transitive relation]].
+\end{theorem}
+
+\begin{proof}
+Let $a < b$ and $b < c$.
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = a < b, b < c
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-a + b}, \map P {-b + c}
+      | c = Definition of $<$
+}}
+{{eqn | o = \leadsto
+      | r = \map P {\paren {-a + b} + \paren {-b + c} }
+      | c = {{Defof|Strict Positivity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-a + c}
+      | c = Properties of $+$ in $D$
+}}
+{{eqn | o = \leadsto
+      | r = a < c
+      | c = Definition of $<$
+}}
+{{end-eqn}}
+And so $<$ is seen to be [[Definition:Transitive Relation|transitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Induced by Strict Positivity Property is Compatible with Addition}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
+Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
+:$\forall a, b \in D: a < b \iff \map P {-a + b}$
+Then $<$ is [[Definition:Relation Compatible with Operation|compatible with $+$]], that is:
+:$\forall x, y, z \in D: x < y \implies \paren {x + z} < \paren {y + z}$
+:$\forall x, y, z \in D: x < y \implies \paren {z + x} < \paren {z + y}$
+\end{theorem}
+
+\begin{proof}
+Let $a < b$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = a < b
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-a + b}
+      | c = by definition of $<$
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-a + b + \paren {-c} + c}
+      | c = for any $c \in D$
+}}
+{{eqn | o = \leadsto
+      | r = \map P {\paren {-a + \paren {-c} } + \paren {b + c} }
+      | c = properties of $+$ in $D$
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-\paren {a + c} + \paren {b + c} }
+      | c = properties of $+$ in $D$
+}}
+{{eqn | o = \leadsto
+      | r = a + c < b + c
+      | c = by definition of $<$
+}}
+{{eqn | o = \leadsto
+      | r = c + a < c + b
+      | c = [[Definition:Commutative Operation|commutativity]] of $+$
+}}
+{{end-eqn}}
+And so $<$ is seen to be [[Definition:Relation Compatible with Operation|compatible with $+$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Induced by Strict Positivity Property is Compatible with Multiplication}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
+Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
+:$\forall a, b \in D: a < b \iff \map P {-a + b}$
+Then $<$ is [[Definition:Relation Compatible with Operation|compatible with $\times$]] in the following sense:
+:$\forall x, y, z \in D: x < y, \map P z \implies \paren {z \times x} < \paren {z \times y}$
+:$\forall x, y, z \in D: x < y, \map P z \implies \paren {x \times z} < \paren {y \times z}$
+\end{theorem}
+
+\begin{proof}
+If $x < y$ then $\map P {-x + y}$.
+Hence:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \map P {-x + y}, \, \map P z
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \map P {z \times \paren {-x + y} }
+      | c = {{Defof|Strict Positivity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-z \times x + z \times y}
+      | c = [[Definition:Distributive Operation|Distributivity]] of $\times$ over $+$
+}}
+{{eqn | o = \leadsto
+      | r = z \times x < z \times y
+      | c = 
+}}
+{{end-eqn}}
+The other result follows from the fact that $\times$ is [[Definition:Commutative Operation|commutative]] in an [[Definition:Integral Domain|integral domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Induced by Strict Positivity Property is Trichotomy}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
+Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
+:$\forall a, b \in D: a < b \iff \map P {-a + b}$
+Then $\forall a, b \in D:$ exactly one of the following conditions applies:
+:$a < b$
+:$a = b$
+:$a > b$
+That is, $<$ is a [[Definition:Trichotomy|trichotomy]].
+\end{theorem}
+
+\begin{proof}
+Take any $a, b \in D$ and consider $-a + b$.
+From the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]], exactly one of the following applies:
+:$(1): \quad \map P {-a + b}$
+:$(2): \quad \map P {-\paren {-a + b} }$
+:$(3): \quad -a + b = 0$
+Taking each of these in turn and taking into account that $\struct {D, +}$ is the [[Definition:Additive Group of Ring|additive group]] of $\struct {D, +, \times}$:
+{{begin-eqn}}
+{{eqn | n = 1
+      | o = 
+      | r = \map P {-a + b}
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = a < b
+      | c = by definition of $<$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | n = 2
+      | o = 
+      | r = \map P {-\paren {-a + b} }
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-b + a}
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = b < a
+      | c = by definition of $<$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | n = 3
+      | o = 
+      | r = -a + b = 0
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = a = b
+      | c = adding $a$ to both sides
+}}
+{{end-eqn}}
+{{qed|lemma}}
+The result has been proved.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] where $P$ is the [[Definition:Strict Positivity Property|(strict) positivity property]].
+Let the [[Definition:Relation|relation]] $<$ be defined on $D$ as:
+:$\forall a, b \in D: a < b \iff \map P {-a + b}$
+Then $<$ is [[Definition:Asymmetric Relation|asymmetric]] and [[Definition:Antireflexive Relation|antireflexive]].
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]], for all $x \in D$ exactly one of the following applies:
+:$(1): \quad \map P x$
+:$(2): \quad \map P {-x}$
+:$(3): \quad x = 0$
+Let $a, b \in D: a < b$.
+Then by definition:
+:$\map P {-a + b}$
+Suppose also that $b < a$.
+Then by definition:
+:$\map P {-b + a}$
+and so:
+:$\map P {-\paren {-a + b} }$
+But then that contradicts the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]]:
+Either:
+:$(1): \quad \map P {-a + b}$
+or:
+:$(2): \quad \map P {-\paren {-a + b} }$
+So $(1)$ and $(2)$ can not both apply, and so if $a < b$ it is not possible that $b < a$.
+Thus $<$ is [[Definition:Asymmetric Relation|asymmetric]].
+Now suppose $\exists a \in D: a < a$.
+Then that would mean $\map P {-a + a}$, that is, $\map P {0_D}$.
+But this also contradicts the conditions of the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law of ordered integral domains]].
+Hence:
+:$\forall a \in D: a \not < a$
+and so $<$ is [[Definition:Antireflexive Relation|antireflexive]].
+{{qed}}
+[[Category:Ordered Integral Domains]]
+7wj0qad5d2yz9f13et7dv0yx2k16dak
+\end{proof}<|endoftext|>
+\section{Properties of Strict Negativity}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]], whose [[Definition:Strict Positivity Property|(strict) positivity property]] is denoted $P$.
+Let $\le$ be the [[Strict Positivity Property induces Total Ordering|total ordering induced by $P$]], and let $<$ be its [[Definition:Strict Total Ordering|strict total ordering]] counterpart.
+Let $N$ be the [[Definition:Strict Negativity Property|(strict) negativity property]] on $D$:
+:$\forall a \in D: \map N a \iff \map P {-a}$
+Then the following properties apply for all $a, b \in D$:
+ 
+=== [[Strict Negativity is equivalent to Strictly Preceding Zero]] ===
+{{:Strict Negativity is equivalent to Strictly Preceding Zero}}
+\end{theorem}<|endoftext|>
+\section{Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero}
+Tags: Absolute Value Function
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
+For all $a \in D$, let $\size a$ denote the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$.
+Then $\size a$ is [[Definition:Strictly Positive|(strictly) positive]] except when $a = 0$.
+\end{theorem}
+
+\begin{proof}
+Let $P$ be the [[Definition:Strict Positivity Property|(strict) positivity property]] on $D$.
+Let $<$ be the [[Definition:Strict Total Ordering|(strict) total ordering]] defined on $D$ as:
+:$a < b \iff a \le b \land a \ne b$
+From the [[Definition:Trichotomy Law (Integral Domain)|trichotomy law]], exactly one of three possibilities holds for any $ a \in D$:
+$(1): \quad \map P a$:
+In this case $0 < a$ and so $\size a = a$.
+So:
+:$\map P a \implies \map P {\size a}$
+$(2): \quad \map P {-a}$:
+In this case $a < 0$ and so $\size a = -a$.
+So:
+:$\map P {-a} \implies \map P {\size a}$
+$(3): \quad a = 0$:
+In this case $\size a = a$.
+So:
+:$a = 0 \implies \size a = 0$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Absolute Values on Ordered Integral Domain}
+Tags: Absolute Value Function
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
+For all $a \in D$, let $\size a$ denote the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$.
+Then:
+:$\size {a + b} \le \size a + \size b$
+\end{theorem}
+
+\begin{proof}
+Let $P$ be the [[Definition:Strict Positivity Property|(strict) positivity property]] on $D$.
+Let $<$ be the [[Definition:Strict Total Ordering|(strict) total ordering]] defined on $D$ as:
+:$a < b \iff a \le b \land a \ne b$
+Let $N$ be the [[Definition:Strict Negativity Property|(strict) negativity property]] on $D$.
+Let $a \in D$.
+If $\map P a$ or $a = 0$ then $a \le \size a$.
+If $\map N a$ then by [[Properties of Strict Negativity|Properties of Strict Negativity: $(1)$]] and definition of [[Definition:Absolute Value on Ordered Integral Domain|absolute value]]:
+:$a < 0 < \size a$
+and hence by [[Definition:Transitive Relation|transitivity]] $<$ we have:
+:$a < \size a$
+By similar reasoning:
+:$-a < \size a$
+Thus for all $a, b \in D$ we have:
+:$a \le \size a, b \le \size b$
+As $<$ is [[Definition:Relation Compatible with Operation|compatible with $+$]], we have:
+:$a + b \le \size a + \size b$
+and:
+:$-\paren {a + b} = \paren {-a} + \paren {-b} \le \size a + \size b$
+But either:
+:$\size {a + b} = a + b$
+or:
+:$\size {a + b} = -\paren {a + b}$
+Hence the result:
+:$\size {a + b} \le \size a + \size b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{One Succeeds Zero in Well-Ordered Integral Domain}
+Tags: Well-Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be a [[Definition:Well-Ordered Integral Domain|well-ordered integral domain]].
+Let $0$ and $1$ be the [[Definition:Ring Zero|zero]] and [[Definition:Unity of Ring|unity]] respectively of $D$.
+Then $0$ is the [[Definition:Immediate Predecessor Element|immediate predecessor]] of $1$:
+:$0 < 1$
+:$\neg \exists a \in D: 0 < a < 1$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there exists an [[Definition:Element|element]] $a \in D$ such that $0 < a < 1$.
+Let us create the [[Definition:Set|set]] $S$ of all such [[Definition:Element|elements]]:
+:$S = \set {x \in D: 0 < x < 1}$
+We know $S$ is not [[Definition:Empty Set|empty]] because it has already been asserted that $a$ is in it.
+And all the [[Definition:Element|elements]] in $S$ are [[Definition:Strictly Positive|strictly positive]] by definition.
+Because $D$ is [[Definition:Well-Ordered Set|well-ordered]] it follows that $S$ has a [[Definition:Minimal Element|minimal element]], which we will call $m$.
+Thus we have $0 < m < 1$.
+Then from [[Square of Element Less than Unity in Ordered Integral Domain]] we have:
+:$0 < m \times m < m$
+Thus $m \times m \in S$ is an [[Definition:Element|element]] of $S$ which is smaller than $m$.
+But $m$ was supposed to be the [[Definition:Minimal Element|minimal element]] of $S$.
+From this contradiction we deduce that there can be no such [[Definition:Element|element]] $a$ such that $0 < a < 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square of Element Less than Unity in Ordered Integral Domain}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
+Let $x \in D$ such that $0 < x < 1$.
+Then:
+: $0 < x \times x < x$
+\end{theorem}
+
+\begin{proof}
+We have that $0 < x < 1$.
+From [[Relation Induced by Strict Positivity Property is Compatible with Multiplication]]:
+:$0 \times x < x \times x < 1 \times x$
+Hence the result.
+{{qed}}
+[[Category:Ordered Integral Domains]]
+payuibiqj3z0g90g32oa6vktwe0b6lr
+\end{proof}<|endoftext|>
+\section{Principle of Mathematical Induction/Well-Ordered Integral Domain}
+Tags: Well-Ordered Integral Domains, Principle of Mathematical Induction
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be a [[Definition:Well-Ordered Integral Domain|well-ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+Let the [[Definition:Unity of Ring|unity]] of $D$ be $1_D$.
+Let $S \subseteq D$ be such that:
+:$1_D \in S$
+:$a \in S \implies a + 1_D \in S$
+Then:
+:$D_{> 0_D} \subseteq S$
+where $D_{> 0_D}$ denotes all the [[Definition:Element|elements]] $d \in D$ such that $\map P d$.
+That is, $D_{> 0_D}$ is the set of all [[Definition:Strictly Positive|(strictly) positive]] elements of $D$.
+\end{theorem}
+
+\begin{proof}
+Let $\overline S$ be the set of all elements of $D_{> 0_D}$ that are not in $S$:
+:$\overline S = D_{> 0_D} \setminus S$
+{{AimForCont}} $\overline S$ is not [[Definition:Empty Set|empty]].
+Then as $D$ is [[Definition:Well-Ordered Integral Domain|well-ordered]], it follows that $\overline S$ has a [[Definition:Minimal Element|minimal element]], which we will call $m$.
+Then $m - 1 \notin \overline S$.
+But $1 < m$ as $1 \in S$, and from [[One Succeeds Zero in Well-Ordered Integral Domain]], $1$ is the minimal positive element of $D$.
+So $0 < m - 1$ and so $m - 1$ is [[Definition:Strictly Positive|strictly positive]].
+Because $m - 1 \notin \overline S$, it follows that $m - 1 \in S$.
+By construction of $S$ it follows that $m \in S$.
+So $m \in S$ and $m \in \overline S$, that is:
+:$m \notin S$
+From this [[Definition:Contradiction|contradiction]] we deduced that there can therefore be no such $m$.
+Hence:
+:$D_{\ge 0_D} \setminus S = \O$.
+From [[Set Difference with Superset is Empty Set]] it follows that $D_{\ge 0_D} \subseteq S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiple of Divisor in Integral Domain Divides Multiple}
+Tags: Divisibility, Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]].
+Let $a, b, c \in D$.
+Let $a \divides b$, where $\divides$ denotes [[Definition:Divisor of Ring Element|divisibility]].
+Then $a \times c$ is a [[Definition:Divisor of Ring Element|divisor]] of $b \times c$.
+\end{theorem}
+
+\begin{proof}
+By definition, if $a \divides b$ then $\exists d \in D: a \times d = b$.
+Then $\paren {a \times d} \times c = b \times c$, that is:
+:$\paren {a \times c} \times d = b \times c$
+which follows because $\times$ is [[Definition:Commutative Operation|commutative]] and [[Definition:Associative|associative]] in an [[Definition:Integral Domain|integral domain]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Divisor Results/One Divides all Integers}
+Tags: Divisors
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = 1
+      | o = \divides
+      | r = n
+}}
+{{eqn | l = -1
+      | o = \divides
+      | r = n
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Integral Domain]], the concept '''[[Definition:Divisor of Integer|divisibility]]''' is fully applicable to the [[Definition:Integer|integers]].
+Therefore this result follows directly from [[Unity Divides All Elements]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Divisor Results/Integer Divides Itself}
+Tags: Divisors, Integer Divisor Results: Integer Divides Itself
+
+\begin{theorem}
+:$n \divides n$
+\end{theorem}<|endoftext|>
+\section{Integer Divisor Results/Integer Divides its Negative}
+Tags: Divisors
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \divides
+      | r = -n
+}}
+{{eqn | l = -n
+      | o = \divides
+      | r = n
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Integral Domain]], the [[Definition:Integer|integers]] are an [[Definition:Integral Domain|integral domain]].
+Hence we can apply [[Product of Ring Negatives]]:
+:$\forall n \in \Z: \exists -1 \in \Z: n = \paren {-1} \times \paren {-n}$
+and [[Product with Ring Negative]]:
+:$\forall n \in \Z: \exists -1 \in \Z: -n = \paren {-1} \times \paren n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Divisor Results/Integer Divides its Absolute Value}
+Tags: Divisors, Absolute Value Function
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \divides
+      | r = \size n
+}}
+{{eqn | l = \size n
+      | o = \divides
+      | r = n
+}}
+{{end-eqn}}
+where:
+:$\size n$ is the [[Definition:Absolute Value|absolute value]] of $n$
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+Let $n > 0$.
+Then $\size n = n$ and [[Integer Divides Itself]] applies.
+Let $n = 0$.
+Then [[Integer Divides Itself]] holds again.
+Let $n < 0$.
+Then $\size n = -n$ and [[Integer Divides its Negative]] applies.
+{{qed}}
+[[Category:Divisors]]
+[[Category:Absolute Value Function]]
+qwokq3o9x7dlgf3p8phr3bj8v8ig1wr
+\end{proof}<|endoftext|>
+\section{Integer Divisor Results/Integer Divides Zero}
+Tags: Divisors, Integer Divisor Results
+
+\begin{theorem}
+:$n \divides 0$
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Integral Domain]], the concept '''[[Definition:Divisor of Ring Element|divides]]''' is fully applicable to the [[Definition:Integer|integers]].
+Therefore this result follows directly from [[Element of Integral Domain Divides Zero]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Divisor Results/Divisors of Negative Values}
+Tags: Divisors
+
+\begin{theorem}
+: $m \mathrel \backslash n \iff -m \divides n \iff m \divides -n \iff -m \divides -n$
+\end{theorem}
+
+\begin{proof}
+Let $m \mathrel \backslash n$.
+From [[Integer Divides its Negative]], we have $-m \divides m$.
+From [[Divisor Relation on Positive Integers is Partial Ordering]] it follows that $-m \divides n$.
+From [[Integer Divides its Negative]], we have $n \divides -n$.
+From [[Divisor Relation on Positive Integers is Partial Ordering]] it follows that $m \divides -n$.
+From above, we have $-m \divides n$.
+From [[Integer Divides its Negative]], we have $n \divides -n$.
+From [[Divisor Relation on Positive Integers is Partial Ordering]] it follows that $-m \divides -n$.
+The reverse implications follow similarly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisors of One}
+Tags: Divisors, 1
+
+\begin{theorem}
+The only [[Definition:Divisor of Integer|divisors]] of $1$ are $1$ and $-1$.
+That is:
+:$a \divides 1 \iff a = \pm 1$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $a = \pm 1$.
+From [[Integer Divides Itself]] we have that $1 \divides 1$.
+From [[Integer Divides its Negative]] we have that $-1 \divides 1$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\exists a \in \Z: a \divides 1$.
+Then $\exists c \in \Z: a c = 1$.
+From [[Absolute Value Function is Completely Multiplicative]] we have that:
+:$\size a \cdot \size c = \size 1$
+Neither $a$ nor $c$ can be zero, from [[Integers form Integral Domain]].
+So $\size a \ge 1$ and $\size c \ge 1$.
+But if $\size a > 1$ then:
+:$\size a \cdot \size c > \size c$
+and so:
+:$\size a \cdot \size c > 1$
+So:
+:$\size a = 1$
+that is:
+:$a = 1$ or $a = -1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisor Relation is Antisymmetric}
+Tags: Divisors, Divisor Relation is Antisymmetric
+
+\begin{theorem}
+''[[Definition:Divisor of Integer|Divides]]'' is a [[Definition:Antisymmetric Relation|antisymmetric relation]] on $\Z_{>0}$, the set of [[Definition:Positive Integer|positive integers]].
+That is:
+:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in \Z_{> 0}$ such that $a \divides b$ and $b \divides a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \divides b
+      | o = \implies
+      | r = \size a \le \size b
+      | c = [[Absolute Value of Integer is not less than Divisors]]
+}}
+{{eqn | l = b \divides a
+      | o = \implies
+      | r = \size b \le \size a
+      | c = [[Absolute Value of Integer is not less than Divisors]]
+}}
+{{eqn | o = \leadsto
+      | r = \size a = \size b
+}}
+{{end-eqn}}
+If we restrict ourselves to the [[Definition:Domain of Relation|domain]] of [[Definition:Positive Integer|positive integers]], we can see:
+:$\forall a, b \in \Z_{>0}: a \divides b \land b \divides a \implies a = b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Continuous at Point iff Oscillation is Zero}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
+Let $x$ be a point in $D$.
+Let $N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhoods]] of $x$.
+Let $\map {\omega_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation]] of $f$ at $x$:
+:$\map {\omega_f} x = \displaystyle \inf \set {\map {\omega_f} I: I \in N_x}$
+where:
+:$\map {\omega_f} I = \displaystyle \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Then $\map {\omega_f} x = 0$ {{iff}} $f$ is [[Definition:Continuous Real Function|continuous]] at $x$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\map {\omega_f} x = 0$.
+Let $\epsilon > 0$.
+Suppose that $\forall I: I \in N_x: \omega_f \left({I} \right) \ge \epsilon$.
+Then by definition, $\map {\omega_f} x \ge \epsilon$. 
+This contradicts $\map {\omega_f} x = 0$.
+From this contradiction we deduce that:
+:$\exists I: I \in N_x: \omega_f \left({I}\right) < \epsilon$
+For this particular $I$, there is an [[Definition:Open Set (Real Analysis)|open set]] $O \subset I$ by the definition of [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhood]].
+Therefore, a $\delta \in \R_{>0}$ exists such that $\openint {x - \delta} {x + \delta}$ is a [[Definition:Subset|subset]] of $O$.
+So for our specific $x$, if $y$ satisfies:
+:$\size {x - y} < \delta$
+Then:
+:$y \in I$
+and, if $y \in D$:
+:$\size {\map f x - \map f y} \le \map {\omega_f} I$
+Since $\map {\omega_f} I < \epsilon$ it follows by the definition of [[Definition:Continuous Real Function|continuity]] that $f$ is [[Definition:Continuous Real Function at Point|continuous]] at $x$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f$ be [[Definition:Continuous Real Function at Point|continuous]] at $x$.
+Then $\forall \epsilon > 0: \exists \delta \in \R_{>0}$ such that, if $y \in D$:
+:$\size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$
+Let the [[Definition:Open Real Interval|interval]] $I_\delta$ be defined as:
+:$I_\delta := \openint {x - \delta} {x + \delta}$
+$I_\delta$ is an [[Definition:Element|element]] of $N_x$ as $I_\delta$ is a [[Definition:Neighborhood (Real Analysis)/Open Subset|neighborhood]] of $x$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {\omega_f} {I_\delta}
+      | r = \sup \set {\size {\map f y - \map f z}: y, z \in I_\delta \cap D}
+      | c = Definition of $\map {\omega_f} {I_\delta}$
+}}
+{{eqn | r = \sup \set {\size {\map f y - \map f x + \map f x - \map f z}: y, z \in I_\delta \cap D}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \sup \set {\size {\map f y - \map f x} + \size {\map f x - \map f z}: y, z \in I_\delta \cap D}
+      | c = [[Triangle Inequality for Vectors in Euclidean Space]] and [[Supremum of Function is less than Supremum of Greater Function]]
+}}
+{{eqn | r = \sup \set {\size {\map f y - \map f x}: y \in I_\delta \cap D} + \sup \set {\size {\map f x - \map f z}: z \in I_\delta \cap D}
+      | c = [[Supremum of Sum equals Sum of Suprema]]
+}}
+{{eqn | r = \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D} + \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D}
+}}
+{{eqn | r = 2 \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D}
+      | c = 
+}}
+{{eqn | o = <
+      | r = 2 \epsilon
+      | c = as $\size {\map f x - \map f y} < \epsilon$ whenever $y \in I_\delta \cap D$
+}}
+{{end-eqn}}
+This gives:
+{{begin-eqn}}
+{{eqn | l = \map {\omega_f} x
+      | r = \inf \set {\map {\omega_f} I: I \in N_x}
+      | c = Definition of $\map {\omega_f} x$
+}}
+{{eqn | o = \le
+      | r = \map {\omega_f} {I_\delta}
+      | c = {{Defof|Infimum of Subset of Real Numbers}} and that $I_\delta \in N_x$
+}}
+{{eqn | o = <
+      | r = 2 \epsilon
+      | c = 
+}}
+{{end-eqn}}
+This holds true for any value of $\epsilon$.
+Thus $\map {\omega_f} x$ must be $0$.
+Hence the result.
+{{qed}}
+[[Category:Real Analysis]]
+5g3boxs69sq27ql4ga0qr7ixu7lklh9
+\end{proof}<|endoftext|>
+\section{Numbers of Type Integer a plus b root 2 are Not a Field}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
+:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
+that is, all [[Definition:Real Number|(real) numbers]] of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
+Then the [[Definition:Algebraic Structure|algebraic structure]]:
+:$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
+where $+$ and $\times$ are [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] on [[Definition:Real Number|real numbers]], is '''not''' a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+We start from the result [[Numbers of Type Integer a plus b root 2 form Subdomain of Reals]].
+We have from that result that $1 + 0 \sqrt 2$ is the [[Definition:Unity of Ring|unity]].
+Now consider a representative element $a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
+From [[Difference of Two Squares]] we have:
+:$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$
+which leads to:
+:$\paren {a + b \sqrt 2} \paren {\dfrac {a - b \sqrt 2} {a^2 - 2 b^2} } = 1$
+so demonstrating that the [[Definition:Product Inverse|product inverse]] of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.
+But for any two given integers $a$ and $b$, the values $\dfrac a {a^2 - 2 b^2}$ and $\dfrac b {a^2 - 2 b^2}$ are not in general [[Definition:Integer|integers]].
+For instance, when $a = 3$ and $b = 4$ we get:
+:$\dfrac a {a^2 - 2 b^2} = \dfrac 3 {23}$ and $\dfrac b {a^2 - 2 b^2} = \dfrac 4 {23}$
+(unless I've got my sums wrong).
+So not every element of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ has a [[Definition:Product Inverse|product inverse]], demonstrating that $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ is not a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Numbers of Type Rational a plus b root 2 form Field}
+Tags: Integral Domains
+
+\begin{theorem}
+Let $\Q \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
+:$\Q \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Q}$
+that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Rational Number|rational numbers]].
+Then the [[Definition:Algebraic Structure|algebraic structure]]:
+:$\struct {\Q \sqbrk {\sqrt 2}, +, \times}$
+where $+$ and $\times$ are conventional [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] on [[Definition:Real Number|real numbers]], is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+By definition, $\Q \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of [[Definition:Real Number|real numbers]].
+Thus $+$ and $\times$ on $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ are [[Definition:Well-Defined Operation|well-defined]].
+=== Closure ===
+Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Q \sqbrk {\sqrt 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a_1 + b_1 \sqrt 2} + \paren {a_2 + b_2 \sqrt 2}
+      | r = \paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \Q \sqbrk {\sqrt 2}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {a_1 + b_1 \sqrt 2} \times \paren {a_2 + b_2 \sqrt 2}
+      | r = \paren {a_1 \times a_2 + 2 \times b_1 \times b_2} + \paren {a_1 \times b_2 + b_1 \times a_2} \sqrt 2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \Q \sqbrk {\sqrt 2}
+      | c = 
+}}
+{{end-eqn}}
+So both $+$ and $\times$ are [[Definition:Closed Algebraic Structure|closed on $\Q \sqbrk {\sqrt 2}$]].
+{{qed|lemma}}
+=== Associativity ===
+As [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] are [[Definition:Associative|associative]] on $\R$ it follows from [[Restriction of Associative Operation is Associative]] that they are also associative on $\Q \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Commutativity ===
+As [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] are [[Definition:Commutative Operation|commutative]] on $\R$ it follows from [[Restriction of Commutative Operation is Commutative]] that they are also [[Definition:Commutative Operation|commutative]] on $\Q \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Identity ===
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b \sqrt 2} + \paren {0 + 0 \sqrt 2}
+      | r = \paren {a + 0} + \paren {b + 0} \sqrt 2
+      | c = 
+}}
+{{eqn | r = a + b \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$.
+So $\paren {0 + 0 \sqrt 2}$ is the [[Definition:Identity Element|identity]] for $+$ on $\Q \sqbrk {\sqrt 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}
+      | r = \paren {a \times 1 + 2 \times b \times 0} + \paren {b \times 1 + a \times 0} \sqrt 2
+      | c = 
+}}
+{{eqn | r = a + b \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {1 + 0 \sqrt 2} \times \paren {a + b \sqrt 2}$.
+So $\paren {1 + 0 \sqrt 2}$ is the [[Definition:Identity Element|identity]] for $\times$ on $\Q \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Inverses ===
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b \sqrt 2} + \paren {-a + \paren {-b} \sqrt 2}
+      | r = \paren {a - a} + \paren {b - b} \sqrt 2
+      | c = 
+}}
+{{eqn | r = 0 + 0 \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {-a + \paren {-b} \sqrt 2} + \paren {a + b \sqrt 2}$.
+So $\paren {-a + \paren {-b} \sqrt 2}$ is the [[Definition:Inverse Element|inverse]] of $\paren {a + b \sqrt 2}$ for $+$ on $\Q \sqbrk {\sqrt 2}$.
+Calculating the [[Definition:Product Inverse|product inverse]] of $\paren {a + b \sqrt 2}$ is less trivial. 
+From [[Difference of Two Squares]] we have:
+:$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$
+which leads to:
+:$\paren {a + b \sqrt 2} \paren {\dfrac {a - b \sqrt 2} {a^2 - 2 b^2} } = 1 = 1 + 0 \sqrt 2$
+so demonstrating that the [[Definition:Product Inverse|product inverse]] of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} - \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.
+As $a$ and $b$ are [[Definition:Rational Number|rational]], it follows that so are $\dfrac a {a^2 - 2 b^2}$ and $\dfrac b {a^2 - 2 b^2}$.
+So the [[Definition:Product Inverse|product inverse]] of $\paren {a + b \sqrt 2}$ is an element of $\Q \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+=== Distributivity ===
+We have that [[Real Multiplication Distributes over Addition]], so by [[Restriction of Operation Distributivity]], $\times$ is [[Definition:Distributive Operation|distributive]] over $+$ on $\Q \sqbrk {\sqrt 2}$.
+{{qed|lemma}}
+The result follows by putting all the pieces together.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic of Ring of Integers Modulo Prime}
+Tags: Ring of Integers Modulo m, Finite Rings
+
+\begin{theorem}
+Let $\struct {\Z_p, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $p$]], where $p$ is a [[Definition:Prime Number|prime number]].
+The [[Definition:Characteristic of Ring|characteristic]] of $\struct {\Z_p, +, \times}$ is $p$.
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Integers Modulo Prime is Field]] we have that $\struct {\Z_p, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+So [[Characteristic of Finite Ring with No Zero Divisors]] applies, and so the [[Definition:Characteristic of Ring|characteristic]] of $\struct {\Z_p, +, \times}$ is [[Definition:Prime Number|prime]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gamma Function Extends Factorial}
+Tags: Gamma Function, Factorials
+
+\begin{theorem}
+:$\forall n \in \N: \map \Gamma {n + 1} = n!$
+\end{theorem}
+
+\begin{proof}
+For $n = 0$:
+{{begin-eqn}}
+{{eqn | l = \map \Gamma 1
+      | r = \int_0^\infty e^{-t} \rd t
+      | c = {{Defof|Integral Form of Gamma Function|Gamma Function}}
+}}
+{{eqn | r = \bigintlimits {-e^{-t} } 0 \infty
+      | c =
+}}
+{{eqn | r = 0 - \paren {-1}
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+Then by [[Gamma Difference Equation]]:
+:$\forall z \in \Z_{> 0}: \map \Gamma {z + 1} = z \, \map \Gamma z$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uniform Limit of Analytic Functions is Analytic}
+Tags: Complex Analysis, Uniform Convergence
+
+\begin{theorem}
+Let $U$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of $\C$.
+Let $\left\{ {f_n}\right\}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Analytic Function|analytic functions]] $f_n : U \to \C$.
+Let $\left\{ {f_n}\right\}$ [[Definition:Locally Uniform Convergence|converge locally uniformly]] to $f$ on $U$.
+Then $f$ is [[Definition:Analytic Function|analytic]].
+\end{theorem}
+
+\begin{proof}
+By [[Equivalence of Local Uniform Convergence and Compact Convergence]], $f_n$ converges to $f$ locally uniformly on $U$.
+Then for any $z \in U$, there is an $\epsilon > 0$ so that:
+:$B_\epsilon \left({z}\right) \subset U$
+and $f_n$ converges uniformly on $B_\epsilon \left({z}\right)$.
+Let $\gamma$ be any simple closed curve in $B_\epsilon \left({z}\right)$.
+Since $f_n \to f$ uniformly on $\gamma$ (because $\gamma \subset B_\epsilon \left({z}\right)$), we have:
+:$\displaystyle \lim_{n \to \infty} \int_\gamma f_n \left({z}\right) \, \mathrm d z = \int_\gamma f \left({z}\right) \, \mathrm d z$
+Since each $f_n$ is analytic, we have that:
+:$\displaystyle \forall n \in \N:  \int_\gamma f_n \left({z}\right) \, \mathrm d z = 0$
+So we conclude also that
+:$\displaystyle \int_\gamma f \left({z}\right) \, \mathrm d z = 0$
+Since $\gamma$ was arbitrary, we have by [[Morera's Theorem]] that $f$ is analytic in $B_\epsilon \left({z}\right)$.
+Since $z$ was arbitrary, $f$ is analytic on all of $U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus of Complex Integral}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right]$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $f : \left[{a \,.\,.\, b}\right] \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex function]].
+Then:
+:$\displaystyle \left|{\int_a^b f \left({t}\right) \rd t}\right| \le \int_a^b \left|{f \left({t}\right)}\right| \rd t$
+where the first integral is a [[Definition:Complex Riemann Integral|complex Riemann integral]], and the second integral is a [[Definition:Definite Integral|definite real integral]].
+\end{theorem}
+
+\begin{proof}
+Define $z \in \C$ as the value of the [[Definition:Complex Riemann Integral|complex Riemann integral]]:
+:$z = \displaystyle \int_a^b f \left({t}\right) \rd t$
+Define $r \in \left[{0 \,.\,.\, \infty }\right)$ as the [[Definition:Complex Modulus|modulus]] of $z$, and $\theta \in \left[{0 \,.\,.\, 2 \pi}\right)$ as the [[Definition:Argument of Complex Number|argument]] of $z$.
+From [[Modulus and Argument of Complex Exponential]], we have $z = re^{i \theta}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = r
+      | r = ze^{-i \theta}
+      | c = [[Reciprocal of Complex Exponential]]
+}}
+{{eqn | r = \int_a^b e^{-i \theta} f \left({t}\right) \rd t
+      | c = [[Linear Combination of Complex Integrals]]
+}}
+{{eqn | r = \int_a^b \operatorname{Re} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t + i \int_a^b \operatorname{Im} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t
+      | c = {{Defof|Complex Riemann Integral}}
+}}
+{{end-eqn}}
+As $r$ is [[Definition:Complex Number#Wholly Real|wholly real]], we have:
+: $\displaystyle 0 = \operatorname{Im}\left({r}\right) = \int_a^b \operatorname{Im} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t$
+Then:
+{{begin-eqn}}
+{{eqn | l = r
+      | r = \int_a^b \operatorname{Re} \left({e^{-i \theta} f \left({t}\right) }\right) \rd t
+}}
+{{eqn | o = \le
+      | r = \int_a^b \left\vert{\operatorname{Re} \left({e^{-i \theta} f \left({t}\right) }\right) }\right\vert \rd t
+      | c = [[Absolute Value of Definite Integral]]
+}}
+{{eqn | o = \le
+      | r = \int_a^b \left\vert{e^{-i \theta} f \left({t}\right) }\right\vert \rd t
+      | c = [[Modulus Larger than Real Part]]
+}}
+{{eqn | r =  \int_a^b \left\vert{e^{-i \theta} }\right\vert \left\vert{f \left({t}\right) }\right\vert \rd t
+}}
+{{eqn | r = \int_a^b \left\vert{f \left({t}\right) }\right\vert \rd t
+      | c = [[Modulus of Exponential of Imaginary Number is One|Modulus of Exponential of Imaginary Number]]
+}}
+{{end-eqn}}
+As $\displaystyle r = \left|{\int_a^b f \left({t}\right) \rd t}\right|$ by its definition, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Triangle Inequality for Contour Integrals}
+Tags: Contour Integration, Named Theorems, Triangle Inequality
+
+\begin{theorem}
+Let $C$ be a [[Definition:Contour (Complex Plane)|contour]].
+Let $f: \Img C \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $\Img C$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C$.
+Then:
+:$\displaystyle \size {\int_C \map f z \rd z} \le \max_{z \mathop \in \Img C} \size {\map f z} \map L C$
+where $\map L C$ denotes the [[Definition:Length of Contour (Complex Plane)|length]] of $C$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Contour (Complex Plane)|contour]], $C$ is a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of a [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$ of [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]].
+Let $C_i$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\int_C \map f z \rd z}
+      | r = \size {\sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \size {\int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}
+      | c = [[Triangle Inequality for Complex Numbers]]
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \size {\map f {\map {\gamma_i} t} } \size {\map {\gamma_i'} t} \rd t
+      | c = [[Modulus of Complex Integral]]
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \max_{t \mathop \in \closedint {a_i} {b_i} } \size {\map f {\map {\gamma_i} t} } \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \max_{z \mathop \in \Img C} \size {\map f z} \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t
+      | c = as $\map {\gamma_i} t \in \Img C$
+}}
+{{eqn | r = \max_{z \mathop \in \Img C} \size {\map f z} \map L C
+      | c = {{Defof|Length of Contour (Complex Plane)}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Jensen's Formula}
+Tags: Complex Analysis, Jensen's Formula
+
+\begin{theorem}
+Let $S$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of the [[Definition:Complex Plane|complex plane]] containing the [[Definition:Closed Disk|closed disk]]:
+:$D_r = \set {z \in \C : \cmod z \le r}$
+of [[Definition:Radius of Closed Ball|radius]] $r$ about $0$.
+Let $f: S \to \C$ be [[Definition:Holomorphic Function|holomorphic]] on $S$. 
+Let $f$ have no [[Definition:Zero of Function|zeroes]] on the circle $\cmod z = r$.
+Let $\map f 0 \ne 0$.
+Let $\rho_1, \ldots, \rho_n$ be the [[Definition:Zero of Function|zeroes]] of $f$ in $D_r$, counted with [[Definition:Multiplicity (Complex Analysis)|multiplicity]].
+Then:
+:$(1): \quad \displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \cmod {\map f {r e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln r - \ln \size {\rho_k} }$
+\end{theorem}
+
+\begin{proof}
+Write $f \left({z}\right) = \left({z - \rho_1}\right) \cdots \left({z - \rho_n}\right) g \left({z}\right)$, so $g \left({z}\right) \ne 0$ for $z \in D_r$.
+It is sufficient to check the equality for each factor of $f$ in this expansion.
+First let $h \left({z}\right) = z - \rho_k$ for some $k \in \left\{{1, \ldots, n}\right\}$.
+Making use of the [[Integration by Substitution|substitution]] $u = r e^{i \theta} - \rho_k$ we find that:
+:$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{h \left({r e^{i \theta} }\right)}\right\vert \ \mathrm d \theta = \frac 1 {2 \pi i} \int_\gamma \frac {\ln \left\vert{u}\right\vert} {u + \rho_k} \, \mathrm d u$
+where $\gamma$ is a circle of radius $r$ centred at $-\rho_k$, traversed [[Definition:Anticlockwise|anticlockwise]].
+{{mistake|I think this last statement is False! The circle is not centered at 0 so $\left\vert{u}\right\vert$ is not constant.}}
+On this circle, $\ln \left\vert{u}\right\vert = \ln r$ is constant, and we have that:
+:$\displaystyle \int_\gamma \frac 1 {u + \rho_k} \ \mathrm d u = \int_{\left\vert{z}\right\vert \mathop = r} \frac {\mathrm d u} u = 2 \pi i$
+Therefore the left hand side of $(1)$ is $\ln r$ as required.
+To show equality for $g \left({z}\right)$, first observe that by the [[Residue Theorem]]:
+:$\displaystyle \int_{\left\vert{z}\right\vert = r} \frac {\ln g \left({z}\right)} z \, \mathrm d z = 2 \pi i \ln g \left({0}\right)$
+Therefore substituting $z = r e^{i \theta}$ we have
+:$\displaystyle 2 \pi i \ln g \left({0}\right) = i \int_0^{2 \pi} \ln g \left({r e^{i \theta} }\right) \, \mathrm d \theta$
+Comparing the imaginary parts of this equality we see that:
+:$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta = \ln \left\vert{g \left({0}\right)}\right\vert$
+as required.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Write
+:$\map f z = \dfrac {r^2 - z \overline {\rho_1} } {r \paren {z - \rho_1} } \cdots \dfrac {r^2 - z \overline {\rho_n} } {r \paren {z - \rho_n} } \map g z$
+so $\map g z \ne 0$ for $z \in D_r$.
+It is sufficient to check equality for each factor of $f$ in this expansion.
+When $\cmod z = r$, we have:
+:$\dfrac 1 z = \dfrac {\overline z} {r^2}$
+and:
+:$\cmod {\dfrac z r} = 1$
+where $\overline z$ denotes the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+So:
+{{begin-eqn}}
+{{eqn | l = \cmod {\frac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} } }
+      | r = \frac {\cmod {r^2 - \overline z \rho_k} } {\cmod {r \paren {z - \rho_k} } }
+      | c = 
+}}
+{{eqn | r = \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {z / r - \rho_k / r} }
+      | c = 
+}}
+{{eqn | r = \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {1 - \rho_k / z} }
+      | c = 
+}}
+{{eqn | r = \frac {\cmod {1 - \overline z \rho_k / r^2} } {\cmod {1 - \overline z \rho_k / r^2} }
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+so the {{LHS}} is $0$.
+Moreover:
+:$\\ln \cmod {\dfrac {r^2 - 0 \overline {\rho_k} } {r \paren {0 - \rho_k} } } = \ln \cmod {\rho_k} - \ln r$
+so the {{RHS}} is $0$.
+Therefore, the formula holds for $\dfrac {r^2 - z \overline {\rho_k} } {r \paren {z - \rho_k} }$.
+Since:
+:$r^2 - z \overline {\rho_i} = 0 \implies \cmod z = r^2 / \cmod {\overline {\rho_i} } > r$
+it follows that $\map g z$ is [[Definition:Holomorphic Function|holomorphic]] without [[Definition:Zero of Function|zeroes]] on $D_r$.
+So the {{RHS}} is $\ln \size {\map g 0}$.
+On the other hand, by the [[Definition:Mean Value Property|mean value property]]:
+:$\displaystyle \dfrac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map g {r e^{i \theta} } } \rd \theta = \ln \map g 0$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue Theorem}
+Tags: Complex Analysis, Named Theorems
+
+\begin{theorem}
+Let $U$ be a [[Definition:Simply Connected|simply connected]] [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Complex Number|complex plane]] $\C$.
+Let $a_1, a_2, \dots, a_n$ be [[Definition:Finite Set|finitely many]] points of $U$.
+Let $f: U \to \C$ be [[Definition:Analytic Function|analytic]] in $U \setminus \set {a_1, a_2, \dots, a_n}$. 
+Let $L = \partial U$ be oriented counterclockwise.
+Then:
+:$\displaystyle \oint_L \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$
+\end{theorem}
+
+\begin{proof}
+Let $\set {U_1, \dotsc, U_n}$ be a set of open subsets of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$.
+Let $U_i \cap U_j = \varnothing$ for all $i \ne j$.
+By [[Existence of Laurent Series]], around each $a_k$ there is an expansion:
+:$\displaystyle \map f z = \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j$ 
+convergent in $U_k$.
+Write: 
+:$\displaystyle X = \bigcup_{i \mathop = 1}^n U_i$
+Then, by [[Contour Integral of Concatenation of Contours]]: 
+:$\displaystyle \oint_L \map f z \rd z = \oint_{\partial \paren {U \setminus X} } \map f z \rd z + \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$ 
+As all poles of $f$ in $U$ are contained in $X$, $f$ is holomorphic on $U \setminus X$, and so by the [[Cauchy-Goursat Theorem]]: 
+:$\displaystyle \oint_{\partial \paren {U \setminus X} } \map f z \rd z = 0$
+Giving: 
+:$\displaystyle \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \oint_{\partial U_k} \map f z \rd z
+	| r = \oint_{\partial U_k} \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j \rd z
+}}
+{{eqn	| r = \oint_{\partial U_k} \paren {\sum_{j \mathop = -\infty}^{-2} c_j \paren {z - a_k}^j + \frac {c_{-1} } {z - a_k} + \sum_{j \mathop = 0}^\infty c_j \paren {z - a_k}^j} \rd z
+}}
+{{eqn	| r = \sum_{j \mathop = -\infty}^{-2} \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z + \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z + \sum_{j \mathop = 0}^\infty \oint_{\partial U_k} c_j \paren {z - a_k}^j \rd z
+}}
+{{eqn	| r = \oint_{\partial U_k} \paren {\frac {c_{-1} } {z - a_k} } \rd z 
+	| c = by [[Contour Integral of Power]], terms where $j \ge 0$ and $j < -1$ vanish
+}}
+{{eqn	| r = 2 \pi i c_{-1}
+	| c = [[Contour Integral of Power]] 
+}}
+{{eqn	| r = 2 \pi i \Res f {a_k}
+	| c = {{Defof|Residue}}
+}}
+{{end-eqn}}
+So:
+:$\displaystyle \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$
+{{qed}}
+[[Category:Complex Analysis]]
+[[Category:Named Theorems]]
+moyv3ydyrhwo2lmkoqzgr51623zznif
+\end{proof}<|endoftext|>
+\section{Cauchy's Integral Formula}
+Tags: Complex Analysis, Cauchy's Integral Formula
+
+\begin{theorem}
+Let $D = \set {z \in \C: \cmod z \le r}$ be the [[Definition:Closed Disk|closed disk]] of [[Definition:Radius of Closed Disk|radius]] $r$ in $\C$.
+Let $f: U \to \C$ be [[Definition:Holomorphic Function|holomorphic]] on some [[Definition:Open Set (Complex Analysis)|open set]] containing $D$.
+Then for each $a$ in the [[Definition:Interior (Complex Analysis)|interior]] of $D$:
+:$\displaystyle \map f a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a} } \rd z$
+where $\partial D$ is the [[Definition:Boundary (Geometry)|boundary]] of $D$, and is traversed [[Definition:Anticlockwise|anticlockwise]].
+\end{theorem}
+
+\begin{proof}
+{{Proofread}}
+Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is [[Definition:Analytic Function|analytic]].
+Let $z_0$ be any point in the region $R$ such that:
+:$\dfrac {\map f z} {z - z_0}$ is [[Definition:Analytic Function|analytic]] everywhere except at $z_0$.
+We draw a circle $C_1$ with center at $z_0$ and radius $r$ such that $r \to 0$.
+This makes $C$ and $C_1$ a multiply connected region.
+{{explain|A diagram at this point would be useful.}}
+According to Cauchy's Integral Theorem for a multiply connected region:
+{{begin-eqn}}
+{{eqn | l = I
+      | o = :=
+      | r = \oint_C \frac {\map f z} {z - z_0} \rd z
+      | c = 
+}}
+{{eqn | r = \oint_{C_1} \frac {\map f z} {z - z_0} \rd z
+      | c = 
+}}
+{{eqn | r = \oint_{C_1} \frac {\map f {z_0} + \paren {\map f z - \map f {z_0} } } {z - z_0} \rd z
+      | c = 
+}}
+{{eqn | r = \map f {z_0} \oint_{C_1} \frac {\rd z} {z - z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z
+      | c = 
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = z - z_0
+      | r = r e^{i \theta}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \d z
+      | r = i r e^{i \theta} \rd \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \oint_{C_1} \frac {\rd z} {z - z_0}
+      | r = \int_0^{2 \pi} \frac {i r e^{i \theta} } {r e^{i \theta} } \rd \theta
+      | c = 
+}}
+{{eqn | r = i \int_0^{2 \pi} \rd \theta
+      | c = 
+}}
+{{eqn | r = 2 \pi i
+      | c = 
+}}
+{{end-eqn}}
+Now:
+:$\displaystyle I = 2 \pi i \map f {z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z$
+According to Epsilon-Delta definition of limit, for every $\left|{z - z_0}\right| < \delta$ there exists a $\epsilon \in \R_{>0}$ such that:
+:$\cmod {\map f z - \map f {z_0} } < \epsilon$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \cmod {\oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z}
+      | o = \le
+      | r = \oint_{C_1} \frac {\cmod {\map f z - \map f {z_0} } } {z - z_0} \cmod {\d z}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac {\epsilon} {\delta} \oint_{C_1} \cmod {\d z}
+      | c = 
+}}
+{{eqn | r = 2 \pi \epsilon
+      | c = 
+}}
+{{end-eqn}}
+{{explain|The precise meaning of $\cmod {\d z}$}}
+As $\epsilon \to 0$:
+:$\displaystyle \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z = 0$
+So:
+{{begin-eqn}}
+{{eqn | l = I
+      | r = \map f {z_0} \oint_{C_1} \frac {\rd z} {z - z_0} + \oint_{C_1} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z
+      | c = 
+}}
+{{eqn | r = 2 \pi i \map f {z_0} + 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \oint_C \frac {\map f z} {z - z_0} \rd z
+      | r = 2 \pi i \, \map f {z_0}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustin Louis Cauchy|cat = Cauchy}}
+\end{proof}<|endoftext|>
+\section{Cauchy's Integral Formula/General Result}
+Tags: Cauchy's Integral Formula
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
+Then for each $a$ in the [[Definition:Interior (Topology)|interior]] of $D$:
+:$\displaystyle f^{\paren n} \paren a = \dfrac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
+where $\partial D$ is the [[Definition:Boundary (Geometry)|boundary]] of $D$, and is traversed [[Definition:Anticlockwise|anticlockwise]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
+=== Basis for the Induction ===
+$\map P 0$ holds, as this is:
+:$\displaystyle \map f a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}} \rd z$
+which is [[Cauchy's Integral Formula]].
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map {f^{\paren k} } a = \frac {k!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z$
+Then we need to show:
+:$\displaystyle \map {f^{\paren {k + 1} } } a = \frac {\paren {k + 1}!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \frac {\rd} {\rd a} \map {f^{\paren k} } a
+      | r = \frac {k!} {2 \pi i} \int_{\partial D} \frac {\rd} {\rd a} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z
+      | c =
+}}
+{{eqn | r = \frac {k!} {2 \pi i}\int_{\partial D} \frac {\paren {k + 1} \map f z} {\paren {z - a}^{k + 2} } \rd z
+      | c =
+}}
+{{eqn | r = \frac {\paren {k + 1}!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z
+      | c =
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \N: \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic of Subfield of Complex Numbers is Zero}
+Tags: Subfields, Complex Numbers
+
+\begin{theorem}
+The [[Definition:Characteristic of Ring|characteristic]] of any [[Definition:Subfield|subfield]] of the [[Definition:Field of Complex Numbers|field of complex numbers]] is $0$.
+\end{theorem}
+
+\begin{proof}
+Suppose to the contrary.
+Let $K$ be a [[Definition:Subfield|subfield]] of $\C$ such that $\Char K = n$ where $n \in \N, n > 0$.
+Then:
+:$\exists a \in K: n \cdot a = 0$
+But as $K$ is a [[Definition:Subfield|subfield]] of $\C$ it follows that $K \subseteq \C$ which means:
+:$\exists a \in \C: n \cdot a = 0$
+Thus, by definition of [[Definition:Characteristic of Ring|characteristic]]:
+:$0 < \Char \C \le n$
+But $\C$ is [[Definition:Infinite Set|infinite]] and so $\Char \C = 0$.
+From that contradiction follows the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic times Ring Element is Ring Zero}
+Tags: Rings with Unity
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
+Let the [[Definition:Ring Zero|zero]] of $R$ be $0_R$ and the [[Definition:Unity of Ring|unity]] of $R$ be $1_R$.
+Let the [[Definition:Characteristic of Ring|characteristic]] of $R$ be $n$.
+Then:
+:$\forall a \in R: n \cdot a = 0_R$
+\end{theorem}
+
+\begin{proof}
+If $a = 0_R$ then $n \cdot a = 0_R$ is immediate.
+So let $a \in R: a \ne 0_R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n \cdot a
+      | r = n \cdot \paren {1_R \circ a}
+      | c = {{Defof|Unity of Ring}}
+}}
+{{eqn | r = \paren {n \cdot 1_R} \circ a
+      | c = [[Powers of Ring Elements]]
+}}
+{{eqn | r = 0_R \circ a
+      | c = {{Defof|Characteristic of Ring}}
+}}
+{{eqn | r = 0_R
+      | c = {{Defof|Ring Zero}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic of Finite Ring is Non-Zero}
+Tags: Finite Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Finite Ring|finite]] [[Definition:Ring with Unity|ring with unity]].
+Then the [[Definition:Characteristic of Ring|characteristic]] of $R$ is not [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+We have that $\struct {R, +, \circ}$ is [[Definition:Finite Ring|finite]], so its [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$ is likewise [[Definition:Finite Group|finite]].
+The result follows by [[Element of Finite Group is of Finite Order]] and the definition of [[Definition:Characteristic of Ring|characteristic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field of Rational Functions is Field}
+Tags: Field Theory, Polynomial Theory
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $K \sqbrk x$ be the [[Ring of Polynomial Forms is Integral Domain|integral domain of polynomial forms]] on $K$.
+Let $\map K x$ be the [[Definition:Field of Rational Functions|field of rational functions]] on $K$.
+Then $\map K x$ forms a [[Definition:Field (Abstract Algebra)|field]].
+If the [[Definition:Characteristic of Ring|characteristic]] of $K$ is $p$, then the [[Definition:Characteristic of Ring|characteristic]] of $\map K x$ is non-[[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Reverted this to the version 9th May 2011 as it's mutated into something that makes less sense.}}
+\end{proof}<|endoftext|>
+\section{Hadamard Factorization Theorem}
+Tags: Entire Functions, Infinite Products, Complex Analysis
+
+\begin{theorem}
+Let $f: \C \to \C$ be an [[Definition:Entire Function|entire function]] of [[Definition:Order of Entire Function|finite order]] $\omega$.
+Let $0$ be a zero of $f$ of [[Definition:Multiplicity (Analytic Function)|multiplicity]] $m\geq0$.
+Let $\left\langle{a_n}\right\rangle$ be the [[Definition:Sequence|sequence]] of nonzero zeroes of $f$, repeated according to [[Definition:Multiplicity (Complex Analysis)|multiplicity]].
+Then $f$ has [[Definition:Rank of Entire Function|finite rank]] $p\leq\omega$ and there exists a [[Definition:Polynomial|polynomial]] $g$ of [[Definition:Degree of Polynomial|degree]] at most $\omega$ such that:
+:$\displaystyle f \left({z}\right) = z^m e^{g(z)} \prod_{n \mathop = 1}^\infty E_p\left( \frac z{a_n} \right)$
+where $E_p$ denotes the $p$th [[Definition:Weierstrass Elementary Factor|Weierstrass elementary factor]].
+\end{theorem}
+
+\begin{proof}
+By [[Convergence Exponent is Less Than Order]], $f$ has finite [[Definition:Exponent of Convergence|exponent of convergence]] $\tau \leq \omega$.
+By [[Relation Between Rank and Exponent of Convergence]], $f$ has [[Definition:Rank of Entire Function|finite rank]] $p\leq\omega$.
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Approximation to Stirling's Formula for Gamma Function}
+Tags: Gamma Function, Stirling's Formula
+
+\begin{theorem}
+Let:
+:$D_\epsilon = \left\{{z \in \C : \left\vert{\operatorname{Arg} \left({z}\right)}\right\vert < \pi - \epsilon,\ \left\vert{z}\right\vert > 1}\right\}$
+where:
+:$\left\vert{\operatorname{Arg} \left({z}\right)}\right\vert$ denotes the [[Definition:Absolute Value|absolute value]] of the [[Definition:Principal Argument|principal argument]] of $z$
+:$\left\vert{z}\right\vert$ denotes the [[Definition:Complex Modulus|modulus]] of $z$
+:$\epsilon \in \R_{>0}$.
+Then for all $z \in D_\epsilon$, the [[Definition:Gamma Function|gamma function]] of $z$ satisfies:
+:$\Gamma \left({z}\right) = \sqrt{\dfrac {2 \pi} z} \left({\dfrac z e}\right)^z\left({1 + \mathcal O \left({z^{-1} }\right)}\right)$
+where $\mathcal O \left({z^{-1} }\right)$ denotes [[Definition:Big-O Notation|big-O]] of $z^{-1}$.
+\end{theorem}
+
+\begin{proof}
+From [[Gamma Function is Unique Extension of Factorial]]:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \left({y + n}\right)^y n!
+      | r = \left({y + n}\right)^y \Gamma \left({n + 1}\right)
+      | c = [[Gamma Function Extends Factorial]]
+}}
+{{eqn | o = \le
+      | r = \Gamma \left({y + n + 1}\right)
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left({n + 1}\right)^y \Gamma \left({n + 1}\right)
+      | c = 
+}}
+{{eqn | r = \left({n + 1}\right)^y n!
+      | c = 
+}}
+{{end-eqn}}
+for $0 < y \le 1$ and $n \in \N$.
+Let $x$ be given.
+Let $n + 1$ be the largest [[Definition:Natural Number|natural number]] such that $n + 1 \le x$.
+Let $x = y + n + 1$, and thus $0 < y \le 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\Gamma \left({x}\right)}{\sqrt{\left({2 \pi}\right)} x^x x^{-1/2} e^{-x} }
+      | o = \le
+      | r = \frac {\left({n + 1}\right)^y n!}{\sqrt{\left({2 \pi}\right)} x^x x^{-1/2} e^{-x} }
+      | c = from $(1)$
+}}
+{{eqn | o = \sim
+      | r = \frac {\left({n + 1}\right)^y n! \sqrt{\left({2 \pi}\right)} n^n n^{1/2} e^{-n} }{\sqrt{\left({2 \pi}\right)} x^x x^{-1/2} e^{-x} }
+      | c = [[Stirling's Formula]]
+}}
+{{eqn | r = \frac {\left({n + 1}\right)^y n! n^n n^{1/2} e^{-n} }{\left({y + n + 1}\right)^{y + n} \left({y + n + 1}\right)^{1/2} e^{-y-n-1} }
+      | c = 
+}}
+{{eqn | r = \left({\frac {n + 1} {y + n + 1} }\right)^y \left({\frac n {y + n + 1} }\right)^{1/2} \left({1 + \frac {y + 1} n }\right)^{-n} e^{y+1}
+      | c = 
+}}
+{{eqn | o = \to
+      | r = 1 \cdot 1 \cdot \frac 1 {e^{y + 1} } e^{y+1} \text { as } n \to \infty
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Similarly for the {{RHS}}.
+The result follows from [[Gamma Function Extends Factorial]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Gamma Function}
+Tags: Gamma Function
+
+\begin{theorem}
+The [[Definition:Gamma Function|gamma function]] $\Gamma \left({z}\right)$ has the following properties:
+\end{theorem}<|endoftext|>
+\section{Poisson Summation Formula}
+Tags: Harmonic Analysis, Fourier Analysis
+
+\begin{theorem}
+Let $f: \R \to \C$ be a [[Definition:Schwarz Function|Schwarz function]].
+Let $\hat f$ be its [[Definition:Fourier Transform|Fourier transform]].
+Then:
+:$\displaystyle \sum_{n \mathop \in \Z} \map f n = \sum_{m \mathop \in \Z} \map {\hat f} m$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\displaystyle \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$
+Then $\map F x$ is $1$-[[Definition:Periodic Function|periodic]] (because of absolute convergence), and has [[Definition:Fourier Coefficient|Fourier coefficients]]:
+{{begin-eqn}}
+{{eqn | l = \hat F_k
+      | r = \int_0^1 \sum_{n \mathop \in \Z} \map f {x + n} e^{-2\pi i k x} \rd x
+      | c = 
+}}
+{{eqn | r = \sum_{n \mathop \in \Z} \int_0^1 \map f {x + n} e^{-2\pi i k x} \rd x
+      | c = because $f$ is [[Definition:Schwarz Function|Schwarz]], so convergence is [[Definition:Uniform Convergence|uniform]]
+}}
+{{eqn | r = \sum_{n \mathop \in \Z} \int_n^{n + 1} \map f x e^{-2\pi i k x} \rd x
+      | c = 
+}}
+{{eqn | r = \int_\R \map f x e^{-2\pi i k x} \rd x
+      | c = 
+}}
+{{eqn | r = \map {\hat f} k
+      | c = where $\hat f$ is the [[Definition:Fourier Transform|Fourier transform]] of $f$
+}}
+{{end-eqn}}
+Therefore by the definition of the [[Definition:Fourier Series|Fourier series]] of $f$:
+:$\displaystyle \map F x = \sum_{k \mathop \in \Z} \map {\hat f} k e^{i k x}$
+Choosing $x = 0$ in this formula:
+:$\displaystyle \sum_{n \mathop \in \Z} \map f n = \sum_{k \mathop \in \Z} \map {\hat f} k$
+as required.
+{{qed}}
+{{Namedfor|Siméon-Denis Poisson|cat = Poisson}}
+\end{proof}<|endoftext|>
+\section{Harmonic Properties of Schwarz Functions}
+Tags: Harmonic Analysis
+
+\begin{theorem}
+Let $f, g : \R \to \C$ be [[Definition:Schwarz Function|Schwarz functions]].
+Let $\hat f$, $\hat g$ be the [[Definition:Fourier Transform|Fourier transforms]] of $f$ and $g$ respectively.
+Then:
+:$(1): \quad \hat f$, $\hat g$ are [[Definition:Schwarz Function|Schwarz functions]].
+:$(2): \quad \map {\widehat {\paren {\hat f} } } x = \map f {-x}$ for all $x \in \R$.
+:$(3): \quad$ If $f * g$ is the [[Definition:Convolution|convolution]] of $f$ and $g$, then:
+{{disambiguate|Definition:Convolution}}
+::::$\widehat {f * g} = \hat f \hat g$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Harmonic Analysis]]
+m37fh6h38xj9zvowgbbbu5ezjkssvwc
+\end{proof}<|endoftext|>
+\section{Euler's Reflection Formula}
+Tags: Gamma Function
+
+\begin{theorem}
+:$\forall z \notin \Z: \map \Gamma z \, \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$
+\end{theorem}
+
+\begin{proof}
+We have the [[Weierstrass Factorization Theorem|Weierstrass products]]:
+:$\displaystyle \map \sin {\pi z} = \pi z \prod_{n \mathop \ne 0} \paren {1 - \frac z n} \map \exp {\frac z n}$
+From the [[Definition:Weierstrass Form of Gamma Function|Weierstrass form of the Gamma function]]:
+:$\displaystyle \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \map \exp {-\frac z n}$
+from which:
+{{begin-eqn}}
+{{eqn   | l = \dfrac 1 {-z \, \map \Gamma z \, \map \Gamma {-z} }
+	| r = \frac {-z^2 \, \map \exp {\gamma z} \, \map \exp {-\gamma z} } {-z} \prod_{n \mathop = 1}^\infty \paren {1 + \frac z n} \paren {1 - \frac z n} \, \map \exp {\frac z n} \, \map \exp {-\frac z n}
+}}
+{{eqn   | r = z \prod_{n \mathop = 1}^\infty \paren {1 - \frac {z^2} {n^2} }
+}}
+{{eqn   | r = \dfrac {\map \sin {\pi z} } \pi 
+	| c = [[Euler Formula for Sine Function]]
+}}
+{{end-eqn}}
+whence:
+{{begin-eqn}}
+{{eqn | l = \map \Gamma z \, \map \Gamma {1 - z}
+      | r = -z \, \map \Gamma z \, \map \Gamma {-z}
+      | c = [[Gamma Difference Equation]]
+}}
+{{eqn | r = \frac \pi {\map \sin {\pi z} }
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Leonhard Paul Euler|cat = Euler}}
+\end{proof}<|endoftext|>
+\section{Legendre's Duplication Formula}
+Tags: Gamma Function, Legendre's Duplication Formula
+
+\begin{theorem}
+:$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \, \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
+where $\N$ denotes the [[Definition:Natural Number|natural numbers]].
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Beta Function|Beta function]]:
+{{begin-eqn}}
+{{eqn | l = \map \Beta {z_1, z_2}
+      | r = \frac {\map \Gamma {z_1} \, \map \Gamma {z_2} } {\map \Gamma {z_1 + z_2} } 
+      | c = 
+}}
+{{eqn | r = \int_0^1 u^{z_1 - 1} \paren {1 - u}^{z_2 - 1} \rd u
+      | c = [[Equivalence of Definitions of Beta Function]]
+}}
+{{end-eqn}}
+Letting $z_1 = z_2 = z$ gives:
+{{begin-eqn}}
+{{eqn | l = \frac {\map \Gamma z \, \map \Gamma z} {\map \Gamma {2 z} }
+      | r = \int_0^1 u^{z - 1} \paren {1 - u}^{z - 1} \rd u
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \int_{-1}^1 \paren {\frac {1 + x} 2 }^{z - 1} \paren {\frac {1 - x} 2}^{z - 1} \rd x
+      | c = [[Integration by Substitution|Substitute]] $u = \dfrac {1 + x} 2$
+}}
+{{eqn | r = \frac 1 {2^{2 z - 1} } \int_{-1}^1 \paren {1 - x^2}^{z - 1} \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | n = 1
+      | l = 2^{2 z - 1} \map \Gamma z \, \map \Gamma z
+      | r = 2 \map \Gamma {2 z} \int_0^1 \paren {1 - x^2}^{z - 1} \rd x
+      | c = by [[Definite Integral of Even Function]]
+}}
+{{end-eqn}}
+Now substituting $u = x^2$ into the [[Definition:Beta Function|Beta function]]:
+:$\displaystyle \map \Beta {z_1, z_2} = \int_0^1 x^{2 z_1 - 2} \paren {1 - x^2}^{z_2 - 1} 2 x \rd x$
+Letting $z_1 = \dfrac 1 2$ and $z_2 = z$ gives:
+:$(2): \quad \displaystyle \map \Beta {\frac 1 2, z} = 2 \int_0^1 \paren {1 - x^2}^{z - 1} \rd x$
+Combining results $(1)$ and $(2)$:
+{{begin-eqn}}
+{{eqn | l = 2^{2 z - 1} \map \Gamma z \, \map \Gamma z 
+      | r = \map \Gamma {2 z} \map \Beta {\frac 1 2, z}
+      | c =
+}}
+{{eqn | r = \map \Gamma {2 z} \frac {\map \Gamma {\frac 1 2} \, \map \Gamma z} {\map \Gamma {\frac 1 2 + z} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \Gamma z \, \map \Gamma {z + \dfrac 1 2} 
+      | r = 2^{1 - 2 z} \map \Gamma {\frac 1 2} \, \map \Gamma {2 z}
+}}
+{{end-eqn}}
+From [[Gamma Function of One Half]]:
+:$\map \Gamma {\dfrac 1 2} = \sqrt \pi$
+It follows that:
+:$\map \Gamma z \, \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Gauss Multiplication Formula]]:
+:$\displaystyle \prod_{k \mathop = 0}^{n - 1} \Gamma \left({z + \frac k n}\right) = \left({2 \pi}\right)^{\left({n - 1}\right) / 2} n^{1/2 - n z} \Gamma \left({n z}\right)$
+Substituting $n = 2$ yields:
+{{begin-eqn}}
+{{eqn | l = \Gamma \left({z}\right) \Gamma \left({z + \frac 1 2}\right)
+      | r = \left({2 \pi}\right)^{1 / 2} 2^{1/2 - 2 z} \Gamma \left({2 z}\right)
+      | c = 
+}}
+{{eqn | r = 2^{1 - 2 z} \sqrt \pi \, \Gamma \left({2 z}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Approximation to Reciprocal times Derivative of Gamma Function}
+Tags: Gamma Function, Reciprocals
+
+\begin{theorem}
+For all $z \in \C$ such that $\cmod {\map \arg z} < \pi - \epsilon, \cmod z > 1$:
+:$\dfrac {\map {\Gamma'} z} {\map \Gamma z} = \ln z + \map {\mathcal O_\epsilon} {z^{-1} }$
+where:
+:$\map {\mathcal O} {z^{-1} }$ denotes [[Definition:Big-O Notation|big-O notation]]
+:the [[Definition:Implied Constant|implied constant]] depends on $\epsilon$.
+\end{theorem}
+
+\begin{proof}
+From [[Stirling's Formula for Gamma Function]]:
+:$\ln \Gamma \left({z}\right) = \left({z - \dfrac 1 2}\right) \ln z - z + \dfrac {\ln 2 \pi} 2 + \mathcal O \left({z^{-1}}\right)$
+Taking the derivative:
+:$(1): \quad \dfrac {\Gamma' \left({z}\right)} {\Gamma \left({z}\right)} = \ln z - \dfrac 1 {2 z} + \dfrac {\mathrm d} {\mathrm dz} \mathcal O \left({z^{-1}}\right)$
+Since there exists $c \left({\epsilon}\right) > 0$ such that:
+:$- \dfrac c {\left\vert{z^{-1} }\right\vert} < \left\vert{\mathcal O \left({z^{-1} }\right)}\right\vert < \dfrac c{\left\vert{z^{-1} }\right\vert}, \quad \left\vert{z}\right\vert > 1$
+{{explain|The above statement needs to be explained and justified.}}
+it follows directly that the third term in $(1)$ is $\mathcal O \left({z^{-1}}\right)$.
+{{qed}}
+[[Category:Gamma Function]]
+[[Category:Reciprocals]]
+0l44q7x1pqd6cadsxsgxeptvnavacjq
+\end{proof}<|endoftext|>
+\section{Complex Conjugate of Gamma Function}
+Tags: Gamma Function, Complex Conjugates
+
+\begin{theorem}
+:$\forall z \in \C \setminus \set {0, -1, -2, \ldots}: \map \Gamma {\overline z} = \overline {\map \Gamma z}$
+\end{theorem}
+
+\begin{proof}
+This is immediate from, say, the [[Definition:Euler Form of Gamma Function|Euler form]] of $\Gamma$ and the fact that [[Definition:Complex Conjugate|complex conjugation]] preserves products and sums.
+[[Category:Gamma Function]]
+[[Category:Complex Conjugates]]
+6vkdpw0tos2m2r2lw5vlcxtn3rmio06
+\end{proof}<|endoftext|>
+\section{Linear Combination of Integers is Ideal}
+Tags: Ideal Theory, Integers
+
+\begin{theorem}
+Let $a, b$ be any [[Definition:Integer|integers]].
+Let $\Bbb S = \set {a x + b y: x, y \in \Z}$.
+Then the [[Definition:Algebraic Structure|algebraic structure]]:
+:$\struct {\Bbb S, +, \times}$
+is an [[Definition:Ideal of Ring|ideal]] of $\Z$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Integer Combination|integer combination]] (or just because it's obvious):
+:$\Bbb S \subseteq \Z$
+and clearly:
+:$\Bbb S \ne \O$
+as $a 0 + b 0 = 0 \in \Bbb S$.
+Let $w_1, w_2 \in \Bbb S$:
+:$w_1 = a x_1 + b y_1, w_2 = a x_2 + b y_2$
+Then $w_1 + \paren {-w_2} = a \paren {x_1 - x_2} + b \paren {y_1 - y_2} \in \Bbb S$ as $x_1 - x_2 \in \Z$ and $y_1 - y_2 \in \Z$.
+Let $r \in \Z$.
+Then:
+:$w_1 \times r = a x_1 r + b y_1 r = a \paren {x_1 r} + b \paren {y_1 r} \in \Bbb S$
+:$r \times w_1 = r a x_1 + r b y_1 = \paren {r x_1} a + \paren {r y_1} b \in \Bbb S$
+The result follows from [[Test for Ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Integer Ideals is Lowest Common Multiple}
+Tags: Ideal Theory, Integers, Lowest Common Multiple
+
+\begin{theorem}
+Let $\ideal m$ and $\ideal n$ be [[Definition:Ideal of Ring|ideals]] of the [[Definition:Integer|integers]] $\Z$.
+Let $\ideal k$ be the [[Definition:Set Intersection|intersection]] of $\ideal m$ and $\ideal n$.
+Then $k = \lcm \set {m, n}$.
+\end{theorem}
+
+\begin{proof}
+By [[Intersection of Ring Ideals is Ideal]] we have that $\ideal k = \ideal m \cap \ideal n$ is an [[Definition:Ideal of Ring|ideal]] of $\Z$.
+By [[Ring of Integers is Principal Ideal Domain]] we have that $\ideal m$, $\ideal n$ and $\ideal k$ are all necessarily [[Definition:Principal Ideal of Ring|principal ideals]].
+By [[Subrings of Integers are Sets of Integer Multiples]] we have that:
+:$\ideal m = m \Z, \ideal n = n \Z$
+Thus:
+:$\ideal k = \set {x \in \Z: n \divides x \land m \divides x}$
+The result follows by [[LCM iff Divides All Common Multiples]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{LCM iff Divides All Common Multiples}
+Tags: Lowest Common Multiple
+
+\begin{theorem}
+Let $a, b \in \Z$ such that $a b \ne 0$.
+Let $m \in \Z: d > 0$.
+Then $m = \lcm \set {a, b}$ {{iff}}:
+: $(1): \quad a \divides m \land b \divides m$
+: $(2): \quad a \divides n \land b \divides n \implies m \divides n$
+That is, in the [[Definition:Set|set]] of [[Definition:Positive Integer|positive integers]], $m$ is the [[Definition:Lowest Common Multiple of Integers|LCM]] of $a$ and $b$ {{iff}} $m$ is a [[Definition:Common Multiple|common multiple]] of $a$ and $b$, and $m$ also [[Definition:Divisor of Integer|divides]] any other [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $m = \lcm \set {a, b}$.
+Then from [[LCM Divides Common Multiple]]:
+: $a \divides n \land b \divides n \implies m \divides n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose $a \divides m \land b \divides m$, and $m$ also [[Definition:Divisor of Integer|divides]] any $n$ that $a$ and $b$ also [[Definition:Divisor of Integer|divide]].
+From $a \divides m \land b \divides m$, we see that $m$ is a [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+From $a \divides n \land b \divides n$, we see that $n$ is also a [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+Also, we have that $m \divides n$.
+From [[Absolute Value of Integer is not less than Divisors]], we see that (in the [[Definition:Domain of Relation|domain]] of $\Z_{>0}$) $m \divides n \implies m \le n$.
+Thus, whatever $m$ may be, it is no larger than $n$.
+Therefore, $m$ must be the least of all the [[Definition:Common Multiple|common multiples]] of $a$ and $b$.
+{{qed}}
+[[Category:Lowest Common Multiple]]
+ojgfu5gmd7pmtf8ef7xaj8tmtiinj36
+\end{proof}<|endoftext|>
+\section{LCM Divides Common Multiple}
+Tags: Lowest Common Multiple
+
+\begin{theorem}
+Let $a, b \in \Z$ such that $a b \ne 0$.
+Let $n$ be any [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+That is, let $n \in \Z: a \divides n, b \divides n$.
+Then:
+:$\lcm \set {a, b} \divides n$
+where $\lcm \set {a, b}$ is the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$.
+{{:Euclid:Proposition/VII/35}}
+\end{theorem}
+
+\begin{proof}
+Let $m = \lcm \set {a, b}$.
+Then $a \divides m$ and $b \divides m$ [[Definition:Lowest Common Multiple of Integers|by definition]].
+Suppose $n$ is some other [[Definition:Common Multiple|common multiple]] of $a$ and $b$ such that $m \nmid n$ ($m$ does not divide $n$).
+Then from the [[Division Theorem]]:
+:$n = k m + r$
+for some integer $k$ and with $0 < r < m$.
+Then since $r = n - k m$, using $a \divides n$ and $a \divides m$:
+: $a \divides r$
+Similarly:
+: $b \divides r$
+Then $r$ is a [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+But we have that $r < m$.
+This [[Proof by Contradiction|contradicts]] the fact that $m$ is the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$.
+So, by contradiction, it follows that $m \divides n$.
+{{qed}}
+{{Euclid Note|35|VII|}}
+\end{proof}<|endoftext|>
+\section{Sum of Integer Ideals is Greatest Common Divisor}
+Tags: Ideal Theory, Integers, Greatest Common Divisor
+
+\begin{theorem}
+Let $\ideal m$ and $\ideal n$ be [[Definition:Ideal of Ring|ideals]] of the [[Definition:Integer|integers]] $\Z$.
+Let $\ideal d = \ideal m + \ideal n$.
+Then $d = \gcd \set {m, n}$.
+\end{theorem}
+
+\begin{proof}
+By [[Sum of Ideals is Ideal]] we have that $\ideal d = \ideal m + \ideal n$ is an [[Definition:Ideal of Ring|ideal]] of $\Z$.
+By [[Ring of Integers is Principal Ideal Domain]] we have that $\ideal m$, $\ideal n$ and $\ideal d$ are all necessarily [[Definition:Principal Ideal of Ring|principal ideals]].
+By [[Subrings of Integers are Sets of Integer Multiples]] we have that:
+:$\ideal m = m \Z, \ideal n = n \Z$
+Thus:
+:$\ideal d = \ideal m + \ideal n = \set {x \in \Z: \exists a, b \in \Z: x = a m + b n}$
+That is, $\ideal d$ is the set of all [[Definition:Integer Combination|integer combinations]] of $m$ and $n$.
+The result follows by [[Bézout's Lemma]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Integers with Principal Ideal}
+Tags: Ideal Theory, Quotient Rings, Integers
+
+\begin{theorem}
+Let $\struct {\Z, +, \times}$ be the [[Integers form Integral Domain|integral domain of integers]].
+Let $n \in \Z$.
+Let $\ideal n$ be the [[Definition:Principal Ideal of Ring|principal ideal of $\struct {\Z, +, \times}$ generated by $n$]].
+The [[Definition:Quotient Ring|quotient ring]] $\struct {\Z, +, \times} / \ideal n$ is [[Definition:Ring Isomorphism|isomorphic]] to $\struct {\Z_n, +_n, \times_n}$, the [[Definition:Ring of Integers Modulo m|ring of integers modulo $n$]].
+Note the special cases where $n = 0$ or $1$:
+:[[Quotient Ring of Integers and Zero]]
+:[[Quotient Ring of Integers and Principal Ideal from Unity]]
+\end{theorem}<|endoftext|>
+\section{Quotient Ring by Null Ideal}
+Tags: Ideal Theory, Quotient Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\struct {\set {0_R}, +, \circ}$ be the [[Definition:Null Ideal|null ideal]] of $\struct {R, +, \circ}$.
+Let $\struct {R / \set {0_R}, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] of $R$ defined by $\set {0_R}$.
+Then $\struct {R / \set {0_R}, +, \circ}$ is [[Definition:Ring Isomorphism|isomorphic]] to $\struct {R, +, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Additive Group of Ring|additive group]] $\struct {R, +}$.
+From [[Trivial Quotient Group is Quotient Group]]:
+:$\struct {R, +} / \set {0_R} \cong \struct {R, +}$
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Ring Homomorphism Preserves Negatives}
+Tags: Ring Homomorphisms
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Then:
+:$\forall x \in R_1: \map \phi {-x} = -\paren {\map \phi x}$
+\end{theorem}
+
+\begin{proof}
+We have that [[Ring Homomorphism of Addition is Group Homomorphism]].
+The result follows from [[Group Homomorphism Preserves Inverses]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]].
+Let $s \in \C$ be a [[Definition:Complex Number|complex number]] with [[Definition:Real Part|real part]] $\sigma>1$.
+Then:
+:$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = - \frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left({x^{s / 2 - 1} + x^{- \left({s + 1}\right) / 2} }\right) \omega \left({x}\right) \rd x$
+where:
+: $\Gamma$ is the [[Definition:Gamma Function|gamma function]]
+: $\displaystyle \omega \left({x}\right) = \sum_{n \mathop = 1}^\infty e^{- \pi n^2 x}$
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Integral Form of Gamma Function|Gamma function]]:
+:$\displaystyle \Gamma \left({\frac s 2}\right) = \int_0^\infty t^{s / 2 - 1} e^{-t} \rd t$
+First substitute $t = \pi n^2 x$ to give:
+:$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) n^{-s} = \int_0^\infty x^{s / 2 - 1} e^{-\pi n^2 x} \rd x$
+By definition, on $\operatorname{Re} \left({s}\right) > 1$:
+:$\displaystyle \zeta \left({s}\right) = \sum_{n \mathop = 1}^\infty n^{-s}$
+Therefore summing over $n \ge 1$:
+:$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = \int_0^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x$
+where:
+:$\displaystyle \omega \left({x}\right) = \sum_{n \mathop = 1}^\infty e^{-\pi n^2 x}$
+Next the integral is split at $x = 1$ as follows:
+{{begin-eqn}}
+{{eqn | l = \pi^{-s / 2}\Gamma \left({\frac s 2}\right) \zeta \left({s}\right)
+      | r = \int_0^1 x^{s / 2 - 1} \omega \left({x}\right) \rd x + \int_1^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x
+      | c =
+}}
+{{eqn | r = \int_1^\infty x^{-s / 2 - 1} \omega \left({\frac 1 x}\right) \rd x + \int_1^\infty x^{s / 2 - 1} \omega \left({x}\right) \rd x
+      | c = [[Integration by Substitution|substituting]] $x \mapsto \dfrac 1 x$ in the first integral.
+}}
+{{end-eqn}}
+Recall the [[Definition:Theta function|Jacobi theta function]]:
+{{begin-eqn}}
+{{eqn | l = \theta \left({x}\right)
+      | r = \sum_{n \mathop \in \Z} e^{- \pi n^2 x}
+      | c = 
+}}
+{{eqn | r = 2 \omega \left({x}\right) + 1
+      | c = 
+}}
+{{end-eqn}}
+Since $e^{-x^2}$ is a [[Definition:Fixed Point|fixed point]] of the [[Definition:Fourier Transform|Fourier transform]], by [[Properties of the Fourier Transform]]:
+:$\mathcal F \left[{e^{-\pi t^2 x} }\right] \left({u}\right) = x^{-1 / 2} e^{-\pi u^2 / x}$
+where $\mathcal F$ denotes the Fourier transform.
+Therefore, by the [[Poisson Summation Formula]]:
+:$\theta \left({x}\right) = \dfrac 1 {\sqrt x} \theta \left({\dfrac 1 x}\right)$
+whence:
+:$\omega \left({\dfrac 1 x}\right) = -\dfrac 1 2 + \dfrac 1 2 \sqrt x + \sqrt x \omega \left({x}\right)$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \int_1^\infty x^{-s / 2 - 1} \omega \left({\frac 1 x}\right) \rd x
+      | r = \int_1^\infty x^{-s / 2 - 1} \left({-\frac 1 2 + \frac 1 2 \sqrt x + \sqrt x \omega \left({x}\right)}\right) \rd x
+      | c = 
+}}
+{{eqn | r = - \frac 1 s + \frac 1 {s - 1} + \int_1^\infty x^{-\left({s + 1}\right)/2} \omega \left({x}\right) \rd x
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\displaystyle \pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = -\frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left[{x^{s / 2} + x^{\left({1 - s}\right) / 2} }\right] \omega \left({x}\right) \frac {\rd x} x$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping on Integers a plus b root 2 to Conjugate is Automorphism}
+Tags: Ring Automorphisms, Integral Domains
+
+\begin{theorem}
+Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
+:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
+that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
+Then the [[Definition:Mapping|mapping]] $\phi: \Z \sqbrk {\sqrt 2} \to \Z \sqbrk {\sqrt 2}$ defined as:
+:$\forall x = a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}: \map \phi {a + b \sqrt 2} = a - b \sqrt 2$
+is a [[Definition:Ring Automorphism|ring automorphism]].
+\end{theorem}
+
+\begin{proof}
+We have [[Numbers of Type Integer a plus b root 2 form Subdomain of Reals]].
+First we note that:
+:$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$
+=== Proof of Bijection ===
+Let $\map \phi {a + b \sqrt 2} = \map \phi {a' + b' \sqrt 2}$.
+Then:
+:$a - b \sqrt 2 = a' - b' \sqrt 2$
+and so:
+:$a + b \sqrt 2 = a' + b' \sqrt 2$
+So $\phi$ is [[Definition:Injection|injective]].
+Now let $y = c + d \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map \phi {c + \paren {-d} \sqrt 2}
+      | r = c - \paren {-d} \sqrt 2
+      | c = 
+}}
+{{eqn | r = c + d \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$\forall y \in \Z \sqbrk {\sqrt 2}: \exists x \in \Z \sqbrk {\sqrt 2}: \map \phi x = y$
+and so $\phi$ is [[Definition:Surjection|surjective]].
+So $\phi$ is a [[Definition:Bijection|bijection]].
+=== Proof of Morphism ===
+Now consider $x = a + b \sqrt 2, y = c + d \sqrt 2$.
+{{begin-eqn}}
+{{eqn | l = \map \phi x + \map \phi y
+      | r = \map \phi {a + b \sqrt 2} + \map \phi {c + d \sqrt 2}
+      | c = 
+}}
+{{eqn | r = \paren {a - b \sqrt 2} + \paren {c - d \sqrt 2}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \paren {a + c} - \paren {b + d} \sqrt 2
+      | c = 
+}}
+{{eqn | r = \map \phi {\paren {a + c} + \paren {b + d} \sqrt 2}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \map \phi {x + y}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi x \, \map \phi y
+      | r = \map \phi {a + b \sqrt 2} \, \map \phi {c + d \sqrt 2}
+      | c = 
+}}
+{{eqn | r = \paren {a - b \sqrt 2} \paren {c - d \sqrt 2}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \paren {a c + 2 b d} - \paren {b c + a d} \sqrt 2
+      | c = 
+}}
+{{eqn | r = \map \phi {\paren {a c + 2 b d} + \paren {b c + a d} \sqrt 2}
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \map \phi {x y}
+      | c = 
+}}
+{{end-eqn}}
+So $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]] which is also a [[Definition:Bijection|bijection]], whose [[Definition:Image of Mapping|image]] equals its [[Definition:Domain of Mapping|domain]].
+Hence the result by definition of [[Definition:Ring Automorphism|ring automorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Epimorphism from Integers to Integers Modulo m}
+Tags: Ring Epimorphisms, Modulo Arithmetic, Integers
+
+\begin{theorem}
+Let $\struct {\Z, +, \times}$ be the [[Definition:Ring of Integers|ring of integers]].
+Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
+Let $\phi: \struct {\Z, +, \times} \to \struct {\Z_m, +_m, \times_m}$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \Z: \map \phi x = \eqclass x m$
+where $\eqclass x m$ is the [[Definition:Residue Class|residue class modulo $m$]].
+Then $\phi$ is a [[Definition:Ring Epimorphism|ring epimorphism]], but specifically not a [[Definition:Ring Monomorphism|ring monomorphism]].
+The [[Definition:Image of Mapping|image]] of $\phi$ is $\struct {\Z_m, +_m, \times_m}$.
+The [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is $m \Z$, the [[Definition:Set of Integer Multiples|set of integer multiples of $m$]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {a + b}
+      | r = \eqclass {a + b} m
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \eqclass a m +_m \eqclass b m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \map \phi a +_m \map \phi b
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map \phi {a \times b}
+      | r = \eqclass {a \times b} m
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \eqclass a m \times_m \eqclass b m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = \map \phi a \times_m \map \phi b
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+Hence $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+Now let $\eqclass a m \in \Z_m$.
+By definition of [[Definition:Residue Class|residue class modulo $m$]]:
+:$\eqclass a m = \set {x \in \Z: \exists k \in \Z: z = a + k m}$
+Setting $k = 0$:
+:$\map \phi a = \eqclass a m$
+and so:
+:$\map {\phi^{-1} } {\eqclass a m} \ne \O$
+Thus $\phi$ is a [[Definition:Surjection|surjection]].
+Now setting $k = 1$, for example, we have that:
+:$\map \phi {a + m} = \eqclass a m$
+and so:
+:$\map \phi a = \map \phi {a + m}$.
+So $\phi$ is specifically not an [[Definition:Injection|injection]].
+It follows by definition that $\phi$ is a [[Definition:Ring Epimorphism|ring epimorphism]], but specifically not a [[Definition:Ring Monomorphism|ring monomorphism]].
+Next we note that:
+:$\forall x \in \Z: \map \phi x \in \Z_m$
+and so:
+:$\Img \phi = \Z_m$
+Finally, we have that the [[Definition:Kernel of Ring Homomorphism|kernel of $\phi$]] is:
+:$\map \ker \phi = \set {x \in \Z: \map \phi x = \eqclass 0 m}$
+Let $\map \phi x = \eqclass 0 m$
+Then $x = 0 + k m$ for some $k \in \Z$.
+That is, $x \in m \Z$ and so:
+:$\map \ker \phi \subseteq m \Z$
+Now let $x \in m \Z$.
+Then:
+:$\exists k \in \Z: x = 0 + k m$
+and so by definition:
+:$\map \phi x = \eqclass 0 m$
+So:
+:$m \Z \subseteq \map \ker \phi$
+Hence:
+:$\map \ker \phi = m \Z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monotone Additive Function is Linear}
+Tags: Analysis, Proofs by Contraposition, Additive Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Monotone Real Function|monotone real function]] which is [[Definition:Additive Function (Conventional)|additive]], that is:
+:$\forall x, y \in \R: \map f {x + y} = \map f x + \map f y$
+Then:
+:$\exists a \in \R: \forall x \in \R: \map f x = a x$
+\end{theorem}<|endoftext|>
+\section{Units of Ring of Polynomial Forms over Field}
+Tags: Polynomial Theory, Units of Rings
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in an [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$ over $F$.
+Then the [[Definition:Unit of Ring|units]] of $F \sqbrk X$ are all the elements of $F \sqbrk X$ whose [[Definition:Degree (Polynomial)|degree]] is $0$.
+\end{theorem}
+
+\begin{proof}
+An element of $F \sqbrk X$ whose [[Definition:Degree (Polynomial)|degree]] is $0$ is merely an element of $F$.
+But note that $0_F$, considered as an element of $F \sqbrk X$, has a [[Definition:Degree (Polynomial)|degree]] which is not defined, so the [[Definition:Null Polynomial over Ring|null polynomial]] is seen to be excluded.
+Any element $a$ of $F$ has an [[Definition:Inverse Element|inverse]] $1_F / a$.
+So all the elements of $F \sqbrk X$ whose [[Definition:Degree (Polynomial)|degree]] is $0$ are [[Definition:Unit of Ring|units]] of $F$ and hence of $F \sqbrk X$.
+Now suppose $\map a X \in F \sqbrk X$ is a [[Definition:Unit of Ring|unit]] of $F \sqbrk X$.
+Then $\map a X \, \map q X = 1_F$ for some $\map q X \in F \sqbrk X$.
+Neither $\map a X$ nor $\map q X$ can be [[Definition:Null Polynomial over Ring|null]].
+So by [[Properties of Degree]] $\map \deg {\map a X} + \map \deg {\map q X} = 0$.
+So $\map \deg {\map a X} = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Units of Numbers of Type Integer a plus b root 2}
+Tags: Integral Domains, Units of Rings
+
+\begin{theorem}
+Let $\Z \sqbrk {\sqrt 2}$ denote the [[Definition:Set|set]]:
+:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
+that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are [[Definition:Integer|integers]].
+Let $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ be the [[Numbers of Type Integer a plus b root 2 form Subdomain of Reals|integral domain]] where $+$ and $\times$ are conventional [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]] on [[Definition:Real Number|real numbers]].
+Then numbers of the form $a + b \sqrt 2$ such that $a^2 - 2 b^2 = \pm 1$ are all [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+For $a + b \sqrt 2$ to be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that:
+:$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$
+In [[Numbers of Type Integer a plus b root 2 are Not a Field]] it is shown that the [[Definition:Product Inverse|product inverse]] of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.
+So if $a^2 - 2 b^2 = \pm 1$ it follows that $c$ and $d$ are [[Definition:Integer|integers]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Units of Gaussian Integers}
+Tags: Gaussian Integers, Units of Gaussian Integers
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
+The [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$.
+\end{theorem}
+
+\begin{proof}
+From [[Gaussian Integer Units are 4th Roots of Unity]]:
+: $\left\{{1, i, -1, -i}\right\}$ constitutes the [[Definition:Set|set]] of the [[Definition:Complex Roots of Unity|$4$th roots of unity]].
+The result follows from [[Roots of Unity under Multiplication form Cyclic Group]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of the [[Definition:Imaginary Unit|imaginary unit]] $i$:
+{{begin-eqn}}
+{{eqn | l = i^2
+      | r = -1
+}}
+{{eqn | l = i^3
+      | r = -i
+}}
+{{eqn | l = i^4
+      | r = 1
+}}
+{{end-eqn}}
+thus demonstrating that $U_\C$ is [[Definition:Generator of Cyclic Group|generated]] by $i$.
+Thus $\left({U_\C, \times}\right)$ is by definition a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $4$]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Units of Gaussian Integers]], $U_\C$ is the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]].
+From [[Group of Units is Group]], $\left({U_\C, \times}\right)$ forms a [[Definition:Group|group]].
+It remains to note that:
+{{begin-eqn}}
+{{eqn | l = i^2
+      | r = -1
+}}
+{{eqn | l = i^3
+      | r = -i
+}}
+{{eqn | l = i^4
+      | r = 1
+}}
+{{end-eqn}}
+thus demonstrating that $U_\C$ is  [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a + b i$ be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk i, +, \times}$.
+Then $a$ and $b$ are not both $0$ as then $a + b i$ would be the [[Definition:Ring Zero|zero]] of $\struct {\Z \sqbrk i, +, \times}$.
+Then:
+:$\exists c, d \in \Z: \paren {a + b i} \paren {c + d i} = 1 + 0 i$
+This leads (after algebra) to:
+:$c = \dfrac a {a^2 + b^2}, d = \dfrac {-b} {a^2 + b^2}$
+Let $a^2 + b^2 = n$.
+We have that $n \in \Z, n > 0$.
+If $c$ and $d$ are [[Definition:Integer|integers]], then $a$ and $b$ must both be [[Definition:Divisor of Integer|divisible]] by $n$.
+Let $a = n x, b = n y$.
+Then:
+:$n^2 x^2 + n^2 y^2 = n$
+and so:
+:$n \paren {x^2 + y^2} = 1$
+Thus $n = a^2 + b^2 = 1$ and so as $a, b \in \Z$ we have:
+:$a^2 = 1, b^2 = 0$
+or:
+:$a^2 = 0, b^2 = 1$
+from which the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\alpha = a + b i$ be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk i, +, \times}$.
+Then by definition of [[Definition:Unit of Ring|unit]]:
+:$\exists\beta = c + d i \in \Z \sqbrk i: \alpha \beta = 1$
+Let $\cmod \alpha$ denote the [[Definition:Complex Modulus|modulus]] of $\alpha$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \cmod \alpha^2 \cdot \cmod \beta^2
+      | r = \cmod {\alpha \beta}^2
+      | c = [[Modulus of Product]]
+}}
+{{eqn | r = \cmod 1^2
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+By [[Divisors of One]]:
+:$\cmod a^2 = 1$ or $-1$
+Since $\cmod \alpha$ and $\cmod \beta$ are [[Definition:Positive Integer|positive integers]]:
+:$\cmod \alpha^2 = a^2 + b^2 = 1$
+and so either:
+:$\cmod a = 1$ and $\cmod b = 0$
+or:
+:$\cmod b = 1$ and $\cmod a = 0$.
+Hence the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk i, +, \times}$ is $\set {\pm 1, \pm i}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers are Euclidean Domain}
+Tags: Integers, Examples of Euclidean Domains
+
+\begin{theorem}
+The [[Definition:Integer|integers]] $\Z$ with the [[Definition:Mapping|mapping]] $\nu: \Z \to \Z$ defined as:
+:$\forall x \in \Z: \map \nu x = \size x$
+form a [[Definition:Euclidean Domain|Euclidean domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Ordered Integral Domain]] we have that $\struct {\Z, +, \times, \le}$ forms an [[Definition:Ordered Integral Domain|ordered integral domain]].
+For all $a \in \Z$, the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$ is defined as:
+:$\size a = \begin{cases}
+a & : 0 \le a \\
+-a & : a < 0
+\end{cases}$
+By [[Product of Absolute Values on Ordered Integral Domain]] we have:
+:$\forall a, b \in \Z: \size a \cdot \size b = \size {a b}$
+Since $\size b > 0$ it follows from [[Relation Induced by Strict Positivity Property is Compatible with Multiplication]] that $\size a \cdot \size b \ge \size a$.
+The second criterion:
+:For any $a, b \in R$, $b \ne 0$, there exist $q, r \in R$ with $\map \nu r < \map \nu b$, or $r = 0$ such that:
+::$ a = q \circ b + r$
+follows from the [[Division Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gaussian Integers form Euclidean Domain}
+Tags: Examples of Euclidean Domains, Complex Numbers, Gaussian Integers form Euclidean Domain
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk i, +, \times}$ be the [[Gaussian Integers form Integral Domain|integral domain of Gaussian Integers]].
+Let $\nu: \Z \sqbrk i \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as: 
+:$\forall a \in \Z \sqbrk i: \map \nu a = \cmod a^2$ 
+where $\cmod a$ is the [[Definition:Modulus of Complex Number|(complex) modulus]] of $a$.
+Then $\nu$ is a [[Definition:Euclidean Valuation|Euclidean valuation]] on $\Z \sqbrk i$.
+Hence $\struct {\Z \sqbrk i, +, \times}$ with $\nu: \Z \sqbrk i \to \Z$ forms a [[Definition:Euclidean Domain|Euclidean domain]].
+\end{theorem}
+
+\begin{proof}
+We have by definition that $\Z \sqbrk i \subseteq \C$.
+Let $a, b \in \Z \sqbrk i$.
+We have from [[Modulus of Product]] that $\cmod a \cdot \cmod b = \cmod {a b}$.
+From [[Complex Modulus is Non-Negative]]:
+:$\forall a \in \C: \cmod a \ge 0$
+and:
+:$\cmod a = 0 \iff a = 0$
+Let $a = x + i y$.
+Suppose $a \in \Z \sqbrk i \ne 0$.
+Then $x \ne 0$ or $y \ne 0$ and $x^2 \ge 1$ or $y^2 \ge 1$.
+So:
+:$\cmod a \ge 1$
+Similarly, $b \in \Z \sqbrk i, b \ne 0 \implies \cmod b \ge 1$.
+Thus it follows that $\cmod {a b} \ge \cmod a$ and so $\nu$ is a [[Definition:Euclidean Valuation|Euclidean valuation]] on $\Z \left[{i}\right]$.
+{{finish|definition of Euclidean valuation requires codomain $\N$, consider modulus squared instead}}
+Now, consider $x, y \in \Z \sqbrk i$.
+We want to find $q, r \in \Z \sqbrk i$ such that $x = q y + r$.
+Note that this means we want $r = y \paren {\dfrac x y - q}$ where $\dfrac x y$ is [[Definition:Complex Number|complex]] but not necessarily [[Definition:Gaussian Integer|Gaussian]].
+We [[Definition:Extension of Mapping|extend]] $\nu$ to the [[Definition:Complex Number|complex numbers]] and define $\nu: \C \to \C$ as:
+:$\forall z \in \C: \map \nu z = \cmod z$
+Then we have:
+:$\map \nu r = \map \nu y \cdot \map \nu {\dfrac x y - q}$
+Thus if $\map \nu {\dfrac x y - q} < 1$ we have:
+:$\map \nu r < \map \nu y$
+Consider the point $P = \dfrac x y$ as a point on the [[Definition:Complex Plane|complex plane]].
+Let $Q$ lie at a point representing the [[Definition:Gaussian Integer|Gaussian integer]] $q$ which lies closest to $P$.
+The distance $P Q$ is at most half the length of a diagonal of a unit square in the complex plane.
+Thus:
+:$\map \nu {\dfrac x y - q} = \cmod {\dfrac x y - q} \le \dfrac {\sqrt 2} 2 = \dfrac 1 {\sqrt 2} < 1$
+This element $q$ and the element $r$, where $\map \nu r < \map \nu y$, are the required values.
+Thus $\Z \sqbrk i$ is a [[Definition:Euclidean Domain|Euclidean domain]].
+{{qed}}
+{{Improve}}
+\end{proof}
+
+\begin{proof}
+We have by definition that $\Z \sqbrk i \subseteq \C$.
+Let $a, b \in \Z \sqbrk i$.
+We have from [[Modulus of Product]] that:
+:$\cmod a \cdot \cmod b = \cmod {a b}$
+From [[Complex Modulus is Norm]] we have that:
+:$\forall a \in \C: \cmod a \ge 0$
+:$\cmod a = 0 \iff a = 0$
+Suppose $a = x + i y \in \Z \sqbrk i$ and $a \ne 0$.
+Then either $x \ne 0$ or $y \ne 0$, so either $x^2 \ge 1$ or $y^2 \ge 1$.
+So $\cmod a^2 \ge 1$.
+If also $b \in \Z \sqbrk i, b \ne 0$, then:
+:$\map \nu {a b} = \cmod {a b}^2 \ge \cmod a^2 = \map \nu a$
+Now, consider $x, y \in Z \sqbrk i$ with $y \ne 0$.
+We want to find $q, r \in Z \sqbrk i$ such that $x = q y + r$ and $\map \nu r < \map \nu y$.
+Note that this means we want $r = y \paren {\dfrac x y - q}$ where $\dfrac x y$ is [[Definition:Complex Number|complex]] but not necessarily [[Definition:Gaussian Integer|Gaussian]].
+Consider the [[Definition:Complex Number|complex number]] $p = \dfrac x y = p_r + i p_i$.
+We [[Definition:Extension of Mapping|extend]] $\nu$ to the [[Definition:Complex Number|complex numbers]] and define $\nu: \C \to \C$ as:
+:$\forall z \in \C: \map \nu z = \cmod z^2$
+Let $q = q_r + i q_i$ be the [[Definition:Gaussian Integer|Gaussian integer]] such that:
+:$\map \nu {p - q} = \cmod {p - q}^2 = \paren {p_r - q_r}^2 + \paren {p_i - q_i}^2$
+is [[Definition:Minimal Element|minimal]].
+That is, $q_r$ is the nearest [[Definition:Integer|integer]] to $p_r$ and $q_i$ is the nearest integer to $p_i$.
+A given real number can be at a [[Definition:Distance between Real Numbers|distance]] at most $1/2$ from an integer, so it follows that:
+:$(1): \quad \map \nu {p - q} \le \paren {\dfrac 1 2}^2 + \paren {\dfrac 1 2}^2 = \dfrac 1 2$
+Now by [[Complex Modulus is Norm]], for any two complex numbers $z_1, z_2$ we have:
+:$\map \nu {z_1 z_2} = \map \nu {z_1} \, \map \nu {z_2}$
+Thus we obtain:
+{{begin-eqn}}
+{{eqn | l = \map \nu {y \paren {p - q} }
+      | r = \map \nu y \, \map \nu {p - q}
+}}
+{{eqn | r = \dfrac {\map \nu y} 2
+      | o = \le
+      | c = by $(1)$
+}}
+{{eqn | r = \map \nu y
+      | o = <
+      | c = since $\map \nu y \ne 0$
+}}
+{{end-eqn}}
+On the other hand:
+{{begin-eqn}}
+{{eqn | l = \map \nu {y \paren {p - q} }
+      | r = \map \nu {y \paren {\frac x y - q} }
+}}
+{{eqn | r = \map \nu {x - y q}
+}}
+{{end-eqn}}
+So letting $r = x - y q$ we have $\map \nu r < \map \nu y$.
+Moreover we trivially have $x = q y + r$.
+Thus $\Z \sqbrk i$ is a [[Definition:Euclidean Domain|Euclidean domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Degree of Product of Polynomials over Ring}
+Tags: Polynomial Theory, Degree of Product of Polynomials over Ring
+
+\begin{theorem}
+:$\forall f, g \in R \left[{X}\right]: \deg \left({f g}\right) \le \deg \left({f}\right) + \deg \left({g}\right)$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Leading Coefficient (Polynomial)|leading coefficient]] of:
+* $f \left({X}\right)$ be $a_n$
+* $g \left({X}\right)$ be $b_n$.
+Then:
+* $f \left({X}\right) = a_n X^n + \cdots + a_0$
+* $g \left({X}\right) = b_n X^n + \cdots + b_0$
+Consider the [[Definition:Leading Coefficient (Polynomial)|leading coefficient]] of the product $f \left({X}\right) g \left({X}\right)$: call it $c$.
+From the definition of [[Definition:Addition of Polynomials|polynomial addition]] and [[Definition:Multiplication of Polynomials|polynomial multiplication]]: 
+* $f \left({X}\right) g \left({X}\right) = c X^{n+m} + \cdots + a_0 b_0$
+Clearly the highest term of $f \left({X}\right) g \left({X}\right)$ can have an index no higher than $n+m$.
+Hence the result:
+:$\deg \left({f g}\right) \not > \deg \left({f}\right) + \deg \left({g}\right)$
+Next, note that the general [[Definition:Ring with Unity|ring with unity]] $(R, +, \circ)$ may have [[Definition:Proper Zero Divisor|proper zero divisors]].
+Therefore it is possible that $X^{n+m}$ may equal $0_R$.
+If that is the case, then the highest term will have an index definitely less than $n+m$.
+That is, in that particular case:
+:$\deg \left({f g}\right) < \deg \left({f}\right) + \deg \left({g}\right)$
+Thus, for a general [[Definition:Ring with Unity|ring with unity]] $(R, +, \circ)$:
+:$\deg \left({f g}\right) \le \deg \left({f}\right) + \deg \left({g}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Degree of Sum of Polynomials}
+Tags: Polynomial Theory
+
+\begin{theorem}
+:$\forall f, g \in R \sqbrk X: \map \deg {f + g} \le \max \set {\map \deg f, \map \deg g}$
+\end{theorem}
+
+\begin{proof}
+First we associate to $f = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$ a formal [[Definition:Vector Space|vector]] $x_f = \tuple {a_0, a_1, \ldots, a_n, 0_R, \ldots} \in R^\infty$.
+Let $x_f^i \in R$ denote the element at the $i$th position. 
+Then:
+:$\displaystyle \map \deg f = \sup \set {i \in \N : x_f^i \ne 0_R}$
+The sum $+$ in the [[Definition:Ring of Polynomials|polynomial ring]] $R \sqbrk X$ gives rise to the following identity in $R^\infty$:
+:$x_{f + g}^i = x_f^i + x_g^i$
+Next, let $f, g \in R \sqbrk X$, and let $d = \max \set {\map \deg f, \map \deg g}$.
+Then $x_f^i = 0_R = x_g^i$ for all $i > d$, so we have:
+:$\displaystyle \map \deg {f + g} = \sup \set {i \in \N : x_{f + g}^i \ne 0_R} = \sup \set {i \in \N : x_f^i + x_g^i \ne 0_R} \le d$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors}
+Tags: Polynomial Theory
+
+\begin{theorem}
+:$\forall f, g \in R \sqbrk X: \map \deg {f g} \ge \map \deg f$
+\end{theorem}
+
+\begin{proof}
+From [[Degree of Product of Polynomials over Integral Domain]], we have:
+:$\map \deg {f g} = \map \deg f + \map \deg g$
+But $\map \deg g \ge 0$ by definition of [[Definition:Degree of Polynomial over Integral Domain|degree]], as $g$ is not [[Definition:Null Polynomial over Ring|null]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial Forms over Field is Euclidean Domain}
+Tags: Field Theory, Examples of Euclidean Domains, Polynomial Theory
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $X$ be [[Definition:Transcendental over Field|transcendental]] in $F$.
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $F$.
+Then $F \sqbrk X$ is a [[Definition:Euclidean Domain|Euclidean domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors]] we have that:
+: $\forall a, b \in F \sqbrk X, a \ne 0_F, b \ne 0_F: \map \deg {a b} \ge \map \deg a$
+where $\map \deg a$ denotes the [[Definition:Degree (Polynomial)|degree]] of $a$.
+From [[Division Theorem for Polynomial Forms over Field]]:
+:$\forall a, b \in F \sqbrk X, b \ne 0_F: \exists q, r \in F \sqbrk X: a = q b + r$
+where $\map \deg r < \map \deg b$ (or $r = 0$), and $q$ and $r$ are unique.
+So $\deg$ is a [[Definition:Euclidean Valuation|Euclidean valuation]] on $F \sqbrk X$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Domain is Principal Ideal Domain}
+Tags: Euclidean Domains, Principal Ideal Domains
+
+\begin{theorem}
+A [[Definition:Euclidean Domain|Euclidean domain]] is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$ and whose [[Definition:Euclidean Valuation|Euclidean valuation]] is $\nu$.
+We need to show that every [[Definition:Ideal of Ring|ideal]] of $\struct {D, +, \times}$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
+Let $U$ be an [[Definition:Ideal of Ring|ideal]] of $\struct {D, +, \times}$ such that $U \ne \set 0$.
+Let $d \in U$ such that $d \ne 0$ and $\map \nu d$ is as small as possible for elements of $U$.
+By definition, $\nu$ is defined as $\nu : D \setminus \set 0 \to \N$, so the [[Definition:Codomain of Mapping|codomain]] of $\nu$ is a subset of the [[Definition:Natural Numbers|natural numbers]].
+By the [[Well-Ordering Principle]], such an element $d$ exists as an element of the [[Definition:Preimage of Element under Mapping|preimage]] of the [[Definition:Smallest Element|least member]] of the [[Definition:Image of Subset under Mapping|image]] of $U$.
+<!--{{explain|Need to ensure that the [[Well-Ordering Principle]] can be applied here.}}-->
+Let $a \in U$.
+Let us write $a = d q + r$ where either $r = 0$ or $\map \nu r < \map \nu d$.
+Then $r = a - d q$ and so $r \in U$.
+Suppose $r \ne 0$.
+That would mean $\map \nu r < \map \nu d$ contradicting $d$ as the element of $U$ with the smallest $\nu$.
+So $r = 0$, which means $a = q d$.
+That is, every element of $U$ is a [[Definition:Divisor of Ring Element|multiple]] of $d$.
+So $U$ is the [[Definition:Principal Ideal of Ring|principal ideal]] generated by $d$.
+This deduction holds for all [[Definition:Ideal of Ring|ideals]] of $D$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial Forms is PID Implies Coefficient Ring is Field}
+Tags: Polynomial Theory, Principal Ideal Domains
+
+\begin{theorem}
+Let $D$ be an [[Definition:Integral Domain|integral domain]].
+Let $D \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
+Let $D sqbrk X$ be a [[Definition:Principal Ideal Domain|principal ideal domain]];
+Then $D$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+Let $y \in D$ be non-zero.
+Then, using the principal ideal property, for some $f \in D \sqbrk X$ we have:
+:$\gen {y, X} = \gen f \subseteq D \sqbrk X$
+Therefore:
+:$\exists p, q \in D \sqbrk X: y = f p, X = f q$
+By [[Properties of Degree]] we conclude that $f = a$ and $q = b + c X$ for some $a, b, c \in D$.
+Substituting into the equation $X = f q$ we obtain:
+:$X = a b + a c X$
+which implies that:
+:$a c = 1$
+That is:
+:$a \in D^\times$
+where $D^\times$ denotes the [[Definition:Group of Units of Ring|group of units]] of $D$.
+Therefore:
+:$\gen f = \gen 1 = D \sqbrk X$
+Therefore:
+:$\exists r, s \in D \sqbrk X: r y + s X = 1$
+If $d$ is the constant term of $r$, then we have $y d = 1$.
+Therefore $y \in D^\times$.
+Our choice of $y$ was arbitrary, so this shows that $D^\times \supseteq D \setminus \set 0$, which says precisely that $D$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elements of Euclidean Domain have Greatest Common Divisor}
+Tags: Euclidean Domains, Ideal Theory
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]].
+Then any two elements $a, b \in D$ have a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] $d$ such that:
+:$d \divides a \land d \divides b$
+:$x \divides a \land x \divides b \implies x \divides d$
+and $d$ is written $\gcd \set {a, b}$.
+For any $a, b \in D$:
+:$\exists s, t \in D: s a + t b = d$
+Any two [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] of any $a, b$ are [[Definition:Associate in Integral Domain|associates]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in D$.
+Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$.
+Then $U$ is an [[Definition:Ideal of Ring|ideal]] of $D$.
+Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are [[Definition:Ideal of Ring|Principal Ideal]].
+By [[Sum of Ideals is Ideal]], $U$ is an ideal.
+By [[Euclidean Domain is Principal Ideal Domain]], $U$ is a [[Definition:Principal Ideal of Ring|principal ideal]], $\ideal d$ say.
+As $a, b \in U$ it follows that $d$ is a [[Definition:Divisor of Ring Element|divisor]] of $a$ and $b$, that is:
+:$d \divides a \land d \divides b$
+Since $d$ itself is in $U$, we have:
+:$\exists s, t \in D: s a + t b = d$
+By [[Common Divisor in Integral Domain Divides Linear Combination]]:
+:$x \divides a \land x \divides b \implies x \divides d$
+So $d$ is a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $a$ and $b$.
+If $d$ and $d'$ are both [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] of $a$ and $b$, then $d \divides a \land d \divides b$ and so $d \divides d'$.
+Similarly $d' \divides d$.
+So $d$ and $d'$ are [[Definition:Associate in Integral Domain|associates]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Associates in Ring of Polynomial Forms over Field}
+Tags: Factorization, Polynomial Theory, Euclidean Domains, Associates
+
+\begin{theorem}
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field (Abstract Algebra)|field]] $F$.
+Let $\map d X$ and $\map {d'} X$ be [[Definition:Polynomial Form|polynomial forms]] in $F \sqbrk X$.
+Then $\map d X$ is an [[Definition:Associate in Integral Domain|associate]] of $\map {d'} X$ {{iff}} $\map d X = c \cdot \map {d'} X$ for some $c \in F, c \ne 0_F$.
+Hence any two [[Definition:Polynomial Form|polynomials]] in $F \sqbrk X$ have a unique [[Definition:Monic Polynomial|monic]] [[Definition:Greatest Common Divisor of Polynomials|GCD]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Associate in Integral Domain|associate]], there exist $\map e X$ and $\map {e'} X$ \in $F \sqbrk X$ such that:
+:$\map d X = \map e X \cdot \map {d'} X$
+:$\map {d'} X = \map {e'} X \cdot \map d X$
+From [[Field is Integral Domain]], $F$ is an [[Definition:Integral Domain|integral domain]].
+From [[Degree of Product of Polynomials over Integral Domain]] it follows that necessarily $\deg e = \deg e' = 0$, as $F$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+Thus for some $c, c' \in F$, it must be that $\map e X = c$ and $\map {e'} X = c'$.
+From the two equations above it follows that $c \cdot c' = 1_F$, where $1_F$ is the [[Definition:Unity of Field|unity]] of $F$.
+Hence, it follows that $c \ne 0_F$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Element Generates Null Ideal}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+For $r \in R$, let $\ideal r$ denote the [[Definition:Ideal of Ring|ideal]] [[Definition:Generator of Ideal|generated]] by $r$.
+Then $\ideal {0_R}$ is the [[Definition:Null Ideal|null ideal]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\ideal {0_R} = \set {r \circ 0_R: r \in R}$
+but for each $r \in R$ we have by [[Ring Product with Zero]] that $r \circ 0_R = 0_R$ for all $r \in R$.
+Therefore $\ideal {0_R}$ is the [[Definition:Null Ideal|null ideal]].
+{{qed}}
+[[Category:Ideal Theory]]
+g0bxzwsul4zh97e4us5incfu0yaqszb
+\end{proof}<|endoftext|>
+\section{Irreducible Elements of Ring of Integers}
+Tags: Integers
+
+\begin{theorem}
+Let $\struct {\Z, +, \times}$ be the [[Integers form Totally Ordered Ring|ring of integers]].
+The [[Definition:Irreducible Element of Ring|irreducible elements]] of $\struct {\Z, +, \times}$ are the [[Definition:Prime Number|prime numbers]] and their [[Definition:Negative|negatives]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Integers form Integral Domain]].
+Therefore the concept of [[Definition:Irreducible Element of Ring|irreducible]] is defined.
+Let $p$ be a [[Definition:Prime Number|prime number]].
+By [[Definition:Prime Number|definition]], the only [[Definition:Divisor of Integer|divisors]] of $p$ are $1, -1, p, -p$.
+From [[Units of Ring of Integers]], $1$ and $-1$ are (the only) [[Definition:Unit of Ring|units]] of $\Z$.
+From [[Associates are Unit Multiples]], $p$ and $-p$ are (the only) [[Definition:Associate of Integer|associates]] of each other.
+Hence the result, from the definition of [[Definition:Irreducible Element of Ring|irreducible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclid's Lemma for Euclidean Domains}
+Tags: Euclidean Domains, Euclid's Lemma
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Unity of Ring|unity]] is $1$.
+Let $a, b, c \in D$.
+Let $a \divides b \times c$, where $\divides$ denotes [[Definition:Divisor of Ring Element|divisibility]].
+Let $a \perp b$, where $\perp$ denotes [[Definition:Coprime Elements of Euclidean Domain|relative primeness]].
+Then $a \divides c$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \perp
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {a, b}
+      | r = 1
+      | c = {{Defof|Coprime Elements of Euclidean Domain}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists x, y \in D:
+      | l = a \times x + b \times y
+      | r = 1
+      | c = [[Bézout's Lemma on Euclidean Domain]]
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | r = c \times \paren {a \times x + b \times y}
+      | c = 
+}}
+{{eqn | r = c \times a \times x + c \times b \times y
+      | c = 
+}}
+{{eqn | r = \paren {a \times c} \times x + \paren {b \times c} \times y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = \divides
+      | r = c \times \paren {a \times x + b \times y}
+      | c = as $a \divides a \times c$ and $a \divides b \times c$
+}}
+{{eqn | r = c \times 1
+      | c = [[Bézout's Lemma on Euclidean Domain]]
+}}
+{{eqn | r = c
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Euclid}}
+\end{proof}<|endoftext|>
+\section{Rational Polynomial is Content Times Primitive Polynomial}
+Tags: Polynomial Theory, Rational Polynomial is Content Times Primitive Polynomial
+
+\begin{theorem}
+Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\map f X \in \Q \sqbrk X$.
+Then:
+:$\map f X = \cont f \, \map {f^*} X$
+where:
+:$\cont f$ is the [[Definition:Content of Rational Polynomial|content]] of $\map f X$
+:$\map {f^*} X$ is a [[Definition:Primitive Polynomial (Ring Theory)|primitive polynomial]].
+For a given polynomial $\map f X$, both $\cont f$ and $\map {f^*} X$ are [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+=== [[Rational Polynomial is Content Times Primitive Polynomial/Existence|Proof of Existence]] ===
+{{:Rational Polynomial is Content Times Primitive Polynomial/Existence}}{{qed|lemma}}
+=== [[Rational Polynomial is Content Times Primitive Polynomial/Uniqueness|Proof of Uniqueness]] ===
+{{:Rational Polynomial is Content Times Primitive Polynomial/Uniqueness}}
+\end{proof}<|endoftext|>
+\section{Product of Rational Polynomials}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\map f X, \map g X \in \Q \sqbrk X$.
+Using [[Rational Polynomial is Content Times Primitive Polynomial]], let these be expressed as:
+:$\map f X = \cont f \cdot \map {f^*} X$
+:$\map g X = \cont g \cdot \map {g^*} X$
+where:
+:$\cont f, \cont g$ are the [[Definition:Content of Rational Polynomial|content]] of $f$ and $g$ respectively
+:$f^*, g^*$ are [[Definition:Primitive Polynomial (Ring Theory)|primitive]].
+Let $\map h X = \map f X \, \map g X$ be the [[Definition:Multiplication of Polynomials|product]] of $f$ and $g$.
+Then:
+:$\map {h^*} X = \map {f^*} X \, \map {g^*} X$
+\end{theorem}
+
+\begin{proof}
+From [[Rational Polynomial is Content Times Primitive Polynomial]]:
+:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \, \map {g^*} X$
+and this expression is [[Definition:Unique|unique]].
+By [[Gauss's Lemma on Primitive Rational Polynomials]] we have that $\map {f^*} X \, \map {g^*} X$ is [[Definition:Primitive Polynomial (Ring Theory)|primitive]].
+From [[Content of Rational Polynomial is Multiplicative]]:
+:$\cont f \cont g = \cont f \cont g > 0$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parallelogram Law}
+Tags: Vector Algebra, Applied Mathematics, Parallelogram Law
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Vector Quantity|vector quantities]].
+Consider a [[Definition:Parallelogram|parallelogram]], two of whose adjacent sides represent $\mathbf a$ and $\mathbf b$ (in [[Definition:Magnitude|magnitude]] and [[Definition:Direction|direction]]).
+:[[File:ParallelogramLaw.png|300px]]
+Then the [[Definition:Diagonal of Quadrilateral|diagonal]] of the [[Definition:Parallelogram|parallelogram]] through that common point represents the [[Definition:Magnitude|magnitude]] and [[Definition:Direction|direction]] of $\mathbf a + \mathbf b$, the [[Definition:Vector Sum|sum]] of $\mathbf a$ and $\mathbf b$.
+\end{theorem}<|endoftext|>
+\section{Equivalence of Definitions of Perfect Set}
+Tags: Perfect Sets
+
+\begin{theorem}
+{{TFAE|def = Perfect Set}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] and let $H \subseteq S$.
+=== Definition 1 implies Definition 2 ===
+Suppose that $H = H'$ where $H'$ is the [[Definition:Derived Set|derived set]] of $H$.
+By definition of [[Definition:Derived Set|derived set]], $H'$ is the set of all [[Definition:Limit Point of Set|limit points]] of $H$.
+So $H$ contains all its [[Definition:Limit Point of Set|limit points]] and so by definition is [[Definition:Closed Set (Topology)|closed]].
+Now we also have that any point not in $H'$ is an [[Definition:Isolated Point of Subset|isolated point]].
+But there are no points in $H$ which are not in $H'$, so $H$ has no [[Definition:Isolated Point of Subset|isolated points]].
+Therefore $H = H'$ implies that $H$ is [[Definition:Closed Set (Topology)|closed]] and has no [[Definition:Isolated Point of Subset|isolated points]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Suppose $H$ is [[Definition:Closed Set (Topology)|closed]] and has no [[Definition:Isolated Point of Subset|isolated points]].
+By [[Closed Set iff Contains all its Limit Points]] we have that $H' \subseteq H$ where $H'$ is the [[Definition:Derived Set|derived set]] of $H$.
+As $S$ has no [[Definition:Isolated Point of Subset|isolated points]], all its elements are elements of its [[Definition:Derived Set|derived set]].
+Therefore $H \subseteq H'$.
+So by definition of [[Definition:Set Equality/Definition 2|set equality]] it follows that $H = H'$.
+{{qed|lemma}}
+=== Definition 2 iff Definition 3 ===
+Let $H$ be [[Definition:Closed Set (Topology)|closed]] with no [[Definition:Isolated Point of Subset|isolated points]].
+By definition, $H$ is [[Definition:Closed Set (Topology)|closed]] {{iff}} it contains all its [[Definition:Limit Point of Set|limit points]].
+Also by definition, $H$ is [[Definition:Dense-in-itself|dense-in-itself]] {{iff}} it has no [[Definition:Isolated Point of Subset|isolated points]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of Interior equals Closure of Complement}
+Tags: Set Closures, Set Interiors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ and $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$.
+Let $\map \complement H$ be the [[Definition:Relative Complement|complement of $H$ in $T$]]:
+:$\map \complement H = T \setminus H$
+Then:
+:$\map \complement {H^\circ} = \paren {\map \complement H}^-$
+and similarly:
+:$\paren {\map \complement H}^\circ = \map \complement {H^-}$
+These can alternatively be written:
+:$T \setminus H^\circ = \paren {T \setminus H}^-$
+:$\paren {T \setminus H}^\circ = T \setminus H^-$
+which, it can be argued, is easier to follow.
+\end{theorem}
+
+\begin{proof}
+Let $\tau$ be the [[Definition:Topology|topology]] on $T$.
+Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = T \setminus H^\circ
+      | r = T \setminus \bigcup_{K \mathop \in \mathbb K} K
+      | c = {{Defof|Interior (Topology)}} of $H$
+}}
+{{eqn | r = \bigcap_{K \mathop \in \mathbb K} \paren {T \setminus K}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{end-eqn}}
+By the definition of [[Definition:Closed Set (Topology)|closed set]], $K$ is [[Definition:Open Set (Topology)|open in $T$]] {{iff}} $T \setminus K$ is [[Definition:Closed Set (Topology)|closed in $T$]].
+Also, from [[Set Complement inverts Subsets]] we have that $T \setminus K \supseteq T \setminus H$.
+Now consider the set $\mathbb K'$ defined as:
+:$\mathbb K' := \set {K' \subseteq T: \paren {T \setminus H} \subseteq K', K' \text { closed in } T}$
+From the above we see that:
+:$K \in \mathbb K \iff T \setminus K \in \mathbb K'$
+Thus:
+{{begin-eqn}}
+{{eqn | l = T \setminus H^\circ 
+      | r = \bigcap_{K' \mathop \in \mathbb K'} K'
+}}
+{{eqn | r = \paren {T \setminus H}^-
+      | c = {{Defof|Closure (Topology)|index = 2|Closure}} of $T \setminus H$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Then we note that:
+{{begin-eqn}}
+{{eqn | l = H^\circ
+      | r = T \setminus \paren {T \setminus H^\circ}
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | r = T \setminus \paren {\paren {T \setminus H}^-}
+      | c = from above
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = \paren {T \setminus H}^\circ
+      | r = T \setminus \paren {\paren {T \setminus \paren {T \setminus H} }^-}
+      | c = from above
+}}
+{{eqn | r = T \setminus H^-
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior of Finite Intersection equals Intersection of Interiors}
+Tags: Set Interiors, Set Intersection
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $n \in \N$.
+Let:
+:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
+Then:
+:$\displaystyle \left({\bigcap_{i \mathop = 1}^n H_i}\right)^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$
+where $H_i^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $H_i$.
+\end{theorem}
+
+\begin{proof}
+In the following, $H_i^-$ denotes the [[Definition:Closure (Topology)|closure]] of the set $H_i$.
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ
+      | r = T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^-
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = T \setminus \paren {\paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i}^-}
+      | c = [[Closure of Finite Union equals Union of Closures]]
+}}
+{{eqn | r = T \setminus \paren {\bigcup_{i \mathop = 1}^n T \setminus H_i^\circ}
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = T \setminus \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n H_i^\circ} }
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \bigcap_{i \mathop = 1}^n H_i^\circ
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Intersection of Regular Open Sets is Regular Open}
+Tags: Regular Open Sets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $n \in \N$.
+Suppose that:
+:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
+where all the $H_i$ are [[Definition:Regular Open Set|regular open]] in $T$.
+That is:
+:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$
+where $H_i^{- \circ}$ denotes the [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure]] of $H_i$.
+Then $\displaystyle \bigcap_{i \mathop = 1}^n H_i$ is [[Definition:Regular Open Set|regular open]] in $T$.
+That is:
+:$\displaystyle \bigcap_{i \mathop = 1}^n H_i = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}
+      | r = \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^\circ
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \paren {\paren {T \setminus \bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^\circ
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^-}^\circ
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n H_i^-}^\circ
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | r = \bigcap_{i \mathop = 1}^n H_i^{- \circ}
+      | c = [[Interior of Finite Intersection equals Intersection of Interiors]]
+}}
+{{eqn | r = \bigcap_{i \mathop = 1}^n H_i
+      | c = as all $H_i$ are [[Definition:Regular Open Set|regular open]] in $T$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Union of Regular Closed Sets is Regular Closed}
+Tags: Regular Closed Sets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $n \in \N$.
+Suppose that:
+:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
+where all the $H_i$ are [[Definition:Regular Closed Set|regular closed]] in $T$.
+That is:
+:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{\circ -}$
+where $H_i^{\circ -}$ denotes the [[Definition:Closure (Topology)|closure]] of the [[Definition:Interior (Topology)|interior]] of $H_i$
+Then $\displaystyle \bigcup_{i \mathop = 1}^n H_i$ is [[Definition:Regular Closed Set|regular closed]] in $T$.
+That is:
+:$\displaystyle \bigcup_{i \mathop = 1}^n H_i = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}
+      | r = \paren {T \setminus \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^-}^-
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^-
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{eqn | r = \paren {\paren {T \setminus \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^-
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^\circ}^-
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \paren {\bigcup_{i \mathop = 1}^n H_i^\circ}^-
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | r = \bigcup_{i \mathop = 1}^n H_i^{\circ -}
+      | c = [[Closure of Finite Union equals Union of Closures]]
+}}
+{{eqn | r = \bigcup_{i \mathop = 1}^n H_i
+      | c = as all $H_i$ are [[Definition:Regular Closed Set|regular closed]] in $T$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary is Intersection of Closure with Closure of Complement}
+Tags: Set Closures, Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $X \subseteq T$.
+Let $\partial X$ denote the [[Definition:Boundary (Topology)|boundary]] of $X$, defined as:
+:$\partial X = X^- \setminus X^\circ$
+Let $\overline X = T \setminus X$ denote the [[Definition:Relative Complement|complement]] of $X$ in $T$.
+Let $X^-$ denote the [[Definition:Closure (Topology)|closure]] of $X$.
+Then:
+:$\partial X = X^- \cap \paren {\overline X}^-$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \partial X
+      | r = X^- \setminus X^\circ
+      | c = {{Defof|Boundary (Topology)|index = 1}}: $X^\circ$ is the [[Definition:Interior (Topology)|interior]] of $X$
+}}
+{{eqn | r = X^- \cap \overline {\paren {X^\circ} }
+      | c = [[Set Difference as Intersection with Relative Complement]]
+}}
+{{eqn | r = X^- \cap \paren {\overline X}^-
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set is Closed iff it Contains its Boundary}
+Tags: Closed Sets, Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]], and let $H \subseteq T$.
+Then $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}}:
+:$\partial H \subseteq H$
+where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\partial H = H^- \cap \paren {T \setminus H}^-$
+where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$.
+Hence from [[Intersection is Subset]] we have that:
+:$\partial H \subseteq H^-$
+Then from [[Closed Set Equals its Closure]], $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}} $H = H^-$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set is Open iff Disjoint from Boundary}
+Tags: Open Sets, Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]], and let $H \subseteq T$.
+Then $H$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}}:
+:$\partial H \cap H = \O$
+where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\partial H = H^- \cap \paren {T \setminus H}^-$
+where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$.
+Hence from [[Intersection is Subset]] we have that:
+:$\partial H \subseteq \left({T \setminus H}\right)^-$
+But from [[Closed Set Equals its Closure]], $\paren {T \setminus H}^- = T \setminus H$ {{iff}} $T \setminus H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+That is, {{iff}} $H$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So $\partial H \subseteq T \setminus H$ {{iff}} $H$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Intersection with Complement is Empty iff Subset]] it follows that $\partial H \cap H = \O$ {{iff}} $H$ is [[Definition:Open Set (Topology)|open]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set is Clopen iff Boundary is Empty}
+Tags: Clopen Sets, Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]], and let $H \subseteq T$.
+Then $H$ is [[Definition:Clopen Set|both closed and open]] in $T$ {{iff}}:
+:$\partial H = \O$
+where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From [[Set is Open iff Disjoint from Boundary]] we have that:
+:$H$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} $\partial H \cap H = \O$
+From [[Set is Closed iff it Contains its Boundary]] we have that:
+:$H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}} $\partial H \subseteq H$
+From the above it follows that $\partial H = \O$ {{iff}} $H$ is [[Definition:Closed Set (Topology)|closed]] and $H$ is [[Definition:Open Set (Topology)|open]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Set is Closed}
+Tags: Closed Sets, Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]], and let $H \subseteq T$.
+Let $\partial H$ denote the [[Definition:Boundary (Topology)|boundary]] of $H$.
+Then $\partial H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\partial H = H^- \cap \paren {T \setminus H}^-$
+where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$
+From [[Topological Closure is Closed]], both $H^-$ and $\paren {T \setminus H}^-$ are [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Topology Defined by Closed Sets]], the [[Definition:Set Intersection|intersection]] of arbitrarily many (in particular two) [[Definition:Closed Set (Topology)|closed sets]] of $T$ is a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+As $\partial H$ is the [[Definition:Set Intersection|intersection]] of $H^-$ and $\left({T \setminus H}\right)^-$ the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Boundary is Contained in Boundary}
+Tags: Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Then:
+:$\map \partial {\partial H} \subseteq \partial H$
+where $\partial H$ is the [[Definition:Boundary (Topology)|boundary]] of $H$.
+That is, the [[Definition:Boundary (Topology)|boundary]] of the [[Definition:Boundary (Topology)|boundary]] of $H$ is contained in the [[Definition:Boundary (Topology)|boundary]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Let $B = \partial H$.
+From [[Boundary of Set is Closed]] we have that $B$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Let $B^-$ denote the [[Definition:Closure (Topology)|closure]] of $B$.
+From [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\partial B = B^- \cap \paren {T \setminus B}^-$
+and so from [[Intersection is Subset]]:
+:$\partial B \subseteq B^-$
+But from [[Closed Set Equals its Closure]]:
+:$B = B^-$
+and so:
+:$\partial B \subseteq B$
+That is:
+:$\map \partial {\partial H} \subseteq \partial H$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Exterior}
+Tags: Set Exteriors
+
+\begin{theorem}
+{{TFAE|def = Exterior (Topology)|view = Exterior|context = Topology (Mathematical Branch)|contextview = Topology}}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+\end{theorem}
+
+\begin{proof}
+Let $H^e$ be defined as:
+:$H^e$ is the [[Definition:Set Complement|complement]] of the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H^e
+      | r = T \setminus H^-
+      | c = 
+}}
+{{eqn | r = \paren {T \setminus H}^\circ
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{end-eqn}}
+Thus:
+:$H^e$ is the [[Definition:Interior (Topology)|interior]] of the [[Definition:Set Complement|complement]] of $H$ in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior is Subset of Exterior of Exterior}
+Tags: Set Interiors, Set Exteriors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $H^e$ denote the [[Definition:Exterior (Topology)|exterior]] of $H$, and let $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$.
+Then:
+:$H^\circ \subseteq \paren {H^e}^e$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {H^e}^e
+      | r = \paren {T \setminus H^e}^\circ
+      | c = {{Defof|Exterior (Topology)|Exterior}}
+}}
+{{eqn | r = \paren {T \setminus \paren {T \setminus H^-} }^\circ
+      | c = [[Equivalence of Definitions of Exterior]]
+}}
+{{eqn | r = \paren {H^-}^\circ
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | o = \supseteq
+      | r = H^\circ
+      | c = [[Interior is Subset of Interior of Closure]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior is Subset of Interior of Closure}
+Tags: Set Closures, Set Interiors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$.
+Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$.
+Then:
+:$H^\circ \subseteq \left({H^-}\right)^\circ$
+\end{theorem}
+
+\begin{proof}
+From [[Set is Subset of its Topological Closure]], we have $H \subseteq H^-$.
+The result follows directly from [[Interior of Subset]].
+{{qed}}
+[[Category:Set Closures]]
+[[Category:Set Interiors]]
+hpi3i0ga2cmk8ylx6ricx5jzzr61qz3
+\end{proof}<|endoftext|>
+\section{Exterior of Finite Union equals Intersection of Exteriors}
+Tags: Set Exteriors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $n \in \N$.
+Let $\forall i \in \closedint 1 n: H_i \subseteq T$.
+Then:
+:$\displaystyle \paren {\bigcup_{i \mathop = 1}^n H_i}^e = \bigcap_{i \mathop = 1}^n H_i^e$
+where $H_i^e$ denotes the [[Definition:Exterior (Topology)|exterior]] of $H_i$.
+\end{theorem}
+
+\begin{proof}
+In the following, $H_i^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of the set $H_i$.
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^e
+      | r = \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^\circ
+      | c = {{Defof|Exterior (Topology)|Exterior}}
+}}
+{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{eqn | r = \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i}^\circ
+      | c = [[Interior of Finite Intersection equals Intersection of Interiors]]
+}}
+{{eqn | r = \bigcap_{i \mathop = 1}^n H_i^e
+      | c = {{Defof|Exterior (Topology)|Exterior}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Union contains Union of Closures}
+Tags: Set Union, Set Closures
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
+Then the [[Definition:Set Union|union]] of the [[Definition:Closure (Topology)|closures]] of the [[Definition:Element|elements]] of $\mathbb H$ is a [[Definition:Subset|subset]] of the [[Definition:Closure (Topology)|closure]] of the [[Definition:Set Union|union]] of $\mathbb H$:
+:$\displaystyle \bigcup_{H \mathop \in \mathbb H} \map \cl H \subseteq \map \cl {\bigcup_{H \mathop \in \mathbb H} H}$
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\displaystyle L = \bigcup_{H \mathop \in \mathbb H} H$.
+We have:
+:$\forall H \in \mathbb H: H \subseteq L$
+so from [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\map \cl H \subseteq \map \cl L$
+It follows from [[Union is Smallest Superset/General Result|Union is Smallest Superset: General Result]] that:
+:$K \subseteq \map \cl L$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set is Subset of its Topological Closure}
+Tags: Set Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $H^-$ be the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
+Then:
+: $H \subseteq H^-$
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Closure (Topology)|definition of closure]], we have:
+:$H^-$ is the [[Definition:Set Union|union]] of $H$ and its [[Definition:Limit Point of Set|limit points]].
+From [[Subset of Union]] it follows directly that:
+:$H \subseteq H^-$
+{{qed}}
+[[Category:Set Closures]]
+43vh957rsenz5hj9tjjazua8jtrws71
+\end{proof}<|endoftext|>
+\section{Set Closure as Intersection of Closed Sets}
+Tags: Set Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let the [[Definition:Closure (Topology)/Definition 1|closure of $H$ (in $T$)]] be defined as:
+:$H^- := H \cup H'$
+where $H'$ is the [[Definition:Derived Set|derived set]] of $H$.
+Let $\mathbb K$ be defined as:
+:$\mathbb K := \left\{{K \supseteq H: K}\right.$ is [[Definition:Closed Set (Topology)|closed]] in $\left.{T}\right\}$
+That is, let $\mathbb K$ be the set of all [[Definition:Closed Set (Topology)|closed sets]] of $T$ which contain $H$.
+Then the [[Definition:Closure (Topology)/Definition 2|closure of $H$ (in $T$)]] can be defined as:
+:$\displaystyle H^- := \bigcap \mathbb K$
+that is, as the [[Definition:Intersection of Set of Sets|intersection]] of all the [[Definition:Closed Set (Topology)|closed sets]] of $T$ which contain $H$.
+\end{theorem}
+
+\begin{proof}
+What needs to be proved here is:
+:$\displaystyle H^- = \bigcap \mathbb K$
+where:
+: $H^- = H \cup H'$
+: $H'$ denotes the [[Definition:Set|set]] of all [[Definition:Limit Point of Set|limit points]] of $H$
+: $\displaystyle \bigcap \mathbb K$ is the [[Definition:Intersection of Set of Sets|intersection]] of all [[Definition:Closed Set (Topology)|closed sets]] of $T$ which contain $H$.
+Let $K \in \mathbb K$.
+From [[Topological Closure of Subset is Subset of Topological Closure]], we have:
+: $H^- \subseteq K^-$
+Thus from [[Closed Set Equals its Closure]], we have:
+: $K^- = K$
+Thus:
+:$\displaystyle \forall K \in \bigcap \mathbb K: H^- \subseteq K$
+since the choice of $K$ is arbitrary. 
+So from [[Intersection is Largest Subset]]:
+: $\displaystyle H^- \subseteq \bigcap \mathbb K$
+{{qed|lemma}}
+Conversely, from [[Topological Closure is Closed]], $H^-$ is [[Definition:Closed Set (Topology)|closed]].
+From [[Set is Subset of its Topological Closure]], $H \subseteq H^-$.
+So $H^-$ is, by definition, a [[Definition:Closed Set (Topology)|closed set]] of $T$ which contains $H$.
+But we have by its definition that $\displaystyle \bigcap \mathbb K$ is the [[Definition:Set Intersection|intersection]] of '''all''' [[Definition:Closed Set (Topology)|closed sets]] in $T$ that contain $H$.
+So from [[Intersection is Subset]] it follows that:
+: $\displaystyle \bigcap \mathbb K \subseteq H^-$
+{{qed|lemma}}
+Finally, we have that:
+: $\displaystyle H^- \subseteq \bigcap \mathbb K$
+: $\displaystyle \bigcap \mathbb K \subseteq H^-$
+So by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+: $\displaystyle H^- = \bigcap \mathbb K$
+which is what we needed to prove.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set Closure is Smallest Closed Set/Topology}
+Tags: Set Closures, Set Closure is Smallest Closed Set
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
+Then $H^-$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Superset|superset]] of $H$ that is [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Define:
+:$\mathbb K := \leftset {K \supseteq H: K}$ is [[Definition:Closed Set (Topology)|closed]] in $\rightset T$
+That is, let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Superset|supersets]] of $H$ that are [[Definition:Closed Set (Topology)|closed]] in $T$.
+The claim is that $H^-$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\mathbb K$.
+From [[Set is Subset of its Topological Closure]]:
+:$H \subseteq H^-$
+From [[Topological Closure is Closed]], $H^-$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Thus $H^- \in \mathbb K$.
+Let $K \in \mathbb K$.
+From [[Set Closure as Intersection of Closed Sets]]:
+:$\displaystyle H^- = \bigcap \mathbb K$
+Therefore, from [[Intersection is Subset/General Result|Intersection is Subset: General Result]]:
+:$H^- \subseteq K$
+Thus by definition $H^-$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\mathbb K$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Interior (Topology)}
+Tags: Set Interiors
+
+\begin{theorem}
+{{TFAE|def = Interior (Topology)|view = interior|context = Topology (Mathematical Branch)|contextview = topology}}
+Let $\struct {T, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbb K$ be defined as:
+:$\mathbb K := \set {K \in \tau: K \subseteq H}$
+That is, let $\mathbb K$ be the set of all [[Definition:Open Set (Topology)|open sets]] of $T$ contained in $H$.
+Then from [[Definition:Interior (Topology)/Definition 1|definition 1]] of the [[Definition:Interior (Topology)|interior]] of $H$, we have:
+:$\displaystyle H^\circ = \bigcup_{K \mathop \in \mathbb K} K$
+That is, $H^\circ$ is the [[Definition:Set Union|union]] of all the [[Definition:Open Set (Topology)|open sets]] of $T$ contained in $H$.
+Let $K \subseteq T$ such that $K$ is [[Definition:Open Set (Topology)|open]] in $T$ and $K \subseteq H$.
+That is, let $K \in \mathbb K$.
+Then from [[Subset of Union]] it follows directly that $K \subseteq H^\circ$.
+So any open set in $T$ contained in $H$ is a subset of $H^\circ$, and so $H^\circ$ is the largest [[Definition:Open Set (Topology)|open set]] of $T$ contained in $H$.
+That is, $H^\circ$ is also the [[Definition:Interior (Topology)/Definition 2|interior of $H$ by definition 2]].
+Hence both definitions are [[Definition:Logical Equivalence|equivalent]].
+{{qed}}
+{{expand|to include definition 3}}
+\end{proof}<|endoftext|>
+\section{Intersection of Interiors contains Interior of Intersection}
+Tags: Set Interiors, Set Intersection
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $T$.
+That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the [[Definition:Power Set|power set]] of $T$.
+Then the [[Definition:Interior (Topology)|interior]] of the [[Definition:Set Intersection|intersection]] of $\mathbb H$ is a [[Definition:Subset|subset]] of the [[Definition:Set Intersection|intersection]] of the [[Definition:Interior (Topology)|interiors]] of the [[Definition:Element|elements]] of $\mathbb H$.
+:$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$
+\end{theorem}
+
+\begin{proof}
+In the following, $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of the set $H$.
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ
+      | r = T \setminus \paren {T \setminus \bigcap_{H \mathop \in \mathbb H} H}^-
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{end-eqn}}
+At this point we note that:
+:$(1): \quad \displaystyle \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \supseteq \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$
+from [[Closure of Union contains Union of Closures]].
+Then we note that:
+:$\displaystyle T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} \subseteq T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}$
+from $(1)$ and [[Set Complement inverts Subsets]].
+Then we continue:
+{{begin-eqn}}
+{{eqn | l = T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}
+      | r = T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} T \setminus H^\circ}
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = T \setminus \paren {T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^\circ} }
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \bigcap_{H \mathop \in \mathbb H} H^\circ
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Exteriors contains Exterior of Union}
+Tags: Set Exteriors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $T$.
+That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the [[Definition:Power Set|power set]] of $T$.
+Then:
+:$\displaystyle \paren {\bigcup_{H \mathop \in \mathbb H} H}^e \subseteq \bigcap_{H \mathop \in \mathbb H} H^e $
+where $H^e$ denotes the [[Definition:Exterior (Topology)|exterior]] of $H$.
+\end{theorem}
+
+\begin{proof}
+In the following, $H^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of the set $H$.
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e
+      | r = \paren {T \setminus \bigcup_{H \mathop \in \mathbb H} H}^\circ
+      | c = {{Defof|Exterior (Topology)|Exterior}}
+}}
+{{eqn | r = \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^\circ
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{eqn | o = \subseteq
+      | r = \bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^\circ
+      | c = [[Intersection of Interiors contains Interior of Intersection]]
+}}
+{{eqn | r = \bigcap_{H \mathop \in \mathbb H} H^e
+      | c = {{Defof|Exterior (Topology)|Exterior}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exterior of Intersection contains Union of Exteriors}
+Tags: Set Exteriors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $T$.
+That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the [[Definition:Power Set|power set]] of $T$.
+Then:
+:$\displaystyle \bigcup_{H \mathop \in \mathbb H} H^e \subseteq \paren {\bigcap_{H \mathop \in \mathbb H} H}^e$
+where $H^e$ denotes the [[Definition:Exterior (Topology)|exterior]] of $H$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \bigcup_{H \mathop \in \mathbb H} H^e
+      | r = \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H^-})
+      | c = {{Defof|Exterior (Topology)|Exterior}}
+}}
+{{eqn | r = T \setminus \bigcap_{H \mathop \in \mathbb H} H^-
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{end-eqn}}
+From [[Closure of Intersection is Subset of Intersection of Closures]]:
+:$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H^-} \subseteq \bigcap_{H \mathop \in \mathbb H} H^-$
+From [[Set Complement inverts Subsets]]:
+:$\displaystyle T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^-} \supseteq T \setminus \bigcap_{H \mathop \in \mathbb H} H^-$
+From the definition of [[Definition:Exterior (Topology)|exterior]]:
+:$\displaystyle T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H}^- = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e$
+Putting this together:
+:$\displaystyle \bigcup_{H \mathop \in \mathbb H} H^e \subseteq \paren {\bigcap_{H \mathop \in \mathbb H} H}^e$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separated Sets are Disjoint}
+Tags: Separated Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A, B \subseteq S$ such that $A$ and $B$ are [[Definition:Separated Sets|separated in $T$]].
+Then $A$ and $B$ are [[Definition:Disjoint Sets|disjoint]]:
+:$A \cap B  = \O$
+\end{theorem}
+
+\begin{proof}
+Let $A$ and $B$ be [[Definition:Separated Sets|separated in $T$]].
+Then:
+{{begin-eqn}}
+{{eqn | l = A^- \cap B
+      | r = \O
+      | c = {{Defof|Separated Sets}}: $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {A \cup A'} \cap B
+      | r = \O
+      | c = {{Defof|Closure (Topology)|Set Closure}}: $A'$ is the [[Definition:Derived Set|derived set]] of $A$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {A \cap B} \cup \paren {A' \cap B}
+      | r = \O
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | ll= \leadsto
+      | l = A \cap B
+      | r = \O
+      | c = [[Union is Empty iff Sets are Empty]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second-Countable Space is First-Countable}
+Tags: First-Countable Spaces, Second-Countable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Second-Countable Space|second-countable]].
+Then $T$ is also [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+By definition $T$ is [[Definition:Second-Countable Space|second-countable]] {{iff}} its [[Definition:Topology|topology]] has a [[Definition:Countable Basis|countable basis]].
+Consider the entire set $S$ as an [[Definition:Open Set (Topology)|open set]].
+From [[Set is Open iff Neighborhood of all its Points]], $S$ has that property.
+As $T$ has a [[Definition:Countable Basis|countable basis]], then (trivially) every point in $T$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+So a [[Definition:Second-Countable Space|second-countable space]] is trivially [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second-Countability is Hereditary}
+Tags: Second-Countable Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Second-Countable Space|second-countable]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Second-Countable Space|second-countable]], $\struct {S, \tau}$ has a [[Definition:Countable Basis|countable basis]].
+That is, $\exists \BB \subseteq \tau$ such that:
+:for all $U \in \tau$, $U$ is a [[Definition:Set Union|union]] of sets from $\BB$
+:$\BB$ is [[Definition:Countable Set|countable]].
+As $H \subseteq S$ it follows that a $H$ itself is a [[Definition:Set Union|union]] of sets from $\BB$.
+The result follows from [[Basis for Topological Subspace]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First-Countability is Hereditary}
+Tags: First-Countable Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:First-Countable Space|first-countable]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:First-Countable Space|first-countable]], every point in $S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T$.
+The [[Definition:Set Intersection|intersection]] of $H$ with the [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] of $S$ provides a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] for $H$.
+As every point in $H$ is also a point in $S$, the result follows from [[Basis for Topological Subspace]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Topological Subspace}
+Tags: Topological Subspaces, Topological Bases
+
+\begin{theorem}
+Let $T = \struct {A, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\O \subseteq H \subseteq A$ and so let $T_H = \struct {H, \tau_H}$ be a [[Definition:Topological Subspace|subspace]] of $T$.
+Let $\BB$ be a [[Definition:Synthetic Basis|(synthetic) basis]] for $T$.
+Let $\BB_H$ be defined as:
+:$B_H = \set {U \cap H: U \in \BB}$
+Then $\BB_H$ is a [[Definition:Synthetic Basis|(synthetic) basis]] for $H$.
+\end{theorem}
+
+\begin{proof}
+$\BB \subseteq \powerset A$ is a [[Definition:Synthetic Basis|synthetic basis]] for $T$ {{iff}}:
+:'''B1:''' $A$ is a [[Definition:Set Union|union]] of sets from $\BB$
+:'''B2:''' If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a [[Definition:Set Union|union]] of sets from $\BB$.
+Let $A = \mathbb S$ is a [[Definition:Set Union|union]] of sets from $\BB$.
+Then:
+:$\displaystyle A = \bigcup_{S \mathop \in \mathbb S} S$
+for some $\mathbb S \subseteq \BB$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap H
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H
+      | c = 
+}}
+{{eqn | r = \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{end-eqn}}
+But if $S \in \BB$, then $S \cap H \in \BB_H$
+So $H$ is a [[Definition:Set Union|union]] of sets from $\BB_H$.
+In the same way we investigate $U_1$ and $U_2$.
+Let $U_1, U_2 \in \BB_H$.
+Then $U_1 = B_1 \cap H, U_2 = B_2 \cap H$ for some $B_1, B_2 \in \BB$.
+Then $U_1 \cap U_2 = B_1 \cap H \cap B_2 \cap H = \paren {B_1 \cap B_2} \cap H$.
+As $\BB$ is a [[Definition:Synthetic Basis|(synthetic) basis]] for $A$, we have that:
+:$\displaystyle B_1 \cap B_2 = \bigcup_{S \mathop \in \mathbb S} S$
+for some $\mathbb S \subseteq \BB$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = U_1 \cap U_2
+      | r = \paren {B_1 \cap B_2} \cap H
+      | c = 
+}}
+{{eqn | r = \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H
+      | c = 
+}}
+{{eqn | r = \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{end-eqn}}
+But if $S \in \BB$, then $S \cap H \in \BB_H$.
+So $U_1 \cap U_2$ is a [[Definition:Set Union|union]] of sets from $\BB_H$.
+So $\BB_H$ fulfils the conditions to be a [[Definition:Synthetic Basis|synthetic basis]] for $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Local Rings}
+Tags: Local Rings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $J \lhd R$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+:$(1): \quad$ If the set $R \setminus J$ is precisely the [[Definition:Group of Units of Ring|group of units]] $R^\times$ of $R$, then $\tuple {R, J}$ is a [[Definition:Local Ring|local ring]].
+:$(2): \quad$ If $1 + x$ is a [[Definition:Unit of Ring|unit]] in $R$ for all $x \in J$ then $\tuple {R, J}$ is [[Definition:Local Ring|local]].
+{{explain|The specific interpretation of the notation $\tuple {R, J}$ is unclear. The usual notation for a ring is something like $\struct {R, +, \times}$.}}
+\end{theorem}
+
+\begin{proof}
+$(1): \quad$ Suppose that $J$ is the set of non-units of $R$.
+Then by [[Ideal of Unit is Whole Ring]], every [[Definition:Ideal of Ring|ideal]] not equal to $R$ is contained in $J$.
+Therefore $J$ is the unique [[Definition:Maximal Ideal of Ring|maximal ideal]] of $R$, so $\tuple {R, J}$ is [[Definition:Local Ring|local]].
+$(2): \quad$ Let $x \in R \setminus J$.
+Then:
+:$J \subsetneq \map I {J \cup \set x} \subseteq R$
+where $\map I S$ is the [[Definition:Generator of Ideal|ideal generated]] by $S$.
+Since $J$ is [[Definition:Maximal Ideal of Ring|maximal]], by definition, $x$ and $J$ [[Definition:Generator of Ideal|generate]] all of $R$.
+Therefore $t x + m = 1$ for some $m \in J$, $t \in R$.
+Thus $t x = 1 - m \in 1 + J$ is a [[Definition:Unit of Ring|unit]] by hypothesis.
+Therefore $x$ is a [[Definition:Unit of Ring|unit]].
+Now use part $(1)$.
+{{qed}}
+[[Category:Local Rings]]
+rndfook9i5ozjv58ox7cg93kkm7xjke
+\end{proof}<|endoftext|>
+\section{Continuity Defined by Closure}
+Tags: Set Closures, Continuous Mappings
+
+\begin{theorem}
+Let $T_1 = \struct {X_1, \tau_1}$ and $T_2 = \struct {X_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] {{iff}}:
+:$\forall H \subseteq X_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
+where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$ in $T_1$.
+That is, {{iff}} the [[Definition:Image of Subset under Mapping|image]] of the [[Definition:Closure (Topology)|closure]] is a [[Definition:Subset|subset]] of the [[Definition:Closure (Topology)|closure]] of the [[Definition:Image of Subset under Mapping|image]].
+\end{theorem}
+
+\begin{proof}
+First we establish some details.
+Let $H \subseteq X_1$.
+Let $\mathbb K_1$ be defined as:
+:$\mathbb K_1 := \set {K \subseteq X_1: H \subseteq K, K \text { closed} }$
+That is, let $\mathbb K_1$ be the set of all [[Definition:Closed Set (Topology)|closed sets]] of $T_1$ which contain $H$.
+Similarly, let $\mathbb K_2$ be defined as:
+:$\mathbb K_2 := \set {K \subseteq X_2: f \sqbrk H \subseteq K, K \text{ closed} }$
+That is, let $\mathbb K_2$ be the set of all [[Definition:Closed Set (Topology)|closed sets]] of $T_2$ which contain $f \sqbrk H$.
+From the definition of [[Definition:Closure (Topology)|closure]], we have that:
+:$\displaystyle H^- = \bigcap \mathbb K_1$
+That is, the [[Definition:Closure (Topology)|closure]] of $H$ is the [[Definition:Set Intersection|intersection]] of all the [[Definition:Closed Set (Topology)|closed sets]] of $T_1$ which contain $H$.
+Similarly:
+:$\displaystyle \paren {f \sqbrk H}^- = \bigcap \mathbb K_2$
+That is, the [[Definition:Closure (Topology)|closure]] of $f \sqbrk H$ is the [[Definition:Set Intersection|intersection]] of all the [[Definition:Closed Set (Topology)|closed sets]] of $T_2$ which contain $f \sqbrk H$.
+We have:
+{{begin-eqn}}
+{{eqn | l = f \sqbrk {H^-}
+      | r = f \sqbrk {\bigcap \mathbb K_1}
+      | c =
+}}
+{{eqn | o = \subseteq
+      | r = \bigcap_{K \mathop \in \mathbb K_1} f \sqbrk K
+      | c = [[Image of Intersection under Mapping]]
+}}
+{{end-eqn}}
+From [[Image of Subset under Relation is Subset of Image/Corollary 2|Image of Subset under Relation is Subset of Image: Corollary 2]]:
+:$H \subseteq K \implies f \sqbrk H \subseteq f \sqbrk K$
+=== Necessary Condition ===
+Suppose $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+From the above we have that :
+:$\displaystyle \paren {f \sqbrk H}^- := \bigcap \mathbb K_2$
+As $f$ is continuous, then:
+:$\forall K \in \mathbb K_2: f^{-1} \sqbrk K$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$
+But as $f \sqbrk H \subseteq K$, it follows from [[Image of Subset under Relation is Subset of Image/Corollary 3|Image of Subset under Relation is Subset of Image: Corollary 3]] that:
+:$H \subseteq f^{-1} \sqbrk K$
+So:
+:$\mathbb K_3 := \set {f^{-1} \sqbrk K: K \text { closed in } T_2, H \subseteq f^{-1} \sqbrk K}$
+consists entirely of [[Definition:Closed Set (Topology)|closed sets]] in $T_1$ which are [[Definition:Superset|supersets]] of $H$.
+That is, $\mathbb K_3 \subseteq \mathbb K_1$.
+So:
+:$\displaystyle \bigcap \mathbb K_1 \subseteq \bigcap \mathbb K_3$
+and so:
+:$\displaystyle f \sqbrk {\bigcap \mathbb K_1} \subseteq f \sqbrk {\bigcap \mathbb K_3}$
+But from [[Image of Intersection under Mapping]]:
+:$\displaystyle f \sqbrk {\bigcap \mathbb K_3} \subseteq \bigcap_{K \mathop \in \mathbb K_3} f \sqbrk K$
+But:
+:$\displaystyle \bigcap_{K \mathop \in \mathbb K_3} f \sqbrk K = \bigcap \mathbb K_2$
+and so:
+:$\displaystyle f \sqbrk {\bigcap \mathbb K_1} \subseteq \bigcap \mathbb K_2$
+which means that:
+:$f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
+as we wanted to show.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $f$ is not [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+From [[Continuity Defined from Closed Sets]], $\exists B \subseteq X_2$ which is [[Definition:Closed Set (Topology)|closed]] in $T_2$ such that $f^{-1} \sqbrk B$ is not [[Definition:Closed Set (Topology)|closed]] in $T_1$.
+By [[Image of Preimage under Mapping]], we have that:
+:$f \sqbrk {f^{-1} \sqbrk B} \subseteq B$
+So from [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B^-$
+From [[Closed Set Equals its Closure]] we have that $B^- = B$.
+Transitively, we get:
+:$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B$
+Because $f^{-1} \sqbrk B$ is not [[Definition:Closed Set (Topology)|closed]] in $T_1$, we have that:
+:$f^{-1} \sqbrk B \subsetneq \paren {f^{-1} \sqbrk B}^-$
+This means there exists an element $x \in \paren {f^{-1} \sqbrk B}^-$ such that $x \notin f^{-1} \sqbrk B$. 
+Therefore $\map f x \notin B$, but $\map f x \in f \sqbrk {\sqbrk {f^{-1} \sqbrk B}^-}$.
+From above, we had:
+:$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B$
+so there exists a set $A \subseteq X_1$, namely $A = f^{-1} \sqbrk B$, such that:
+:$f \sqbrk {A^-} \nsubseteq \paren {f \sqbrk A}^-$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ring is Local}
+Tags: Valuation Rings, Local Rings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Valuation Ring|valuation ring]].
+Then $R$ is a [[Definition:Local Ring|local ring]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Valuation Rings]]
+[[Category:Local Rings]]
+80qyrb0gwzowczadyz86mtm5a7a5um9
+\end{proof}<|endoftext|>
+\section{Bijection is Open iff Inverse is Continuous}
+Tags: Open Mappings, Continuous Mappings
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ be a [[Definition:Bijection|bijection]].
+Then $f$ is [[Definition:Open Mapping|open]] {{iff}} $f^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be a [[Definition:Bijection|bijection]].
+Let $g := f^{-1}$.
+By [[Bijection iff Inverse is Bijection]] we have that $g$ is a [[Definition:Bijection|bijection]] and that $g^{-1} = f$.
+Let $f$ be [[Definition:Open Mapping|open]].
+Then by definition of [[Definition:Open Mapping|open mapping]]:
+:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
+taking $H \in \tau_1$ by definition of [[Definition:Open Set (Topology)|open in $T_1$]].
+But $f = g^{-1}$ and so:
+:$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$
+which is exactly the definition for $g$ to be [[Definition:Continuous Mapping (Topology)|continuous]].
+The argument works the other way.
+Let $g$ be [[Definition:Continuous Mapping (Topology)|continuous]].
+Then by definition of [[Definition:Continuous Mapping (Topology)|continuous mapping]]:
+:$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$
+But $g^{-1} = f$ and so:
+:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
+which is exactly the definition for $f$ to be [[Definition:Open Mapping|open]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bijection is Open iff Closed}
+Tags: Open Mappings, Closed Mappings, Bijections
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ be a [[Definition:Bijection|bijection]].
+Then $f$ is [[Definition:Open Mapping|open]] {{iff}} $f$ is [[Definition:Closed Mapping|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be a [[Definition:Bijection|bijection]].
+Suppose $f$ is an [[Definition:Open Mapping|open mapping]].
+From the definition of [[Definition:Open Mapping|open mapping]]:
+:$\forall H \in \tau_1: f \sqbrk H \in \tau_2$
+As $f$ is a [[Definition:Bijection|bijection]]:
+:$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$
+By definition of [[Definition:Closed Set (Topology)|closed set]]:
+:$S_1 \setminus H$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$
+and as $f$ is an [[Definition:Open Mapping|open mapping]]:
+:$f \sqbrk {S_1 \setminus H} = S_2 \setminus f \sqbrk H$
+is [[Definition:Closed Set (Topology)|closed]] in $T_2$.
+Hence by definition $f$ is a [[Definition:Closed Mapping|closed mapping]].
+A similar argument demonstrates that if $f$ is [[Definition:Closed Mapping|closed]] then it is [[Definition:Open Mapping|open]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T2 Space is T1 Space}
+Tags: T1 Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]]:
+:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
+As $U \cap V = \O$ it follows from the definition of [[Definition:Disjoint Sets|disjoint sets]] that:
+:$x \in U \implies x \notin V$
+:$y \in V \implies y \notin U$
+So if $x \in U, y \in V$ then:
+:$\exists U \in \tau: x \in U, y \notin U$
+:$\exists V \in \tau: y \in V, x \notin V$
+which is precisely the definition of a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1 Space is T0 Space}
+Tags: T0 Spaces, T1 Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Fréchet Space (Topology)|Fréchet ($T_1$) space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Kolmogorov Space|Kolmogorov ($T_0$) space]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S, \tau}$ be a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+Let $x, y \in S: x \ne y$.
+From the definition of [[Definition:Fréchet Space (Topology)|$T_1$ space]]:
+:'''Both'''
+::$\exists U \in \tau: x \in U, y \notin U$
+:'''and''':
+::$\exists V \in \tau: y \in V, x \notin V$
+From the [[Rule of Simplification]]:
+: $\exists U \in \tau: x \in U, y \notin U$
+From the [[Rule of Addition]]:
+:'''Either'''
+::$\exists U \in \tau: x \in U, y \notin U$
+:'''or''':
+::$\exists V \in \tau: y \in V, x \notin V$
+which is precisely the definition of a [[Definition:Kolmogorov Space|Kolmogorov ($T_0$) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T5 Space is T4 Space}
+Tags: T4 Spaces, T5 Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:T5 Space|$T_5$ space]].
+Then $\struct {S, \tau}$ is also a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S, \tau}$ be a [[Definition:T5 Space|$T_5$ space]].
+From the definition of [[Definition:T5 Space|$T_5$ space]]:
+:$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
+where $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$ in $T$.
+Let $C, D \subseteq S$ be [[Definition:Disjoint Sets|disjoint sets]] which are [[Definition:Closed Set (Topology)|closed]] in $T$.
+Thus $C, D \in \map \complement \tau$ from the definition of [[Definition:Closed Set (Topology)|closed set]].
+From [[Topological Closure is Closed]]:
+:$C^- = C, D^- = D$
+and so from $C \cap D = \O$:
+:$C^- \cap D = C \cap D^- = \O$
+Thus from the definition of [[Definition:T5 Space|$T_5$ space]]:
+:$\forall C, D \in \map \complement \tau, C \cap D = \O: \exists U, V \in \tau: C \subseteq U, D \subseteq V$ 
+which is precisely the definition of a [[Definition:T4 Space|$T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Normal Space is Normal Space}
+Tags: Completely Normal Spaces, Normal Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Completely Normal Space|completely normal space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Normal Space|normal space]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S, \tau}$ be a [[Definition:Completely Normal Space|completely normal space]].
+From the definition, $\struct {S, \tau}$ is a [[Definition:Completely Normal Space|completely normal space]] {{iff}}:
+:$\struct {S, \tau}$ is a [[Definition:T5 Space|$T_5$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+We have that a [[T5 Space is T4 Space|$T_5$ space is a $T_4$ space]].
+So:
+:$\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+which is precisely the definition of a [[Definition:Normal Space|normal space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Regular Space is T2 Space}
+Tags: Regular Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+From the definition of [[Definition:Regular Space|regular space]]:
+:$\struct {S, \tau}$ is a [[Definition:T3 Space|$T_3$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Let $x, y \in S$.
+As $T$ is [[Definition:Kolmogorov Space|$T_0$]], it follows that either:
+:$\exists V \in \tau: x \in V, y \notin V$
+or:
+:$\exists V \in \tau: y \in V, x \notin V$
+that is, there exists $V$, an [[Definition:Open Set (Topology)|open set]], containing one but not the other.
+{{WLOG}}, suppose that $\exists V \in \tau: y \in V, x \notin V$.
+Then by definition of [[Definition:Relative Complement|relative complement]]:
+:$x \in \relcomp S V$
+Let $F := \relcomp S V$.
+As $V$ is [[Definition:Open Set (Topology)|open]], by definition of [[Definition:Closed Set (Topology)|closed set]] we have that $F = \relcomp S V$ is [[Definition:Closed Set (Topology)|closed]].
+That is:
+:$\relcomp S V \in \tau$
+As $y \in V$ it follows that $y \notin F$, that is:
+:$y \in \relcomp S F$
+Now $\struct {S, \tau}$ is a [[Definition:T3 Space|$T_3$ space]], and so:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
+So we have that:
+:$x \in F \subseteq U \implies x \in U$
+:$y \notin F, y \in V$
+such that $U \cap V = \O$.
+So:
+:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
+which is precisely the definition of a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Space is T3 Space}
+Tags: T3 Spaces, Normal Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+Then $\struct {S, \tau}$ is also a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+From the definition of [[Definition:Normal Space|normal space]]:
+:$\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|Fréchet ($T_1$) space]].
+Let $F$ be any [[Definition:Closed Set (Topology)|closed set]] in $T$.
+Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
+As $T$ is a [[Definition:Fréchet Space (Topology)|Fréchet ($T_1$) space]] it follows from [[Equivalence of Definitions of T1 Space|Equivalence of Definitions of $T_1$ Space]] that $\set y$ is [[Definition:Closed Set (Topology)|closed]].
+As $T = \struct {S, \tau}$ is a [[Definition:Normal Space|normal space]], we have that:
+:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
+That is, for any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $A$ and $B$ respectively.
+But $F$ and $\set y$ are [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]].
+So:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
+which is precisely the definition of a [[Definition:T3 Space|$T_3$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Hausdorff Space is Hausdorff Space}
+Tags: Completely Hausdorff Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Completely Hausdorff Space|completely Hausdorff space]].
+From the [[Definition:Completely Hausdorff Space|definition]]:
+:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
+We have that [[Set is Subset of its Topological Closure]] and so $U \subseteq U^-$ and $V \subseteq V^-$.
+This leads to:
+:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
+which is precisely the definition of a [[Definition:Hausdorff Space|Hausdorff ($T_2$) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Regular Space is Completely Hausdorff Space}
+Tags: Regular Spaces, Completely Hausdorff Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+Let $x, y \in S$ with $x \ne y$.
+From the definition of a [[Definition:Completely Hausdorff Space/Definition 2|completely Hausdorff Space]]:
+:$T$ is [[Definition:Completely Hausdorff Space/Definition 2|completely Hausdorff]] if $x$ and $y$ are separated by [[Definition:Closed Neighborhood|closed neighborhoods]].
+From the [[Definition:Regular Space|definition]]:
+:$\struct {S, \tau}$ is a [[Definition:T3 Space|$T_3$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+From [[Regular Space is T2 Space|Regular Space is $T_2$ Space]], there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] containing $x$ and $y$ respectively:
+:$\exists U_x, U_y \in \tau: x \in U_x, y \in U_y, U_x \cap U_y = \O$
+From [[Definition:T3 Space/Definition 2|definition of $T_3$ space]], each [[Definition:Open Set (Topology)|open set]] contains a [[Definition:Closed Neighborhood|closed neighborhood]] around each of its points, specifically for $x$ and $y$:
+:$\exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U_x$
+:$\exists N_y: \relcomp S {N_y} \in \tau: \exists V_y \in \tau: y \in V_y \subseteq N_y \subseteq U_y$
+From [[Subsets of Disjoint Sets are Disjoint]]:
+:$N_x \cap N_y = \O$
+Hence $x$ and $y$ are separated by [[Definition:Closed Neighborhood|closed neighborhoods]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of T0 Space}
+Tags: T0 Spaces
+
+\begin{theorem}
+{{TFAE|def = T0 Space|view = $T_0$ (Kolmogorov) space}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+=== Definition by Open Sets implies Definition by Limit Points ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall x, y \in S$, either:
+::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
+::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
+Let $x, y \in S$ such that $x \ne y$.
+By the definition of [[Definition:Limit Point of Point|limit point of a point]]:
+:$(3): \quad x$ is a [[Definition:Limit Point of Point|limit point of $y$]] if every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $x \in U$ contains $y$.
+:$(4): \quad y$ is a [[Definition:Limit Point of Point|limit point of $x$]] if every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $y \in U$ contains $x$.
+Thus:
+:If $(3)$ holds then $(1)$ is false.
+:If $(4)$ holds then $(2)$ is false.
+So both $(3)$ and $(4)$ can not both hold at once.
+Hence $T = \struct {S, \tau}$ is a [[Definition:Topological Space|topological space]] for which no two points can be [[Definition:Limit Point of Point|limit points]] of each other.
+{{qed|lemma}}
+=== Definition by Limit Points implies Definition by Open Sets ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which no two points can be [[Definition:Limit Point of Point|limit points]] of each other.
+Let $x, y \in S$ such that $x \ne y$.
+Suppose that:
+: $(1): \quad \nexists U \in \tau: x \in U, y \notin U$
+and
+: $(2): \quad \nexists U \in \tau: y \in U, x \notin U$
+Expressing this in different terms:
+:$(3): \quad$ Every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $x \in U$ contains $y$.
+:$(4): \quad$ Every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $y \in U$ contains $x$.
+That is, by definition:
+:$x$ is a [[Definition:Limit Point of Point|limit point of $y$]] and $y$ is a [[Definition:Limit Point of Point|limit point of $x$]].
+From this contradiction we deduce that:
+:$\forall x, y \in S$, either:
+::$\exists U \in \tau: x \in U, y \notin U$
+:or:
+::$\exists U \in \tau: y \in U, x \notin U$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of T1 Space}
+Tags: T1 Spaces
+
+\begin{theorem}
+{{TFAE|def = T1 Space|view = $T_1$ (Fréchet) space}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+=== Definition by Open Sets implies Definition by Closed Points ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall x, y \in S$, both:
+::$(1): \quad \exists U \in \tau: x \in U, y \notin U$
+and:
+::$(2): \quad \exists U \in \tau: y \in U, x \notin U$
+Let $x, y \in S$.
+By the definition of [[Definition:Limit Point of Point|limit point of a point]], the above condition means:
+:$(3): \quad y$ is a [[Definition:Limit Point of Point|limit point of $x$]] if every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $y \in U$ contains $x$.
+Thus $(2)$ and $(3)$ give us that $y$ is ''not'' a [[Definition:Limit Point of Point|limit point of $x$]].
+As $x$ and $y$ are any two points in $S$, it follows that $x$ has no limit points.
+Thus it holds [[Definition:Vacuous Truth|vacuously]] that all the [[Definition:Limit Point of Point|limit point of $x$]] are in $\set x$.
+By [[Closed Set iff Contains all its Limit Points]], we have that $\set x$ is a [[Definition:Closed Set (Topology)|closed set]] in $T$.
+This holds for all $x \in S$.
+Hence $T = \struct {S, \tau}$ is a [[Definition:Topological Space|topological space]] for which all points are [[Definition:Closed Point|closed]].
+{{qed|lemma}}
+=== Definition by Closed Points implies Definition by Open Sets ===
+$T = \struct {S, \tau}$ is a [[Definition:Topological Space|topological space]] for which all points are [[Definition:Closed Point|closed]].
+Let $x, y \in S$.
+By [[Closed Set iff Contains all its Limit Points]], we have that:
+:all the [[Definition:Limit Point of Point|limit points of $x$]] are in $\set x$
+:all the [[Definition:Limit Point of Point|limit points of $y$]] are in $\set y$.
+It follows by [[Definition:Limit Point of Point|definition of limit point]] (reversing the [[Equivalence of Definitions of T1 Space#Definition by Open Sets implies Definition by Closed Points|above argument]]) that:
+:$\exists U \in \tau: x \in U, y \notin U$
+and
+:$\exists U \in \tau: y \in U, x \notin U$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T3 1/2 Space is T3 Space}
+Tags: T3 1: 2 Spaces, T3 Spaces
+
+\begin{theorem}
+Let $T$ be a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Then $T$ is also a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+From the definition of [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]:
+:For any [[Definition:Closed Set (Topology)|closed set]] $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$.
+Let $F \subseteq S$ be a [[Definition:Closed Set (Topology)|closed set]] in $T$ and let $y \in \relcomp S F$.
+An [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]] $f: S \to \closedint 0 1$ where:
+:$\forall a \in F: \map f a = 0$
+:$\map f y = 1$
+Let:
+:$U = \map {f^{-1} } 0$
+:$V = \map {f^{-1} } 1$
+As $f$ is [[Definition:Continuous Mapping (Topology)|continuous]], both $U$ and $V$ are [[Definition:Open Set (Topology)|open]] in $T$.
+Suppose $x \in U \cap V$.
+Then we would have:
+:$x \in U \implies \map f x = 0$
+:$x \in V \implies \map f x = 1$
+which contradicts the [[Definition:Many-to-One Relation|many-to-one nature of a mapping]].
+So $U \cap V = \O$.
+Thus we have:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
+That is, for any [[Definition:Closed Set (Topology)|closed set]] $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
+which is precisely the definition of a [[Definition:T3 Space|$T_3$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tychonoff Space is Regular, T2 and T1}
+Tags: Tychonoff Spaces, Regular Spaces, Hausdorff Spaces, T1 Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Tychonoff Space|Tychonoff space]].
+Then $\struct {S, \tau}$ is also:
+:a [[Definition:Regular Space|regular space]]
+:a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]]
+:a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Tychonoff Space|Tychonoff space]].
+From the definition of [[Definition:Tychonoff Space|Tychonoff space]]:
+:$\struct {S, \tau}$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+We have that a [[T3 1/2 Space is T3 Space|$T_{3 \frac 1 2}$ space is a $T_3$ space]].
+From the definition, a [[Definition:Regular Space|regular space]] is:
+:a [[Definition:T3 Space|$T_3$ space]]
+:a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+So a [[Definition:Tychonoff Space|Tychonoff space]] is a [[Definition:Regular Space|regular space]].
+Then we have a [[Regular Space is T2 Space|regular space is a $T_2$ (Hausdorff) space]].
+Then we have a [[T2 Space is T1 Space|$T_2$ (Hausdorff) space is a $T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Space is Tychonoff Space}
+Tags: Tychonoff Spaces, Normal Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Tychonoff Space|Tychonoff (completely regular) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+From the definition of [[Definition:Normal Space|normal space]]:
+:$\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Let $F$ be any [[Definition:Closed Set (Topology)|closed set]] in $T$.
+Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
+As $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]] it follows from [[Equivalence of Definitions of T1 Space|Equivalence of Definitions of $T_1$ Space]] that $\set y$ is [[Definition:Closed Set (Topology)|closed]].
+As $T = \struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]], we have that for any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$ there exists an [[Definition:Urysohn Function|Urysohn function]] for $A$ and $B$.
+But $F$ and $\set y$ are [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]].
+So there exists an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$.
+This is the definition of a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Next we note that as $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], from [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]] it follows that $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+So $T$ is both a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]] and [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+So, by definition, $T$ is a [[Definition:Tychonoff Space|Tychonoff space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T4 and T3 Space is T 3 1/2}
+Tags: T4 Spaces, T3 Spaces, T3 1: 2 Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be:
+:a [[Definition:T4 Space|$T_4$ space]]
+and also:
+:a [[Definition:T3 Space|$T_3$ space]].
+Then $T$ is also a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:T4 Space|$T_4$ space]] which is also a [[Definition:T3 Space|$T_3$ space]].
+From it being [[Definition:T3 Space|$T_3$]]:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
+Consider this $U \in \tau$, which is [[Definition:Disjoint Sets|disjoint]] from $\set y$.
+Then $\relcomp S U$ is a [[Definition:Closed Set (Topology)|closed set]] which is [[Definition:Disjoint Sets|disjoint]] from $F$ but such that $\set y \subseteq \relcomp S U$.
+As $T$ is a [[Definition:T4 Space|$T_4$ space]], we have that from [[Urysohn's Lemma]] there exists an [[Definition:Urysohn Function|Urysohn function]] $f$ for $F$ and $\relcomp S U$.
+As $\set y \subseteq \relcomp S U$, this function $f$ is a [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$ as well.
+So:
+:For any [[Definition:Closed Set (Topology)|closed set]] $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$.
+which is precisely the definition of a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Axioms Preserved under Homeomorphism}
+Tags: Separation Axioms
+
+\begin{theorem}
+The [[Definition:Separation Axioms|separation axioms]] are preserved under [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+Let $T_A = \left({S_A, \tau_A}\right), T_B = \left({S_B, \tau_B}\right)$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+\end{theorem}<|endoftext|>
+\section{T0 Space is Preserved under Homeomorphism}
+Tags: T0 Spaces, Homeomorphisms
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Homeomorphism/Topological Spaces/Definition 4|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
+The result follows from [[T0 Space is Preserved under Closed Bijection|$T_0$ (Kolmogorov) Space is Preserved under Closed Bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1 Space is Preserved under Homeomorphism}
+Tags: T1 Spaces, Homeomorphisms
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Homeomorphism/Topological Spaces/Definition 4|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]  [[Definition:Bijection|bijection]].
+The result follows from [[T1 Space is Preserved under Closed Bijection|$T_1$ (Fréchet) Space is Preserved under Closed Bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T2 Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, Hausdorff Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Homeomorphism/Topological Spaces/Definition 4|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
+The result follows from [[T2 Space is Preserved under Closed Bijection|$T_2$ (Hausdorff) Space is Preserved under Closed Bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Hausdorff Space is Preserved under Homeomorphism}
+Tags: Completely Hausdorff Spaces, Homeomorphisms
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Homeomorphism/Topological Spaces/Definition 4|homeomorphism]], $\phi$ is a [[Definition:Closed Mapping|closed]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
+The result follows from [[Completely Hausdorff Space is Preserved under Closed Bijection|$T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Closed Bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Regular Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, Regular Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Regular Space|regular space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+We have that $\struct {S, \tau}$ is a [[Definition:Regular Space|regular space]] {{iff}}:
+:$\struct {S, \tau}$ is a [[Definition:T3 Space|$T_3$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+From [[T3 Space is Preserved under Homeomorphism|$T_3$ Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:T3 Space|$T_3$ space]], then so is $T_B$.
+From [[T0 Space is Preserved under Homeomorphism|$T_0$ (Kolmogorov) Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]], then so is $T_B$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T3 Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, T3 Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:T3 Space|$T_3$ space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Suppose that $T_A$ is a [[Definition:T3 Space|$T_3$ space]].
+Let $F$ be [[Definition:Closed Set (Topology)|closed]] in $T_B$.
+Let $y \in S_B$ such that $y \notin F$.
+From [[Preimage of Intersection under Mapping]] it follows that $\phi^{-1} \sqbrk F$ and $\map {\phi^{-1}} y$ are [[Definition:Disjoint Sets|disjoint]].
+Also, as $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], it is [[Definition:A Fortiori|a fortiori]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Thus by [[Continuity Defined from Closed Sets]], $\phi^{-1} \sqbrk F$ is [[Definition:Closed Set (Topology)|closed]].
+Now as $T_A$ is a [[Definition:T3 Space|$T_3$ space]], we find [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U_1$ containing $\phi^{-1} \sqbrk F$ and $U_2$ containing $\map {\phi^{-1}} y$.
+From [[Image of Subset is Subset of Image]], we have $F = \phi \sqbrk {\phi^{-1} \sqbrk F} \subseteq \phi \sqbrk {U_1}$.
+Here, the first equality follows from [[Subset equals Image of Preimage iff Mapping is Surjection]], as $\phi$ is [[Definition:A Fortiori|a fortiori]] [[Definition:Surjection|surjective]], being a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], we deduce also $\set y \subseteq \phi \sqbrk {U_2}$.
+From [[Image of Intersection under Injection]] it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are [[Definition:Disjoint Sets|disjoint]].
+Since $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], they are also both [[Definition:Open Set (Topology)|open]] in $T_B$.
+Therewith, we have construed two [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] in $T_B$, one containing $F$, and the other containing $y$.
+Hence $T_B$ is shown to be a [[Definition:T3 Space|$T_3$ space]] as well.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T3 1/2 Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, T3 1: 2 Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Let $F \subseteq S_B$ be [[Definition:Closed Set (Topology)|closed]], and let $y \in S_B$ such that $y \notin F$.
+Let $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$.
+Since $T_A$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]], there exists an [[Definition:Urysohn Function|Urysohn function]] $f: X_A \to \closedint 0 1$ for $G$ and $\set z$.
+Define $g: S_B \to \closedint 0 1$ by:
+:$\map g x = \map f {\map {\phi^{-1} } x}$
+By [[Composite of Continuous Mappings is Continuous]], $g$ is [[Definition:Continuous Mapping|continuous]].
+Also, for all $x \in F$:
+:$\map g x = 0$
+as $\map {\phi^{-1} } x \in G$.
+Similarly, because $\map {\phi^{-1} } y = z$:
+:$\map g y = 1$
+Hence $g$ is an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$.
+Since $F$ and $y$ were arbitrary, it follows that $T_B$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tychonoff Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, Tychonoff Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Tychonoff Space|Tychonoff (completely regular) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+We have that $\struct {S, \tau}$ is a [[Definition:Tychonoff Space|Tychonoff space]] {{iff}}:
+:$\struct {S, \tau}$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+From [[T3 1/2 Space is Preserved under Homeomorphism|$T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]], then so is $T_B$.
+From [[T0 Space is Preserved under Homeomorphism|$T_0$ (Kolmogorov) Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]], then so is $T_B$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Space is Preserved under Homeomorphism}
+Tags: Normal Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Normal Space|normal space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+We have that $\struct {S, \tau}$ is a [[Definition:Normal Space|normal space]] {{iff}}:
+:$\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+From [[T4 Space is Preserved under Homeomorphism|$T_4$ Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:T4 Space|$T_4$ space]], then so is $T_B$.
+From [[T1 Space is Preserved under Homeomorphism|$T_1$ Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], then so is $T_B$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T4 Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, T4 Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:T4 Space|$T_4$ space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Suppose that $T_A$ is a [[Definition:T4 Space|$T_4$ space]].
+Let $B_1$ and $B_2$ be [[Definition:Closed Set (Topology)|closed]] in $T_B$ that are [[Definition:Disjoint Sets|disjoint]].
+Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the [[Definition:Preimage of Subset under Mapping|preimages]] of $B_1$ and $B_2$ under $\phi$, respectively.
+From [[Preimage of Intersection under Mapping]] it follows that $A_1$ and $A_2$ are [[Definition:Disjoint Sets|disjoint]].
+Also, as $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], it is [[Definition:A Fortiori|a fortiori]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Thus [[Continuity Defined from Closed Sets]] applies to yield that both $A_1$ and $A_2$ are [[Definition:Closed Set (Topology)|closed]].
+Now as $T_A$ is a [[Definition:T4 Space|$T_4$ space]], we find [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U_1$ containing $A_1$ and $U_2$ containing $A_2$.
+From [[Image of Subset is Subset of Image]], we have $B_1 = \phi \sqbrk {\phi^{-1} \sqbrk {B_1} } = \phi \sqbrk {A_1} \subseteq \phi \sqbrk {U_1}$.
+Here, the first equality follows from [[Subset equals Image of Preimage iff Mapping is Surjection]], as $\phi$ is [[Definition:A Fortiori|a fortiori]] [[Definition:Surjection|surjective]], being a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], we deduce also $B_2 \subseteq \phi \sqbrk {U_2}$.
+From [[Image of Intersection under Injection]] it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are [[Definition:Disjoint Sets|disjoint]].
+Since $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], they are also both [[Definition:Open Set (Topology)|open]] in $T_B$.
+Therewith, we have construed two [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] in $T_B$, one containing $B_1$, and the other containing $B_2$.
+Hence $T_B$ is shown to be a [[Definition:T4 Space|$T_4$ space]] as well.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Normal Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, Completely Normal Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:Completely Normal Space|completely normal space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+We have that $\struct {S, \tau}$ is a [[Definition:Completely Normal Space|completely normal space]] {{iff}}:
+:$\struct {S, \tau}$ is a [[Definition:T5 Space|$T_5$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+From [[T5 Space is Preserved under Homeomorphism|$T_5$ Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:T5 Space|$T_5$ space]], then so is $T_B$.
+From [[T1 Space is Preserved under Homeomorphism|$T_1$ Space is Preserved under Homeomorphism]]:
+:If $T_A$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], then so is $T_B$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T5 Space is Preserved under Homeomorphism}
+Tags: Homeomorphisms, T5 Spaces
+
+\begin{theorem}
+If $T_A$ is a [[Definition:T5 Space|$T_5$ space]], then so is $T_B$.
+\end{theorem}<|endoftext|>
+\section{Inverse of Homeomorphism is Homeomorphism}
+Tags: Homeomorphisms
+
+\begin{theorem}
+Let $T, T'$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T \to T'$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+Then $f^{-1}: T' \to T$ is also a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] is a [[Definition:Bijection|bijection]] such that both $f$ and $f^{-1}$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+As $f$ is a [[Definition:Bijection|bijection]] then by [[Bijection iff Inverse is Bijection]], so is $f^{-1}$.
+So by definition $f^{-1}$ is a [[Definition:Bijection|bijection]] such that both $f^{-1}$ and $\left({f^{-1}}\right)^{-1}$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+The result follows from [[Inverse of Inverse of Bijection]].
+{{qed}}
+[[Category:Homeomorphisms]]
+47c5e8g6whe2mnurj633c2axmzllsxc
+\end{proof}<|endoftext|>
+\section{T0 Space is Preserved under Closed Bijection}
+Tags: T0 Spaces
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Closed Mapping|closed]] [[Definition:Bijection|bijection]].
+If $T_A$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Let $T_A$ be a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+By [[Bijection is Open iff Closed]], $\phi$ is an [[Definition:Open Mapping|open]] [[Definition:Bijection|bijection]].
+By [[Bijection is Open iff Inverse is Continuous]], it follows that $\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By definition: 
+:$\forall x, y \in S_A$, either:
+::$\exists U \in \tau_A: x \in U, y \notin U$
+:or:
+::$\exists U \in \tau_A: y \in U, x \notin U$
+Suppose that:
+:$\exists x, y \in S_B: \forall V \in \tau_B: x, y \in V \lor x, y \notin V$
+That is:
+:$\exists x, y \in S_B: \forall V \in \tau_B: \set {x, y} \subseteq V \lor \set {x, y} \cap V = \O$
+From [[Image of Subset under Relation is Subset of Image/Corollary 3|Image of Subset under Relation is Subset of Image: Corollary 3]] it follows that:
+:$\forall V \in \tau_B: \phi^{-1} \sqbrk {\set {x, y} } \subseteq \phi^{-1} \sqbrk V \lor \phi^{-1} \sqbrk {\set {x, y} } \cap \phi^{-1} \sqbrk V = \O$
+that is:
+:$\forall V \in \tau_B: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in \phi^{-1} \sqbrk V \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin \phi^{-1} \sqbrk V$
+But by definition of [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]], $U = \phi^{-1} \sqbrk V$ is [[Definition:Open Set (Topology)|open]] in $T_A$ {{iff}} $V$ is [[Definition:Open Set (Topology)|open]] in $T_B$.
+Thus:
+:$\forall U \in \tau_A: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in U \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin U$
+This contradicts the condition that $T_A$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+It follows that $T_B$ must also be a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Hausdorff Space is Preserved under Closed Bijection}
+Tags: Completely Hausdorff Spaces
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Closed Mapping|closed]] [[Definition:Bijection|bijection]].
+If $T_A$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Let $T_A$ be a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+Then:
+:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A^- \cap V_A^- = \O$
+where $U_A^-$ signifies the [[Definition:Closure (Topology)|closure]] of $U_A$.
+Suppose that $T_B$ is not [[Definition:Completely Hausdorff Space|completely Hausdorff]].
+Then:
+:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B^- \cap V_B^- \ne \O$
+That is, there exists at least one pair of points $a$ and $b$ for which the [[Definition:Closure (Topology)|closures]] of all the [[Definition:Open Set (Topology)|open sets]] containing $a$ and $b$ are not [[Definition:Disjoint Sets|disjoint]].
+From [[Topological Closure is Closed]], both $U_B^-$ and $V_B^-$ are [[Definition:Closed Set (Topology)|closed]].
+From [[Topology Defined by Closed Sets]], it follows that $W_B = U_B^- \cap V_B^-$ is also a [[Definition:Closed Set (Topology)|closed set]].
+From [[Preimage of Intersection under Mapping]], we have:
+:$\phi^{-1} \sqbrk {U_B^- \cap V_B^-} = \phi^{-1} \sqbrk {U_B^-} \cap \phi^{-1} \sqbrk {V_B^-}$
+Let $H_A = \phi^{-1} \sqbrk {U_B^-}, K_A = \phi^{-1} \sqbrk {V_B^-}, W_A = \phi^{-1} \sqbrk {W_B}$.
+Then we have:
+:$H_A \cap K_A = W_A \ne \O$
+By [[Bijection is Open iff Closed]], $\phi$ is an [[Definition:Open Mapping|open]] [[Definition:Bijection|bijection]].
+By [[Bijection is Open iff Inverse is Continuous]], it follows that $\phi^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+From [[Continuity Defined from Closed Sets]], we have that $H_A$, $K_A$ and $W_A$ are all [[Definition:Closed Set (Topology)|closed]].
+Also, let $U_A = \phi^{-1} \sqbrk {U_B}$ and $V_A = \phi^{-1} \sqbrk {V_B}$.
+From [[Continuity Defined by Closure]], we have that:
+:$\phi^{-1} \sqbrk {U_B^-} \subseteq \paren {\phi^{-1} \sqbrk {U_B} }^-$
+:$\phi^{-1} \sqbrk {V_B^-} \subseteq \paren {\phi^{-1} \sqbrk {V_B} }^-$
+that is:
+:$H_A \subseteq U_A^-$
+:$K_A \subseteq V_A^-$
+As $H_A \cap K_A \ne \O$ it follows that $U_A^- \cap V_A^- \ne \O$.
+From the [[Definition:Bijection|bijective]] nature of $\phi$, we have that:
+:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
+:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
+Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
+Thus we have that:
+:$\exists x, y \in X_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A^- \cap V_A^- \ne \O$
+contradicting the fact that $T_A$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+Hence $T_B$ must after all be a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1 Space is Preserved under Closed Bijection}
+Tags: T1 Spaces
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Closed Mapping|closed]] [[Definition:Bijection|bijection]].
+If $T_A$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Let $T_A$ be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+By definition, all points in $T_A$ are [[Definition:Closed Point|closed]].
+Let $a \in S_A$.
+Then $\set a$ is a [[Definition:Closed Set (Topology)|closed set]].
+As $\phi$ is a [[Definition:Closed Mapping|closed mapping]] it follows directly that $\phi \sqbrk {\set a}$ is [[Definition:Closed Set (Topology)|closed]].
+As $\phi$ is a [[Definition:Bijection|bijection]] it follows that every point in $S_B$ is the [[Definition:Image of Element under Mapping|image]] under $\phi$ of a single point in $S_A$.
+Hence every point in $S_B$ is [[Definition:Closed Point|closed]].
+That is, $T_B$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T2 Space is Preserved under Closed Bijection}
+Tags: Hausdorff Spaces
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Closed Mapping|closed]] [[Definition:Bijection|bijection]].
+If $T_A$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]], then so is $T_B$.
+\end{theorem}
+
+\begin{proof}
+Let $T_A$ be a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Then:
+:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$
+Suppose that $T_B$ is not [[Definition:Hausdorff Space|Hausdorff]].
+Then:
+:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B \cap V_B \ne \O$
+That is, there exists at least one pair of points $a$ and $b$ for which all the [[Definition:Open Set (Topology)|open sets]] containing $a$ and $b$ are not [[Definition:Disjoint Sets|disjoint]].
+As $\tau_B$ is a [[Definition:Topology|topology]], it follows that $W_B = U_B \cap V_B$ is also an [[Definition:Open Set (Topology)|open set]].
+Let $U_A = \phi^{-1} \sqbrk {U_B}, V_A = \phi^{-1} \sqbrk {V_B}, W_A = \phi^{-1} \sqbrk {W_B}$.
+From [[Preimage of Intersection under Mapping]], we have:
+:$\phi^{-1} \sqbrk {U_B \cap V_B} = \phi^{-1} \sqbrk {U_B} \cap \phi^{-1} \sqbrk {V_B}$
+that is:
+:$U_A \cap V_A = W_A$
+By [[Bijection is Open iff Closed]], $\phi$ is an [[Definition:Open Mapping|open]] [[Definition:Bijection|bijection]].
+By [[Bijection is Open iff Inverse is Continuous]], it follows that $\phi^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+As $\phi$ is [[Definition:Continuous Mapping (Topology)|continuous]], all of $U_A, V_A, W_A$ are [[Definition:Open Set (Topology)|open in $T_A$]].
+From the [[Definition:Bijection|bijective]] nature of $\phi$, we have that:
+:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
+:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
+Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
+Thus we have that:
+:$\exists x, y \in S_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A \cap V_A \ne \O$
+contradicting the fact that $T_A$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Hence $T_B$ must after all be a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Properties Preserved by Expansion}
+Tags: Separation Axioms
+
+\begin{theorem}
+These [[Definition:Separation Axioms|separation properties]] are preserved under [[Definition:Expansion of Topology|expansion]]:
+:[[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) Space]]
+:[[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) Space]]
+:[[Definition:Hausdorff Space|$T_2$ (Hausdorff) Space]]
+:[[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (Completely Hausdorff) Space]]
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Set|set]].
+Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be [[Definition:Topological Space|topological spaces]] based on $S$ such that $\tau_2$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_1$.
+That is, let $\tau_1$ and $\tau_2$ be [[Definition:Topology|topologies]] on $S$ such that $\tau_1 \subseteq \tau_2$.
+Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the [[Definition:Identity Mapping|identity mapping]] from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$.
+From [[Identity Mapping to Expansion is Closed]], we have that $I_S$ is [[Definition:Closed Mapping|closed]].
+We also have [[Identity Mapping is Bijection]].
+So we can directly apply:
+:[[T0 Space is Preserved under Closed Bijection|$T_0$ (Kolmogorov) Space is Preserved under Closed Bijection]]
+:[[T1 Space is Preserved under Closed Bijection|$T_1$ (Fréchet) Space is Preserved under Closed Bijection]]
+:[[T2 Space is Preserved under Closed Bijection|$T_2$ (Hausdorff) Space is Preserved under Closed Bijection]]
+:[[Completely Hausdorff Space is Preserved under Closed Bijection|$T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Closed Bijection]]
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping to Expansion is Closed}
+Tags: Closed Mappings, Identity Mappings
+
+\begin{theorem}
+Let $S$ be a set on which $\tau_1$ and $\tau_2$ are [[Definition:Topology|topologies]] such that:
+:$\tau_1 \subseteq \tau_2$
+that is, such that $\tau_2$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_1$.
+Let $I_X: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the [[Definition:Identity Mapping|identity mapping]] from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$.
+Then $I_S$ is [[Definition:Closed Mapping|closed]].
+\end{theorem}
+
+\begin{proof}
+$\tau_1 \subseteq \tau_2$ means that every [[Definition:Open Set (Topology)|open set]] of $\struct {S, \tau_1}$ is also an [[Definition:Open Set (Topology)|open set]] of $\struct {S, \tau_2}$.
+Let $A \subseteq S$ be [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau_1}$
+Then by definition $S \setminus A$ is [[Definition:Open Set (Topology)|open]] in $\struct {S, \tau_1}$.
+Then $\map {I_S} {S \setminus A} = S \setminus A$ is [[Definition:Open Set (Topology)|open]] in $\struct {S, \tau_2}$.
+So, by definition, $\map {I_S} A = A$ is [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau_2}$.
+So for all $A \subseteq S$ [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau_1}$, it holds that $\map {I_S} A$ is [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau_2}$.
+So by definition of [[Definition:Closed Mapping|closed mapping]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Properties Preserved under Topological Product}
+Tags: Separation Axioms, Product Spaces
+
+\begin{theorem}
+Let $\mathbb S = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $\displaystyle T = \struct {S, \tau} = \prod_{i \mathop \in I} \struct{S_i, \tau_i}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\mathbb S$.
+Then $T$ has one of the following properties {{iff}} each of $\struct {S_i, \tau_i}$ has the same property:
+:[[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) Property]]
+:[[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) Property]]
+:[[Definition:Hausdorff Space|$T_2$ (Hausdorff) Property]]
+:[[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (Completely Hausdorff) Property]]
+:[[Definition:T3 Space|$T_3$ Property]]
+:[[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ Property]]
+If $T = \struct {S, \tau}$ has one of the following properties then each of $\struct {S_i, \tau_i}$ has the same property:
+:[[Definition:T4 Space|$T_4$ Property]]
+:[[Definition:T5 Space|$T_5$ Property]]
+but the [[Definition:Converse Statement|converse]] does not necessarily hold.
+\end{theorem}
+
+\begin{proof}
+:[[Product Space is T0 iff Factor Spaces are T0|Product Space is $T_0$ iff Factor Spaces are $T_0$]]
+:[[Product Space is T1 iff Factor Spaces are T1|Product Space is $T_1$ iff Factor Spaces are $T_1$]]
+:[[Product Space is T2 iff Factor Spaces are T2|Product Space is $T_2$ iff Factor Spaces are $T_2$]]
+:[[Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff]]
+:[[Product Space is T3 iff Factor Spaces are T3|Product Space is $T_3$ iff Factor Spaces are $T_3$]]
+:[[Product Space is T3 1/2 iff Factor Spaces are T3 1/2|Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$]]
+:[[Factor Spaces are T4 if Product Space is T4|Factor Spaces are $T_4$ if Product Space is $T_4$]]
+:[[Factor Spaces are T5 if Product Space is T5|Factor Spaces are $T_5$ if Product Space is $T_5$]]
+\end{proof}<|endoftext|>
+\section{Separation Properties Preserved in Subspace}
+Tags: Topological Subspaces, Separation Axioms
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T_H$ be a [[Definition:Topological Subspace|subspace]] of $T$.
+If $T$ has one of the following properties then $T_H$ has the same property:
+:[[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) Property]]
+:[[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) Property]]
+:[[Definition:Hausdorff Space|$T_2$ (Hausdorff) Property]]
+:[[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (Completely Hausdorff) Property]]
+:[[Definition:T3 Space|$T_3$ Property]]
+:[[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ Property]]
+:[[Definition:T5 Space|$T_5$ Property]]
+That is, the above properties are all [[Definition:Hereditary Property (Topology)|hereditary]].
+\end{theorem}
+
+\begin{proof}
+:[[T0 Property is Hereditary|$T_0$ Property is Hereditary]]
+:[[T1 Property is Hereditary|$T_1$ Property is Hereditary]]
+:[[T2 Property is Hereditary|$T_2$ Property is Hereditary]]
+:[[Completely Hausdorff Property is Hereditary]]
+:[[T3 Property is Hereditary|$T_3$ Property is Hereditary]]
+:[[T3 1/2 Property is Hereditary|$T_3 \frac 1 2$ Property is Hereditary]]
+:[[T5 Property is Hereditary|$T_5$ Property is Hereditary]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T4 Property Preserved in Closed Subspace}
+Tags: Closed Sets, T4 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T_K$ be a [[Definition:Topological Subspace|subspace]] of $T$ such that $K$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+If $T$ is a [[Definition:T4 Space|$T_4$ space]] then $T_K$ is also a [[Definition:T4 Space|$T_4$ space]].
+That is, the property of being a [[Definition:T4 Space|$T_4$ space]] is [[Definition:Weakly Hereditary Property|weakly hereditary]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:T4 Space|$T_4$ space]].
+Then:
+:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
+That is, for any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $A$ and $B$ respectively.
+We have that the set $\tau_K$ is defined as:
+:$\tau_K := \set {U \cap K: U \in \tau}$
+where $K$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Let $A, B \subseteq K$ be [[Definition:Closed Set (Topology)|closed in $K$]] such that $A \cap B = \O$.
+From [[Intersection with Subset is Subset]] we have that $A \subseteq K \iff A \cap K = A$.
+As $K$ is itself [[Definition:Closed Set (Topology)|closed in $T$]], it follows that so is $A \cap K = A$ from [[Topology Defined by Closed Sets]].
+Similarly, so is $B \cap K = B$.
+Because $T$ is a [[Definition:T4 Space|$T_4$ space]], we have that:
+:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
+As $A, B \subseteq K$ such that  we have that:
+:$A \subseteq U \cap K, B \subseteq V \cap K: \paren {U \cap K} \cap \paren {V \cap K} = \O$
+From the definition of [[Definition:Topological Subspace|topological subspace]], both $U \cap K$ and $V \cap K$ are [[Definition:Open Set (Topology)|open]] in $K$.
+Thus the [[Definition:T4 Space|$T_4$ axiom]] is satisfied in $K$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Normal iff Every Subspace is Normal}
+Tags: Completely Normal Spaces, Normal Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $T$ is a [[Definition:Completely Normal Space|completely normal space]] {{iff}} every [[Definition:Topological Subspace|subspace]] of $T$ is [[Definition:Normal Space|normal]].
+\end{theorem}
+
+\begin{proof}
+From the definitions, we have that:
+:$T$ is a [[Definition:Completely Normal Space|completely normal space]] {{iff}}:
+::$\struct {S, \tau}$ is a [[Definition:T5 Space|$T_5$ space]]
+::$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+:$T$ is a [[Definition:Normal Space|normal space]] {{iff}}:
+::$\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]]
+::$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+From [[T1 Property is Hereditary|$T_1$ Property is Hereditary]], any [[Definition:Topological Subspace|subspace]] of a [[Definition:Fréchet Space (Topology)|$T_1$ space]] is also a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+Then we have that a [[T5 Space iff Every Subspace is T4|space is $T_5$ iff every subspace is $T_4$]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tychonoff Space is Urysohn Space}
+Tags: Urysohn Spaces, Tychonoff Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Tychonoff Space|Tychonoff space]].
+Then $\struct {S, \tau}$ is also an [[Definition:Urysohn Space|Urysohn space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Tychonoff Space|Tychonoff space]].
+Let $x, y \in S: x \ne y$.
+From the definition of [[Definition:Tychonoff Space|Tychonoff space]]:
+:$\struct {S, \tau}$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+From [[Tychonoff Space is Regular, T2 and T1|Tychonoff Space is Regular, $T_2$ and $T_1$]] we use the fact that $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Clearly $\set x$ and $\set y$ are [[Definition:Disjoint Sets|disjoint]].
+From the definition of [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], we have that $\set x$ and $\set y$ are [[Definition:Closed Set (Topology)|closed]].
+So by definition of a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]], there exists an [[Definition:Urysohn Function|Urysohn function]] for $\set x$ and $\set y$.
+This is exactly the definition of an an [[Definition:Urysohn Space|Urysohn space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Urysohn Space is Completely Hausdorff Space}
+Tags: Completely Hausdorff Spaces, Urysohn Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be an [[Definition:Urysohn Space|Urysohn space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be an [[Definition:Urysohn Space|Urysohn space]].
+Then for any [[Definition:Distinct Elements|distinct points]] $x, y \in S$ (i.e. $x \ne y$), there exists an [[Definition:Urysohn Function|Urysohn function]] for $\set x$ and $\set y$.
+{{proof wanted|Then we do some stuff.}}
+Thus:
+:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
+which is precisely the definition of a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Perfectly Normal Space is Completely Normal Space}
+Tags: Completely Normal Spaces, Perfectly Normal Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Perfectly Normal Space|perfectly normal space]].
+Then $T$ is also a [[Definition:Completely Normal Space|completely normal space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Perfectly Normal Space|perfectly normal space]].
+From the [[Definition:Perfectly Normal Space|definition]]:
+:$T$ is a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]
+:$T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{ProofWanted|Don't know where to start at the moment. I'm tired, it's been a busy day.}}
+\end{proof}<|endoftext|>
+\section{T3 Space is Semiregular}
+Tags: T3 Spaces, Semiregular Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:T3 Space|$T_3$ space]].
+Then $T$ is a [[Definition:Semiregular Space|semiregular space]].
+\end{theorem}
+
+\begin{proof}
+Let $\BB = \set {B \subseteq S: B^{- \circ} = B}$.
+In other words, $\BB$ is the collection of [[Definition:Regular Open Set|regular open sets]] contained in $S$.
+It follows immediately from the definition of a [[Definition:Synthetic Basis|basis]] that our theorem is proved if we can show that:
+:$\BB$ is a [[Definition:Cover of Set|cover]] for $S$
+:$\forall U, V \in \BB: \forall x \in U \cap V: \exists W \in \BB: x \in W \subseteq U \cap V$
+First we show that $\BB$ is a [[Definition:Cover of Set|cover]] for $S$. 
+Since $S$ is [[Definition:Open Set (Topology)|open]]:
+:$S^\circ = S$
+Since $S$ is [[Definition:Closed Set (Topology)|closed]]:
+:$S^- = S$
+Therefore:
+:$S^{- \circ} = S$
+and so:
+:$S \in \BB$
+From [[Set is Subset of Union/General Result|Set is Subset of Union]]:
+:$S \subseteq \bigcup \BB$
+Thus $\BB$ [[Definition:Cover of Set|covers]] $S$.
+{{qed|lemma}}
+Next we demonstrate the second condition for $\BB$ to be a [[Definition:Basis (Topology)/Synthetic Basis/Definition 2|basis]].
+Let $U, V \in \BB$, $x \in U \cap V$.
+We will show that for some $W \in \BB$:
+:$x \in W \subseteq U \cap V$
+By [[General Intersection Property of Topological Space]], $U \cap V \in \tau$. 
+Since $T$ is [[Definition:T3 Space/Definition 2|$T_3$]], there is a [[Definition:Closed Neighborhood|closed neighborhood]] $N_x$ around $x$ that is contained in $U \cap V$:
+:$\exists N_x: \relcomp S {N_x} \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \paren {U \cap V}$
+Let $W := {N_x}^{- \circ}$. 
+Then by [[Interior of Closure is Regular Open]], $W^{- \circ} = W$ and $W \in \BB$.
+On the other hand, since $N_x$ is [[Definition:Closed Set (Topology)|closed]]:
+:${N_x}^- = N_x$, and thus $W = {N_x}^\circ$
+By definition of [[Definition:Interior (Topology)|interior]], we have $Q \subseteq W \subseteq N_x$.
+Therefore we have $x \in W \subseteq N_x \subseteq U \cap V$, as desired.
+It follows that $\BB$ is a [[Definition:Synthetic Basis|basis]] for $\tau$.
+Hence the result by definition of [[Definition:Semiregular Space|semiregular space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Regular Space is Semiregular Space}
+Tags: Semiregular Spaces, Regular Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Semiregular Space|semiregular space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+From the [[Definition:Regular Space|definition]]:
+:$\struct {S, \tau}$ is a [[Definition:T3 Space|$T_3$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+We also have that a [[T3 Space is Semiregular|$T_3$ Space is Semiregular]].
+Hence the result, by definition of [[Definition:Semiregular Space|semiregular space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Space is Sigma-Compact}
+Tags: Compact Spaces, Sigma-Compact Spaces
+
+\begin{theorem}
+Every [[Definition:Compact Topological Space|compact space]] is [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Sigma-Compact Space|$\sigma$-compact space]] is the [[Definition:Set Union|union]] of [[Definition:Countable Set|countably]] many [[Definition:Compact Topological Space|compact]] sets.
+A [[Definition:Compact Topological Space|compact space]] is the [[Definition:Set Union|union]] of exactly one [[Definition:Compact Topological Space|compact space]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sigma-Compact Space is Lindelöf}
+Tags: Sigma-Compact Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Every [[Definition:Sigma-Compact Space|$\sigma$-compact space]] is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+By definition:
+:$T$ is a [[Definition:Lindelöf Space|Lindelöf space]] {{iff}} every [[Definition:Open Cover|open cover]] of $X$ has a [[Definition:Countable Subcover|countable subcover]].
+By definition of [[Definition:Sigma-Compact Space|$\sigma$-compact space]], $T = \bigcap \TT$ where $\TT$ is the [[Definition:Set Union|union]] of [[Definition:Countable Set|countably]] many [[Definition:Compact Topological Subspace|compact subspaces]].
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Each [[Definition:Element|element]] of $\TT$ is [[Definition:Cover of Set|covered]] by a [[Definition:Finite Set|finite]] number of [[Definition:Element|elements]] of $\CC$.
+Hence $T$ is [[Definition:Cover of Set|covered]] by a [[Definition:Countable Union|countable union]] of a [[Definition:Finite Set|finite]] number of [[Definition:Element|elements]] of $\CC$.
+Hence $\CC$ has a [[Definition:Countable Subcover|countable subcover]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Space is Countably Compact}
+Tags: Sequentially Compact Spaces, Countably Compact Spaces
+
+\begin{theorem}
+A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Topological Space|topological space]] is also [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Topological Space|topological space]].
+By definition of [[Definition:Countably Compact Space/Definition 3|countably compact space]], it suffices to show that every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has an [[Definition:Accumulation Point of Sequence|accumulation point]] in $S$.
+By the definition of a [[Definition:Sequentially Compact Space|sequentially compact space]], every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Subsequence|subsequence]] which [[Definition:Convergent Sequence (Topology)|converges]] in $S$.
+The result follows from [[Limit Point of Sequence is Accumulation Point]].
+{{qed}}
+{{ACC|Equivalence of Definitions of Countably Compact Space}}
+\end{proof}<|endoftext|>
+\section{Countably Compact Space is Weakly Countably Compact}
+Tags: Countably Compact Spaces, Weakly Countably Compact Spaces
+
+\begin{theorem}
+Every [[Definition:Countably Compact Space|countably compact space]] is a [[Definition:Weakly Countably Compact Space|weakly countably compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact space]].
+By definition:
+:$T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Limit Point of Set|limit point]] in $S$.
+By definition of [[Definition:Countably Compact Space/Definition 5|countably compact space]]:
+:every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $S$.
+By definition, an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] is a [[Definition:Limit Point of Set|limit point]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1 Space is Weakly Countably Compact iff Countably Compact}
+Tags: T1 Spaces, Countably Compact Spaces, Weakly Countably Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) topological space]].
+Then $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} $T$ is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+We have that a [[Countably Compact Space is Weakly Countably Compact]] whatever the [[Definition:Separation Axioms|separation properties]].
+{{qed|lemma}}
+=== Necessary Condition ===
+We need to show that a [[Definition:Fréchet Space (Topology)|$T_1$ space]] which is [[Definition:Weakly Countably Compact Space|weakly countably compact]] is also necessarily [[Definition:Countably Compact Space|countably compact]].
+So, let $T = \struct {S, \tau}$ be a [[Definition:Fréchet Space (Topology)|$T_1$ space]] which is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+Let $A \subseteq S$.
+Suppose $x$ to be a [[Definition:Limit Point of Set|limit point]] but not an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $A$.
+Then there exists some [[Definition:Open Set (Topology)|open set]] $U_x \in \tau$ such that $x \in U_x$ which contains only a [[Definition:Finite Set|finite number]] of points of $A$, say:
+:$U_x = \set {a_1, a_2, \ldots, a_n}$
+But this, we have specified, is a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+So this would imply that $x$ has an [[Definition:Open Neighborhood|open neighborhood]] which contains no points of $A$.
+That is, $x$ is not a [[Definition:Limit Point of Set|limit point]] of $A$ after all.
+Hence if $x$ is a [[Definition:Limit Point of Set|limit point]] of $A$, it is an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $A$.
+Thus $S$ is [[Definition:Countably Compact Space|countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Compact Space is Pseudocompact}
+Tags: Countably Compact Spaces, Pseudocompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact space]].
+Then $T$ is a [[Definition:Pseudocompact Space|pseudocompact space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact space]].
+Then every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+By definition:
+:$T$ is [[Definition:Pseudocompact Space|pseudocompact]] {{iff}} every [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Real-Valued Function|real-valued function]] on $S$ is [[Definition:Bounded Real-Valued Function|bounded]].
+Let $f: S \to \R$ be such that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Consider the sets:
+:$S_n = \set {x \in S: \size {\map f x} < n}$
+$S_n$ forms a [[Definition:Countable Cover|countable cover]] of $S$ whose [[Definition:Finite Subcover|finite subcover]] yields a [[Definition:Bound of Real-Valued Function|bound]] for the [[Definition:Absolute Value|absolute value]] of $f$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pseudocompact Normal Space is Countably Compact}
+Tags: Countably Compact Spaces, Pseudocompact Spaces, Normal Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+Then $T$ is [[Definition:Pseudocompact Space|pseudocompact]] {{iff}} $T$ is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+By [[Countably Compact Space is Pseudocompact]] we already have that if $T$ is [[Definition:Countably Compact Space|countably compact]] then $T$ is [[Definition:Pseudocompact Space|pseudocompact]].
+It remains to prove that if $T$ is [[Definition:Pseudocompact Space|pseudocompact]] then $T$ is [[Definition:Countably Compact Space|countably compact]].
+{{AimForCont}} $T$ is not [[Definition:Countably Compact Space|countably compact]].
+By definition of [[Definition:Countably Compact Space/Definition 5|countably compact space]], $S$ contains an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] $H = \set {x_n}$ with no [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $T$.
+Since, by definition of [[Definition:Normal Space|normal]], $S$ is a [[Definition:T1 Space|$T_1$ space]], $H$ is [[Definition:Closed Set (Topology)|closed]] and [[Definition:Discrete Space|discrete]] in the [[Definition:Topological Subspace|subspace topology]].
+{{Explain|$H$ is [[Definition:Closed Set (Topology)|closed]] and [[Definition:Discrete Space|discrete]] in the [[Definition:Topological Subspace|subspace topology]].}}
+By definition, $T$ is [[Definition:Pseudocompact Space|pseudocompact]] {{iff}} every [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Real-Valued Function|real-valued function]] on $S$ is [[Definition:Bounded Real-Valued Function|bounded]].
+By definition of [[Definition:Normal Space|normal]], we have that $S$ is a [[Definition:T4 Space|$T_4$ space]].
+Hence the [[Tietze Extension Theorem]] guarantees a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Extension of Mapping|extension]] to $S$ of the [[Definition:Unbounded Mapping|unbounded]] [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] $f: H \to \R$ defined by $\map f {x_n} = n$.
+This [[Definition:Contradiction|contradicts]] our hypothesis that $S$ is [[Definition:Pseudocompact Space|pseudocompact]].
+It follows by [[Proof by Contradiction]] that $T$ is [[Definition:Countably Compact Space|countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Compact Lindelöf Space is Compact}
+Tags: Countably Compact Spaces, Compact Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Lindelöf Space|Lindelöf space]] which is also [[Definition:Countably Compact Space|countably compact]].
+Then $T$ is [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+By the definitions:
+:If $T = \struct {S, \tau}$ is a [[Definition:Lindelöf Space|Lindelöf space]] then every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+:If $T = \struct {S, \tau}$ is a [[Definition:Countably Compact Space|countably compact space]] then every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+It follows trivially that every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+Hence the result by definition of [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Space is Weakly Locally Compact}
+Tags: Compact Spaces, Weakly Locally Compact Spaces, Compact Space is Weakly Locally Compact
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Compact Topological Space|compact space]].
+Then $T$ is a [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be a [[Definition:Compact Topological Space|compact space]].
+By definition, $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]] {{iff}} each point of $T$ has a [[Definition:Compact Topological Space|compact]] [[Definition:Neighborhood of Point|neighborhood]] in $T$.
+Let $x \in S$ be an arbitrary point of $T$.
+By [[Space is Neighborhood of all its Points]], $T$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+Thus $x$ has a [[Definition:Compact Topological Space|compact]] [[Definition:Neighborhood of Point|neighborhood]] in $T$.
+Hence the result by definition of [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows directly from:
+:[[Compact Space is Weakly Sigma-Locally Compact|Compact Space is Weakly $\sigma$-Locally Compact]]
+:a [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space]] is [[Definition:Weakly Locally Compact Space|weakly locally compact]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completed Riemann Zeta Function has Order One}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+The [[Definition:Completed Riemann Zeta Function|completed Riemann zeta function]] $\xi$ has [[Definition:Order of Entire Function|order]] at most $1$.
+\end{theorem}
+
+\begin{proof}
+We are required to prove that:
+:$\xi \left({s}\right) = \dfrac 1 2 s \left({s - 1}\right) \pi^{-s/2} \Gamma \left({\dfrac s 2}\right) \zeta \left({s}\right) \ll \exp \left({|s|^\beta}\right)$
+for all $\beta > 1$, where $\ll$ is the [[Definition:Big-O Notation|order notation]].
+Note that by the [[Functional Equation for Riemann Zeta Function]], it is sufficient to check this for $\Re \left({s}\right) \ge \dfrac 1 2$.
+We simply check this fact for each factor.
+Evidently:
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 s \left({s - 1}\right) \pi^{-s/2}
+      | o = \ll
+      | r = \exp\left({2 \log s - \frac s 2 \log \pi}\right)
+}}
+{{eqn | o = \ll
+      | r = \exp\left({c_1 \left\vert{s}\right\vert}\right)
+}}
+{{end-eqn}}
+for some $c_1 > 0$.
+For the [[Definition:Gamma Function|gamma]] factor, we have [[Stirling's Formula for Gamma Function]]:
+:$\displaystyle \log \Gamma \left({s}\right) = \left({s - \frac 1 2}\right)\log s - s + \frac {\log 2 \pi} 2 + \sum_{n \mathop = 1}^{d-1} \frac{B_{2 n}}{ 2 n \left({2 n - 1}\right)s^{2 n-1}} + \mathcal O \left({s^{1 - 2 d} }\right)$
+This is valid only away from the [[Definition:Pole|poles]] of $\Gamma$ at $s = 0, -1, -2, \ldots$
+However, it is assumed that $\Re \left({s}\right) \ge \dfrac 1 2$, so this is not a problem.
+The error term $\mathcal O \left({s^{1 - 2d} }\right)$ is small for large $s$.
+More generally, the largest contribution is the term $\left({s - \dfrac 1 2}\right)\log s$, so we have:
+:$\log \Gamma\left({\dfrac s 2}\right) \ll \left\vert{s}\right\vert \log \left\vert{s}\right\vert$
+That is:
+:$\Gamma \left({\dfrac s 2}\right) \ll \exp \left({\left\vert{s}\right\vert \log \left\vert{s}\right\vert}\right)$
+Finally, from [[Integral Representation of Riemann Zeta Function in terms of Fractional Part]], for $\Re \left({s}\right) > \dfrac 1 2$:
+:$\displaystyle \zeta \left({s}\right) = \frac s {s-1} - s \int_1^\infty \left\{{x}\right\} x^{-s-1} \ \mathrm d x$
+It is seen that for $\Re \left({s}\right) > \dfrac 1 2$, the integral is bounded, and therefore:
+:$\left({1 - s}\right) \zeta \left({s}\right) \ll \mathcal O (\left\vert{s}\right\vert^2) \ll \exp \left({\left\vert{s}\right\vert}\right)$
+Combining these facts, and using that $\log s \ll s^\epsilon$ for all $\epsilon > 0$ (shown by [[Upper Bound of Natural Logarithm]]), we have:
+:$\left|{\xi \left({s}\right)}\right| \ll \exp\left({\left\vert{s}\right\vert^{1 + \epsilon} }\right)$ 
+for all $\epsilon > 0$, and the proof is complete.
+{{qed}}
+[[Category:Riemann Zeta Function]]
+jk1z74v284tro7oyhtwcf4lspax9183
+\end{proof}<|endoftext|>
+\section{Product Equation for Riemann Zeta Function}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+There exists a constant $B$ such that
+:$\displaystyle \frac{\zeta'(s)}{\zeta(s)} = B - \frac 1{s-1} +\frac12 \log \pi  - \frac12 \frac{\Gamma'(s/2+1)}{\Gamma(s/2+1)} + \sum_\rho \left( \frac 1{s-\rho} + \frac1\rho \right)$
+where $\zeta$ is the [[Definition:Riemann Zeta Function|Riemann zeta function]], $\rho$ runs over the non-trivial zeros of $\zeta$, and $\Gamma$ is the [[Definition:Gamma Function|gamma function]].
+\end{theorem}
+
+\begin{proof}
+Let $\xi$ be the [[Definition:Completed Riemann Zeta Function|completed Riemann zeta function]]
+:$\displaystyle \xi(s) = \frac 12 s(s-1) \pi^{-s/2} \Gamma\left(\frac s2\right) \zeta(s)$
+By [[Gamma Difference Equation]], this gives 
+:$\displaystyle \xi(s) = (s-1) \pi^{-s/2} \Gamma\left(\frac s2 + 1\right) \zeta(s)$
+So taking the logarithm,
+:$\displaystyle \log \xi(s) = \log (s - 1) - \frac s2 \log\pi + \log\Gamma\left(\frac s2 + 1\right) + \log \zeta(s)$
+And taking the derivative we obtain
+:$\displaystyle \frac{\xi'(s)}{\xi(s)} = \frac1{s-1} -\frac12 \log \pi + \frac{\Gamma'\left(s/2 + 1\right)}{\Gamma\left(s/2 + 1\right)} + \frac{\zeta'(s)}{\zeta(s)} \qquad (1)$
+We have that the [[Completed Riemann Zeta Function has Order One]], so by the [[Hadamard Factorisation Theorem]], there exist constants $A$, $B$ such that
+:$\displaystyle \xi(s) = \exp(A+Bs)\: \prod_\rho \left( 1 - \frac s{\rho} \right) \exp\left( \frac s{\rho} \right)$
+where $\rho$ runs over the zeros of $\xi$, that is, the nontrivial zeros of $\zeta$.
+Taking the logarithm, we obtain
+:$\displaystyle \log \xi(s) = A+Bs + \sum_\rho \left[ \log \left( 1 - \frac s{\rho} \right) + \frac s{\rho} \right]$
+which has derivative
+:$\displaystyle \frac{\xi'(s)}{\xi(s)} = B + \sum_\rho \left( \frac 1{s - \rho} + \frac 1\rho\right) \qquad (2)$
+Combining $(1)$ and $(2)$ we have
+:$\displaystyle B + \sum_\rho \left( \frac 1{s - \rho} + \frac 1\rho\right) =  \frac1{s-1} -\frac12 \log \pi + \frac{\Gamma'\left(s/2 + 1\right)}{\Gamma\left(s/2 + 1\right)} + \frac{\zeta'(s)}{\zeta(s)}$
+from which the result follows by rearranging the terms.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zeros of Functions of Finite Order}
+Tags: Entire Functions
+
+\begin{theorem}
+Let $\map f z$ be an [[Definition:Entire Function|entire function]] which satisfies:
+:$\map f 0 \ne 0$
+:$\cmod {\map f z} \ll \map \exp {\map \alpha {\cmod z} }$
+for all $z \in \C$ and some function $\alpha$, where $\ll$ is the [[Definition:Big-O Notation|order notation]].
+For $T \ge 1$, let:
+:$\map N T = \# \set {\rho \in \C: \map f r = 0, \ \cmod \rho < T}$
+where $\#$ denotes the [[Definition:Cardinality|cardinality]] of a set.
+Then:
+:$\map N T \ll \map \alpha {2 T}$
+\end{theorem}
+
+\begin{proof}
+Fix $T \ge 1$ and let $\rho_1, \rho_2, \ldots, \rho_n$ be an enumeration of the zeros of $f$ with modulus less than $T$, counted with [[Definition:Multiplicity (Complex Analysis)|multiplicity]].
+By [[Jensen's Formula]]:
+:$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln T - \ln \size {\rho_k} }$
+Let $\rho_0 = 1$, $\rho_{n + 1} = T$, $r_k = \size {\rho_k}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_0^T \map N t \frac {\d t} t
+      | r = \sum_{k \mathop = 0}^n \int_{r_k}^{r_{k + 1} } \map N t \frac {\d t} t
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n k \, \map \ln {\frac{r_{k + 1} } {r_k} }
+      | c = as by the definition of $N$, it is constant value $k$ on each interval $\openint {\size {\rho_k} }  {\size {\rho_{k + 1} } }$
+}}
+{{eqn | r = \map \ln {\frac {T^n} {r_1 \dotsm r_n} }
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {\ln T - \ln r_k}
+}}
+{{end-eqn}}
+and
+{{begin-eqn}}
+{{eqn | l = \int_0^T \map N t \frac {\d t} t
+      | r = \int_0^2 \map N {\frac {T \theta} 2} \frac {\d \theta} \theta
+}}
+{{eqn | o = \ge
+      | r = \map N {\frac T 2} \int_1^2 \frac {\d \theta} \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \map N {\frac T 2} \ln 2
+      | c = {{Defof|Logarithm}}
+}}
+{{end-eqn}}
+Moreover, by hypothesis we have that:
+:$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta \le \sup_{\theta \mathop \in \closedint 0 {2 \pi} } \ln \size {\map f {T e^{i \theta} } } \ll \map \alpha T$
+Putting these facts into [[Jensen's Formula]] we have:
+:$\map N {\dfrac T 2} \ln 2 + \cmod {\map f 0} \ll \map \alpha T$
+which implies:
+:$\map N T \ll \map \alpha {2 T}$
+{{qed}}
+=== Proof of Corollary ===
+Let $\epsilon > 0$, $\map N 0 = 0$, so that:
+:$\displaystyle \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le \sum_{T \mathop \ge 1} \paren {\map N T - \map N {T - 1} } T^{-1 - \epsilon}$
+We have $\map N T \ll 2 T$, so $\map N T - \map N {T - 1}$ is bounded in $T$, say by $C > 0$.
+Therefore:
+:$\displaystyle \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le C \ \sum_{T \mathop \ge 1} \frac 1 {T^{1 + \epsilon} }$
+and the sum on the {{RHS}} converges absolutely for $\epsilon > 0$.
+{{qed}}
+[[Category:Entire Functions]]
+jndqk6h2kh48t5fq924xgeqplpf53wv
+\end{proof}<|endoftext|>
+\section{Strongly Locally Compact Space is Weakly Locally Compact}
+Tags: Weakly Locally Compact Spaces, Strongly Locally Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Strongly Locally Compact Space|strongly locally compact space]].
+Then $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+Let $x \in S$.
+By definition, there exists an [[Definition:Open Set (Topology)|open set]] $U_x$ of $T$ such that:
+:$x \in U_x$
+:${U_x}^-$ (the [[Definition:Closure (Topology)|closure]] of $U_x$) is [[Definition:Compact Topological Subspace|compact]].
+From [[Set is Subset of its Topological Closure]], $U_x \subseteq {U_x}^-$ and so $x \in {U_x}^-$.
+Thus $x$ is contained in a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Point|neighborhood]].
+As this holds for all $x$, $T$ is a [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weakly Locally Compact Hausdorff Space is Strongly Locally Compact}
+Tags: Hausdorff Spaces, Weakly Locally Compact Spaces, Strongly Locally Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Let $T$ be [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+Then $T$ is [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+As $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]], $x$ is contained in a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Point|neighborhood]] $N_x$.
+As $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]], we can use the result [[Compact Subspace of Hausdorff Space is Closed]].
+Thus the [[Definition:Interior (Topology)|interior]] of $N_x$ has a [[Definition:Closure (Topology)|closure]] which is [[Definition:Compact Topological Subspace|compact]].
+Hence the result, from definition of [[Definition:Strongly Locally Compact Space|strongly locally compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weakly Sigma-Locally Compact iff Weakly Locally Compact and Lindelöf}
+Tags: Weakly Locally Compact Spaces, Weakly Sigma-Locally Compact Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+{{TFAE}}
+:$(1): \quad T$ is [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact]]
+:$(2): \quad T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]] and [[Definition:Lindelöf Space|Lindelöf]]
+\end{theorem}
+
+\begin{proof}
+=== 1 implies 2 ===
+Let $T = \struct {S, \tau}$ be [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact]].
+Then [[Definition:Weakly Sigma-Locally Compact Space|by definition]]:
+:$T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]]
+:$T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+From [[Sigma-Compact Space is Lindelöf]] it follows directly that:
+:$T$ is [[Definition:Lindelöf Space|Lindelöf]]
+:$T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed|lemma}}
+=== 2 implies 1 ===
+Now let $T = \struct {S, \tau}$ be [[Definition:Weakly Locally Compact Space|weakly locally compact]] and [[Definition:Lindelöf Space|Lindelöf]].
+By definition:
+:$T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]] {{iff}} every point of $S$ is contained in a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Set|neighborhood]].
+:$T$ is [[Definition:Lindelöf Space|Lindelöf]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+Thus the [[Definition:Interior (Topology)|interiors]] of the [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Set|neighborhoods]] are an [[Definition:Open Cover|open cover]] of $S$.
+As $T$ is [[Definition:Lindelöf Space|Lindelöf]], this cover has a [[Definition:Countable Subcover|countable subcover]].
+Thus $T$ is the [[Definition:Set Union|union]] of [[Definition:Countable Set|countably]] many [[Definition:Compact Topological Subspace|compact subspaces]].
+That is, $T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]], and so [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second-Countable Space is Lindelöf}
+Tags: Second-Countable Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Second-Countable Space|second-countable]].
+Then $T$ is also a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be [[Definition:Second-Countable Space|second-countable]].
+Then by [[Definition:Second-Countable Space|definition]] its [[Definition:Topology|topology]] has a [[Definition:Countable Basis|countable basis]].
+Let $\BB$ be this [[Definition:Countable Basis|countable basis]].
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Every set in $\CC$ is the [[Definition:Set Union|union]] of a subset of $\BB$.
+So $\CC$ itself is the [[Definition:Set Union|union]] of a subset of $\BB$.
+This [[Definition:Set Union|union]] of a subset of $\BB$ is therefore a [[Definition:Countable Subcover|countable subcover]] of $\CC$.
+That is, $T$ is by definition [[Definition:Lindelöf Space|Lindelöf]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separable Space satisfies Countable Chain Condition}
+Tags: Separable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Separable Space|separable space]].
+Then $T$ satisfies the [[Definition:Countable Chain Condition|countable chain condition]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Separable Space|separable space]].
+Then by [[Definition:Separable Space|definition]] there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $X$ which is [[Definition:Everywhere Dense|everywhere dense]].
+We need to show that every [[Definition:Disjoint Sets|disjoint set]] of [[Definition:Open Set (Topology)|open sets]] of $T$ is [[Definition:Countable Set|countable]].
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Logarithmic Derivative of Riemann Zeta Function}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]]:
+:$\displaystyle \forall s \in \C: \map \Re s > 1: \map \zeta s = \sum_{n \mathop \ge 1} n^{-s}$
+Then for all $s$ with $\map \Re s > 1$:
+:$\displaystyle -\frac {\map {\zeta'} s} {\map \zeta s} = \sum_{n \mathop \ge 1} \map \Lambda n n^{-s}$
+where $\Lambda$ is [[Definition:Von Mangoldt Function|von Mangoldt's function]].
+\end{theorem}
+
+\begin{proof}
+By [[Sum of Reciprocals of Powers as Euler Product]]:
+:$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
+where $p$ ranges over the [[Definition:Prime Number|primes]].
+From [[Laws of Logarithms]]:
+:$\displaystyle \ln \map \zeta s = - \sum_p \map \ln {1 - p^{-s} }$
+Taking the derivative:
+{{begin-eqn}}
+{{eqn | l = \frac {\map {\zeta'} s} {\map \zeta s}
+      | r = -\sum_p \frac {\paren {\ln p} p^{-s} } {1 - p^{-s} }
+      | c = [[Logarithmic Derivative of Infinite Product of Analytic Functions]]
+}}
+{{eqn | r = -\sum_p \paren {\ln p} \paren {\frac 1 {1 - p^{-s} } - 1}
+      | c = 
+}}
+{{eqn | r = -\sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{end-eqn}}
+Also, by the [[Definition:Von Mangoldt Function|definition of $\Lambda$]]:
+:$\displaystyle \sum_{n \mathop \ge 1} \map \Lambda n n^{-s} = \sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}$
+and the proof is complete.
+{{qed}}
+[[Category:Riemann Zeta Function]]
+7qdyq1f0w6xbecwodhq440lfqs8qgvh
+\end{proof}<|endoftext|>
+\section{Riemann Zeta Has No Zeros With Real Part One}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]].
+Then for all $t \in \R$, $\zeta(1 + it) \neq 0$
+\end{theorem}
+
+\begin{proof}
+Throughout, the complex variable $s$ is $s = \sigma + it$.
+We have, for $\sigma > 1$,
+{{begin-eqn}}
+{{eqn|l= -\frac{\zeta'(s)}{\zeta(s)}
+     |r= \sum_{n\geq 1} \Lambda(n) n^{-s}
+     |c= By [[Logarithmic Derivative of Riemann Zeta Function]]
+}}
+{{eqn|l= 
+     |r=\sum_{n\geq 1} \Lambda(n) n^{-\sigma} \exp\left( -it\log n\right)
+     |c=Where $s = \sigma + it$
+}}
+{{eqn|l= 
+     |r=\sum_{n\geq 1} \Lambda(n) n^{-\sigma} \left( \cos(t\log n) - i \sin(t\log n) \right)
+     |c=By [[Euler's Formula]]
+}}
+{{end-eqn}}
+Therefore,
+:$\displaystyle -\Re\left( \frac{\zeta'(s)}{\zeta(s)} \right) = \sum_{n\geq 1} \Lambda(n) n^{-\sigma} \cos(t\log n) \qquad (1)$
+Now observe that
+:$3 + 4\cos\theta + \cos(2\theta) = 2(1+ \cos\theta)^2 \geq 0$
+Because for all $n \geq 1$ we have $\Lambda(n)n^{-\sigma} \geq 0$, we have
+{{begin-eqn}}
+{{eqn|l=0
+     |r=\sum_{n\geq 1} \Lambda(n) n^{-\sigma} \left\{ 3 + 4\cos(t\log n) + \cos(2t\log n) \right\}
+     |c=
+     |o=\leq
+}}
+{{eqn|l=
+     |r=-\Re\left( 3\frac{\zeta'(\sigma)}{\zeta(\sigma)} + 4\frac{\zeta'(\sigma + it)}{\zeta(\sigma + it)} + \frac{\zeta'(\sigma + 2it)}{\zeta(\sigma + 2it)} \right)
+     |c=By $(1)$
+}}
+{{end-eqn}}
+Now let 
+:$\eta(s) = \zeta(s)^3\cdot \zeta(s+it)^4\cdot \zeta(s+2it)$
+Then the above computation has shown that
+:$\displaystyle\Re\left( \frac{\eta'(s)}{\eta(s)} \right) \leq 0$
+By [[Poles of Riemann Zeta Function]] we know that $\zeta$ has a [[Definition:Simple Pole|simple pole]] at $s=1$ with [[Definition:Residue (Complex Analysis)|residue]] $1$.
+Suppose that $1+it$ is a zero of $\zeta$ of [[Definition:Order of a Zero|order]] $d \geq 1$.
+Therefore, at $s = 1$, $\eta$ has a zero of order $4d - 3 \geq 0$, that is,
+:$\displaystyle \eta(s) \sim (s-1)^{4d -3}$
+as $s \to 1^+$, where $\sim$ indicates asymptotic equality, and superscript $+$ denotes a limit from the right along the real line. Therefore
+:$\displaystyle \frac{\eta'(s)}{\eta(s)} \sim \frac{4d -3}{s-1}$
+as $s \to 1^+$. Since $\displaystyle \Re\left( \frac{4d -3}{s-1}\right) \to + \infty$ as $s \to 1^+$, it follows that
+:$\displaystyle \Re\left(\frac{\eta'(s)}{\eta(s)}\right) \to \infty$
+as $s \to 1^+$. But we have already shown that
+:$\displaystyle\Re\left( \frac{\eta'(s)}{\eta(s)} \right) \leq 0$
+a contradiction.
+{{qed}}
+[[Category:Riemann Zeta Function]]
+fo6o84yn30601r2nq1x3org4ioua0l4
+\end{proof}<|endoftext|>
+\section{Compact Space is Countably Compact}
+Tags: Compact Spaces, Countably Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact space]].
+Then $T$ is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact space]].
+Then [[Definition:Compact Topological Space|by definition]] every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+So every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+Hence [[Definition:Countably Compact Space/Definition 1|by definition]] $T$ is [[Definition:Countably Compact Space|countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second-Countable Space is Compact iff Countably Compact}
+Tags: Compact Spaces, Countably Compact Spaces, Second-Countable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Second-Countable Space|second-countable space]].
+Then $T$ is [[Definition:Compact Topological Space|compact]] {{iff}} $T$ is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+We have that a [[Compact Space is Countably Compact|compact space is countably compact]] (whether the [[Definition:Topological Space|space]] is [[Definition:Second-Countable Space|second-countable]] or not).
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $T$ be [[Definition:Countably Compact Space|countably compact]].
+Since $T$ is [[Definition:Second-Countable Space|second-countable]], we can choose a [[Definition:Countable Basis|countable (analytic) basis]] $\BB$ for $\tau$.
+Let $\CC$ be an [[Definition:Open Cover|open cover]] for $S$.
+Define:
+:$\CC' = \set {B \in \BB: \exists C \in \CC: B \subseteq C}$
+Then $\CC' \subseteq \BB$, and so $\CC'$ is [[Definition:Countable Set|countable]].
+By the definition of an [[Definition:Analytic Basis|analytic basis]], we have $\BB \subseteq \tau$; since [[Subset Relation is Transitive|$\subseteq$ is a transitive relation]], it therefore follows that $\CC' \subseteq \tau$.
+From [[Equivalence of Definitions of Analytic Basis]]:
+:$\forall C \in \CC: \forall x \in C: \exists B \in \CC': x \in B$
+That is, by the definition of [[Definition:Union of Set of Sets|set union]]:
+:$\displaystyle \forall C \in \CC: C \subseteq \bigcup \CC'$
+Hence, by [[Union is Smallest Superset/General Result|Union is Smallest Superset]]:
+:$\displaystyle S \subseteq \bigcup \CC \subseteq \bigcup \CC'$
+Since [[Subset Relation is Transitive|$\subseteq$ is a transitive relation]], it follows that $\CC'$ is an [[Definition:Open Cover|open cover]] for $S$.
+Since $T$ is [[Definition:Countably Compact Space|countably compact]], there exists a [[Definition:Finite Subcover|finite subcover]] $\FF$ of $\CC'$ for $S$.
+By the [[Principle of Finite Choice]], there exists an [[Definition:Indexed Family|indexed family]] $\family {C_B}_{B \mathop \in \FF}$ such that:
+:$\forall B \in \FF: B \subseteq C_B \in \CC$
+Note that $\set {C_B: B \in \FF}$ is a [[Definition:Finite Subcover|finite subcover]] of $\CC$ for $X$.
+Hence, $T$ is [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsymmetric Functional Equation for Riemann Zeta Function}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]].
+Let $\Gamma$ be the [[Definition:Gamma Function|gamma function]].
+Then for all $s \in \C$,
+:$\displaystyle \map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \, \map \cos {\frac {\pi s} 2} \map \Gamma s \, \map \zeta s$
+\end{theorem}
+
+\begin{proof}
+We have for $s \notin \Z$ [[Euler's Reflection Formula]]:
+:$\map \Gamma s \, \map \Gamma {1 - s} = \dfrac \pi {\map \sin {\pi s} }$
+Replacing $s \mapsto \dfrac {1 + s} 2$ we deduce:
+{{begin-eqn}}
+{{eqn | l = \map \Gamma {\frac {1 + s} 2} \, \map \Gamma {\frac {1 - s} 2}
+      | r = \frac \pi {\map \sin {\pi \paren {1 + s} / 2} }
+      | c = substituting $s \mapsto \dfrac {1 + s} 2$
+}}
+{{eqn | r = \frac \pi {\map \cos {\pi s / 2} }
+      | c = [[Sine and Cosine are Periodic on Reals]]
+}}
+{{end-eqn}}
+Also, we have [[Legendre's Duplication Formula]] for $z \notin -\dfrac 1 2 \N_0$:
+:$\map \Gamma s \, \map \Gamma {s + \dfrac 1 2} = 2^{1 - 2 s} \sqrt \pi \map \Gamma {2 s}$
+Replacing $s \mapsto s / 2$ this yields:
+:$\map \Gamma {\dfrac s 2} \, \map \Gamma {\dfrac {1 + s } 2} = 2^{1 - s} \sqrt \pi \, \map \Gamma s$
+Together these give:
+:$(1): \quad \dfrac {\map \Gamma {s / 2} } {\map \Gamma {\paren {1 - s} / 2} } = 2^{1 - s} \pi^{-1/2} \, \map \Gamma s \, \map \cos {\pi s / 2}$
+Now we take the [[Functional Equation for Riemann Zeta Function]]:
+:$\pi^{-s/2} \, \map \zeta s \, \map \Gamma {s / 2} \, \map \Gamma {\dfrac {1 - s} 2}^{-1} = \pi^{\paren {s - 1} / 2} \, \map \zeta {1 - s}$
+and substitute $(1)$ to give:
+:$\pi^{\paren {s - 1} / 2} \, \map \zeta {1 - s} = \pi^{-\paren {s + 1} / 2} \, \map \zeta s 2^{1 - s} \, \map \Gamma s \, \map \cos {\pi s / 2}$
+Multiplying by $\pi^{\paren {s - 1} / 2}$ this becomes:
+:$\map \zeta {1 - s} = \pi^{-s} 2^{1 - s} \, \map \cos {\pi s / 2} \, \map \Gamma s \, \map \zeta s$
+as desired.
+{{qed}}
+[[Category:Riemann Zeta Function]]
+e0b4dqr0ckvuouvjabqz7z7qo1p7ry1
+\end{proof}<|endoftext|>
+\section{Countably Compact First-Countable Space is Sequentially Compact}
+Tags: Sequentially Compact Spaces, Countably Compact Spaces, First-Countable Spaces, Countably Compact First-Countable Space is Sequentially Compact
+
+\begin{theorem}
+A [[Definition:Countably Compact Space|countably compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]] is also [[Definition:Sequentially Compact Space|sequentially compact]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+: [[Infinite Sequence in Countably Compact Space has Accumulation Point]]
+: [[Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
+By definition of [[Definition:Sequentially Compact Space|sequentially compact]], it is sufficient to show that every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Convergent Sequence (Topology)|convergent]] [[Definition:Subsequence|subsequence]].
+Let $\sequence {s_n}$ be any [[Definition:Infinite Sequence|sequence]] in $S$.
+By [[Infinite Sequence in Countably Compact Space has Accumulation Point]], $\sequence {s_n}$ has an [[Definition:Accumulation Point of Sequence|accumulation point]] $p \in S$.
+As $T$ is [[Definition:First-Countable Space|first-countable]], $p$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]], say:
+:$\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$
+Then a [[Definition:Subsequence|subsequence]] $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, [[Definition:Convergent Sequence (Topology)|converges]] to $p$.
+{{handwaving|This last statement is "by intimidation"; justification needed}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue at Simple Pole}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: \C \to \C$ be a [[Definition:Complex Function|function]] [[Definition:Meromorphic Function|meromorphic]] on some [[Definition:Region (Complex Analysis)|region]], $D$, containing $a$. 
+Let $f$ have a [[Definition:Simple Pole|simple pole]] at $a$. 
+Then the [[Definition:Residue (Complex Analysis)|residue]] of $f$ at $a$ is given by:
+:$\displaystyle \operatorname{Res} \left({f, a}\right) = \lim_{z \mathop \to a} \left({z - a}\right) f \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+By [[Existence of Laurent Series]], there exists a [[Definition:Laurent Series|Laurent series]]:
+:$\displaystyle f\left({z}\right) = \sum_{n \mathop = -\infty}^\infty c_n \left({z - a}\right)^n$
+which is [[Definition:Convergent Series of Numbers|convergent]] in $D \setminus \left\{a\right\}$, where $\left({c_n}\right)$ is a doubly infinite sequence of complex coefficients. 
+We are given that $f$ has only a [[Definition:Simple Pole|simple pole]] at $a$.
+Thus $c_n = 0$ for $n < -1$.
+So we can write: 
+:$\displaystyle f\left({z}\right) = \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n + \frac {c_{-1} } {z - a}$
+Then:
+{{begin-eqn}}
+{{eqn	| l = \lim_{z \mathop \to a} \left({z - a}\right) f\left({z}\right)
+	| r = \lim_{z \mathop \to a} \left({z - a}\right) \left({ \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n + \frac {c_{-1} } {z - a} } \right)
+}}
+{{eqn	| r = \lim_{z \mathop \to a} \left({ \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^{n + 1} + c_{-1} }\right)
+}}
+{{eqn	| r = \sum_{n \mathop = 0}^\infty c_n \left({a - a}\right)^{n+1} + c_{-1}
+}}
+{{eqn	| r = 0 \sum_{n \mathop = 0}^\infty c_n + c_{-1} 
+}}
+{{eqn	| r = c_{-1}
+}}
+{{eqn	| r = \operatorname{Res} \left({f, a}\right)
+	| c = {{Defof|Residue (Complex Analysis)|Residue}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Complex Analysis]]
+0zekccwvg5l1wwblefgy8gxori8tg9q
+\end{proof}<|endoftext|>
+\section{Poles of Riemann Zeta Function}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]].
+Then $\zeta$ has a [[Definition:Simple Pole|simple pole]] at $s = 1$ with [[Definition:Residue (Complex Analysis)|residue]] $1$, and no other poles.
+\end{theorem}
+
+\begin{proof}
+By [[Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part]]:
+:$\displaystyle \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
+is [[Definition:Meromorphic Function|meromorphic]] for $\map \Re s > 0$, and the integral [[Definition:Convergence|converges]] to a finite value for fixed $s$ in this region.
+Therefore in this region the only pole of $\zeta$ is at $s = 1$, with residue:
+{{begin-eqn}}
+{{eqn | l = \Res \zeta 1
+      | r = \lim_{s \mathop \to 1} \paren {s - 1} \paren {\frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x}
+      | c = [[Residue at Simple Pole]]
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+By [[Unsymmetric Functional Equation for Riemann Zeta Function]]:
+:$\displaystyle \map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\frac 1 2 s \pi} \, \map \Gamma s \, \map \zeta s$
+Therefore, for $\map \Re s \le 0$:
+{{begin-eqn}}
+{{eqn | l = \map \zeta s
+      | r = \map \zeta {1 - t}
+      | c = for some $t$ with $\map \Re t \ge 1$
+}}
+{{eqn | r = 2^{1 - t} \pi^{-t} \map \cos {\frac 1 2 t \pi} ,\ \map \Gamma t \,\map \zeta t
+}}
+{{end-eqn}}
+By [[Complex Exponential Function is Entire]], the factor $\displaystyle 2^{1 - t} \pi^{-t}$ has no poles when $\map \Re t \ge 1$.
+By [[Poles of Gamma Function]], $\map \Gamma t$ has no poles when $\map \Re t \ge 1$.
+By [[Complex Cosine Function is Entire]], $\map \cos {\dfrac 1 2 t \pi}$ also has no poles in this region.
+Therefore, the only possible pole is a simple pole at $t = 1$ from the factor $\map \zeta t$.
+But at this point:
+:$\map \cos {\dfrac 1 2 t \pi} = \map \cos {\dfrac \pi 2} = 0$
+which cancels the pole of $\zeta$.
+{{qed}}
+[[Category:Riemann Zeta Function]]
+jsk325rcv2jct7le98xlkhivv8t9y96
+\end{proof}<|endoftext|>
+\section{Properties of Dirichlet Convolution}
+Tags: Dirichlet Convolution
+
+\begin{theorem}
+Let $f, g, h$ be [[Definition:Arithmetic Function|arithmetic functions]].
+Let $*$ denote [[Definition:Dirichlet Convolution|Dirichlet convolution]].
+Let $\iota$ be the [[Definition:Identity Arithmetic Function|identity arithmetic function]].
+Then the following properties hold:
+\end{theorem}<|endoftext|>
+\section{Ring of Arithmetic Functions is Ring with Unity}
+Tags: Arithmetic Functions, Examples of Rings
+
+\begin{theorem}
+Let $\AA$ be the set of all [[Definition:Arithmetic Function|arithmetic functions]].
+Let $*$ denote [[Definition:Dirichlet Convolution|Dirichlet convolution]], and $+$ the [[Definition:Pointwise Addition|pointwise sum]] of functions.
+The [[Definition:Ring of Arithmetic Functions|ring of arithmetic functions]] $\struct {\AA, +, *}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+\end{theorem}
+
+\begin{proof}
+By [[Structure Induced by Abelian Group Operation is Abelian Group]], $\struct {\AA, +}$ is an [[Definition:Abelian Group|abelian group]].
+By [[Properties of Dirichlet Convolution]], $*$ is [[Definition:Commutative Operation|commutative]], [[Definition:Associative|associative]] and has a [[Definition:Unity of Ring|unity]].
+Therefore $\struct {\AA, +, *}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+{{finish|link to a proof for distributive}}
+[[Category:Arithmetic Functions]]
+[[Category:Examples of Rings]]
+jy35b34ylhbf2lw3bll4jlig6smlf56
+\end{proof}<|endoftext|>
+\section{Units of Ring of Arithmetic Functions}
+Tags: Arithmetic Functions
+
+\begin{theorem}
+Let $f$ be an [[Definition:Arithmetic Function|arithmetic function]].
+Then $f$ is a [[Definition:Unit of Ring|unit]] in the [[Definition:Ring of Arithmetic Functions|ring of arithmetic functions]] {{iff}}:
+:$\map f 1 \ne 0$
+\end{theorem}
+
+\begin{proof}
+Follows immediately from [[Invertibility of Arithmetic Functions]] and the definition of [[Definition:Ring of Arithmetic Functions|ring of arithmetic functions]].
+{{qed}}
+[[Category:Arithmetic Functions]]
+ms9ywwi9foyyqyxreu2xt3go953vc5z
+\end{proof}<|endoftext|>
+\section{Möbius Inversion Formula}
+Tags: Dirichlet Convolution, Möbius Function
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Arithmetic Function|arithmetic functions]].
+Then:
+:$(1): \quad \displaystyle \map f n = \sum_{d \mathop \divides n} \map g d$
+{{iff}}:
+:$(2): \quad \displaystyle \map g n = \sum_{d \mathop \divides n} \map f d \, \map \mu {\frac n d}$
+where:
+: $d \divides n$ denotes that $d$ is a [[Definition:Divisor of Integer|divisor]] of $n$
+: $\mu$ is the [[Definition:Möbius Function|Möbius function]].
+\end{theorem}
+
+\begin{proof}
+Let $u$ be the [[Definition:Unit Arithmetic Function|unit arithmetic function]] and $\iota$ the [[Definition:Identity Arithmetic Function|identity arithmetic function]].
+Let $*$ denote [[Definition:Dirichlet Convolution|Dirichlet convolution]].
+Then equation $(1)$ states that $f = g * u$ and  $(2)$ states that $g = f * \mu$.
+The proof rests on the following facts:
+By the [[Sum of Möbius Function over Divisors/Lemma|lemma to Sum of Möbius Function over Divisors]]:
+:$\mu * u = \iota$
+By [[Properties of Dirichlet Convolution]], [[Definition:Dirichlet Convolution|Dirichlet convolution]] is [[Definition:Commutative Operation|commutative]], [[Definition:Associative|associative]] and $h * \iota = h$ for all $h$.
+We have:
+{{begin-eqn}}
+{{eqn | l = f = g * u
+      | o = \leadsto
+      | r = f * \mu = \paren {g * u} * \mu
+}}
+{{eqn | o = \leadsto
+      | r = f * \mu = g * \paren {u * \mu}
+}}
+{{eqn | o = \leadsto
+      | r = f * \mu = g
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = g = f * \mu
+      | o = \leadsto
+      | r = g * u = \paren {f * \mu} * u
+}}
+{{eqn | o = \leadsto
+      | r = g * u = f * \paren {\mu * u}
+}}
+{{eqn | o = \leadsto
+      | r = g * u = f
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+{{Namedfor|August Ferdinand Möbius|cat = Möbius}}
+\end{proof}<|endoftext|>
+\section{Fully Normal Space is Normal Space}
+Tags: Fully Normal Spaces, Normal Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Fully Normal Space|fully normal space]].
+Then $T$ is a [[Definition:Normal Space|normal space]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Fully Normal Space|fully normal]] {{iff}}:
+:$T$ is [[Definition:Fully T4 Space|fully $T_4$]]
+:$T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+We have that a [[Fully T4 Space is T4 Space|fully $T_4$ space is also a $T_4$ space]].
+So:
+:$T$ is a [[Definition:T4 Space|$T_4$ Space]]
+:$T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]]
+which is precisely the definition of a [[Definition:Normal Space|normal space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Paracompact Space is Countably Paracompact}
+Tags: Paracompact Spaces, Countably Paracompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Paracompact Space|paracompact space]].
+Then $T$ is a [[Definition:Countably Paracompact Space|countably paracompact space]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Paracompact Space|paracompact space]] {{iff}} every [[Definition:Open Cover|open cover]] of $T$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+This also applies to all [[Definition:Countable Set|countable]] [[Definition:Open Cover|open covers]].
+So every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $T$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+This is precisely the definition for a [[Definition:Countably Paracompact Space|countably paracompact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metacompact Space is Countably Metacompact}
+Tags: Metacompact Spaces, Countably Metacompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Metacompact Space|metacompact space]].
+Then $T$ is a [[Definition:Countably Metacompact Space|countably metacompact space]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Metacompact Space|metacompact space]] iff every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]].
+This also applies to all [[Definition:Countable Set|countable]] [[Definition:Open Cover|open covers]].
+So every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]].
+This is precisely the definition for a [[Definition:Countably Metacompact Space|countably metacompact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fully T4 Space is T4 Space}
+Tags: Fully T4 Spaces, T4 Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Fully T4 Space|fully $T_4$ space]].
+Then $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+Recall that $\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]] {{iff}}:
+:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
+That is, for any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq X$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $A$ and $B$ respectively.
+Consider the [[Definition:Open Cover|open cover]] $\UU = \set {\map \complement A, \map \complement B}$.
+Since $T$ is [[Definition:Fully T4 Space|fully $T_4$]], there exists a [[Definition:Star Refinement|star refinement]] $\VV$ of $\UU$.
+Define $\VV_A$ and $\VV_B$ by:
+:$\VV_A := \set {V \in \VV: A \cap V \ne \O}$
+:$\VV_B := \set {V \in \VV: B \cap V \ne \O}$
+and subsequently $U_A$ and $U_B$ by:
+:$U_A := \bigcup \VV_A$
+:$U_B := \bigcup \VV_B$
+Now given $a \in A$, since $\VV$ is a [[Definition:Open Cover|cover]], there exists $V \in \VV$ such that $a \in V$.
+But then $V \in \VV_A$, so that:
+:$a \in U_A$
+Hence $A \subseteq U_A$, and similarly $B \subseteq U_B$.
+{{AimForCont}} that $x \in U_A \cap U_B$.
+Then there exist $V_A \in \VV_A, V_B \in \VV_B$ such that:
+:$x \in V_A, x \in V_B$
+By definition of $\VV_A$ and $\VV_B$, there exist:
+:$a \in A \cap V_A$
+:$b \in B \cap V_B$
+Now recall the definition of the [[Definition:Star (Topology)|star]] $x^*$ of $x$:
+:$x^* := \bigcup \set {V \in \VV: x \in V}$
+Since $T$ is [[Definition:Fully T4 Space|fully $T_4$]], either $x^* \subseteq \map \complement A$ or $x^* \subseteq \map \complement B$.
+But $a \in V_A \subseteq x^*$, so $x^* \nsubseteq \map \complement A$.
+Similarly $b \in V_B \subseteq x^*$, so $x^* \nsubseteq \map \complement B$.
+From this [[Definition:Contradiction|contradiction]] it is inferred that $x \in U_A \cap U_B$ cannot exist.
+That is:
+:$U_A \cap U_B = \O$
+and it follows that $T$ is a [[Definition:T4 Space|$T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fully Normal Space is Paracompact}
+Tags: Fully Normal Spaces, Paracompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Fully Normal Space|fully normal space]].
+Then $T$ is [[Definition:Paracompact Space|paracompact]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Fully Normal Space|fully normal]] {{iff}}:
+:$T$ is [[Definition:Fully T4 Space|fully $T_4$]]
+:$T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Then $T$ is [[Definition:Fully T4 Space|fully $T_4$]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Star Refinement|star refinement]].
+Let $\UU$ be an [[Definition:Open Cover|open cover]] for $T$.
+Then from the definition, there exists a [[Definition:Cover of Set|cover]] $\VV$ for $T$ such that:
+:$\displaystyle \forall x \in S: \exists U \in \UU: \paren {\bigcup \set {V \in \VV: x \in V} } \subseteq U$
+Recall from the definition of [[Definition:Paracompact Space|paracompact]]:
+:$T$ is [[Definition:Paracompact Space|paracompact]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+{{proof wanted|It remains to fill in the gap.}}
+\end{proof}<|endoftext|>
+\section{Subcover is Refinement of Cover}
+Tags: Covers
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\mathcal C$ be a [[Definition:Cover of Set|cover]] for $S$.
+Let $\mathcal V$ be a [[Definition:Subcover|subcover]] of $\mathcal C$.
+Then $\mathcal V$ is a [[Definition:Refinement of Cover|refinement]] of $\mathcal C$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Subcover|subcover]]:
+: $\mathcal V \subseteq \mathcal C$
+That is, every element of $\mathcal V$ is an element of $\mathcal C$.
+From the definition of [[Definition:Subset|subset]], every element of $\mathcal V$ is the [[Definition:Subset|subset]] of some element of $\mathcal C$.
+This is precisely the definition of [[Definition:Refinement of Cover|refinement]].
+{{qed}}
+[[Category:Covers]]
+jctjtgc73oevd6f4pktjiwyjrpblx12
+\end{proof}<|endoftext|>
+\section{Compact Space is Paracompact}
+Tags: Compact Spaces, Paracompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact space]].
+Then $T$ is [[Definition:Paracompact Space|paracompact]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Compact Topological Space|compact]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+From [[Subcover is Refinement of Cover]], it follows that every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+This is precisely the definition of [[Definition:Paracompact Space|paracompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Paracompact Space is Metacompact}
+Tags: Paracompact Spaces, Metacompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Paracompact Space|paracompact space]].
+Then $T$ is also [[Definition:Metacompact Space|metacompact]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Paracompact Space|paracompact]] {{iff}} every [[Definition:Open Cover|open cover]] of $T$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+Consider some [[Definition:Open Cover|open cover]] $\UU$ of $T$.
+Let $x \in S$.
+Then there exists some [[Definition:Neighborhood of Point|neighborhood]] $\exists N_x$ of $x$ which intersects only [[Definition:Finite Set|finitely many]] [[Definition:Element|elements]] of $\UU$.
+Thus $x$ itself can be in only [[Definition:Finite Set|finitely many]] elements of $\UU$.
+Hence $T$ must be, by definition, [[Definition:Metacompact Space|metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Paracompact Space is Countably Metacompact}
+Tags: Countably Paracompact Spaces, Countably Metacompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Paracompact Space|countably paracompact space]].
+Then $T$ is [[Definition:Countably Metacompact Space|countably metacompact]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Countably Paracompact Space|countably paracompact]] {{iff}} every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $X$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+Consider some [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] $\mathcal U$ of $X$.
+Let $x \in X$.
+Then there exists some [[Definition:Neighborhood of Point|neighborhood]] $\N_x$ of $x$ which intersects only [[Definition:Finite Set|finitely many]] [[Definition:Element|elements]] of $\mathcal U$.
+Thus $x$ itself can be in only [[Definition:Finite Set|finitely many]] [[Definition:Element|elements]] of $\mathcal U$.
+Hence $T$ must be, by definition, [[Definition:Countably Metacompact Space|countably metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Compact Space is Countably Paracompact}
+Tags: Countably Compact Spaces, Countably Paracompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact space]].
+Then $T$ is [[Definition:Countably Paracompact Space|countably paracompact]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T$ is [[Definition:Countably Compact Space|countably compact]] {{iff}} every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+From [[Subcover is Refinement of Cover]], it follows that every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+This is precisely the definition of [[Definition:Countably Paracompact Space|countably paracompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metacompact Countably Compact Space is Compact}
+Tags: Compact Spaces, Countably Compact Spaces, Metacompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact space]] which is also [[Definition:Metacompact Space|metacompact]].
+Then $T$ is [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Countably Compact Space|countably compact space]], every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Finite Subcover|finite subcover]].
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact space]] which is also [[Definition:Metacompact Space|metacompact]].
+Let $\UU_\alpha$ be an [[Definition:Open Cover|open cover]] of $S$.
+Then let $\VV_\beta$ be the [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]], guaranteed by its [[Definition:Metacompact Space|metacompactness]].
+Let $x \in S$.
+We have that $x$ is an [[Definition:Element|element]] of only a [[Definition:Finite Set|finite number]] of the [[Definition:Element|elements]] of $\VV_\beta$, as $\VV_\beta$ is [[Definition:Point Finite|point finite]].
+Consider all the [[Definition:Subcover|subcovers]] of $\VV_\beta$, and [[Definition:Ordering|order]] them by [[Definition:Subset|subset]].
+Consider any [[Definition:Chain (Set Theory)|chain]] $\CC$ of such [[Definition:Subcover|subcovers]].
+Suppose $x$ is not in the [[Definition:Set Intersection|intersection]] of $\CC$.
+Then it would fail to be [[Definition:Cover of Set|covered]] by one of the [[Definition:Element|elements]] of $\CC$.
+This would be a [[Definition:Contradiction|contradiction]] of the fact that every [[Definition:Element|element]] of $\CC$ is a [[Definition:Cover of Set|cover]] of $\VV_\beta$. 
+So the [[Definition:Set Intersection|intersection]] of a [[Definition:Chain (Set Theory)|chain]] of such [[Definition:Subcover|subcovers]] is itself a [[Definition:Subcover|subcover]].
+Hence it must be that $\VV$ has a [[Definition:Minimal Set|minimal]] [[Definition:Subcover|subcover]], $\VV_\gamma$, say.
+Now each element $V_\gamma$ of $\VV_\gamma$ must itself contain a unique element $x_\gamma \in V_\gamma$ which belongs to no element of $\VV_\gamma$.
+This is because $\VV_\gamma$ is minimal.
+If $\VV_\gamma$ were [[Definition:Infinite|infinite]], the set of all $\set {x_\gamma}$ would be an infinite set without an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]].
+This can not be the case.
+{{explain|Why can this not be the case?}}
+Thus $\VV_\gamma$ is a [[Definition:Finite Cover|finite cover]].
+So we have constructed a [[Definition:Finite Subcover|finite subcover]] for $\UU_\alpha$, demonstrating that $T$ is [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Metacompact Lindelöf Space is Metacompact}
+Tags: Countably Metacompact Spaces, Metacompact Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Lindelöf Space|Lindelöf space]] which is also [[Definition:Countably Metacompact Space|countably metacompact]].
+Then $T$ is [[Definition:Metacompact Space|metacompact]].
+\end{theorem}
+
+\begin{proof}
+By the definitions:
+:If $T = \struct {S, \tau}$ is a [[Definition:Lindelöf Space|Lindelöf space]] then every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+:If $T = \struct {S, \tau}$ is a [[Definition:Countably Metacompact Space|countably metacompact space]] then every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]].
+It follows trivially that every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]].
+Hence the result by definition of [[Definition:Metacompact Space|metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Paracompact Lindelöf Space is Paracompact}
+Tags: Countably Paracompact Spaces, Paracompact Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Lindelöf Space|Lindelöf space]] which is also [[Definition:Countably Paracompact Space|countably paracompact]].
+Then $T$ is [[Definition:Paracompact Space|paracompact]].
+\end{theorem}
+
+\begin{proof}
+By the definitions:
+:If $T = \struct {S, \tau}$ is a [[Definition:Lindelöf Space|Lindelöf space]] then every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+:If $T = \struct {S, \tau}$ is a [[Definition:Countably Paracompact Space|countably paracompact space]] then every [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+It follows trivially that every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Locally Finite Cover|locally finite]].
+Hence the result by definition of [[Definition:Paracompact Space|paracompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dirichlet's Approximation Theorem}
+Tags: Diophantine Equations
+
+\begin{theorem}
+Let $\alpha, x \in \R$.
+Then there exist [[Definition:Integer|integers]] $a, q$ such that:
+: $\gcd \left\{{a, q}\right\} = 1$, $1 \le q \le x$
+and:
+:$\left|{\alpha - \dfrac a q}\right| \le \dfrac 1 {q x}$
+\end{theorem}
+
+\begin{proof}
+It is sufficient to find $a, q$ not necessarily [[Definition:Coprime Integers|coprime]].
+Once such an $a, q$ have been found, then a [[Definition:Coprime Integers|coprime]] pair can be found by dividing [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] of $\dfrac a q$ by the [[Definition:Greatest Common Divisor of Integers|GCD]] of $a$ and $q$.
+Let $X = \left\lfloor{x}\right\rfloor$ be the [[Definition:Floor Function|integer part]] of $x$.
+Let $\alpha_q = \alpha q - \left\lfloor{\alpha q}\right\rfloor \in \left[{0 \,.\,.\, 1}\right)$ for $q = 1, \ldots, X$.
+Consider the [[Definition:Pairwise Disjoint|disjoint]] [[Definition:Real Interval|real intervals]]:
+:$I_k = \left[{\dfrac{k-1} {X + 1} \,.\,.\, \dfrac k {X+1}} \right), \quad k=1,\ldots, X + 1$
+Suppose there exists $q$ such that:
+:$\alpha_q \in \left[{0 \,.\,.\, \dfrac 1{X+1}} \right)$
+Then:
+:$0 \le \alpha q - \left\lfloor{\alpha q}\right\rfloor < \dfrac 1 {X+1}$
+Hence:
+:$\left|{\alpha - \dfrac{\left\lfloor{\alpha q}\right\rfloor} q}\right| < \dfrac 1 {q \left({X + 1}\right)} < \dfrac 1 {q x}$
+Taking $a = \left\lfloor{\alpha q}\right\rfloor$, the construction is complete.
+{{qed|lemma}}
+Suppose that not to be the case, but suppose that there exists $q$ such that:
+: $\alpha_q \in \left[{\dfrac X {X + 1} \,.\,.\, 1}\right)$
+Then similarly:
+:$\dfrac X {X+1} < \alpha q - \left\lfloor{\alpha q}\right\rfloor < 1$
+so:
+:$-\dfrac 1 {X+1} < \alpha q - \left\lfloor{\alpha q}\right\rfloor - 1 < 0$
+and:
+:$\left|{\alpha  - \dfrac {\left\lfloor{\alpha q}\right\rfloor + 1} q}\right| < \dfrac 1 {q \left({X+1}\right)} < \dfrac 1 {q x}$
+Taking $a = \left\lfloor{\alpha q}\right\rfloor + 1$, again the construction is complete.
+{{qed|lemma}}
+If neither of the above cases holds, then the $X - 1$ remaining intervals $I_k$ contain the $X$ values of $\alpha_q$.
+Therefore, by the [[Pigeonhole Principle]], there are $k_0 \in \left\{{2, \ldots, X}\right\}$ and $q_1 < q_2$ such that: :$\alpha_{q_1}, \alpha_{q_2} \in I_{k_0}$
+Then for $i = 1, 2$:
+:$\dfrac{k_0 - 1} {X + 1} \le \alpha q_i - \left\lfloor{\alpha q_i}\right\rfloor < \dfrac {k_0} {X+1}$
+Therefore:
+:$\left|{\alpha q_2 - \alpha q_1 - \left\lfloor{\alpha q_2}\right\rfloor + \left\lfloor{\alpha q_1}\right\rfloor}\right| < \dfrac 1 {X + 1}$
+and therefore:
+:$\left|{\alpha - \dfrac{\left\lfloor{\alpha q_2}\right\rfloor - \left\lfloor{\alpha q_1}\right\rfloor} {q_2 - q_1}}\right| < \dfrac 1 {\left({X + 1}\right) \left({q_2 - q_1}\right)} < \dfrac 1 {x \left({q_2 - q_1}\right)}$
+Taking $a = \left\lfloor{\alpha q_2}\right\rfloor - \left\lfloor{\alpha q_1}\right\rfloor$ and $q = q_2 - q_1$ the construction is complete.
+{{qed}}
+{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
+[[Category:Diophantine Equations]]
+1tm1fhvu4mf49qcj6b5cwklvhc05tit
+\end{proof}<|endoftext|>
+\section{Convergence of Dirichlet Series with Bounded Partial Sums}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\left\langle{a_n}\right\rangle_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $\C$.
+Suppose that there exists $B > 0$ such that for all $n, m \in \N$:
+:$\displaystyle \left\vert{\sum_{k \mathop = m}^n a_n}\right\vert \le B$
+Then the [[Definition:Dirichlet Series|Dirichlet series]]:
+: $\displaystyle f \left({s}\right) = \sum_{n \mathop \ge 1} a_n n^{-s}$
+[[Definition:Locally Uniform Convergence|converges locally uniformly]] to an [[Definition:Analytic Function|analytic function]] on $\Re \left({s}\right) > 0$.
+\end{theorem}
+
+\begin{proof}
+By [[Exponential is Entire]], the [[Definition:Partial Sum|partial sums]]:
+:$\displaystyle f_N \left({s}\right) = \sum_{n \mathop = 1}^N a_n n^{-s}$
+are [[Definition:Analytic Function|analytic]].
+So by [[Uniform Limit of Analytic Functions is Analytic]] it is sufficient to show [[Definition:Locally Uniform Convergence|locally uniform convergence]].
+For $0 < A < \pi / 2$, $\delta > 0$ we let:
+:$D_{A, \delta} = \left\{{s \in \C : -A < \arg \left({s}\right) < A,\ \Re \left({s}\right) > \delta}\right\}$
+Then for any $s \in \C$ such that $\Re \left({s}\right) > 0$, we can choose $A$, $\delta$ such that $s \in D_{A,\delta}$.
+So it is sufficient to prove [[Definition:Locally Uniform Convergence|locally uniform convergence]] in this region.
+Also note that if $\lambda_n = \log n$, then:
+:$\displaystyle f \left({s}\right) = \sum_{n \mathop \ge 1} a_n e^{-\lambda_n s}$
+By [[Abel's Lemma: Formulation 2]], for $N, M \in \N$ we have:
+:$\displaystyle \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} = \sum_{n \mathop = M}^{N - 1} \left({\sum_{k \mathop = M}^n a_n}\right) \left[{e^{-\lambda_n s} - e^{-\lambda_{n + 1} s}}\right] + e^{-\lambda_N s} \left({\sum_{n \mathop = M}^N a_n}\right)$
+Let $\epsilon > 0$ be arbitrary, and choose $N_0 \in \N$ such that:
+:$\displaystyle \forall n \ge N_0: \left|{e^{-\lambda_n s}}\right| < \epsilon$
+Now for $N, M \ge N_0$, by the above we obtain:
+:$(1): \quad \displaystyle \left\vert \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} \right\vert \le B \sum_{n \mathop = M}^{N - 1} \left\vert e^{-\lambda_n s} - e^{-\lambda_{n+1} s} \right\vert + B \epsilon$
+For any $\alpha, \beta \in \R$ we have:
+{{begin-eqn}}
+{{eqn | l = \left\vert{e^{-\alpha s} - e^{-\beta s} }\right\vert
+      | r = \left\vert{s \int_\alpha^\beta e^{-x s} \ \mathrm d x}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{s}\right\vert \int_\alpha^\beta e^{-x \sigma} \ \mathrm d x
+      | c = where $\sigma = \Re \left({s}\right)$
+}}
+{{eqn | r = \frac{\left\vert{s}\right\vert} \sigma \left({e^{-\alpha \sigma} - e^{-\beta \sigma} }\right)
+      | c = 
+}}
+{{end-eqn}}
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \left\vert{\sum_{n \mathop = M}^N a_n e^{-\lambda_n s} }\right\vert
+      | o = \le
+      | r = \frac {B \left\vert{s}\right\vert} \sigma \sum_{n \mathop = M}^{N - 1} \left({e^{-\lambda_n \sigma} - e^{-\lambda_{n + 1} \sigma} }\right) + B \epsilon
+      | c = by $(1)$
+}}
+{{eqn | r = \frac {B \left\vert{s}\right\vert} \sigma \left({e^{-\lambda_N \sigma} - e^{-\lambda_M \sigma} }\right) + B \epsilon
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac {B \epsilon \left\vert{s}\right\vert} \sigma + B \epsilon
+      | c = 
+}}
+{{end-eqn}}
+Finally letting $\Im \left({s}\right) = t$:
+:$\dfrac{\left\vert{s}\right\vert} \sigma = \dfrac{\sqrt{\sigma^2 + t^2} } \sigma = \sqrt{1 + \dfrac {t^2} {\sigma^2} }$
+and:
+:$\dfrac {\pi} 2 > A \ge \arg \left({s}\right) = \arctan \left({\dfrac t \sigma}\right)$
+so $\dfrac t \sigma < \tan A$ is [[Definition:Bounded Uniformly|bounded uniformly]] in $D_{A, \delta}$.
+Thus letting $N \to \infty$, we have shown that:
+:$\displaystyle \left\vert{\sum_{n \mathop = M}^\infty a_n e^{-\lambda_n s} }\right\vert \to 0$
+as $M \to \infty$ [[Definition:Locally Uniform Convergence|uniformly]] in $D_{A, \delta}$.
+{{qed}}
+[[Category:Analytic Number Theory]]
+lqwt0v3yzx14ccjck17qzdu4uz9jwvo
+\end{proof}<|endoftext|>
+\section{Convergence of Dirichlet Series with Bounded Coefficients}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\left\langle{a_n}\right\rangle_{n \mathop \in \N}$ be a [[Definition:Bounded Sequence|bounded sequence]] in $\C$.
+Then the [[Definition:Dirichlet Series|Dirichlet series]]:
+:$\displaystyle f \left({s}\right) = \sum_{n \mathop \ge 1} a_n n^{-s}$
+[[Definition:Absolutely Convergent Series|converges absolutely]] and [[Definition:Locally Uniform Convergence|locally uniformly]] to an [[Definition:Analytic Function|analytic function]] on $\Re \left({s}\right) > 1$.
+\end{theorem}
+
+\begin{proof}
+By [[Exponential is Entire]], the [[Definition:Partial Sum|partial sums]]:
+:$\displaystyle f_N \left({s}\right) = \sum_{n \mathop = 1}^N a_n n^{-s}$
+are [[Definition:Analytic Function|analytic]].
+So by [[Uniform Limit of Analytic Functions is Analytic]] it is sufficient to show [[Definition:Locally Uniform Convergence|locally uniform convergence]].
+Let $B$ be a [[Definition:Bound of Sequence|bound]] for the $a_n$:
+:$\forall n \in \N: \left\vert{a_n}\right\vert \le B$
+Let $D$ be any [[Definition:Open Set (Complex Analysis)|open subset]] of $\Re \left({s}\right) > 1$.
+So for some $\kappa > 0$:
+:$\forall s \in D: \Re \left({s}\right) \ge 1 + \kappa$
+Now:
+{{begin-eqn}}
+{{eqn | l = \left\vert{f_N \left({s}\right)}\right\vert
+      | o = \le
+      | r = \sum_{n \mathop = 1}^N \left\vert{a_n}\right\vert \left\vert{n^{-s} }\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = B \sum_{n \mathop = 1}^N \frac 1 {n^{1 + \kappa} }
+      | c = 
+}}
+{{end-eqn}}
+which we know to be finite (!).
+{{MissingLinks|I would imagine the statement (!) is proved somewhere, but can't think what it'd be called}}
+[[Category:Analytic Number Theory]]
+jeq3skah0g8mnus3u7r39tc3pmc1d7q
+\end{proof}<|endoftext|>
+\section{Dirichlet L-Function from Trivial Character}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\chi_0$ be the [[Definition:Trivial Character|trivial]] [[Definition:Dirichlet Character|Dirichlet character]] modulo $q$.
+{{explain|Trivial character we got, Dirichlet character we got, we still need a page for trivial Dirichlet character. There exists on another page a link to [[Definition:Dirichlet Character/Trivial Character]] which ought to be straightforward to construct.}}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]].
+Then:
+:$\displaystyle \map L {s, \chi_0} = \map \zeta s \cdot \prod_{p \mathop \divides q} \paren {1 - p^{-s} }$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\map {\chi_0} a = \begin{cases} 1 & : \gcd \set {a, q} = 1 \\ 0 & : \text{otherwise} \end{cases}$
+{{explain|Worth specifying the domain of $a$ for added clarity here.}}
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \map L {s, \chi_0}
+      | r = \prod_p \paren {1 - \map \chi p p^{-s} }^{-1}
+      | c = [[Euler Product]]: $p$ ranges over the [[Definition:Prime Number|primes]]
+}}
+{{eqn | r = \prod_{p \mathop \nmid q} \paren {1 - p^{-s} }^{-1}
+      | c = [[Fundamental Theorem of Arithmetic]]: $\gcd \set {p, q} > 1$ {{iff}} $p \divides q$
+}}
+{{eqn | r = \prod_{p \mathop \divides q} \paren {1 - p^{-s} } \prod_p \paren {1 - p^{-s} }^{-1}
+      | c =
+}}
+{{eqn | r = \map \zeta s \prod_{p \mathop \divides q} \paren {1 - p^{-s} }
+      | c = [[Euler Product]]
+}}
+{{end-eqn}}
+{{explain|Not sure whether [[Fundamental Theorem of Arithmetic]] is where you need to go for $\gcd \set {p, q} > 1$ {{iff}} $p \divides q$ -- I'm fairly sure there's a result on {{ProofWiki}} demonstrating it more directly}}
+Hence the result.
+{{Qed}}
+[[Category:Analytic Number Theory]]
+1o3ubcvfr7e7dkorv40cltzcfylb59q
+\end{proof}<|endoftext|>
+\section{Separable Metacompact Space is Lindelöf}
+Tags: Separable Spaces, Metacompact Spaces, Lindelöf Spaces, Separable Metacompact Space is Lindelöf
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Separable Space|separable topological space]] which is also [[Definition:Metacompact Space|metacompact]].
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+$T$ is [[Definition:Separable Space|separable]] {{iff}} there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]].
+$T$ is [[Definition:Metacompact Space|metacompact]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]].
+$T$ is a [[Definition:Lindelöf Space|Lindelöf space]] if every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+Having established the definitions, we proceed.
+Let $T = \left({S, \tau}\right)$ be [[Definition:Separable Space|separable]] and [[Definition:Metacompact Space|metacompact]].
+{{AimForCont}} there exists an [[Definition:Open Cover|open cover]] $\UU$ of $S$ which has no [[Definition:Countable Subcover|countable subcover]].
+As $T$ is [[Definition:Metacompact Space|metacompact]], $\UU$ has an [[Definition:Open Refinement|open refinement]] $\VV$ which is [[Definition:Point Finite|point finite]].
+By nature of $\UU$, which has no [[Definition:Countable Subcover|countable subcover]], $\VV$ is [[Definition:Uncountable Set|uncountable]].
+{{explain|How would $\VV$ being countable lead to a countable subcover of $\UU$? ACC is helpful for this.}}
+[[Definition:By Hypothesis|By hypothesis]], $T$ is [[Definition:Separable Space|separable]].
+Let $\SS$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]].
+Then each $V \in \VV$ contains some $s \in \SS$.
+{{explain|why? Also, this is not true if $\VV$ contains the empty set.}}
+So some $s \in \SS$ is contained in an [[Definition:Uncountable Set|uncountable]] number of elements of $\VV$.
+Thus, by definition, $\VV$ is not [[Definition:Point Finite|point finite]].
+Thus no [[Definition:Uncountable Set|uncountable]] [[Definition:Open Refinement|open refinement]] $\VV$ of $\UU$ exists which is [[Definition:Point Finite|point finite]].
+It follows that $\VV$ must be [[Definition:Countable Set|countable]].
+Thus $\UU$ has a [[Definition:Countable Subcover|countable subcover]].
+That is, by definition, $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+$T$ is [[Definition:Metacompact Space|metacompact]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has an [[Definition:Open Refinement|open refinement]] which is [[Definition:Point Finite|point finite]].
+$T$ is a [[Definition:Lindelöf Space|Lindelöf space]] if every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+Having established the definitions, we proceed.
+Let $\mathcal U$ be an [[Definition:Open Cover|open cover]] of $S$.
+Let $\mathcal V$ be a [[Definition:Point Finite|point finite]] [[Definition:Open Refinement|open refinement]] of $\mathcal U$.
+By [[Point Finite Set of Open Sets in Separable Space is Countable]], $\mathcal V$ is [[Definition:Countable Set|countable]].
+Define a mapping $H$ on $\mathcal V$ thus:
+:$\forall V \in \mathcal V: H \left({V}\right) = \left\{{U \in \mathcal U: V \subseteq U}\right\}$
+By [[Image of Countable Set under Mapping is Countable]], the [[Definition:Image of Mapping|image]] of $H$ is [[Definition:Countable Set|countable]].
+Call this [[Definition:Image of Mapping|image]] $I$.
+Since $\mathcal V$ is a refinement of $\mathcal U$, $\varnothing \notin I$.
+By the [[Axiom:Axiom of Countable Choice|Axiom of Countable Choice]], $I$ has a [[Definition:Choice Function|choice function]] $c$.
+Then $G = c \circ H: \mathcal V \to \mathcal U$ is a mapping such that:
+:$\forall V \in \mathcal V: V \subseteq G \left({V}\right)$
+Then $\mathcal Q = G \left({\mathcal V}\right)$ is [[Definition:Countable Set|countable]] by [[Image of Countable Set under Mapping is Countable]].
+Each element of $\mathcal Q$ is an element of $\mathcal U$ by the definition of $G$.
+Let $x \in S$.
+Then since $\mathcal V$ is a [[Definition:Cover of Set|cover]] for $S$:
+: $\exists V \in \mathcal V: x \in V$
+Then $x \in V \subseteq G \left({V}\right) \in \mathcal Q$.
+Thus $\mathcal Q$ is a [[Definition:Countable Subcover|countable subcover]] of $\mathcal U$.
+Thus each [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]], so $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Functional Equation for Dirichlet L-Functions}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\chi$ be a [[Definition:Dirichlet Character#Primitive Characters|primitive Dirichlet character]] to the modulus $q \geq 1$.
+Let $\map \Lambda {s, \chi}$ be the [[Definition:Completed Dirichlet L-Function|completed $L$-function]] for $\chi$.
+Let $\map \tau \chi$ denote the [[Definition:Gauss Sum|Gaussian sum]].
+Then for all $s \in \C$:
+:$\map \Lambda {s, \chi} = i^{-\kappa} \dfrac {\map \tau \chi} {\sqrt q} \map \Lambda {1 - s, \overline \chi}$
+where $\kappa = \dfrac 1 2 \paren {1 - \map \chi {-1} }$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Analytic Number Theory]]
+dcfnpn6q6rk91zwth2b331fnv9zb9ge
+\end{proof}<|endoftext|>
+\section{Orthogonality Relations for Characters}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] with [[Definition:Identity Element|identity]] $e$.
+Let $G^*$ be the dual group of [[Definition:Character (Number Theory)|characters]] $\chi : G \to \C_{\ne 0}$.
+Let $\chi_0$ be the [[Definition:Trivial Character|trivial character]] on $G$.
+Let $\psi: G \to \C_{\ne 0}$ be any [[Definition:Character (Number Theory)|character]].
+Let $y \in G$ be arbitrary.
+Then:
+:$\displaystyle \sum_{x \mathop \in G} \psi \left({x}\right) = \begin{cases}
+\left\lvert{G}\right\rvert & : \psi = \chi_0 \\
+0 & : \psi \ne \chi_0 \end{cases}$
+and:
+:$\displaystyle \sum_{\chi \mathop \in G^*} \chi \left({y}\right) = \begin{cases}
+\left\lvert{G^*}\right\rvert & : y = e \\
+0 & : y \ne e \end{cases}$
+\end{theorem}
+
+\begin{proof}
+If $\psi = \chi_0$, then it is straightforward that:
+{{begin-eqn}}
+{{eqn | l = \sum_{x \mathop \in G} \psi \left({x}\right)
+      | r = \sum_{x \mathop \in G} \chi_0 \left({x}\right)
+      | c = Assumption
+}}
+{{eqn | r = \sum_{x \mathop \in G} 1
+      | c = {{Defof|Trivial Character}}
+}}
+{{eqn | r = \vert G \vert
+      | c = 
+}}
+{{end-eqn}}
+If $\psi \neq \chi_0$, then $\exists y \in G$ such that $\psi \left({y}\right) \neq 1$.
+As $x$ runs through $G$ in the summation, $yx$ also runs through $G$.
+So:
+{{begin-eqn}}
+{{eqn | l = \sum_{x \mathop \in G} \psi \left({x}\right)
+      | r = \sum_{x \mathop \in G} \psi \left({y x}\right)
+      | c = by the above claim
+}}
+{{eqn | r = \sum_{x \mathop \in G} \psi \left({y}\right) \psi \left({x}\right)
+      | c = $\psi$ as a [[Definition:Character (Number Theory)|character]] is by definition a [[Definition:Group Homomorphism|homomorphism]]
+}}
+{{eqn | r = \psi \left({y}\right) \sum_{x \mathop \in G} \psi \left({x}\right)
+      | c = 
+}}
+{{end-eqn}}
+Since by assumption $\psi \left({y}\right) \ne 1$, it must be true that:
+:$\displaystyle \sum_{x \mathop \in G} \psi \left({x}\right) = 0$
+{{qed}}
+{{refactor|Suggestion: Move the dual space version into a corollary, as it is a direct application of the main theorem on the dual space, with identifying $x \in G$ as a character: $x: \chi \mapsto \chi \left({x}\right)$}}
+[[Category:Analytic Number Theory]]
+abb5loovdpeici0lkr1ik5j3bcfxspw
+\end{proof}<|endoftext|>
+\section{Analytic Continuation of Dirichlet L-Function}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\chi : G := \left({\Z / q \Z}\right)^\times \to \C^\times$ be a [[Definition:Dirichlet Character|Dirichlet character]] modulo $q$.
+{{explain|$\C^\times$}}
+Let $L \left({s, \chi}\right)$ be the [[Definition:Dirichlet L-function|Dirichlet $L$-function]] for $\chi$.
+Let $\chi$ be the [[Definition:Dirichlet Character/Trivial Character|trivial character]].
+Then $L \left({s, \chi}\right)$ has an analytic continuation to $\C$ except for a [[Definition:Simple Pole|simple pole]] at $s = 1$.
+Let $\chi$ be non-trivial.
+Then $L \left({s, \chi}\right)$ is analytic on $\Re \left({s}\right) > 0$.
+\end{theorem}
+
+\begin{proof}
+Let $\chi$ be the trivial character.
+Then by [[Dirichlet L-Function from Trivial Character]]:
+:$\displaystyle L \left({s, \chi}\right) = \zeta \left({s}\right) \cdot \prod_{p \mathop \backslash q} \left(1 - p^{-s}\right)$
+where $\backslash$ denotes [[Definition:Divisor of Integer|divisibility]].
+Also, by [[Poles of Riemann Zeta Function]], $\zeta$ is analytic on $\C$ except for a simple pole at $s = 1$.
+Since $L \left({s, \chi}\right)$ is just $\zeta$ times some constant, the same holds for this function.
+If $\chi$ is nontrivial, then by the [[Orthogonality Relations for Characters]]:
+:$\displaystyle \sum_{x \mathop \in G} \chi \left({x}\right) = 0$
+By definition, $\chi$ is $q$-[[Definition:Periodic Function|periodic]], and zero on integers not [[Definition:Coprime Integers|coprime]] to $q$.
+So for any $M \in \N$:
+:$\displaystyle \sum_{n \mathop = M + 1}^{M + Q} \chi \left({n}\right) = 0$
+Let $M, N \in \N$ be arbitrary.
+Let $d$ be such that $M + q d \le N \le M + q \left({d + 1}\right)$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = M}^N \chi \left({n}\right)
+      | r = \sum_{k \mathop = 0}^{d-1} \sum_{n \mathop = 0}^{q-1} \chi \left({M + kq + n}\right) + \sum_{n \mathop = M + q d}^N \chi \left({n}\right)
+      | c =
+}}
+{{eqn | r = \sum_{n \mathop = M \mathop + q d}^N \chi \left({n}\right)
+      | c = because $\chi$ is $q$-[[Definition:Periodic Function|periodic]], and zero on integers not [[Definition:Coprime Integers|coprime]] to $q$
+}}
+{{eqn | o = \le
+      | r = q
+      | c = because $\left\vert{N - M + q d}\right\vert \le q$
+}}
+{{end-eqn}}
+So the coefficients $\chi \left({n}\right)$ of $L \left({s, \chi}\right)$ have [[Definition:Bounded Sequence|bounded]] [[Definition:Sequence of Partial Sums|partial sums]].
+Therefore, by [[Convergence of Dirichlet Series with Bounded Partial Sums]], $L \left({s, \chi}\right)$ [[Definition:Locally Uniform Convergence|converges locally uniformly]] to an analytic function on $\Re \left({s}\right) > 0$.
+{{qed}}
+{{expand|Should be extended to all of $\C$, put the above down to get to Dirichlet's theorem}}
+[[Category:Analytic Number Theory]]
+75jlpx5ar0zsewjbqdfhqqn3sri6k4q
+\end{proof}<|endoftext|>
+\section{L-Function does not Vanish at One}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\psi$ be a [[Definition:Dirichlet Character#Trivial Character|non-trivial Dirichlet charater]] modulo $q$.
+Let $\map L {s, \chi}$ be the [[Definition:Dirichlet L-function|Dirichlet $L$-function associated to $\chi$]].
+Then $\map L {1, \chi} \ne 0$.
+\end{theorem}
+
+\begin{proof}
+Let $G^*$ be the [[Definition:Group|group]] of characters modulo $q$.
+{{explain|"[[Definition:Group|group]] of characters modulo $q$" - we have the definition for character, and we have (somewhere) the definition of the ring of integers modulo $q$, but not this entity.}}
+Let $\displaystyle \map {\zeta_q} s = \prod_{\chi \mathop \in G^*} \map L {s, \chi} R$.
+Among the factors of $\zeta_q$ is the the $L$-function associated to the trivial character, which by [[Analytic Continuation of Dirichlet L-Function]] we know to have a [[Definition:Simple Pole|simple pole]] at $s = 1$.
+{{AimForCont}} $\map L {1, \psi} = 0$. 
+Then the zero of this factor kills the [[Definition:Pole (Complex Analysis)|pole]] of the principal $L$-function.
+So by [[Analytic Continuation of Dirichlet L-Function]] $\zeta_q$ is analytic on $\map \Re s > 0$.
+For $\map \Re s > 1$ we have:
+{{begin-eqn}}
+{{eqn | l = \map {\zeta_q} s
+      | r = \prod_{\chi \mathop \in G^*} \prod_p \frac 1 {1 - \map \chi p p^{-s} }
+      | c = [[Euler Product]], where $p$ ranges over the [[Definition:Prime Number|primes]]
+}}
+{{eqn | r = \prod_{p \mathop \nmid q} \prod_{\chi \mathop \in G^*}  \frac 1 {1 - \map \chi p p^{-s} }
+      | c = because the [[Euler Product]] is [[Definition:Absolutely Convergent Series|absolutely convergent]]
+}}
+{{end-eqn}}
+For any prime $p$, let $f_p$ be the order of $p$ mod $q$.
+{{explain|order}}
+Then:
+:$\map \chi p^{f_p} = \map \chi {p^{f_p} } = \map \chi 1 = 1$
+So $\map \chi p$ is an $f_p$th root of unity.
+Moreover by the [[Orthogonality Relations for Characters]] each distinct such root occurs $\map \phi q / f_p$ times among the numbers $\map \chi p$ where $\chi \in G^*$.
+Also, letting $\xi$ be a primitive $f_p$th root of unity we find that for any $u \in \C$:
+:$\displaystyle \prod_{i \mathop = 0}^{f_p} \paren {1 - \xi^i u} = 1 - u^\xi$
+Putting these facts together:
+:$\displaystyle \prod_{\chi \mathop \in G^*} \frac 1 {1 - \map \chi p p^{-s} } = \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \map {\zeta_q} s
+      | r = \prod_{p \mathop \divides q} \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}
+      | c =
+}}
+{{eqn | r = \prod_{p \mathop \divides q} \paren {1 + p^{-f_p s} + p^{-2 f_p s} + \cdots}^{\map \phi q / f_p}
+      | c =
+}}
+{{end-eqn}}
+Also, if $\chi_0$ is the trivial character modulo $q$, by [[Euler Product]] we have:
+:$\displaystyle \map L {\map \phi q s, \chi_0} = \prod_{p \mathop \nmid q} \paren {1 + p^{-\map \phi q s} + p^{-2 \map \phi q s} + \cdots}$
+from which we see that for $s \in \R_{\ge 0}$:
+:$\map {\zeta_q} s \ge \map L {\map \phi q s, \chi_0}$
+However, by [[Analytic Continuation of Dirichlet L-Function]], $\map L {\map \phi q s, \chi_0}$ diverges for $s = \map \phi q^{-1}$, and therefore so does $\map {\zeta_q} s$.
+But we showed above that $\map {\zeta_q} s$ converges for $\map \Re s > 0$, a [[Definition:Contradiction|contradiction]].
+{{qed}}
+[[Category:Analytic Number Theory]]
+9xcwrmj6yiktscec0jn9tgxixhzz0vh
+\end{proof}<|endoftext|>
+\section{Logarithm of Dirichlet L-Functions}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\chi$ be a [[Definition:Dirichlet Character|Dirichlet character]] modulo $q$.
+The [[Definition:Dirichlet Series|Dirichlet series]]
+:$\map f s = \displaystyle \sum_{n \mathop \ge 1} \sum_p \frac {\map \chi p^n} {n p^{n s} }$
+[[Definition:Absolutely Convergent Series|converges absolutely]] to an [[Definition:Analytic Function|analytic function]], where $p$ ranges over the primes.
+Moreover, $\map f s$ defines a [[Definition:Branch (Complex Analysis)|branch]] of $\ln \map L {s, \chi}$
+\end{theorem}
+
+\begin{proof}
+By [[Convergence of Dirichlet Series with Bounded Coefficients]], $\map f s$ converges absolutely on $\map \Re s > 1$ to an analytic function.
+For fixed $s \in \set {\map \Re s > 1}$:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop \ge 1} \sum_p \frac {\map \chi p^n} {n p^{n s} }
+      | r = \sum_p \paren {\frac {\map \chi p} {p^s} + \frac {\map \chi p^2} {2 p^{2 s} } + \cdots}
+      | c = [[Manipulation of Absolutely Convergent Series]]
+}}
+{{eqn | r = \sum_p \map \ln {\frac 1 {1 - \map \chi p p^{-s} } }
+      | c = [[Taylor Series of Logarithm]]; this is the branch of $\ln$ with $\map \ln {1 + x} = x - \dfrac {x^2} 2 + \cdots$ for small $x$
+}}
+{{eqn | r = \ln \prod_p \paren {\frac 1 {1 - \map \chi p p^{-s} } }
+      | c = [[Logarithm is Continuous]]; this not necessarily the same branch of $\ln$
+}}
+{{eqn | r = \ln \map L {s, \chi}
+      | c = [[Euler Product]]
+}}
+{{end-eqn}}
+Hence $\map f s$ is an analytic branch of $\ln$ on $\map \Re s > 1$.
+{{qed}}
+[[Category:Analytic Number Theory]]
+pfys7o7jkgvmengnjn3eetzutl6ls03
+\end{proof}<|endoftext|>
+\section{Dirichlet's Theorem on Arithmetic Sequences}
+Tags: Analytic Number Theory, Prime Numbers, Arithmetic Sequences, Dirichlet's Theorem on Arithmetic Sequences
+
+\begin{theorem}
+Let $a, q$ be [[Definition:Coprime Integers|coprime integers]].
+Let $\PP_{a, q}$ be the set of [[Definition:Prime Number|primes]] $p$ such that $p \equiv a \pmod q$.
+Then $\PP_{a, q}$ has [[Definition:Dirichlet Density|Dirichlet density]]:
+:$\map \phi q^{-1}$
+where $\phi$ is [[Definition:Euler Phi Function|Euler's phi function]].
+In particular, $\PP_{a, q}$ is [[Definition:Infinite Set|infinite]].
+\end{theorem}
+
+\begin{proof}
+=== [[Dirichlet's Theorem on Arithmetic Sequences/Lemma 1|Lemma 1]] ===
+{{:Dirichlet's Theorem on Arithmetic Sequences/Lemma 1}}{{qed|lemma}}
+Define:
+:$\eta_{a, q} : n \mapsto \begin{cases}
+1 & : n \equiv a \pmod q\\
+0 & : \text{otherwise}
+\end{cases}$
+=== [[Dirichlet's Theorem on Arithmetic Sequences/Lemma 2|Lemma 2]] ===
+{{:Dirichlet's Theorem on Arithmetic Sequences/Lemma 2}}{{qed|lemma}}
+We have:
+{{begin-eqn}}
+{{eqn | l = \sum_{p \mathop \in \PP_{a, q} } p^{-s}
+      | r = \sum_p \map {\eta_{a, q} } p p^{-s}
+      | c = 
+}}
+{{eqn | r = \sum_p \sum_{\chi \mathop \in G^*} \frac {\map {\overline \chi} a} {\map \phi q} \, \map \chi p p^{-s}
+      | c = [[Dirichlet's Theorem on Arithmetic Sequences/Lemma 2|Lemma 2]]
+}}
+{{eqn | r = \frac 1 {\map \phi q} \sum_p \frac {\map {\chi_0} p} {p^s} + \sum_{\substack {\chi \mathop \in G^* \\ \chi \mathop \ne \chi_0} } \frac {\map {\overline \chi} a} {\map \phi q} \sum_p \map \chi p p^{-s}
+      | c = where $\chi_0$ is the trivial character on $G$
+}}
+{{end-eqn}}
+By [[Dirichlet's Theorem on Arithmetic Sequences/Lemma 1|Lemma 1]], the first term grows like $\dfrac 1 {\map \phi q} \ln \dfrac 1 {s - 1}$ as $s \to 1$, while all other terms are bounded.
+That is:
+:$\displaystyle \sum_{p \mathop \in \PP_{a, q} } \frac 1 {p^s} \sim \frac 1 {\map \phi q} \, \map \ln {\dfrac 1 {s - 1} }$
+as $s \to 1$. 
+{{qed}}
+{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
+\end{proof}<|endoftext|>
+\section{Discrete Fourier Transform on Abelian Group}
+Tags: Fourier Analysis, Representation Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
+Let $G^*$ be the dual group of [[Definition:Character (Number Theory)|characters]] $G \to \C^\times$.
+Let $\eta: G \to \C$ be a [[Definition:Mapping|mapping]] from $G$ to the set of [[Definition:Complex Number|complex numbers]].
+Then for all $x \in G$:
+:$\displaystyle \eta \left({x}\right) = \frac 1 {\phi \left({q}\right)} \sum_{\chi \mathop \in G^*} \langle \eta, \chi \rangle_G \chi(x)$
+where:
+:$\displaystyle \langle \eta, \chi \rangle_G = \sum_{x \mathop \in G} \eta \left({x}\right) \overline{\chi} \left({x}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn|l = \frac 1 {\phi \left({q}\right)} \sum_{\chi \mathop \in G^*} \langle \eta, \chi \rangle_G \chi \left({y}\right)
+     |r = \frac 1 {\phi \left({q}\right)} \sum_{\chi \mathop \in G^*} \sum_{x \mathop \in G} \eta \left({x}\right) \overline \chi \left({x}\right) \chi \left({y}\right)
+     |c =
+}}
+{{eqn|r = \frac 1 {\phi \left({q}\right)} \sum_{x \mathop \in G} \eta(x) \sum_{\chi \mathop \in G^*} \overline \chi \left({x}\right) \chi \left({y}\right)
+     |c =
+}}
+{{eqn|r = \frac 1 {\phi \left({q}\right)} \eta \left({y}\right) \phi \left({y}\right)
+     |c = [[Orthogonality Relations for Characters]]
+}}
+{{eqn|r = \eta \left({y}\right)
+     |c =
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Fourier Analysis]]
+[[Category:Representation Theory]]
+dlnm1gr781v18q9tma21hb0389hk2s3
+\end{proof}<|endoftext|>
+\section{Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods}
+Tags: Hausdorff Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
+Let $V_1$ and $V_2$ be [[Definition:Compact Topological Space|compact sets]] in $T$.
+Then $V_1$ and $V_2$ have [[Definition:Disjoint Sets|disjoint]] [[Definition:Neighborhood of Set|neighborhoods]].
+\end{theorem}
+
+\begin{proof}
+Let $\FF$ be the [[Definition:Set|set]] of all [[Definition:Ordered Pair|ordered pairs]] $\tuple {Z, W}$ such that:
+:$Z, W \in \tau$
+:$V_1 \subseteq Z$
+:$Z \cap W = \O$
+By the [[Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma|lemma]], $\Img \FF$ [[Definition:Cover of Set|covers]] $V_2$.
+{{explain|It is not clear what mapping or relation $\Img \FF$ refers to.}}
+By the definition of [[Definition:Compact Topological Space|compact space]], there exists a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $K$ of $\Img \FF$ which also [[Definition:Finite Cover|covers]] $V_2$.
+By the definition of [[Definition:Topology|topology]], $\displaystyle \bigcup K$ is [[Definition:Open Set (Topology)|open]].
+By the [[Principle of Finite Choice]], there exists a [[Definition:Bijection|bijection]] $\GG \subseteq \FF$ such that $\Img \GG = K$.
+{{explain}}
+Then $\GG$, and hence its [[Definition:Preimage of Mapping|preimage]], will be [[Definition:Finite Set|finite]].
+Let $\displaystyle J = \bigcap \Preimg \GG$.
+By [[Intersection is Largest Subset]], $V_1 \subseteq J$.
+By the definition of a [[Definition:Topology|topology]], $J$ is [[Definition:Open Set (Topology)|open]].
+Then $\displaystyle \bigcup K$ and $J$ are [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] such that $\displaystyle V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Subsets of T3 Spaces}
+Tags: T3 Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:T3 Space|$T_3$ space]].
+Let $A \subseteq S$ be [[Definition:Compact Topological Space|compact]] in $T$.
+Then for each $U \in \tau$ such that $A \subseteq U$:
+:$\exists V \in \tau: A \subseteq V \subseteq V^- \subseteq U$
+where $V^-$ denotes the [[Definition:Closure (Topology)|closure]] of $V$.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq S$ be [[Definition:Compact Topological Space|compact]] in $T$.
+Let $U \in \tau$ such that $A \subseteq U$.
+Since $T$ is [[Definition:T3 Space/Definition 2|$T_3$]]:
+:Each [[Definition:Open Set (Topology)|open set]] contains a [[Definition:Closed Neighborhood|closed neighborhood]] around each of its points:
+::$\forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U$
+Note that $\set {V_x: x \in A}$ forms an [[Definition:Open Cover|open cover]] for $A$.
+By [[Definition:Compact Topological Space|compactness]] of $A$, $\set {V_x: x \in A}$ has a [[Definition:Finite Subcover|finite subcover]].
+Therefore there is a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $I \subseteq A$ such that: 
+:$\set {V_x: x \in I}$ is an [[Definition:Open Cover|open cover]] for $A$.
+Let $V = \displaystyle \bigcup_{i \mathop \in I} V_x$, $N = \displaystyle \bigcup_{i \mathop \in I} N_x$.
+By definition of a [[Definition:Topology|topology]], the set $\displaystyle \bigcup_{i \mathop \in I} V_x$ is [[Definition:Open Set|open]] since it is a  [[Definition:Set Union|union]] of [[Definition:Open Set|open sets]].
+By [[Finite Union of Closed Sets is Closed in Topological Space]], the set $\displaystyle \bigcup_{i \mathop \in I} N_x$ is [[Definition:Closed Set|closed]] since it is a [[Definition:Finite|finite]] [[Definition:Set Union|union]] of [[Definition:Closed Set|closed sets]].
+By [[Set Union Preserves Subsets/Families of Sets|Set Union Preserves Subsets]], since $V_x \subseteq N_x \; \forall x \in I \subseteq U$:
+:$V \subseteq N$
+By [[Set Closure is Smallest Closed Set in Topological Space]]:
+:$V \subseteq V^- \subseteq N$
+Thus:
+{{begin-eqn}}
+{{eqn | l = A
+      | o = \subseteq
+      | r = V
+      | c = {{Defof|Open Cover}}
+}}
+{{eqn | o = \subseteq
+      | r = V^-
+}}
+{{eqn | o = \subseteq
+      | r = N
+}}
+{{eqn | o = \subseteq
+      | r = U
+      | c = [[Union of Subsets is Subset]]: $\forall x \in I \subseteq U: N_x \subseteq U$
+}}
+{{end-eqn}}
+satisfies our claim.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Hausdorff Space is T4}
+Tags: T4 Spaces, Hausdorff Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Hausdorff Space|Hausdorff space]].
+Then $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Compact Subspace of Hausdorff Space is Closed]].
+We also have that a [[Closed Subspace of Compact Space is Compact]].
+We also have that [[Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods]].
+$T$ is a [[Definition:T4 Space|$T_4$ space]] when any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed subsets]] of $S$ are [[Definition:Separated by Neighborhoods|separated by neighborhoods]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Hausdorff Topology is Minimal Hausdorff}
+Tags: Hausdorff Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]] which is [[Definition:Compact Topological Space|compact]].
+Then $\tau$ is the [[Definition:Minimal Set|minimal]] [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] $\powerset S$ such that $T$ is a [[Definition:Hausdorff Space|Hausdorff space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there exists a [[Definition:Topology|topology]] $\tau'$ on $S$ such that:
+:$\tau' \subseteq \tau$ but $\tau' \ne \tau$
+:$\tau'$ is a [[Definition:Hausdorff Space|Hausdorff space]].
+From [[Equivalence of Definitions of Finer Topology]]:
+:the [[Definition:Identity Mapping|identity mapping]] $I_S: \struct {S, \tau} \to \struct {S, \tau'}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Let $A \in \tau$.
+Then $S \setminus A \subseteq S$ is [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau}$.
+By [[Closed Subspace of Compact Space is Compact]], $S \setminus A$ is [[Definition:Compact Topological Space|compact]] in $\struct {S, \tau}$.
+From [[Continuous Image of Compact Space is Compact]]:
+:$I_S \sqbrk {S \setminus A}$ is also [[Definition:Compact Topological Space|compact]].
+[[Definition:By Hypothesis|By hypothesis]], $\struct {S, \tau'}$ is a [[Definition:Hausdorff Space|Hausdorff space]].
+From [[Compact Subspace of Hausdorff Space is Closed]]:
+:$I_S \sqbrk {S \setminus A}$ is [[Definition:Closed Set (Topology)|closed]] in $\struct {S, \tau'}$.
+Hence:
+:$A = S \setminus I_S \sqbrk {S \setminus A} \in \tau'$
+By definition of [[Definition:Subset|subset]]:
+:$\tau \subseteq \tau'$
+Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.
+Hence by definition of [[Definition:Equality of Sets|set equality]]:
+:$\tau' = \tau$
+But this [[Definition:Contradiction|contradicts]] our hypothesis that $\tau' \ne \tau$.
+By [[Proof by Contradiction]], it follows that no [[Definition:Topology|topology]] which is [[Definition:Strictly Coarser Topology|strictly coarser]] than $\tau$ can be [[Definition:Hausdorff Space|Hausdorff]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Hausdorff Topology is Maximally Compact}
+Tags: Hausdorff Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]] which is [[Definition:Compact Topological Space|compact]].
+Then $\tau$ is [[Definition:Maximal Set|maximally]] [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\tau'$ be a [[Definition:Topology|topology]] on $S$ such that $\tau \subseteq \tau'$ but that $\tau \ne \tau'$.
+Consider the [[Definition:Identity Mapping|identity mapping]] $I_S: \struct {S, \tau'} \to \struct {S, \tau}$.
+From [[Separation Properties Preserved in Subspace]], $I_S$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]] from a [[Definition:Hausdorff Space|Hausdorff space]] to a [[Definition:Compact Topological Space|compact]] [[Definition:Hausdorff Space|Hausdorff space]].
+{{AimForCont}} $\struct {S, \tau'}$ is [[Definition:Compact Topological Space|compact]].
+Then $I_S$ is also [[Definition:Open Set (Topology)|open]], and so $\tau' \subseteq \tau$.
+{{explain|Why does the fact that $\struct {S, \tau'}$ is compact mean $I_S$ is [[Definition:Open Set (Topology)|open]]?}}
+Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.
+Hence by definition of [[Definition:Equality of Sets|set equality]]:
+:$\tau' = \tau$
+But this [[Definition:Contradiction|contradicts]] our hypothesis that $\tau' \ne \tau$.
+By [[Proof by Contradiction]], it follows that no [[Definition:Topology|topology]] which is [[Definition:Finer Topology|strictly finer]] than $\tau$ can be [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximum Cardinality of Separable Hausdorff Space}
+Tags: Hausdorff Spaces, Separable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]] which is [[Definition:Separable Space|separable]].
+Then $S$ can have a [[Definition:Cardinality|cardinality]] no greater than $2^{2^{\aleph_0} }$.
+\end{theorem}
+
+\begin{proof}
+Let $D$ be an [[Definition:Everywhere Dense|everywhere dense]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Countable Set|countable]], as is guaranteed as $T$ is [[Definition:Separable Space|separable]].
+Consider the [[Definition:Mapping|mapping]] $\Phi: S \to 2^{\powerset D}$ defined as:
+:$\forall x \in S: \map {\map \Phi x} A = 1 \iff A = D \cap U_x$ for some [[Definition:Neighborhood of Point|neighborhood]] $U_x$ of $x$
+{{explain|It is not clear in Steen & Seeabch what is meant by $\Phi: S \to 2^{\powerset D}$ -- presumably $2^{\powerset D}$ is the ordinal which is the power set of the power set of $D$. It is also not clear what the notation $\map {\map \Phi x} A$ means -- in fact is may be the case that a transcription error has been committed.  Hence the proof cannot be attempted until these points have been cleared up.}}
+It is seen that if $T$ is a [[Definition:Hausdorff Space|Hausdorff space]], then $\Phi$ is an [[Definition:Injection|injection]].
+It follows that:
+:$\card S \le \card {2^{\powerset D} } = 2^{2^{\aleph_0} }$
+{{explain|the chain of reasoning leading to the above}}
+\end{proof}<|endoftext|>
+\section{Cluster Point of Ultrafilter is Unique}
+Tags: Filter Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $S$.
+Let $x \in S$ be a [[Definition:Cluster Point of Filter|cluster point]] of $\FF$.
+Then there is no point $y \in S: y \ne x$ such that $y$ is also a [[Definition:Cluster Point of Filter|cluster point]] of $\FF$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y$ both be [[Definition:Cluster Point of Filter|cluster points]] of an [[Definition:Ultrafilter on Set|ultrafilter]] $\FF$ on a [[Definition:Set|set]] $S$.
+Then by definition:
+:$\forall U \in \FF: \set {x, y} \subseteq U$
+But then we can create a [[Definition:Finer Filter on Set|finer filter]] $\FF \cup \set {\set x}$ on $S$.
+So $\FF$ is not an [[Definition:Ultrafilter on Set|ultrafilter]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Separation Axioms}
+Tags: Separation Axioms
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Separation Axioms|separation axioms]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Topological Space|topological space]] $T$ satsifies [[Definition:Separation Axioms|separation axiom]] $P_1$, then $T$ also satisfies [[Definition:Separation Axioms|separation axiom]] $P_2$.
+Then the following sequence of [[Definition:Separation Axioms|separation axioms]] holds:
+{|
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Perfectly Normal Space|Perfectly Normal]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Perfectly T4 Space|Perfectly $T_4$]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T4 Space|$T_4$]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Completely Normal Space|Completely Normal]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T5 Space|$T_5$]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Normal Space|Normal]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T4 Space|$T_4$]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$]] ||
+| align="center" | $\impliedby$ ||
+| align="center" | [[Definition:Completely Regular Space|Completely Regular (Tychonoff)]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Urysohn Space|Urysohn]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:T3 Space|$T_3$]] ||
+| align="center" | $\impliedby$ ||
+| align="center" | [[Definition:Regular Space|Regular]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (Completely Hausdorff)]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Semiregular Space|Semiregular]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Hausdorff Space|$T_2$ (Hausdorff)]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet)]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov)]] ||
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+* [[Definition:Perfectly Normal Space|Perfectly Normal]] implies [[Definition:Perfectly T4 Space|Perfectly $T_4$]] implies [[Definition:T4 Space|$T_4$]] by definition.
+* [[Perfectly Normal Space is Completely Normal Space]].
+* [[Definition:Completely Normal Space|Completely Normal]] implies [[Definition:T5 Space|$T_5$]] by definition.
+* [[Completely Normal Space is Normal Space]].
+* [[T5 Space is T4 Space|$T_5$ space is $T_4$ space]].
+* [[Definition:Normal Space|Normal]] implies [[Definition:T4 Space|$T_4$]] by definition.
+* [[Normal Space is Tychonoff Space|Normal Space is Tychonoff (Completely Regular) Space]].
+* [[Definition:Completely Regular Space|Completely Regular (Tychonoff)]] implies [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$]] by definition.
+* [[Tychonoff Space is Regular, T2 and T1|Completely Regular (Tychonoff) Space is Regular Space]].
+* [[T3 1/2 Space is T3 Space|$T_{3 \frac 1 2}$ Space is $T_3$ Space]].
+* [[Definition:Regular Space|Regular]] implies [[Definition:T3 Space|$T_3$]] by definition.
+* [[Tychonoff Space is Urysohn Space|Completely Regular (Tychonoff) Space is Urysohn Space]].
+* [[Urysohn Space is Completely Hausdorff Space]].
+* [[Regular Space is Completely Hausdorff Space]].
+* [[Regular Space is Semiregular Space]].
+* [[Completely Hausdorff Space is Hausdorff Space|Completely Hausdorff Space is $T_2$ (Hausdorff) Space]].
+* [[Definition:Semiregular Space|Semiregular]] implies [[Definition:Hausdorff Space|$T_2$ (Hausdorff)]] by definition.
+* [[T2 Space is T1 Space|$T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space]].
+* [[T1 Space is T0 Space|$T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Global Compactness Properties}
+Tags: Compact Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Compact Topological Space|compactness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Topological Space|topological space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Sequentially Compact Space|Sequentially Compact]] ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:Compact Topological Space|Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Countably Compact Space|Countably Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Pseudocompact Space|Pseudocompact]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:Sigma-Compact Space|$\sigma$-Compact]] ||
+| align="center" | ||
+| align="center" | [[Definition:Weakly Countably Compact Space|Weakly Countably Compact]] ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:Lindelöf Space|Lindelöf Space]] ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+* [[Sequentially Compact Space is Countably Compact]].
+* [[Compact Space is Countably Compact]].
+* [[Compact Space is Sigma-Compact|Compact Space is $\sigma$-Compact]].
+* [[Sigma-Compact Space is Lindelöf|$\sigma$-Compact Space is Lindelöf]].
+* [[Countably Compact Space is Weakly Countably Compact]].
+* [[Countably Compact Space is Pseudocompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Local Compactness Properties}
+Tags: Compact Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Compact Topological Space|compactness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Topological Space|topological space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | [[Definition:Compact Topological Space|Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Strongly Locally Compact Space|Strongly Locally Compact]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Weakly Sigma-Locally Compact Space|Weakly $\sigma$-Locally Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Weakly Locally Compact Space|Weakly Locally Compact]] ||
+| align="center" | $\Longleftarrow$ ||
+| align="center" | [[Definition:Locally Compact Space|Locally Compact]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:Sigma-Compact Space|$\sigma$-Compact]] ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:Lindelöf Space|Lindelöf Space]] ||
+| align="center" | ||
+| align="center" | ||
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+* [[Compact Space is Strongly Locally Compact]].
+* [[Strongly Locally Compact Space is Weakly Locally Compact]].
+* [[Locally Compact Space is Weakly Locally Compact]].
+* [[Compact Space is Weakly Sigma-Locally Compact|Compact Space is Weakly $\sigma$-Locally Compact]].
+* A [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact]] is both [[Definition:Weakly Locally Compact Space|weakly locally compact]] and [[Definition:Sigma-Compact Space|$\sigma$-compact]] by definition.
+* [[Sigma-Compact Space is Lindelöf|$\sigma$-Compact Space is Lindelöf]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Integral Dependence}
+Tags: Algebraic Number Theory, Commutative Algebra
+
+\begin{theorem}
+Let $A$ be an [[Definition:Ring Extension|extension]] of a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] $R$.
+For $x \in A$, the following are equivalent:
+{{begin-eqn}}
+{{eqn | n = 1
+      | o =
+      | r = \) $x$ is [[Definition:Integral Element of Ring Extension|integral]] over $R$ \(
+}}
+{{eqn | n = 2
+      | o =
+      | r = \) The $R$-[[Definition:Module|module]] $R \left[{x}\right]$ is [[Definition:Finitely Generated Module|finitely generated]] \(
+}}
+{{eqn | n = 3
+      | o =
+      | r = \) There exists a [[Definition:Subring|subring]] $B$ of $A$ such that $x \in B$, $R \subseteq B$ and $B$ is a [[Definition:Finitely Generated Module|finitely generated $R$-module]] \(
+}}
+{{eqn | n = 4
+      | o =
+      | r = \) There exists a [[Definition:Subring|subring]] $B$ of $A$ such that $x B \subseteq B$ and $B$ is [[Definition:Finitely Generated Module|finitely generated]] over $R$ \(
+}}
+{{eqn | n = 5
+      | o =
+      | r = \) There exists a [[Definition:Faithful Module|faithful]] $R \left[{x}\right]$-[[Definition:Module|module]] $B$ that is [[Definition:Finitely Generated Module|finitely generated]] as an $R$-[[Definition:Module|module]] \(
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+=== $(1) \implies (2)$ ===
+By hypothesis, there exist $r_0, \ldots, r_{n-1} \in R$ such that:
+:$x^n + r_{n-1} x^{n-1} + \cdots + r_1 x + r_0 = 0$
+So the powers $x^k$, $k \ge n$ can be written as an $R$-[[Definition:Linear Combination|linear combination]] of:
+:$\left\{ {1, \ldots, x^{n-1} }\right\}$
+Therefore this [[Definition:Set|set]] [[Definition:Generator|generates]] $R \left[{x}\right]$.
+{{qed|lemma}}
+=== $(2) \implies (3)$ ===
+$B = R \left[{x}\right]$ trivially satisfies the required conditions.
+{{qed|lemma}}
+=== $(3) \implies (4)$ ===
+By $(3)$ we have an $R$-module $B$ such that $R \subseteq B$, $B$ is finitely generated over $R$.
+Also, $x \in B$, so $x B \subseteq B$ as required.
+{{qed|lemma}}
+=== $(4) \implies (5)$ ===
+By $(4)$ we have an $R \left[{x}\right]$-module $B$ that is finitely generated over $R$.
+Let $y$ lie in the [[Definition:Annihilator|annihilator]] $\operatorname{Ann}_{R \left[{x}\right]} \left({B}\right)$
+We have that $1 \in B$.
+Then in particular $y \cdot 1 = 0$, and $y = 0$.
+Therefore, $B$ is [[Definition:Faithful Module|faithful]] over $R \left[{x}\right]$.
+{{qed|lemma}}
+=== $(5) \implies (1)$ ===
+Let $B$ be as in $(5)$, say generated by $m_1, \ldots, m_n \in B$.
+Then there are $r_{i j} \in R$, $i, j = 1,\ldots, n$ such that:
+:$\displaystyle x \cdot m_i = \sum_{j \mathop = 1}^n r_{i j} m_j$
+Let $b_{i j} = x \delta_{i j} - r_{i j}$ where $\delta_{i j}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Then:
+:$\displaystyle \sum_{j \mathop = 1}^n b_{i j} m_j = 0, \quad i = 1, \ldots, n$
+So, let $M = \left({b_{i j} }\right)_{1 \le i, j \le n}$.
+Then by [[Cramer's Rule]]:
+:$\left({\det M}\right) m_i = 0$, $i = 1, \ldots, n$
+Since $\det M \in R \left[{x}\right]$, also $\det M \in \operatorname{Ann}_{R \left[{x}\right]} \left({B}\right)$.
+So $\det M = 0$ by hypothesis.
+But $\det M = 0$ is a [[Definition:Monic Polynomial|monic polynomial]] in $x$ with coefficients in $R$.
+Thus $x$ is integral over $R$.
+{{qed}}
+[[Category:Algebraic Number Theory]]
+[[Category:Commutative Algebra]]
+4kaiy1bq732m22rz278dj5jzit8839p
+\end{proof}<|endoftext|>
+\section{Transitivity of Finite Generation}
+Tags: Commutative Algebra, Algebraic Number Theory
+
+\begin{theorem}
+Let $A \subseteq B \subseteq C$ be [[Definition:Ring (Abstract Algebra)|rings]].
+Suppose $B$ is a [[Definition:Finitely Generated Module|finitely generated]] $A$-[[Definition:Module|module]], and $C$ is a finitely generated $B$-module.
+Then $C$ is a finitely generated $A$-module.
+\end{theorem}
+
+\begin{proof}
+Let $b_1,\ldots, b_n$ [[Definition:Generator|generate]] $B$ over $A$.
+Let $c_1,\ldots, c_m$ generate $C$ over $B$.
+Then for any $x \in C$ there are $\beta_k \in B$, $k=1,\ldots, m$ such that:
+:$\displaystyle x = \sum_{k = 1}^m \beta_k c_k$
+For any $i \in \{1,\ldots,m\}$ there exists $\alpha_{ij} \in A$, $j = 1,\ldots, n$ such that
+:$\displaystyle \beta_i = \sum_{j = 1}^n \alpha_{ij}b_j$
+So
+:$\displaystyle x = \sum_{k = 1}^m \sum_{j = 1}^n \alpha_{kj} b_j c_k$
+Therefore, $\{b_jc_k : j = 1,\ldots,n;\ k = 1,\ldots, m\}$ generates $C$ over $A$.
+{{qed}}
+[[Category:Commutative Algebra]]
+[[Category:Algebraic Number Theory]]
+sz9wj70oearyvyybtmztbg84dxkwuw5
+\end{proof}<|endoftext|>
+\section{Integral Closure is Subring}
+Tags: Commutative Algebra, Algebraic Number Theory
+
+\begin{theorem}
+Let $A$ be an [[Definition:Ring Extension|extension]] of a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] $\struct {R, +, \circ}$.
+Let $C$ be the [[Definition:Integral Closure|integral closure]] of $R$ in $A$.
+Then $C$ is a [[Definition:Subring|subring]] of $A$.
+\end{theorem}
+
+\begin{proof}
+For each $r \in R$, we have $r$ is a zero of $a - r = 0$.
+{{MissingLinks|zero}}
+Hence $R \subseteq C$, and in particular, $C$ is [[Definition:Non-Empty Set|non-empty]].
+By [[Subring Test]] it is sufficient to prove that if $x, y \in C$ then $x + \paren {-y}, x \circ y \in C$.
+Let $x, y \in C$.
+By [[Equivalent Definitions of Integral Dependence|Equivalent Definitions of Integral Dependence: $(1) \implies (2)$]], $R \sqbrk x$ and $R \sqbrk y$ are [[Definition:Finitely Generated Module|finitely generated]] over $R$.
+Because $R \subseteq R \sqbrk x$, $R \sqbrk x \sqbrk y$ is also [[Definition:Finitely Generated Module|finitely generated]] over $R \sqbrk x$.
+Therefore, we have [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Ring Extension|extensions]]:
+:$R \hookrightarrow R \sqbrk x \hookrightarrow R \sqbrk x \sqbrk y = R \sqbrk {x, y}$
+Hence, by [[Transitivity of Finite Generation]], $R \sqbrk {x, y}$ is [[Definition:Finitely Generated Module|finitely generated]] over $R$.
+So by [[Equivalent Definitions of Integral Dependence|Equivalent Definition of Integral Dependence: $(2) \implies (1)$]], every element of $R \sqbrk {x, y}$ is [[Definition:Integral Element of Ring Extension|integral]] over $R$.
+In particular, $x + \paren {-y}$ and $x \circ y$ are [[Definition:Integral Element of Ring Extension|integral]] over $R$.
+{{qed}}
+[[Category:Commutative Algebra]]
+[[Category:Algebraic Number Theory]]
+a4tpxtnv2uundr184ymd3rw6xox1tln
+\end{proof}<|endoftext|>
+\section{Localization of Ring is Unique}
+Tags: Commutative Algebra
+
+\begin{theorem}
+Let $A$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $S \subseteq A$ be a [[Definition:Multiplicatively Closed Subset of Ring|multiplicatively closed subset]].
+Let $\left({A_S, \iota}\right)$ and $\left({\tilde A_S, \tilde \iota}\right)$ both satisfy the definition of the [[Definition:Localization of Ring|localization]] of $A$ at $S$.
+Then there is a [[Definition:Canonical Isomorphism|canonical isomorphism]] $\phi: A_S \to \tilde A_S$.
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Localization of Ring|localization]], there exist unique [[Definition:Ring Homomorphism|homomorphisms]]:
+:$g : A_S \to \tilde A_S$
+:$h : \tilde A_S \to A_S$
+such that:
+:$h \circ \iota = \tilde \iota$
+and:
+:$g \circ \tilde \iota = \iota$
+Therefore:
+:$\tilde \iota = h \circ \iota = h \circ \left({g \circ \tilde \iota}\right) = \left({h \circ g}\right) \circ \tilde \iota$
+The [[Definition:Identity Mapping|identity mapping]] also satisfies this equality (that is, $\tilde \iota = \left({h \circ g}\right) \circ \tilde \iota$).
+Therefore by the uniqueness of $h$ and $g$ we have:
+:$h \circ g = I_{\tilde A_S}$
+where $I_{\tilde A_S}$ denotes the [[Definition:Identity Mapping|identity mapping]] on $\tilde A_S$.
+Similarly we have that:
+:$g \circ h = I_{A_S}$
+where $I_{A_S}$ denotes the [[Definition:Identity Mapping|identity mapping]] on $A_S$.
+Therefore by [[Bijection iff Left and Right Inverse]] $g = \phi$ is a [[Definition:Bijection|bijective]] [[Definition:Ring Homomorphism|homomorphisms]], that is, an [[Definition:Ring Isomorphism|isomorphism]].
+{{qed}}
+[[Category:Commutative Algebra]]
+q003p2eryv948l1wioz91exvyyozyl6
+\end{proof}<|endoftext|>
+\section{Spectrum of Ring is Nonempty}
+Tags: Commutative Algebra
+
+\begin{theorem}
+Let $A$ be a [[Definition:Non-Trivial Ring|non-trivial]] [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Then its [[Definition:Spectrum of Ring|prime spectrum]] is [[Definition:Non-Empty Set|non-empty]]:
+:$\Spec A \ne \O$
+\end{theorem}
+
+\begin{proof}
+This is a reformulation of [[Ring with Unity has Prime Ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Space is Strongly Locally Compact}
+Tags: Compact Spaces, Strongly Locally Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact space]].
+Then $T$ is a [[Definition:Strongly Locally Compact Space|strongly locally compact space]].
+\end{theorem}
+
+\begin{proof}
+From [[Underlying Set of Topological Space is Clopen]], $S$ is [[Definition:Clopen Set|clopen]] in $T$.
+From [[Closed Set Equals its Closure]], $S = S^-$.
+So every point of $S$ is contained in an [[Definition:Open Set (Topology)|open set]] (that is, $S$) whose [[Definition:Closure (Topology)|closure]] (that is, $S$ again) is [[Definition:Compact Topological Subspace|compact]] (as  $T = \struct {S, \tau}$ itself is [[Definition:Compact Topological Space|compact]]).
+That is precisely the definition of a [[Definition:Strongly Locally Compact Space|strongly locally compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Space is Weakly Sigma-Locally Compact}
+Tags: Weakly Sigma-Locally Compact Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact space]].
+Then $T$ is a [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Topological Space|compact space]].
+We have that:
+:[[Compact Space is Sigma-Compact]]
+:[[Compact Space is Weakly Locally Compact]]
+Hence by definition $T$ is [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Paracompactness Properties}
+Tags: Compact Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Paracompact Space|paracompactness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Topological Space|topological space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Compact Topological Space|Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Countably Compact Space|Countably Compact]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Fully Normal Space|Fully Normal]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Paracompact Space|Paracompact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Countably Paracompact Space|Countably Paracompact]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Fully T4 Space|Fully $T_4$]] ||
+| align="center" | ||
+| align="center" | [[Definition:Metacompact Space|Metacompact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Countably Metacompact Space|Countably Metacompact]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+|-
+| align="center" | [[Definition:T4 Space|$T_4$]] ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+* [[Compact Space is Countably Compact]].
+* [[Compact Space is Paracompact]].
+* [[Fully Normal Space is Paracompact]].
+* [[Countably Compact Space is Countably Paracompact]].
+* [[Paracompact Space is Countably Paracompact]].
+* [[Definition:Fully Normal Space|Fully Normal Space]] is [[Definition:Fully T4 Space|Fully $T_4$]] by definition.
+* [[Fully T4 Space is T4 Space|Fully $T_4$ Space is $T_4$]].
+* [[Paracompact Space is Metacompact]].
+* [[Metacompact Space is Countably Metacompact]].
+* [[Countably Paracompact Space is Countably Metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compactness Properties in Hausdorff Spaces}
+Tags: Hausdorff Spaces, Compact Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Compact Topological Space|compactness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | [[Definition:Fully T4 Space|Fully $T_4$]] ||
+| align="center" | $\iff$ ||
+| align="center" | [[Definition:Paracompact Space|Paracompact]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Weakly Sigma-Locally Compact Space|Weakly $\sigma$-Locally Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T4 Space|$T_4$]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Weakly Locally Compact Space|Weakly Locally Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:First-Countable Space|First-Countable]] and [[Definition:Countably Compact Space|Countably Compact]]||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T3 Space|$T_3$]] ||
+|}
+\end{theorem}
+
+\begin{proof}
+From [[T2 Space is T1 Space|$T_2$ Space is $T_1$ Space]] and [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]], we note that in a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]]:
+:A [[Definition:Fully T4 Space|fully $T_4$ space]] is a [[Definition:Fully Normal Space|fully normal space]]
+:A [[Definition:T4 Space|$T_4$ space]] is a [[Definition:Normal Space|normal space]]
+:A [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]] is a [[Definition:Tychonoff Space|Tychonoff space]]
+:A [[Definition:T3 Space|$T_3$ space]] is a [[Definition:Regular Space|regular space]]
+all by definition.
+From [[Hausdorff Space is Fully T4 iff Paracompact|Hausdorff Space is Fully $T_4$ iff Paracompact]] we have that:
+:[[Definition:Fully T4 Space|Fully $T_4$]] $\iff$ [[Definition:Paracompact Space|Paracompact]].
+From that same result and [[Fully T4 Space is T4 Space|Fully $T_4$ Space is $T_4$ Space]]:
+:[[Definition:Paracompact Space|Paracompact]] $\implies$ [[Definition:T4 Space|$T_4$]]
+From [[Normal Space is Tychonoff Space]] it follows that:
+:[[Definition:T4 Space|$T_4$]] $\implies$ [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$]]
+From [[First-Countable Space is Sequentially Compact iff Countably Compact]] and [[Sequentially Compact Hausdorff Space is Regular]] it follows that:
+:[[Definition:First-Countable Space|First-Countable]] and [[Definition:Countably Compact Space|Countably Compact]] $\implies$ [[Definition:T3 Space|$T_3$]]
+The further justifications are listed as follows:
+:[[Sigma-Locally Compact Hausdorff Space is T4|$\sigma$-Locally Compact Hausdorff Space is $T_4$]]
+:[[Definition:Weakly Sigma-Locally Compact Space|Weakly $\sigma$-Locally Compact]] implies [[Definition:Weakly Locally Compact Space|Weakly Locally Compact]] by definition.
+:[[T3 1/2 Space is T3 Space|$T_{3 \frac 1 2}$ Space is $T_3$ Space]]
+{{ProofWanted|Some results still to be written.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Locus of Set is Zero Locus of Generated Ideal}
+Tags: Algebraic Geometry
+
+\begin{theorem}
+Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $n\geq1$ be a [[Definition:Natural Number|natural number]].
+Let $A = k \left[{X_1, \ldots, X_n}\right]$ be the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] in $n$ variables over $k$.
+Let $T \subseteq A$ be a [[Definition:Subset|subset]], and $V \left({T}\right)$ the [[Definition:Zero Locus of Set of Polynomials|zero locus]] of $T$.
+Let $J = \left({T}\right)$ be the [[Definition:Ideal of Ring|ideal]] [[Definition:Generator|generated]] by $T$.
+Then:
+: $V \left({T}\right) = V \left({J}\right)$
+\end{theorem}
+
+\begin{proof}
+Let $x \in V \left({T}\right)$, so $f \left({x}\right) = 0$ for all $f \in T$.
+By definition, $J$ is the set of [[Definition:Linear Combination|linear combinations]] of elements of $T$ over $k$.
+So any $g \in J$ is of the form
+:$g = k_1 t_1 + \cdots + k_r t_r$
+with $k_i \in k$ and $t_i \in T$.
+Therefore:
+{{begin-eqn}}
+{{eqn|l=g \left({x}\right)
+     |r=k_1 t_1 \left({x}\right) + \cdots + k_r t_r \left({x}\right)
+}}
+{{eqn|l=
+     |r=0
+     |c=because $f \left({x}\right) = 0$ for all $f \in T$
+}}
+{{end-eqn}}
+Therefore $x \in V \left({J}\right)$.
+Conversely, if $x \in V \left({J}\right)$, then $f \left({x}\right) = 0$ for all $f \in J$.
+But $T \subseteq J$, so in particular $f \left({x}\right) = 0$ for all $f \in T$.
+So $x \in V \left({T}\right)$.
+{{qed}}
+[[Category:Algebraic Geometry]]
+n4qyz7rgoqcle2fezabo58byr75biyn
+\end{proof}<|endoftext|>
+\section{Compactness Properties in T3 Spaces}
+Tags: T3 Spaces, Compact Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Compact Topological Space|compactness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:T3 Space|$T_3$ space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | [[Definition:Second-Countable Space|Second-Countable]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Lindelöf Space|Lindelöf]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | $\Downarrow$||
+| align="center" | ||
+| align="center" | [[Definition:Fully T4 Space|Fully $T_4$]] $\iff$ [[Definition:Paracompact Space|Paracompact]] ||
+|-
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:T5 Space|$T_5$]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:T4 Space|$T_4$]] ||
+|-
+|}
+\end{theorem}
+
+\begin{proof}
+The justifications are listed as follows:
+:[[Second-Countable T3 Space is T5|Second-Countable $T_3$ Space is $T_5$]]
+:[[Second-Countable Space is Lindelöf]]
+:[[Lindelöf T3 Space is Paracompact|Lindelöf $T_3$ Space is Paracompact]]
+:[[T3 Space is Fully T4 iff Paracompact|$T_3$ Space is Fully $T_4$ iff Paracompact]]
+:[[Fully T4 Space is T4 Space|Fully $T_4$ Space is $T_4$ Space]]
+:[[T5 Space is T4 Space|$T_5$ Space is $T_4$ Space]]
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Finite Product of Sigma-Compact Spaces is Sigma-Compact}
+Tags: Sigma-Compact Spaces, Product Spaces
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a [[Definition:Finite Set|finite set]] of [[Definition:Topological Space|topological spaces]].
+Let $\displaystyle \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the [[Definition:Product Space (Topology)|product space]] of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.
+Let each of $\struct {S_i, \tau_i}$ be [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+Then $\struct {S, \tau}$ is also [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+Then $\displaystyle S_1 = \bigcup_{n \mathop = 1}^\infty C_n$ and $\displaystyle S_2 = \bigcup_{n \mathop = 1}^\infty D_n$ where all the $C_n$ and $D_n$ are [[Definition:Compact Topological Subspace|compact sets]].
+From [[Tychonoff's Theorem]], the set $\KK = \set {C_n \times D_m: n, m \in \N}$ is a [[Definition:Countable Set|countable set]] of [[Definition:Compact Topological Subspace|compact sets]] of $S_1 \times S_2$ with the [[Definition:Product Topology|product topology]].
+Consider $\tuple {a, b} \in S_1 \times S_2$.
+Then $a \in S_1$ and $b \in S_2$.
+Using the [[Definition:Sigma-Compact Space|$\sigma$-compactness]] of $S_1$ and $S_2$, $\exists n, m$ such that $a \in C_n$ and $b \in D_m$.
+So $\tuple {a, b} \in C_n \times D_m \in \KK$.
+Thus, $\bigcup \KK = S_1 \times S_2$ and then $S_1 \times S_2$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+Using induction on the number of factors of the product we get the result.
+{{finish|Complete the inductive proof.}}
+\end{proof}<|endoftext|>
+\section{Countable Product of Sequentially Compact Spaces is Sequentially Compact}
+Tags: Sequentially Compact Spaces, Product Spaces
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] with [[Definition:Countable Set|countable cardinality]].
+Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexed Family|indexed]] by $I$.
+Let $\displaystyle \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
+Let each of $\struct {S_\alpha, \tau_\alpha}$ be [[Definition:Sequentially Compact Space|sequentially compact]].
+Then $\struct {S, \tau}$ is also [[Definition:Sequentially Compact Space|sequentially compact]].
+\end{theorem}<|endoftext|>
+\section{Finite Product of Weakly Locally Compact Spaces is Weakly Locally Compact}
+Tags: Weakly Locally Compact Spaces, Product Spaces
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a [[Definition:Finite Set|finite set]] of [[Definition:Topological Space|topological spaces]].
+Let $\displaystyle \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the [[Definition:Product Space (Topology)|product space]] of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.
+Let each of $\struct {S_i, \tau_i}$ be [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+Then $\struct {S, \tau}$ is also [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+\end{theorem}
+
+\begin{proof}
+Pick any $\tuple {x_1, \dots, x_n} \in S$.
+Since each of $\struct {S_i, \tau_i}$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]], $x_i$ has a [[Definition:Compact Space|compact]] [[Definition:Neighborhood of Point|neighborhood]].
+That is, there is a [[Definition:Compact Space|compact set]] $N_i$ such that:
+:$\exists U_i \in \tau_i: x_i \in U_i \subseteq N_i$
+By [[Cartesian Product of Subsets/Family of Subsets]], we have:
+:$\displaystyle \tuple {x_1, \dots, x_n} \in \prod_{i \mathop = 1}^n U_i \subseteq \prod_{i \mathop = 1}^n N_i$
+By definition of [[Definition:Product Space (Topology)|product space]]:
+:$\displaystyle \prod_{i \mathop = 1}^n U_i \in \tau$
+By [[Topological Product of Compact Spaces/Finite Product]]:
+:$\displaystyle \prod_{i \mathop = 1}^n N_i$ is [[Definition:Compact Space|compact]]
+Therefore $\displaystyle \prod_{i \mathop = 1}^n N_i$ is a [[Definition:Compact Space|compact]] [[Definition:Neighborhood of Point|neighborhood]] of $\tuple {x_1, \dots, x_n}$.
+As $\tuple {x_1, \dots, x_n}$ is arbitrary, $\struct {S, \tau}$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Product of First-Countable Spaces is First-Countable}
+Tags: First-Countable Spaces, Product Spaces
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] with [[Definition:Countable Set|countable cardinality]].
+Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexed Family|indexed]] by $I$.
+Let $\displaystyle \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
+Let each of $\struct {S_\alpha, \tau_\alpha}$ be [[Definition:First-Countable Space|first-countable]].
+Then $\struct {S, \tau}$ is also [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $x = \family {x_\alpha} \in S$.
+Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] for $x_\alpha$ in $\struct {S_\alpha, \tau_\alpha}$ for each $\alpha \in I$.
+Let $\pr_\alpha$ denote the [[Definition:Projection|projection]] of $S$ onto $S_\alpha$.
+Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.
+Let $\KK_J = \displaystyle \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$.
+From [[Product Space Local Basis Induced from Factor Spaces Local Bases]]:
+:$\displaystyle \BB_x = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a [[Definition:Local Basis|local basis]] of $x$ in the [[Definition:Product Space (Topology)|product space]] $\struct {S_\alpha, \tau_\alpha}$
+Now, each of the $\LL_\alpha$'s is [[Definition:Countable Set|countable]]. 
+Since the $\KK_J$'s can be identified with a [[Definition:Finite Set|finite]] [[Definition:Cartesian Product|product]] of [[Definition:Countable Set|countable sets]], they are each [[Definition:Countable Set|countable]]
+From [[Countable Union of Countable Sets is Countable]], $\BB_x$ forms a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] of $x$ in the [[Definition:Product Space (Topology)|product space]] $\struct {S_\alpha, \tau_\alpha}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Product of Second-Countable Spaces is Second-Countable}
+Tags: Second-Countable Spaces, Product Spaces
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] with [[Definition:Countable Set|countable cardinality]].
+Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexed Family|indexed]] by $I$.
+Let $\displaystyle \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
+Let each of $\struct {S_\alpha, \tau_\alpha}$ be [[Definition:Second-Countable Space|second-countable]].
+Then $\struct {S, \tau}$ is also [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a [[Definition:Countable Basis|countable basis]] for $\tau_\alpha$ for each $\alpha \in I$.
+Let $\pr_\alpha$ denote the [[Definition:Projection|projection]] of $S$ onto $S_\alpha$.
+Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.
+Let $\KK_J = \displaystyle \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$.
+From [[Product Space Basis Induced from Factor Space Bases]]:
+:$\displaystyle \BB = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a [[Definition:Basis (Topology)|basis]] of the [[Definition:Product Space (Topology)|product space]].
+Now, each of the $\LL_\alpha$'s is [[Definition:Countable Set|countable]]. 
+Since the $\KK_J$'s can be identified with a [[Definition:Finite Set|finite]] [[Definition:Cartesian Product|product]] of [[Definition:Countable Set|countable sets]], they are each [[Definition:Countable Set|countable]]
+From [[Countable Union of Countable Sets is Countable]], $\BB$ forms a [[Definition:Countable Basis|countable basis]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Product of Separable Spaces is Separable}
+Tags: Separable Spaces, Product Spaces
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] with [[Definition:Countable Set|countable cardinality]].
+Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexed Family|indexed]] by $I$.
+Let $\displaystyle \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
+Let each of $\struct {S_\alpha, \tau_\alpha}$ be [[Definition:Separable Space|separable]].
+Then $\struct {S, \tau}$ is also [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+For each $\alpha \in I$, let $D_\alpha$ be a [[Definition:Countable Set|countable]] [[Definition:Everywhere Dense|everywhere dense subset]] of $\struct {S_\alpha, \tau_\alpha}$.
+Let $D = \displaystyle \prod_{\alpha \mathop \in I} D_\alpha$.
+From [[Countable Union of Countable Sets is Countable]], $D$ is a [[Definition:Countable Set|countable set]] of $S$.
+From [[Natural Basis of Tychonoff Topology]], the [[Definition:Set|set]] $\BB$ of [[Definition:Cartesian Product|cartesian products]] of the form $\displaystyle \prod_{\alpha \mathop \in I} U_\alpha$ where:
+:for all $\alpha \in I : U_\alpha \in \tau_\alpha$
+:for all but finitely many [[Definition:Index|indices]] $\alpha : U_\alpha = X_\alpha$
+is a [[Definition:Analytic Basis|basis]] for $\tau$.
+From [[Analytic Basis Characterization of Denseness]], to show that $D$ is an [[Definition:Everywhere Dense|everywhere dense subset]] in $\struct {S, \tau}$ it is sufficient to show that:
+:$\forall U \in \BB : U \cap D \ne \O$
+Let $U = \displaystyle \prod_{\alpha \mathop \in I} U_\alpha \in \BB$.
+Then:
+:$\forall \alpha \in I : U_\alpha \cap D_\alpha \ne \O$
+For all $\alpha \in I$, let $x_\alpha \in U_\alpha \cap D_\alpha$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \family{x_\alpha}_{\alpha \in I}
+      | o = \in 
+      | r = \prod_{\alpha \mathop \in I} \paren {U_\alpha \cap D_\alpha}
+}}
+{{eqn | r = \paren {\prod_{\alpha \mathop \in I} U_\alpha} \bigcap \paren {\prod_{\alpha \mathop \in I}  D_\alpha}
+      | c = [[General Case of Cartesian Product of Intersections]]
+}}
+{{eqn | r = U \cap D
+}}
+{{end-eqn}}
+Thus:
+:$U \cap D \ne \O$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compactness Properties Preserved under Continuous Surjection}
+Tags: Compact Spaces, Continuous Mappings, Surjections
+
+\begin{theorem}
+Let $T_A = \strict {S_A, \tau_A}$ and $T_B = \strict {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
+If $T_A$ has one of the following properties, then $T_B$ has the same property:
+:[[Definition:Compact Topological Space|Compact Space]]
+:[[Definition:Sigma-Compact Space|$\sigma$-Compact Space]]
+:[[Definition:Countably Compact Space|Countable Compact Space]]
+:[[Definition:Sequentially Compact Space|Sequential Compact Space]]
+:[[Definition:Lindelöf Space|Lindelöf Space]]
+\end{theorem}
+
+\begin{proof}
+=== [[Compactness is Preserved under Continuous Surjection|Proof for Compactness]] ===
+{{:Compactness is Preserved under Continuous Surjection}}
+=== [[Sigma-Compactness is Preserved under Continuous Surjection|Proof for $\sigma$-Compactness]] ===
+{{:Sigma-Compactness is Preserved under Continuous Surjection}}
+=== [[Countable Compactness is Preserved under Continuous Surjection|Proof for Countable Compactness]] ===
+{{:Countable Compactness is Preserved under Continuous Surjection}}
+=== [[Sequential Compactness is Preserved under Continuous Surjection|Proof for Sequential Compactness]] ===
+{{:Sequential Compactness is Preserved under Continuous Surjection}}
+=== [[Lindelöf Property is Preserved under Continuous Surjection|Proof for Lindelöf Property]] ===
+{{:Lindelöf Property is Preserved under Continuous Surjection}}
+\end{proof}<|endoftext|>
+\section{Projection from Product Topology is Open}
+Tags: Tychonoff Topology, Open Mappings, Projections, Projection from Product Topology is Open and Continuous
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $T = \struct {T_1 \times T_2, \tau}$ be the [[Definition:Product Space (Topology)|product space]] of $T_1$ and $T_2$, where $\tau$ is the [[Definition:Tychonoff Topology|Tychonoff topology]] on $S$.
+Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the [[Definition:Projection (Mapping Theory)|first and second projections]] from $T$ onto its [[Definition:Factor Space|factors]].
+Then both $\pr_1$ and $\pr_2$ are [[Definition:Open Mapping|open]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$.
+It follows from the definition of [[Definition:Tychonoff Topology|Tychonoff topology]] that $U$ can be expressed as:
+:$\displaystyle U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$
+where $J$ is an arbitrary [[Definition:Indexing Set|index set]], $n_j \in \N$, $i_{k,j} \in \set {1,2}$, and $U_{k,j} \in \tau_{i_{k,j} }$.
+For all $i \in \set {1, 2}$, define $V_{i, k, j} \in \tau_i$ by:
+:$V_{i, k, j} = \begin {cases} U_{k, j} & : i = i_{k, j} \\ S_i & : i \ne i_{k, j} \end {cases}$
+By definition of [[Definition:Projection (Mapping Theory)|projection]]:
+:$\map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } = \displaystyle \prod_{i \mathop = 1}^2 V_{i, k, j}$
+{{WLOG}}, we consider $\pr_1$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {\pr_1} U
+      | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } }
+      | c = [[Image of Union under Relation/Family of Sets|Image of Union under Relation: Family of Sets]]
+}}
+{{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i \mathop = 1}^2 V_{i, k, j} }
+}}
+{{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\prod_{i \mathop = 1}^2 \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} }
+      | c = [[Cartesian Product of Intersections]]
+}}
+{{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j}
+      | c = {{Defof|Projection (Mapping Theory)|Projection}}
+}}
+{{end-eqn}}
+As $\displaystyle \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j} \in \tau_1$, it follows that $\pr_1$ is [[Definition:Open Mapping|open]].
+The proof for $\pr_2$ is the same, [[Definition:Mutatis Mutandis|mutatis mutandis]].
+\end{proof}<|endoftext|>
+\section{Weak Local Compactness is Preserved under Open Continuous Surjection}
+Tags: Weakly Locally Compact Spaces, Open Mappings, Continuous Mappings
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] which is also an [[Definition:Open Mapping|open mapping]] and a [[Definition:Surjection|surjection]].
+If $T_A$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]], then $T_B$ is also [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\phi$ be a [[Definition:Mapping|mapping]] which is [[Definition:Surjection|surjective]], [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and [[Definition:Open Mapping|open]].
+Let $T_A$ be [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+Take $b \in S_B$.
+Let $V$ be a [[Definition:Neighborhood of Point|neighbourhood]] of $b$.
+Since $\phi$ is [[Definition:Surjection|surjective]]:
+:$\forall y \in S_B: \exists x \in S_A: x \in \map {\phi^{-1} } y$
+From the [[Definition:Weakly Locally Compact Space|weak local compactness]] of $T_A$ and the [[Definition:Everywhere Continuous Mapping (Topology)|continuity]] of $\phi$, there exists a [[Definition:Compact Subspace|compact]] [[Definition:Neighborhood of Point|neighbourhood]] $K$ of $x$ such that $\phi \sqbrk K \subseteq V$.
+Since $K$ is a [[Definition:Neighborhood of Point|neighbourhood]] of $x$, then $x \in K^\circ$ and $y \in \phi \sqbrk {K^\circ} \subseteq \phi \sqbrk K$, where $K^\circ$ is the [[Definition:Interior (Topology)|interior]] of $K$.
+$\phi$ is an [[Definition:Open Mapping|open mapping]] and $K^\circ$ is an [[Definition:Open Set (Topology)|open set]], so $\phi \sqbrk {K^\circ}$ is also [[Definition:Open Set (Topology)|open]].
+Finally we get that $y \in \phi \sqbrk K \subseteq V$, where $\phi \sqbrk K$ is a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Point|neighbourhood]].
+Thus $T_B$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compactness Properties Preserved under Projection Mapping}
+Tags: Compact Spaces, Continuous Mappings, Projections
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] with [[Definition:Countable Set|countable cardinality]].
+Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexed Family|indexed]] by $I$.
+Let $\displaystyle \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
+Let $\pr_\alpha: \struct {S, \tau} \to \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Projection (Mapping Theory)|projection on the $\alpha$ coordinate]].
+Then $\pr_\alpha$ preserves the following [[Definition:Compact Topological Space|compactness]] properties.
+That is, if $\struct {S, \tau}$ has one of the following properties, then each of $\struct {S_\alpha, \tau_\alpha}$ has the same property:
+:[[Definition:Compact Topological Space|Compact Space]]
+:[[Definition:Sigma-Compact Space|$\sigma$-Compact Space]]
+:[[Definition:Countably Compact Space|Countably Compact Space]]
+:[[Definition:Sequentially Compact Space|Sequentially Compact Space]]
+:[[Definition:Lindelöf Space|Lindelöf Space]]
+:[[Definition:Locally Compact Space|Locally Compact Space]]
+:[[Definition:Weakly Locally Compact Space|Weakly Locally Compact Space]]
+:[[Definition:Paracompact Space|Paracompact Space]]
+\end{theorem}
+
+\begin{proof}
+First note that [[Projection from Product Topology is Continuous]].
+Also note that [[Projection is Surjection]].
+It follows from [[Compactness Properties Preserved under Continuous Surjection]] that:
+:[[Definition:Compact Topological Space|Compact Space]]
+:[[Definition:Sigma-Compact Space|$\sigma$-Compact Space]]
+:[[Definition:Countably Compact Space|Countably Compact Space]]
+:[[Definition:Sequentially Compact Space|Sequentially Compact Space]]
+:[[Definition:Lindelöf Space|Lindelöf Space]]
+are all preserved under [[Definition:Projection (Mapping Theory)|projections]].
+Next note that [[Projection from Product Topology is Open]].
+It follows from [[Local Compactness is Preserved under Open Continuous Surjection]] that [[Definition:Locally Compact Space|local compactness]] is preserved under [[Definition:Projection (Mapping Theory)|projections]].
+It follows from [[Weak Local Compactness is Preserved under Open Continuous Surjection]] that [[Definition:Weakly Locally Compact Space|weak local compactness]] is preserved under [[Definition:Projection (Mapping Theory)|projections]].
+The final result is that [[Paracompactness is Preserved under Projections]].
+{{finish|Finish these results off.}}
+\end{proof}<|endoftext|>
+\section{Countability Properties Preserved under Projection Mapping}
+Tags: Countability Axioms, Continuous Mappings, Projections
+
+\begin{theorem}
+Let $\sequence{\struct{S_\alpha, \tau_\alpha}}$ be a [[Definition:Sequence|sequence]] of [[Definition:Topological Space|topological spaces]].
+Let $\displaystyle \struct{S, \tau} = \prod \struct{S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\sequence {\struct{S_\alpha, \tau_\alpha}}$.
+Let $\pr_\alpha: \struct{S, \tau} \to \struct{S_\alpha, \tau_\alpha}$ be the [[Definition:Projection (Mapping Theory)|projection on the $\alpha$ coordinate]].
+Then $\pr_\alpha$ preserves the following [[Definition:Countability Axioms|countability properties]].
+That is, if $\struct{S, \tau}$ has one of the following properties, then each of $\struct{S_\alpha, \tau_\alpha}$ has the same property.
+:[[Definition:Separable Space|Separability]]
+:[[Definition:First-Countable Space|First-Countability]]
+:[[Definition:Second-Countable Space|Second-Countability]]
+\end{theorem}
+
+\begin{proof}
+First note that [[Projection from Product Topology is Continuous]].
+It follows from [[Continuous Image of Separable Space is Separable]] that [[Definition:Separable Space|separability]] is preserved under [[Definition:Projection (Mapping Theory)|projections]].
+Next note that [[Projection from Product Topology is Open]].
+It follows from [[Countability Axioms Preserved under Open Continuous Surjection]] that:
+:[[Definition:First-Countable Space|First-Countability]]
+:[[Definition:Second-Countable Space|Second-Countability]]
+are preserved under [[Definition:Projection (Mapping Theory)|projections]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countability Axioms Preserved under Open Continuous Surjection}
+Tags: Countability Axioms, Open Mappings, Continuous Mappings, Surjections
+
+\begin{theorem}
+Let $T_A = \left({S_A, \tau_A}\right)$ and $T_B = \left({S_B, \tau_B}\right)$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Surjection|surjective]] [[Definition:Open Mapping|open mapping]] which is also [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+If $T_A$ has one of the following properties, then $T_B$ has the same property:
+:[[Definition:First-Countable Space|First-Countability]]
+:[[Definition:Second-Countable Space|Second-Countability]]
+\end{theorem}
+
+\begin{proof}
+=== [[First-Countability is Preserved under Open Continuous Surjection|Proof for First-Countability]] ===
+{{:First-Countability is Preserved under Open Continuous Surjection}}
+=== [[Second-Countability is Preserved under Open Continuous Surjection|Proof for Second-Countability]] ===
+{{:Second-Countability is Preserved under Open Continuous Surjection}}
+\end{proof}<|endoftext|>
+\section{Connectedness of Points is Equivalence Relation}
+Tags: Equivalence Relations, Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $a \sim b $ denote the [[Definition:Relation|relation]]:
+:$a \sim b \iff a$ is [[Definition:Connected Points (Topology)|connected]] to $b$
+where $a, b \in S$.
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+We have that $\set a \subseteq S$ is a [[Definition:Degenerate Connected Set|(degenerate) connected set]] of $S$.
+So $a$ is in the same [[Definition:Connected Set (Topology)|connected set]] as itself and so $a \sim a$.
+So $\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+If $x \sim y$ then $x$ is in the same [[Definition:Connected Set (Topology)|connected set]] as $y$ by definition.
+Trivially it follows that $y$ is in the same [[Definition:Connected Set (Topology)|connected set]] as $x$ and so $y \sim x$.
+So $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $x \sim y$ and $y \sim z$.
+Then by definition:
+:$x$ is in the same [[Definition:Connected Set (Topology)|connected set]] as $y$
+:$y$ is in the same [[Definition:Connected Set (Topology)|connected set]] as $z$
+By [[Union of Connected Sets with Non-Empty Intersections is Connected]] it follows that $x$ is in the same [[Definition:Connected Set (Topology)|connected set]] as $z$.
+Thus so $x \sim z$.
+So $\sim$ has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+$\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Hence by definition it is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Atlas Belongs to Unique Differentiable Structure}
+Tags: Manifolds
+
+\begin{theorem}
+Let $M$ be a [[Definition:Locally Euclidean Space|locally Euclidean space of dimension $d$]].
+Let $\mathcal A$ be an [[Definition:Atlas|atlas]] on $M$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Differentiable Structure|differentiable structure]] $\mathcal F$ on $M$ with $\mathcal A \in \mathcal F$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal F$ be the [[Definition:Equivalence Class|equivalence class]] of $\mathcal A$ under the [[Definition:Relation|relation]] of [[Definition:Compatible Atlases|compatibility]].
+By [[Compatibility of Atlases is Equivalence Relation]], this is indeed an [[Definition:Equivalence Relation|equivalence relation]].
+By definition we have $\mathcal A \in \mathcal F$.
+By [[Relation Partitions Set iff Equivalence]], $\mathcal F$ is an [[Definition:Element|element]] of the [[Definition:Partition (Set Theory)|partition]] of [[Definition:Equivalence Class|equivalence classes]].
+By definition, the [[Definition:Element|elements]] of a [[Definition:Partition (Set Theory)|partition]] are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+Therefore if $\mathcal G \ne \mathcal F$ is an [[Definition:Element|element]] of the [[Definition:Partition (Set Theory)|partition]], we must have:
+: $\mathcal A \notin \mathcal G$
+Therefore $\mathcal A$ belongs to exactly one [[Definition:Differentiable Structure|differentiable structure]].
+{{qed}}
+{{explain|Clarification is needed as to why this result should be categorised in [[:Category:Manifolds]].}}
+[[Category:Manifolds]]
+1d6a163jw7q2sy421hs3o2njg2xs4ya
+\end{proof}<|endoftext|>
+\section{Surgery for Rings}
+Tags: Ring Isomorphisms, Ring Monomorphisms
+
+\begin{theorem}
+Let $R$ and $S$ be [[Definition:Commutative and Unitary Ring|commutative rings with unity]], and $\phi: R \to S$ a [[Definition:Ring Monomorphism|ring monomorphism]].
+Then there is a ring $T$ [[Definition:Ring Isomorphism|isomorphic]] to $S$ that contains $R$ as a [[Definition:Subring|subring]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be the [[Definition:Disjoint Union (Set Theory)|disjoint union]] $T = R \cup \paren {S \setminus \Img \phi}$.
+Define $\theta : T \to S$ as follows:
+:If $x \in R$, then $\map \theta x = \map \phi x$
+:If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
+We claim that $\theta$ is an isomorphism.
+''Injectivity'': Let $\map \theta x = \map \theta y$.
+Since $\theta \sqbrk R = \phi \sqbrk R$, we have $\theta \sqbrk R \cap \paren {S \setminus \Img \phi} = \O$.
+Therefore either $x, y \in R$ or $x, y \in \paren {S \setminus \Img \phi}$.
+If $x, y \in R$ then $\map \phi x = \map \phi y$, so $\theta$ is injective because $\phi$ is.
+If $x, y \in \paren {S \setminus \Img \phi}$ then $\theta$ acts as the identity and $x = y$.
+''Surjectivity'': Under $\theta$, $R$ surjects onto $\Img \phi = \phi \sqbrk R$ by definition.
+Moreover:
+:$\theta \sqbrk {S \setminus \Img \phi} = I \sqbrk {S \setminus \Img \phi} = \paren {S \setminus \Img \phi}$
+where $I$ denotes the [[Definition:Identity Mapping|identity mapping]].
+Therefore $\theta$ is surjective everywhere.
+Now we endow $T$ with addition and multiplication: for $x, y \in T$, let
+:$x + y = \map {\theta^{-1} } {\map \theta x + \map \theta y}$
+:$x y = \map {\theta^{-1} } {\map \theta x \, \map \theta y}$
+so trivially we have
+:$\map \theta {x + y} = \map \theta x + \map \theta y$
+:$\map \theta {x y} = \map \theta x \, \map \theta y$
+Therefore $T$ is the isomorphic image of a ring, and an isomorphism is precisely a map that preserves the ring axioms.
+Thus $T$ is a ring [[Definition:Ring Isomorphism|isomorphic]] to $S$ containing $R$ as a subring.
+{{qed}}
+{{proof wanted|I wonder whether this page might be no more than an application of the [[Embedding Theorem]].}}
+[[Category:Ring Isomorphisms]]
+[[Category:Ring Monomorphisms]]
+orqom74s8aylmkd8ikavss1u965boik
+\end{proof}<|endoftext|>
+\section{Clopen Set contains Components of All its Points}
+Tags: Clopen Sets, Connected Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$ be [[Definition:Clopen Set|both closed and open]] in $T$.
+Then $H$ contains the [[Definition:Component (Topology)|components]] of all of its points.
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a [[Definition:Clopen Set|clopen set]] in $T$.
+By definition, $H$ is [[Definition:Open Set (Topology)|open]] and so $H \in \tau$.
+But as $H$ is also [[Definition:Closed Set (Topology)|closed]], by definition $\relcomp S H \in \tau$ where $\complement_S$ denotes [[Definition:Relative Complement|complement relative to $S$]].
+Thus $H$ and $\relcomp S H$ are both [[Definition:Open Set (Topology)|open]] such that:
+:$H \cap \relcomp S H = \O$ from [[Intersection with Relative Complement is Empty]]
+:$H \cup \relcomp S H = S$ from [[Union with Relative Complement]]
+and so forming a [[Definition:Partition (Topology)|partition]] of $T$.
+As $H$ and $\relcomp S H$ are both [[Definition:Closed Set (Topology)|closed]], it follows from [[Closed Set Equals its Closure]] that:
+:$H \cap \paren {\relcomp S H}^- = \O = H^- \cap \relcomp S H$
+and so by definition $H$ and $\relcomp S H$ are [[Definition:Separated Sets|separated]].
+{{AimForCont}} that $H$ does not contain the [[Definition:Component (Topology)|components]] of all of its points.
+That is:
+:$\exists x \in H: \map {\operatorname {Comp}_x} T \nsubseteq H$
+where $\map {\operatorname {Comp}_x} T$ is the [[Definition:Component (Topology)|component]] of $x$ in $T$.
+Then:
+:$\exists z \in \map {\operatorname {Comp}_x} T: z \in H \cup \relcomp S H$
+This means that $H \cup \relcomp S H$ is [[Definition:Connected Set (Topology)|connected]].
+This [[Definition:Contradiction|contradicts]] the fact that $H$ and $\relcomp S H$ are [[Definition:Separated Sets|separated]].
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Connectedness Between Two Points is an Equivalence Relation}
+Tags: Equivalence Relations, Connectedness Between Two Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $a \sim b $ denote the [[Definition:Relation|relation]]:
+:$a \sim b \iff \exists S \subseteq T: S$ is [[Definition:Connected Between Two Points|connected between the two points ]] $a$ and $b$
+where $a, b \in X$.
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+Trivially, any [[Definition:Separation (Topology)|separation]] of $T$ is a [[Definition:Partition (Set Theory)|partition]] of $S$.
+So any $a \in S$ will be in exactly one of the [[Definition:Open Set (Topology)|open sets]] forming the [[Definition:Partition (Set Theory)|partition]].
+So $a \sim a$ and $\sim$ is seen to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+If $a \sim b$ then each [[Definition:Separation (Topology)|separation]] of $T$ includes a single [[Definition:Open Set (Topology)|open set]] $U \ne \O$ which contains both $a$ and $b$.
+Trivially it follows that $b \sim a$ and so $\sim$ is seen to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $a \sim b$ and $b \sim c$.
+Then by definition:
+:Each [[Definition:Separation (Topology)|separation]] of $T$ includes a single [[Definition:Open Set (Topology)|open set]] $U_1 \ne \O$ which contains both $a$ and $b$.
+:Each [[Definition:Separation (Topology)|separation]] of $T$ includes a single [[Definition:Open Set (Topology)|open set]] $U_2 \ne \O$ which contains both $b$ and $c$.
+So $b \in U_1$ and $b \in U_2$ and so $U_1 \cap U_2 \ne \O$.
+So by the definition of [[Definition:Partition (Set Theory)|partition]] it follows that $U_1 = U_2$ and so $a \sim c$.
+So $\sim$ has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+$\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Hence by definition it is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quasicomponent is Intersection of Clopen Sets}
+Tags: Clopen Sets, Connectedness Between Two Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $p \in S$.
+Then the [[Definition:Quasicomponent|quasicomponent]] containing $p$ equals the [[Definition:Set Intersection|intersection]] of all [[Definition:Clopen Set|sets which are both open and closed]] which contain $p$.
+\end{theorem}
+
+\begin{proof}
+Let $C$ be the [[Definition:Quasicomponent|quasicomponent]] of $p$.
+Let $Q$ be the [[Definition:Set|set]] of [[Definition:Clopen Set|clopen sets]] containing $p$ that are not equal to $S$.
+First let $x \in C$.
+Let $U \in Q$.
+By definition of $Q$:
+:$p \in U$
+Let $V = \relcomp S U$ be the [[Definition:Relative Complement|complement of $U$ relative to $S$]].
+By [[Complement of Clopen Set is Clopen]] $V$ is also a [[Definition:Clopen Set|clopen set]] of $T$.
+By [[Clopen Set and Complement form Separation]] $U$ and $V$ [[Definition:Separation (Topology)|separate]] $S$.
+By the definition of [[Definition:Quasicomponent|quasicomponent]], $U \mid V$ is a [[Definition:Separation (Topology)|separation]] of $S$ which contains a single [[Definition:Open Set (Topology)|open set]] $U$ containing both $x$ and $p$.
+Thus:
+:$x \in U$
+As this holds for all such $U$:
+:$x \in \displaystyle \bigcap Q$
+As this holds for all such $x$:
+:$C \subseteq Q$
+{{qed|lemma}}
+Let $x \in \displaystyle \bigcap Q$.
+Let $U \mid V$ be a [[Definition:Separation (Topology)|separation]] of $S$.
+Let $p \in U$.
+Then $U \in Q$.
+Thus $x \in U$.
+Thus $x$ lies on the same side as $p$ of any [[Definition:Separation (Topology)|separation]] of $X$.
+So $x \in C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Space with One Quasicomponent is Connected}
+Tags: Connectedness Between Two Points, Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which has one [[Definition:Quasicomponent|quasicomponent]].
+Then $T$ is [[Definition:Connected Topological Space|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+[[Definition:By Hypothesis|By hypothesis]], the [[Definition:Quasicomponent|quasicomponent]] of $x$ is $S$ itself.
+Thus by definition of [[Definition:Quasicomponent|quasicomponent]]:
+:$\forall y \in S: y \sim x$
+where $\sim$ is the [[Definition:Relation|relation]] defined on $T$ as:
+:$x \sim y \iff T$ is [[Definition:Connected Between Two Points|connected between the two points]] $x$ and $y$
+Let $K = \displaystyle \bigcap_{x \mathop \in U} U: U$ is [[Definition:Clopen Set|clopen]] in $T$.
+By [[Quasicomponent is Intersection of Clopen Sets]]:
+:$\displaystyle \bigcap K = S$
+Thus there is no [[Definition:Non-Empty Set|non-empty]] [[Definition:Clopen Set|clopen set]] of $T$ apart from $S$.
+The result follows by definition of [[Definition:Connected (Topology)/Topological Space/Definition 4|connected space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Localization of Ring Exists}
+Tags: Commutative Algebra
+
+\begin{theorem}
+Let $A$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $S \subseteq A$ be a [[Definition:Multiplicatively Closed Subset|multiplicatively closed subset]] with $0 \notin S$.
+Then there exists a [[Definition:Localization of Ring|localization]] $\struct {A_S, \iota}$ of $A$ at $S$.
+\end{theorem}
+
+\begin{proof}
+Define a [[Definition:Relation|relation]] $\sim$ on the [[Definition:Cartesian Product|Cartesian product]] $A \times S$ by:
+:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
+=== [[Localization of Ring Exists/Lemma 1|Lemma 1]] ===
+{{:Localization of Ring Exists/Lemma 1}}{{qed|lemma}}
+Let $A_S$ be used to denote the  $\paren {A \times S} / \sim$.
+Let $a / s$ be the [[Definition:Equivalence Class|equivalence class]] of $\tuple {a, s}$ in $\paren {A \times S} / \sim$.
+For $\dfrac a s, \dfrac b t \in A_S$, let the following be defined:
+:$\dfrac a s + \dfrac b t = \dfrac {a t + b s} {s t}$
+:$\dfrac a s \cdot \dfrac b t = \dfrac {a b} {s t}$
+=== [[Localization of Ring Exists/Lemma 2|Lemma 2]] ===
+{{:Localization of Ring Exists/Lemma 2}}{{qed|lemma}}
+Now define $\iota: A \to A_S$ by:
+:$\map \iota a := \dfrac a 1$
+It is to be shown that $\struct {A_S, \iota}$ satisfy the universal property for localization.
+{{MissingLinks|universal property for localization}}
+Let $B$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $g: A \to B$ be a [[Definition:Mapping|mapping]] such that:
+:$\map g S \subseteq B^\times$
+{{explain|$B^\times$}}
+Suppose that $h: A_S \to B$ is a [[Definition:Ring Homomorphism|ring homomorphism]] with $h \circ \iota = g$.
+Then we must have:
+:$\map h {\dfrac a 1} = \map g a$
+and:
+:$\map h {\dfrac 1 s} \cdot \map h {\dfrac s 1} = 1$
+Therefore:
+:$\map h {\dfrac 1 s} \map g s = 1$
+so:
+:$\map h {\dfrac 1 s} = \map g s^{-1}$
+Therefore:
+:$\map h {\dfrac a s} = \map h {\dfrac a 1} \cdot \map h {\dfrac 1 s} = \map g a \map g s^{-1}$.
+So if such $h$ exists it must equal $\map g a \map g s^{-1}$, so is unique.
+{{MissingLinks|Links to results demonstrating the validity of the steps in the above.}}
+Therefore, to conclude the proof we pull out:
+=== [[Localization of Ring Exists/Lemma 3|Lemma 3]] ===
+{{:Localization of Ring Exists/Lemma 3}}{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transitivity of Integrality}
+Tags: Algebraic Number Theory, Commutative Algebra, Transitivity of Integrality
+
+\begin{theorem}
+Let $A \subseteq B \subseteq C$ be [[Definition:Ring Extension|extensions]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
+Suppose that $C$ is [[Definition:Integral Ring Extension|integral]] over $B$, and $B$ is [[Definition:Integral Ring Extension|integral]] over $A$.
+Then $C$ is [[Definition:Integral Ring Extension|integral]] over $A$.
+\end{theorem}
+
+\begin{proof}
+First, a lemma:
+=== [[Transitivity of Integrality/Lemma|Lemma]] ===
+{{:Transitivity of Integrality/Lemma}}{{qed|lemma}}
+Now let $x \in C$.
+Thus $x$ is supposed [[Definition:Integral Element of Ring Extension|integral]] over $B$.
+That is, we can find an expression:
+:$(1): \quad x^n + b_{n - 1} x^{n - 1} + \dotsb + b_1 x + b_0 = 0, \quad b_i \in B, \ i = 0, \dotsc, n - 1$
+Let $D$ be the [[Definition:Subring|subring]] of $C$ generated by $A \cup \set {b_0, \dotsc, b_{n - 1} }$.
+By the lemma, $D$ is finitely generated over $A$.
+Moreover, $D \sqbrk x$ is finitely generated over $D$ because of the equation $(1)$.
+Therefore by [[Transitivity of Finite Generation]]:
+:$D \sqbrk x$ is a finitely generated $A$-module.
+Finally by [[Equivalent Definitions of Integral Dependence]], $x$ is [[Definition:Integral Element of Ring Extension|integral]] over $A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Closure is Integrally Closed}
+Tags: Algebraic Number Theory, Commutative Algebra
+
+\begin{theorem}
+Let $A \subseteq B$ be an [[Definition:Ring Extension|extension]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
+Let $C$ be the [[Definition:Integral Closure|integral closure]] of $A$ in $B$.
+Then $C$ is [[Definition:Integrally Closed|integrally closed]].
+\end{theorem}
+
+\begin{proof}
+Suppose $x \in B$ is integral over $C$.
+Certainly $C$ is integral over $A$, so by [[Transitivity of Integrality]], $C[x]$ is integral over $A$.
+In particular, $x$ is integral over $A$, so $x \in C$.
+{{Qed}}
+[[Category:Algebraic Number Theory]]
+[[Category:Commutative Algebra]]
+1k8rp56mvkw2r0m8ykdktnr7yke9yul
+\end{proof}<|endoftext|>
+\section{Universal Property for Field of Quotients}
+Tags: Fields of Quotients, Integral Domains, Universal Properties
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
+Let $\struct {F, \oplus, \cdot}$ be a [[Definition:Field of Quotients|field of quotients]] of $D$.
+Then $F$ satisfies the following [[Definition:Universal Property|universal property]]:
+There exists a [[Definition:Ring Homomorphism|(ring) homomorphism]] $\iota : D \to F$ such that:
+::for every [[Definition:Field (Abstract Algebra)|field]] $\tilde F$ and
+:and:
+::for every [[Definition:Ring Homomorphism|(ring) homomorphism]] $\phi: D \to \tilde F$
+:there exists a unique [[Definition:Field Homomorphism|field homomorphism]] $\psi: F \to \tilde F$ satisfying:
+:::$\psi \iota = \phi$
+That is, the following diagram [[Definition:Commutative Diagram|commutes]]:
+:[[File:FieldFracComDiag.jpg|Universal Property]]
+Namely we may take:
+:$\psi: a / b \mapsto \map \phi a \map \phi b^{-1}$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Fields of Quotients]]
+[[Category:Integral Domains]]
+[[Category:Universal Properties]]
+qpbvhwzhxjrhrmcmdi13a29p3ot2g5j
+\end{proof}<|endoftext|>
+\section{Constant Mapping is Continuous}
+Tags: Continuous Mappings, Constant Mappings
+
+\begin{theorem}
+Let $T_A = \left({A, \tau_A}\right)$ and $T_B = \left({B, \tau_B}\right)$ be [[Definition:Topological Space|topological spaces]].
+Let $b \in B$ be any point in $B$.
+Let $f_b: A \to B$ be the [[Definition:Constant Mapping|constant mapping]] defined by:
+:$\forall x \in A: f_b \left({x}\right) = b$
+Then $f_b$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+We have by definition that:
+:$\forall x \in A: f_b \left({x}\right) = b$
+So:
+$f^{-1} \left({b}\right) = A$
+For $c \in B: c \ne B$ we have that:
+$f^{-1} \left({c}\right) = \varnothing$
+Let $U \in \tau_B$ such that $b \in U$.
+Then $f^{-1} \left({U}\right) = A$
+Let $V \in \tau_B$ such that $b \notin V$.
+Then $f^{-1} \left({V}\right) = \varnothing$
+From the definition of [[Definition:Topology|topology]], $A$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Empty Set is Element of Topology]], $\varnothing$ is also [[Definition:Open Set (Topology)|open]] in $T$.
+So $f_b$ fulfils the conditions for it to be [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Path-Connectedness is Equivalence Relation}
+Tags: Equivalence Relations, Path-Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $a \sim b $ denote the [[Definition:Relation|relation]]:
+:$a \sim b \iff a$ is [[Definition:Path-Connected Points|path-connected]] to $b$
+where $a, b \in S$.
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+From [[Point is Path-Connected to Itself]], we have that $a \sim a$.
+So $\sim$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+If $a \sim b$ then $a$ is is [[Definition:Path-Connected Points|path-connected]] to $b$ by definition.
+We form the mapping $g: \closedint 0 1 \to \closedint 0 1$:
+:$\map g x = 1 - x$
+which is trivially [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By [[Composite of Continuous Mappings is Continuous]] $f \circ g$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Putting it together we see that $f \circ g$ maps $0$ to $b$ and $1$ to $a$.
+So $b \sim a$ and $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Follows directly from [[Joining Paths makes Another Path]].
+{{qed|lemma}}
+$\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Hence by definition it is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arc-Connectedness is Equivalence Relation}
+Tags: Equivalence Relations, Arc-Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $a \sim b $ denote the [[Definition:Relation|relation]]:
+:$a \sim b \iff a$ is [[Definition:Arc-Connected Points|arc-connected]] to $b$
+where $a, b \in S$.
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+The definition of [[Definition:Arc-Connected Points|arc-connected]] specifically makes provision that:
+:$\forall a \in S: a \sim a$
+specifically to ensure that $\sim$ be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+If $x \sim y$ then $x$ is is [[Definition:Arc-Connected Points|arc-connected]] to $y$ by definition.
+In the trivial case of $x = y$, $y \sim x$ by [[Definition:Reflexive Relation|reflexivity]].
+Suppose that $x \ne y$.
+We form the [[Definition:Mapping|mapping]] $g: \R \to \R$:
+:$\map g x = 1 - x$
+which is trivially [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and [[Definition:Injection|injective]].
+By [[Composite of Continuous Mappings is Continuous]], $g \circ f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By [[Composite of Injections is Injection]], $g \circ f$ is also [[Definition:Injection|injective]].
+Putting it together we see that $g \circ f$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Injection|injection]] which maps $0$ to $y$ and $1$ to $x$.
+So $y \sim x$ and $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Follows directly from [[Joining Arcs makes Another Arc]].
+{{qed|lemma}}
+$\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Hence by definition it is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Joining Arcs makes Another Arc}
+Tags: Topology
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $\mathbb I \subseteq \R$ be the [[Definition:Closed Unit Interval|closed unit interval]] $\left[{0 \,.\,.\, 1}\right]$.
+Let $a,b,c$ be three distinct points of $T$.
+Let $f, g: \mathbb I \to T$ be [[Definition:Arc (Topology)|arcs]] in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
+Let $h: \mathbb I \to T$ be the [[Definition:Mapping|mapping]] given by:
+:$h \left({x}\right) = \begin{cases}
+f \left({2 x}\right) & : x \in \left[{0 \,.\,.\, \dfrac 1 2}\right] \\
+g \left({2 x - 1}\right) & : x \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]
+\end{cases}$
+Then either:
+:$h$ is an [[Definition:Arc (Topology)|arc]] in $T$
+or
+:There exists some [[Definition:Restriction of Mapping|restriction]] of $h$ which, possibly after reparametrisation, is an [[Definition:Arc (Topology)|arc]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Arc in Topological Space is Path]], $f$ and $g$ are also [[Definition:Path (Topology)|paths]] in $T$.
+So by [[Joining Paths makes Another Path]] it follows that $h$ is a [[Definition:Path (Topology)|path]] in $T$.
+Now if $\operatorname{Im} \left({f}\right) \cap \operatorname{Im} \left({g}\right) = b$ it can be seen that:
+:$\displaystyle \forall x \in \operatorname{Im} \left({h}\right): x = \begin{cases}
+f \left({y}\right) & \text{for some } y \in \left[{0 \,.\,.\, \dfrac 1 2}\right], \text{ or} \\
+g \left({z}\right) & \text{for some } z \in \left[{\dfrac 1 2 \,.\,.\, 1}\right] \\
+\end{cases}$
+and it follows that $h$ is an [[Definition:Injection|injection]], and therefore an [[Definition:Arc (Topology)|arc]].
+{{mistake|that should be an XOR, really. And the intervals for $y$ and $z$ are simply $I$}}
+On the other hand, suppose:
+: $\exists y \in \left[{0 \,.\,.\, \dfrac 1 2}\right]: \exists z \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]: f \left({y}\right) = g \left({z}\right)$
+such that $f \left({y}\right) \ne b$.
+{{finish|It remains to show that you can built an arc out of the bits up till where the arcs cross. It's intuitively obvious but requires some analysis work.}}
+{{qed}}
+[[Category:Topology]]
+dd45te87wsxmhkn1yz0eacd2javv00r
+\end{proof}<|endoftext|>
+\section{Arc in Topological Space is Path}
+Tags: Topology
+
+\begin{theorem}
+Let $T = \struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $f$ be an [[Definition:Arc (Topology)|arc]] in $T$.
+Then $f$ is a [[Definition:Path (Topology)|path]] in $T$.
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Arc (Topology)|arc]] from $a$ to $b$ is a [[Definition:Path (Topology)|path]] $f: I \to T$ such that $f$ is [[Definition:Injection|injective]].
+Hence the result, by definition.
+{{qed}}
+[[Category:Topology]]
+k0foahpy900i24trhw6en2y1gfckc4o
+\end{proof}<|endoftext|>
+\section{Vector Times Magnitude Same Length As Magnitude Times Vector}
+Tags: Vector Algebra
+
+\begin{theorem}
+Let $\mathbf u$ and $\mathbf v$ be two [[Definition:Vector (Linear Algebra)|vectors]] in the [[Definition:Vector Space|vector space]] $\struct {G, +_G, \circ}_K$ over a [[Definition:Division Ring|division ring]] $\struct {K, +_K, \times}$ with subfield $\R$ such that $\R \subseteq \map Z K$ with $\map Z K$ the [[Definition:Center (Abstract Algebra)/Ring|center]] of $K$
+Let $\norm {\mathbf u}$ and $\norm {\mathbf v}$ be the [[Definition:Vector Length|lengths]] of $\mathbf u$ and $\mathbf v$ respectively.
+Then:
+:$\norm {\paren {\norm {\mathbf v} \circ \mathbf u} } = \norm {\paren {\norm {\mathbf u} \circ \mathbf v} }$
+\end{theorem}<|endoftext|>
+\section{Relationship between Component Types}
+Tags: Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $p \in S$.
+Let:
+: $A$ be the [[Definition:Arc Component|arc component]] of $p$
+: $P$ be the [[Definition:Path Component|path component]] of $p$
+: $C$ be the [[Definition:Component (Topology)|component]] of $p$
+: $Q$ be the [[Definition:Quasicomponent|quasicomponent]] of $p$.
+Then:
+:$A \subseteq P \subseteq C \subseteq Q$
+In general, the [[Definition:Subset|inclusions]] do not hold in the other direction.
+\end{theorem}
+
+\begin{proof}
+Let $f \in A$.
+By [[Arc in Topological Space is Path]] we have that $f \in P$.
+That is, $A \subseteq P$.
+{{qed|lemma}}
+----
+Let $f \in P$.
+From [[Path-Connected Space is Connected]] we have directly that $P \subseteq C$.
+{{qed|lemma}}
+----
+Let $f \in C$.
+From [[Connected Space is Connected Between Two Points]] we have directly that $C \subseteq Q$.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+{{finish|Plenty to do.}}
+\end{proof}<|endoftext|>
+\section{Increasing Sequence in Ordered Set Terminates iff Maximal Element}
+Tags: Increasing Mappings, Sequences
+
+\begin{theorem}
+Let $\left({P, \le}\right)$ be an [[Definition:Ordering|ordered set]].
+{{TFAE}}
+: $(1): \quad$ Every increasing [[Definition:Sequence|sequence]] $x_1 \leq x_2 \leq x_3 \leq \cdots$ with $x_i \in P$ eventually terminates: there is $n \in \N$ such that $x_n = x_{n+1} = \cdots$.
+: $(2): \quad$ Every non-empty subset of $P$ has a [[Definition:Maximal|maximal]] element.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Suppose $(1)$ holds.
+Pick $\varnothing \ne S \subseteq P$.
+Let $x_1 \in S$ be arbitrary.
+Given $x_k \in S$, pick $x_{k+1} \in S$ strictly bigger than $x_k$.
+By hypothesis the process must eventually terminate, say $x_n$ is the last element.
+Then by construction there are no larger elements than $x_n$, i.e. $x_n$ is a maximal element of $S$.
+{{qed|lemma}}
+{{improve|Last three sentences can be made more explicit and less handwaving}}
+=== $(2)$ implies $(1)$ ===
+Suppose $(2)$ holds.
+Let $\left({x_k}\right)_{k \in \N}$ be an [[Definition:Increasing Sequence|increasing sequence]] of elements of $P$.
+By hypothesis, the set $\left\{{x_k: k \in \N}\right\}$ has a [[Definition:Maximal Element|maximal element]], say $x_n$.
+Then since $x_m \ge x_n$ for all $m \geq n$, we must have $x_n = x_{n+1} = \cdots$.
+{{qed}}
+[[Category:Increasing Mappings]]
+[[Category:Sequences]]
+26tn5iv2h0yqpm30p23o2v3z3lvjqoz
+\end{proof}<|endoftext|>
+\section{Characterisation of UFDs}
+Tags: Unique Factorization Domains, Factorization
+
+\begin{theorem}
+Let $A$ be an [[Definition:Integral Domain|integral domain]].
+{{TFAE}}
+: $(1): \quad A$ is a [[Definition:Unique Factorization Domain|unique factorisation domain]] 
+: $(2): \quad A$ is a [[Definition:GCD Domain|GCD domain]] satisfying the [[Definition:Ascending Chain Condition|ascending chain condition]] on [[Definition:Principal Ideal of Ring|principal ideals]].
+: $(3): \quad A$ satisfies the [[Definition:Ascending Chain Condition|ascending chain condition]] on [[Definition:Principal Ideal of Ring|principal ideals]] and every [[Definition:Irreducible Element of Ring|irreducible element]] of $A$ is a [[Definition:Prime Element|prime element]] of $A$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Unique Factorization Domains]]
+[[Category:Factorization]]
+2y7r9urx0npae77nocb10s6j20xxxss
+\end{proof}<|endoftext|>
+\section{Ultraconnected Space is Path-Connected}
+Tags: Path-Connected Spaces, Ultraconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Ultraconnected Space|ultraconnected]].
+Then $T$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Ultraconnected Space|ultraconnected]].
+Let $a, b \in S$.
+Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the [[Definition:Closure (Topology)|closure]] of $\set a$.
+Such a $p$ can be chosen, as $T$ being [[Definition:Ultraconnected Space|ultraconnected]] guarantees that $\set a^- \cap \set b^- \ne \O$.
+Consider the [[Definition:Mapping|mapping]] $f: \closedint 0 1 \to X$ such that:
+:$\map f x = \begin{cases}
+a & : x \in \hointr 0 {\dfrac 1 2} \\
+p & : x = \dfrac 1 2 \\
+b & : x \in \hointl {\dfrac 1 2} 1 \\
+\end{cases}$
+Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+{{explain|The above is a nontrivial claim}}
+The result follows from the definition of [[Definition:Path-Connected Space|path-connectedness]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Real-Valued Function on Irreducible Space is Constant}
+Tags: Irreducible Spaces, Constant Mappings
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
+Let $f: S \to \R$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Real-Valued Function|real-valued function]].
+Then $f$ is [[Definition:Constant Mapping|constant]], that is:
+:$\exists a \in \R: \forall x \in S: \map f x = a$
+\end{theorem}
+
+\begin{proof}
+By definition of $T = \struct {S, \tau}$ being [[Definition:Irreducible Space|irreducible]]:
+:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O\implies U_1 \cap U_2 \ne \O$
+For every $x \in \R$, define:
+:$L_x := f^{-1} \sqbrk {\openint {-\infty} x}$
+:$U_x := f^{-1} \sqbrk {\openint x \infty}$
+By [[Definition:Everywhere Continuous Mapping (Topology)|continuity]] of $f$, these are [[Definition:Open Set (Topology)|open]] in $T$.
+They are also [[Definition:Disjoint Sets|disjoint]], because for each $s \in S$, $\map f s < x$ and $\map f s > x$ cannot both hold simultaneously.
+Suppose now that $\map f s = a$ and $\map f {s'} = b$, for some $s, s' \in S$ and $a, b \in \R$ with $a < b$.
+Define $c = \dfrac {a + b} 2$.
+Then $a < c < b$, whence:
+:$s \in L_c$ and $s' \in U_c$
+However, because $T$ is [[Definition:Irreducible Space|irreducible]], it follows that either $L_c$ or $U_c$ must be [[Definition:Empty Set|empty]].
+This contradicts the existence of such $a$ and $b$.
+Hence $f$ must be [[Definition:Constant Mapping|constant]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Irreducible Space is Pseudocompact}
+Tags: Pseudocompact Spaces, Irreducible Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
+Then $T$ is [[Definition:Pseudocompact Space|pseudocompact]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Continuous Real-Valued Function on Irreducible Space is Constant]].
+A [[Definition:Constant Mapping|constant mapping]] is trivially [[Definition:Bounded Mapping|bounded]].
+Hence the result by definition of [[Definition:Pseudocompact Space|pseudocompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Connectedness Properties}
+Tags: Connected Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Connected Topological Space|connectedness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Topological Space|topological space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Ultraconnected Space|Ultraconnected]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Arc-Connected Space|Arc-Connected]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Path-Connected Space|Path-Connected]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Irreducible Space|Irreducible]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Connected Topological Space|Connected]] ||
+|-
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+:[[Ultraconnected Space is Path-Connected]]
+:[[Arc-Connected Space is Path-Connected]]
+:[[Path-Connected Space is Connected]]
+:[[Irreducible Space is Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arc-Connected Space is Path-Connected}
+Tags: Path-Connected Spaces, Arc-Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Arc-Connected Space|arc-connected]].
+Then $T$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Arc-Connected Space|arc-connected]].
+Then $\forall x, y \in S$, there exists a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Injection|injection]] $f: \closedint 0 1 \to S$, such that $\map f 0 = x$ and $\map f 1 = y$.
+As $f$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Injection|injection]], it is also simply a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+The result follows from the definition of [[Definition:Path-Connected Space|path-connectedness]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Irreducible Space is Connected}
+Tags: Irreducible Spaces, Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
+Then $T$ is [[Definition:Connected Topological Space|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Irreducible Space|irreducible]].
+Then:
+:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
+So trivially there are no two [[Definition:Open Set (Topology)|open sets]] that can form a [[Definition:Separation (Topology)|separation]] of $T$.
+The result follows from definition of [[Definition:Connected Topological Space|connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Ultraconnected Space is not T1}
+Tags: T1 Spaces, Ultraconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Ultraconnected Space|ultraconnected]].
+If $S$ has more than one [[Definition:Element|element]], then $T$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+That is, if $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]] with more than one [[Definition:Element|element]], it is not [[Definition:Ultraconnected Space|ultraconnected]].
+\end{theorem}
+
+\begin{proof}
+$T = \struct {S, \tau}$ be [[Definition:Ultraconnected Space|ultraconnected]].
+Thus by definition:
+:$(1): \quad \forall x, y \in S: \set x^- \cap \set y^- \ne \O$
+Let $a, b \in S$ such that $a \ne b$.
+{{AimForCont}} $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+By definition of [[Definition:Fréchet Space (Topology)/Definition 2|$T_1$ Space]], $\set a$ and $\set b$ are [[Definition:Closed Set (Topology)|closed]].
+From [[Closed Set Equals its Closure]] we have that $\set a^- = \set a$ and $\set b^- = \set b$.
+It immediately follows that:
+:$\set a^- \cap \set b^- = \O$
+But that [[Definition:Contradiction|contradicts]] $(1)$ above. 
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ultraconnected Space is T4}
+Tags: T4 Spaces, Ultraconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Ultraconnected Space|ultraconnected]].
+Then $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+Recall the definition of a [[Definition:T4 Space|$T_4$ space]]:
+$T = \struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]] {{iff}}:
+:for any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$, there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $A$ and $B$ respectively.
+As no two [[Definition:Closed Set (Topology)|closed sets]] of an [[Definition:Ultraconnected Space|ultraconnected space]] are actually [[Definition:Disjoint Sets|disjoint]], it follows that [[Definition:T4 Space|$T_4$-ness]] follows [[Definition:Vacuous Truth|vacuously]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dot Product of Vector with Itself}
+Tags: Dot Product, Dot Product of Vector with Itself
+
+\begin{theorem}
+:$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf u \cdot \mathbf u
+      | r = u_1 u_1 + u_2 u_2 + \cdots + u_n u_n
+      | c = {{Defof|Dot Product}}
+}}
+{{eqn | r = u_1^2 + u_2^2 + \cdots + u_n^2
+      | c = 
+}}
+{{eqn | r = \paren {\sqrt {\sum_{i \mathop = 1}^n u_i^2} }^2
+      | c = 
+}}
+{{eqn | r = \norm {\mathbf u}^2
+      | c = {{Defof|Vector Length|subdef = Real Vector Space|Vector Length in $\R^n$}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf u \cdot \mathbf u
+      | r = \left\Vert{\mathbf u}\right\Vert \left\Vert{\mathbf u}\right\Vert \cos \angle \mathbf u, \mathbf u
+      | c = [[Cosine Formula for Dot Product]]
+}}
+{{eqn | r = \left\Vert{\mathbf u}\right\Vert^2 \cos 0
+      | c = since the angle between a vector and itself is $0$
+}}
+{{eqn | r = \left\Vert{\mathbf u}\right\Vert^2
+      | c = [[Cosine of Zero is One]]
+}}
+{{end-eqn}}
+{{qed}}
+=== Note ===
+Because this theorem is used to prove the general ($n$-dimensional) case of [[Cosine Formula for Dot Product]], this proof is circular the way we have defined the dot product.
+However, some texts use the cosine formula as the definition of the dot product and derive the sum of products form as a consequence.
+The two definitions are equivalent, so we have included this proof to show how the statement would be proved from the cosine definition.
+\end{proof}<|endoftext|>
+\section{Cosine Formula for Dot Product}
+Tags: Dot Product
+
+\begin{theorem}
+Let $\mathbf v,\mathbf w$ be two non-[[Definition:Zero Vector|zero]] [[Definition:Vector (Euclidean Space)|vectors in $\R^n$]].
+The [[Definition:Dot Product|dot product]] of $\mathbf v$ and $\mathbf w$ can be calculated by:
+:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
+where:
+:$\norm {\, \cdot \,}$ denotes [[Definition:Vector Length|vector length]] and
+:$\theta$ is the [[Definition:Angle Between Vectors|angle between $\mathbf v$ and $\mathbf w$]].
+\end{theorem}
+
+\begin{proof}
+There are two cases, the first where the two vectors are not scalar multiples of each other, and the second where they are.
+=== Case 1 ===
+:[[File:AngleBetweenTwoVectors.png|500px]]
+Let the two vectors $\mathbf v$ and $\mathbf w$ not be scalar multiples of each other. 
+Then by the definition of [[Definition:Angle Between Vectors|angle between vectors]], we have $\theta$ defined as in the [[Definition:Triangle (Geometry)|triangle]] as shown above.
+(Note that from [[Angle Between Non-Zero Vectors Always Defined]], such a triangle is guaranteed to exist).
+By the [[Law of Cosines]]:
+:$\norm {\mathbf v - \mathbf w}^2 = \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
+Now, observe that:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf v - \mathbf w}^2
+      | r = \paren {\mathbf v - \mathbf w} \cdot \paren {\mathbf v - \mathbf w}
+      | c = [[Dot Product of Vector with Itself]]
+}}
+{{eqn | r = \mathbf v \cdot \mathbf v - 2 \paren {\mathbf v \cdot \mathbf w} + \mathbf w \cdot \mathbf w
+      | c = [[Properties of Dot Product]]
+}}
+{{eqn | r = \norm {\mathbf v}^2 + \norm {\mathbf w}^2 - 2 \paren {\mathbf v \cdot \mathbf w}
+      | c = [[Dot Product of Vector with Itself]]
+}}
+{{end-eqn}}
+Equating these two expressions for $\norm {\mathbf v - \mathbf w}^2$ gives:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta
+      | r = \norm {\mathbf v}^2 + \norm {\mathbf w}^2 - 2 \paren {\mathbf v \cdot \mathbf w}
+      | c =
+}}
+{{eqn | l = -2 \paren {\norm {\mathbf v} \norm {\mathbf w} \cos \theta}
+      | r = -2 \paren {\mathbf v \cdot \mathbf w}
+      | c =
+}}
+{{eqn | l = \norm {\mathbf v} \norm {\mathbf w} \cos \theta
+      | r = \mathbf v \cdot \mathbf w
+      | c =
+}}
+{{end-eqn}}
+which is exactly the desired result.
+{{qed|lemma}}
+=== Case 2 ===
+Let $\mathbf v = \paren {v_1, v_2, \ldots, v_n}$ and $\mathbf w = \paren {w_1, w_2, \ldots, w_n}$.
+{{WLOG}}, let $\mathbf v = c \mathbf w$, where $c$ is some scalar. 
+If $c > 0$, then by the definition of [[Definition:Angle Between Vectors|angle between vectors]]:
+:$\theta = 0 \implies \cos \theta = 1$
+If $c < 0$, then by the definition of [[Definition:Angle Between Vectors|angle between vectors]]:
+:$\theta = \pi \implies \cos \theta = -1$
+(Note that $c$ cannot be $0$ because we have stipulated $\mathbf v$ and $\mathbf w$ to be non-zero).
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf v \cdot \mathbf w
+      | r = \paren {c \mathbf w} \cdot \mathbf w
+      | c =
+}}
+{{eqn | r = c \paren {\mathbf w \cdot \mathbf w}
+      | c = [[Properties of Dot Product]]
+}}
+{{eqn | r = c \paren {\norm {\mathbf w}^2}
+      | c = [[Dot Product of Vector with Itself]]
+}}
+{{eqn | r = \paren {c \norm {\mathbf w} } \norm {\mathbf w}
+      | c =
+}}
+{{eqn | r = c \sqrt {\sum_{i \mathop = 1}^n w_i^2} \norm {\mathbf w}
+      | c ={{Defof|Vector Length}} in $\R^n$
+}}
+{{eqn | r = \sgn \paren c \sqrt{c^2 \sum_{i \mathop = 1}^n w_i^2} \norm {\mathbf w}
+      | c = where $\sgn \paren c$ is the [[Definition:Signum Function|signum function]], that is the algebraic sign of $c$
+}}
+{{eqn | r = \sgn \paren c \sqrt {\sum_{i \mathop = 1}^n \paren {c w_i}^2} \norm {\mathbf w}
+      | c =
+}}
+{{eqn | r = \sgn \paren c \sqrt {\sum_{i \mathop = 1}^n v_i^2} \norm {\mathbf w}
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = \sgn \paren c \norm {\mathbf v} \norm {\mathbf w}
+      | c = Definition of $\norm {\mathbf u}$
+}}
+{{eqn | r = \norm {\mathbf v} \norm {\mathbf w} \cos \theta
+      | c = $\sgn \paren c = \cos \theta$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quasicomponents and Components are Equal in Locally Connected Space}
+Tags: Locally Connected Spaces, Components, Quasicomponents
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Connected Space|locally connected]].
+Then $A \subseteq S$ is a [[Definition:Component (Topology)|component]] of $T$ {{iff}} $A \subseteq S$ is a [[Definition:Quasicomponent|quasicomponent]] of $T$.
+\end{theorem}<|endoftext|>
+\section{Irreducible Space is Locally Connected}
+Tags: Irreducible Spaces, Locally Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]].
+Then $T$ is [[Definition:Locally Connected Space|locally connected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Irreducible Space|irreducible]].
+Then:
+:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
+So trivially there are no two [[Definition:Open Set (Topology)|open sets]] that can form a [[Definition:Separation (Topology)|separation]] of $T$.
+As a [[Definition:Basis (Topology)|basis]] consists of [[Definition:Open Set (Topology)|open sets]], this applies to all sets in a [[Definition:Basis (Topology)|basis]] for $T$.
+The result follows from definition of [[Definition:Locally Connected Space|locally connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quasicomponents and Path Components are Equal in Locally Path-Connected Space}
+Tags: Locally Path-Connected Spaces, Quasicomponents, Path Components
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Path-Connected Space|locally path-connected]].
+Then $A \subseteq S$ is a [[Definition:Path Component|path component]] of $T$ {{iff}} $A \subseteq S$ is a [[Definition:Quasicomponent|quasicomponent]] of $T$.
+\end{theorem}<|endoftext|>
+\section{Quasicomponents and Arc Components are Equal in Locally Arc-Connected Space}
+Tags: Locally Arc-Connected Spaces, Quasicomponents, Arc Components
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+Then $A \subseteq S$ is an [[Definition:Arc Component|arc component]] of $T$ {{iff}} $A$ is a [[Definition:Quasicomponent|quasicomponent]] of $T$.
+\end{theorem}<|endoftext|>
+\section{Locally Arc-Connected Space is Locally Path-Connected}
+Tags: Locally Arc-Connected Spaces, Locally Path-Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+Then $T$ is also [[Definition:Locally Path-Connected Space|locally path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Locally Arc-Connected Space|arc-connected]].
+Then $T$ has a [[Definition:Basis (Topology)|basis]] consisting entirely of [[Definition:Arc-Connected Set|arc-connected sets]].
+From [[Arc-Connected Space is Path-Connected]], this [[Definition:Basis (Topology)|basis]] consisting entirely of [[Definition:Path-Connected Set (Topology)|path-connected sets]].
+The result follows from definition of [[Definition:Locally Path-Connected Space|locally path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Locally Path-Connected Space is Locally Connected}
+Tags: Locally Path-Connected Spaces, Locally Connected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Locally Path-Connected Space|locally path-connected]].
+Then $T$ is also [[Definition:Locally Connected Space|locally connected]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$ be any point of $T$.
+Let $\BB$ be a [[Definition:Local Basis|local basis]] of [[Definition:Path-Connected Set (Topology)|path-connected sets]] for $x$.
+From [[Path-Connected Space is Connected]], $\BB$ is a [[Definition:Local Basis|local basis]] of [[Definition:Connected Set (Topology)|connected sets]].
+Thus, $T$ is [[Definition:Locally Connected Space|locally connected]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Locally Euclidean Space is Locally Compact}
+Tags: Manifolds
+
+\begin{theorem}
+Let $M$ be a [[Definition:Locally Euclidean Space|locally Euclidean space]] of some dimension $d$.
+Then $M$ is [[Definition:Locally Compact Space|locally compact]].
+\end{theorem}
+
+\begin{proof}
+Let $m \in M$ be arbitrary.
+By definition there is an [[Definition:Open Neighborhood|open neighborhood]] $U$ of $m$, [[Definition:Homeomorphic Metric Spaces|homeomorphic]] to an [[Definition:Open Set of Metric Space|open subset]] of $\R^d$.
+By the definition of an [[Definition:Open Set of Metric Space|open set]], there is some [[Definition:Open Ball of Metric Space|open ball]]:
+:$B = B_\delta \left({\phi \left({m}\right)}\right) = \left\{{x \in \R^d: \left|{x - \phi \left({m}\right)}\right| < \delta}\right\}$
+of [[Definition:Radius of Open Ball|radius]] $\delta$ containing $\phi \left({m}\right)$, contained in $U$.
+By [[Closure of Open Ball in Metric Space]] and [[Topological Closure is Closed]], the set:
+:$C = \left\{{x \in \R^d: \left|{x - \phi \left({m}\right)}\right| \le \dfrac \delta 2}\right\}$
+is [[Definition:Closed Set (Topology)|closed]], and $C \subseteq B \subseteq U$.
+Moreover, $C$ is trivially [[Definition:Bounded Metric Space|bounded]], hence [[Definition:Compact Topological Space|compact]] by the [[Heine-Borel Theorem]].
+Now if $\phi$ is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]] $U \to \R^d$, then by definition $\phi^{-1}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Therefore by [[Continuous Image of Compact Space is Compact]], $\phi^{-1} \left({C}\right) \subseteq M$ is [[Definition:Compact Topological Space|compact]].
+Furthermore $m \in \phi^{-1} \left({C}\right)$ because $\phi \left({m}\right) \in C$.
+Thus every point of $M$ has a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood (Topology)|neighborhood]].
+{{Qed}}
+{{explain|Clarification is needed as to why this result should be categorised in [[:Category:Manifolds]].}}
+[[Category:Manifolds]]
+atj03zp0r9bpm24yi9z708tx91dy20y
+\end{proof}<|endoftext|>
+\section{UFD is GCD Domain}
+Tags: Unique Factorization Domains, Factorization
+
+\begin{theorem}
+Let $A$ be a [[Definition:Unique Factorization Domain|unique factorisation domain]].
+Then $A$ is a [[Definition:GCD Domain|GCD domain]].
+\end{theorem}
+
+\begin{proof}
+Let $x \divides y$ denote [[Definition:Divisor of Ring Element|$x$ divides $y$]].
+Let $x, y \in A$, with [[Definition:Complete Factorization|complete factorizations]]:
+:$x = u x_1 \cdots x_r$
+:$y = v y_1 \cdots y_s$
+where:
+:$u, v$ are [[Definition:Unit of Ring|units]]
+:the $x_i$, $y_i$ [[Definition:Irreducible Element of Ring|irreducible]].
+We arrange the [[Definition:Complete Factorization|complete factorizations]] as follows:
+:$x = u \paren {x_1 \cdots x_t} x_{t + 1} \cdots x_r$
+:$y = v \paren {y_1 \cdots y_t} y_{t + 1} \cdots y_s$
+where:
+:$t \le \min \set {r, s}$
+:For $i = 1, \ldots, t$, $x_i$ and $y_i$ are [[Definition:Associate in Integral Domain|associates]]
+:For any $i \in \set {t + 1, \ldots, r}$, $j \in \set {t + 1, \ldots, s}$, $x_i$ and $y_j$ are not [[Definition:Associate in Integral Domain|associates]].
+Let $d = x_1 \cdots x_t$ (recall that the [[Definition:Vacuous Product|empty product]] is $1$, i.e. $d = 1$ when $t = 0$).
+We claim that $d$ is a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] for $x$ and $y$.
+Certainly $d \divides x$ and $d \divides y$.
+So, let $f$ be another [[Definition:Common Divisor of Ring Elements|common divisor]] of $x$ and $y$.
+We can find $w, z \in A$ such that $x = f w$, and $y = f z$.
+If $f$ is a [[Definition:Unit of Ring|unit]], then $f \divides d$ by definition.
+{{AimForCont}} $f \nmid d$.
+Then the [[Definition:Complete Factorization|complete factorization]] of $f$ must contain an [[Definition:Irreducible Element of Ring|irreducible element]] that does not [[Definition:Divisor of Ring Element|divide]] $d$.
+Call this [[Definition:Irreducible Element of Ring|irreducible element]] $g$.
+We have that:
+:$g$ must [[Definition:Divisor of Ring Element|divide]] some $x_j$ where $j > t$
+and
+:$g$ must [[Definition:Divisor of Ring Element|divide]] some $y_k$ where $k > t$.
+Either:
+:$g$ is a [[Definition:Unit of Ring|unit]], [[Definition:Contradiction|contradicting]] its [[Definition:Irreducible Element of Ring|irreducibility]]
+or:
+:$x_j$ and $y_k$ are not [[Definition:Irreducible Element of Ring|irreducible]], which is a [[Definition:Contradiction|contradiction]] also.
+Hence by [[Proof by Contradiction]]:
+:$f \divides d$
+and so $x$ and $y$ have a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]].
+{{qed}}
+[[Category:Unique Factorization Domains]]
+[[Category:Factorization]]
+k6llamsyc7acqtb6vxxrbunq6tnjqvp
+\end{proof}<|endoftext|>
+\section{Unique Factorization Domain is Integrally Closed}
+Tags: Algebraic Number Theory, Factorization, Unique Factorization Domains
+
+\begin{theorem}
+Let $A$ be a [[Definition:Unique Factorization Domain|unique factorization domain (UFD)]].
+Then $A$ is [[Definition:Integrally Closed|integrally closed]].
+\end{theorem}
+
+\begin{proof}
+Let $K$ be the [[Definition:Field of Quotients|field of quotients]] of $A$.
+Let $x \in K$ be [[Definition:Integral Element of Ring Extension|integral]] over $A$.
+Let:
+$x = a / b$
+for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$.
+This makes sense because a [[UFD is GCD Domain]].
+There is an equation:
+:$\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$
+{{explain|This is not actually an equation, it has no equals sign in it.}}
+with $a_i \in A$, $i = 0, \dotsc, n - 1$.
+Multiplying by $b^n$, we obtain:
+:$a^n + b c = 0$
+with $c \in A$.
+Therefore:
+:$b \divides a^n$
+{{AimForCont}} $b$ is not a [[Definition:Unit of Ring|unit]].
+Then:
+:$\gcd \set {a, b} \notin A^\times$
+which is a [[Definition:Contradiction|contradiction]].
+So $b$ is a [[Definition:Unit of Ring|unit]], and:
+:$a b^{-1} \in A$
+{{explain|Make it clear why this fulfils the conditions for the statement of the result}}
+{{qed}}
+[[Category:Algebraic Number Theory]]
+[[Category:Factorization]]
+[[Category:Unique Factorization Domains]]
+hc338lvrefaby5nmas6fbieb7ik5n10
+\end{proof}<|endoftext|>
+\section{Totally Disconnected Space is T1}
+Tags: T1 Spaces, Totally Disconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Totally Disconnected Space|totally disconnected]].
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Totally Disconnected Space|totally disconnected]].
+Then as its [[Definition:Component (Topology)|components]] are [[Definition:Singleton|singletons]], it follows that each of its points is [[Definition:Closed Point|closed]].
+From [[Equivalence of Definitions of T1 Space|Equivalence of Definitions of $T_1$ Space]], it follows that $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Linear Operator is Well Defined}
+Tags: Linear Operators, Determinants
+
+\begin{theorem}
+Let $V$ be a [[Definition:Trivial Vector Space|nontrivial]] [[Definition:Finite Dimensional Vector Space|finite dimensional]] [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $A: V \to V$ be a [[Definition:Linear Operator|linear operator]] of $V$.
+Then the [[Definition:Determinant of Linear Operator|determinant]] $\det A$ of $A$ is [[Definition:Well Defined|well defined]], that is, does not depend on the choice of a [[Definition:Basis of Vector Space|basis]] of $V$.
+\end{theorem}
+
+\begin{proof}
+Let $A_\mathcal B$ and $A_\mathcal C$ be the [[Definition:Matrix|matrices]] of $A$ [[Definition:Relative Matrix|relative]] to $\mathcal B$ and $\mathcal C$ respectively.
+Let $\det$ also denote the [[Definition:Determinant of Matrix|determinant]] of a [[Definition:Matrix|matrix]].
+We are required to show that $\det A_\mathcal B = \det A_\mathcal C$.
+Let $P$ be the [[Definition:Change of Basis Matrix|change of basis]] matrix from $\mathcal B$ to $\mathcal C$.
+By [[Change of Coordinate Vectors Under Linear Mapping]] and since $A_\mathcal B$ and $A_\mathcal C$ represent the same [[Definition:Linear Operator|linear operator]] with respect to different bases, the following diagram commutes:
+:[[File:Determinant Independent of Basis.png|250px]]
+where $u \in V$, and $\left[{u}\right]_\mathcal B$ indicates the [[Definition:Coordinate Vector|coordinate vector]] of $u$ with repect to $\mathcal B$, and similarly for $\left[{u}\right]_\mathcal C$.
+That is, $P A_\mathcal B = A_\mathcal C P$.
+Therefore, because a [[Change of Basis is Invertible]] we have:
+: $A_\mathcal B = P^{-1} A_\mathcal C P$
+So:
+{{begin-eqn}}
+{{eqn | l = \det \left({A_\mathcal B}\right)
+      | r = \det \left({P^{-1} A_\mathcal C P}\right)
+}}
+{{eqn | r = \det \left({P^{-1} }\right) \det \left({A_\mathcal C}\right) \det \left({P}\right)
+      | c = [[Determinant of Matrix Product]]
+}}
+{{eqn | r = \det \left({P}\right)^{-1} \det \left({A_\mathcal C}\right) \det \left({P}\right)
+      | c = [[Determinant of Inverse Matrix]]
+}}
+{{eqn | r = \det \left({A_\mathcal C}\right)
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Linear Operators]]
+[[Category:Determinants]]
+k7z62jj2gyon3luhlhzgyqlbikf8elv
+\end{proof}<|endoftext|>
+\section{Localization Preserves Integral Closure}
+Tags: Algebraic Number Theory
+
+\begin{theorem}
+Let $A \subseteq B$ be an [[Definition:Ring Extension|extension]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
+Let $C$ be the [[Definition:Integral Closure|integral closure]] of $A$ in $B$.
+Let $S \subseteq A$ be a [[Definition:Multiplicatively Closed Subset of Ring|multiplicatively closed subset]].
+Then $C_S$ is the [[Definition:Integral Closure|integral closure]] of $A_S$ in $B_S$, where subscript $S$ indicates the [[Definition:Localization of Ring|localization]] at $S$.
+\end{theorem}
+
+\begin{proof}
+First we show that $C_S$ is an integral extension of $A_S$.
+Let $x \in C_S$, and $\iota$ be the canonical [[Definition:Inclusion Mapping|inclusion]] from a [[Definition:Ring (Abstract Algebra)|ring]] to its [[Definition:Localization of Ring|localization]].
+There exists $s \in S$ such that $sx \in \iota(C)$, say $sx = \iota(c)$.
+Since $c \in C$ is [[Definition:Integral Element of Ring Extension|integral]], there is an equation:
+:$c^n + a_{n-1}c^{n-1} + \cdots + a_1 c + a_0 = 0$
+with the $a_i \in A$. Since $\iota$ is a [[Definition:Ring Homomorphism|homomorphism]], we have:
+:$\left({s x}\right)^n + \iota \left({a_{n-1} }\right) \left({s x}\right)^{n-1} + \cdots + \iota \left({a_1}\right) \left({s x}\right) + \iota \left({a_0}\right) = 0$
+{{explain|Just a link, probably, to indicate the nature of $\iota$ - so as to confirm that it is a (ring) homomorphism.}}
+Multiplying by $s^{-n}$ this gives:
+:$x^n + s^{-1}\iota \left({a_{n-1} }\right)x^{n-1} + \cdots + x^{1-n} \iota \left({a_1}\right) x + s^{-n}\iota \left({a_0}\right) = 0$
+This is a [[Definition:Monic Polynomial|monic polynomial]] with coefficients in $A_S$, so $x$ is integral over $A_S$.
+Now let $x \in B_S$ be integral over $A_S$.
+So we have an equation
+:$x^n + r_{n-1}x^{n-1} + \cdots + r_1 x + r_0 = 0$
+Let:
+:$t \in S$ be such that $t x \in \iota \left({B}\right)$
+:$u_i$ be such that $u_i r_i \in \iota \left({A}\right)$ for $i = 0, \ldots, n-1$
+:$s = t u_0 \cdots u_{n-1}$.
+Then we have:
+:$\left({s x}\right)^n + r_{n-1} s \left({s x}\right)^{n-1} + \cdots + r_1 s^{n-1} \left({s x}\right) + r_0 s^n = 0$
+Now let:
+: $s x = \iota \left({y}\right)$ for $y \in B$
+: $r_{n-i} s^i = \iota \left({q_i}\right)$ for $i = 1, \ldots, n$
+Then:
+:$\iota \left({y^n + q_1 y^{n-1} + \cdots + q_{n-1} y + q_n}\right) = 0$
+Now:
+: $\ker \left({\iota: B \to B_S}\right) = \left\{{x \in B : s x = 0 \text{ for some } s \in S}\right\}$
+So there exists $v \in S$ such that:
+:$\left({v y}\right)^n + q_1 v (\left({v y}\right)^{n-1} + \cdots + q_{n-1} v^{n-1} \left({v y}\right) + q_n v^n = 0$
+Therefore $v y \in B$ is [[Definition:Integral Element of Ring Extension|integral]] over $A$, so $vy \in C$.
+Therefore:
+:$\iota \left({v y}\right) \in C_S$
+and:
+:$\iota \left({y}\right) \in C_S$
+Thus:
+:$x = s^{-1} \iota \left({y}\right) \in C_S$
+{{qed}}
+{{MissingLinks}}
+[[Category:Algebraic Number Theory]]
+fn2hcx9gs0fpovp7mrjceq789a5r8vh
+\end{proof}<|endoftext|>
+\section{Integrally Closed is Local Property}
+Tags: Algebraic Number Theory
+
+\begin{theorem}
+Let $A$ be an [[Definition:Integral Domain|integral domain]].
+For a [[Definition:Prime Ideal of Ring|prime ideal]] $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the [[Definition:Localization of Ring|localization]] at $S = A \divides \mathfrak p$.
+Then the following are equivalent:
+:$(1): \quad A$ is [[Definition:Integrally Closed|integrally closed]]
+:$(2): \quad A_{\mathfrak p}$ is [[Definition:Integrally Closed|integrally closed]] for all [[Definition:Prime Ideal of Ring|prime ideals]] $\mathfrak p$.
+:$(3): \quad A_{\mathfrak m}$ is [[Definition:Integrally Closed|integrally closed]] for all [[Definition:Maximal Ideal of Ring|maximal ideals]] $\mathfrak m$.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Let $\map Q R$ denote the [[Definition:Field of Quotients|field of quotients]] of an [[Definition:Integral Domain|integral domain]] $R$.
+We have by [[Localization Preserves Integral Closure]] that:
+:$\map Q {A_{\mathfrak p}} = \map Q A$
+Hence $A_{\mathfrak p}$ is [[Definition:Integrally Closed|integrally closed]] for all [[Definition:Prime Ideal of Ring|prime ideals]] $\mathfrak p$.
+{{qed|lemma}}
+=== $(2)$ implies $(3)$ ===
+This is true because a [[Maximal Ideal of Commutative and Unitary Ring is Prime Ideal]].
+{{qed|lemma}}
+=== $(3)$ implies $(1)$ ===
+{{ProofWanted|Need some more results on localization first}}
+[[Category:Algebraic Number Theory]]
+qtf8nxkecuh3zjh3gb4g70co6oyzuoh
+\end{proof}<|endoftext|>
+\section{Totally Disconnected and Locally Connected Space is Discrete}
+Tags: Discrete Topology, Locally Connected Spaces, Totally Disconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is both [[Definition:Totally Disconnected Space|totally disconnected]] and [[Definition:Locally Connected Space|locally connected]].
+Then $T$ is the [[Definition:Discrete Space|discrete space]] on $S$.
+\end{theorem}
+
+\begin{proof}
+So, let $T = \struct {S, \tau}$ be that [[Definition:Topological Space|topological space]] which is both [[Definition:Totally Disconnected Space|totally disconnected]] and [[Definition:Locally Connected Space|locally connected]].
+As $T$ is [[Definition:Totally Disconnected Space|totally disconnected]], every point is a [[Definition:Component (Topology)|component]] and therefore [[Definition:Closed Point|closed]].
+As $T$ is [[Definition:Locally Connected Space|locally connected]], there exists a [[Definition:Basis (Topology)|basis]] $\BB$ of $T$ such that every element of $\BB$ is a [[Definition:Component (Topology)|component]] of $T$.
+In order for $T$ to be covered by $\BB$, every [[Definition:Singleton|singleton]] subset of $T$ must be in $\BB$.
+From the definition of [[Definition:Topology|topology]], the union of any number of these [[Definition:Singleton|singleton]] sets is an [[Definition:Open Set (Topology)|open set]] of $T$.
+That is, every subset of $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
+That is, every element of the [[Definition:Power Set|power set]] of $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
+This is precisely the definition of the [[Definition:Discrete Space|discrete space]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Totally Disconnected but Connected Set must be Singleton}
+Tags: Totally Disconnected Spaces, Connected Sets, Singletons
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$ be both [[Definition:Totally Disconnected Space|totally disconnected]] and [[Definition:Connected Set (Topology)|connected]].
+Then $H$ is a [[Definition:Singleton|singleton]].
+\end{theorem}
+
+\begin{proof}
+If $H$ is [[Definition:Totally Disconnected Space|totally disconnected]] then its individual points are [[Definition:Separated Points|separated]].
+If $H$ is [[Definition:Connected Set (Topology)|connected]] it can not be represented as the [[Definition:Set Union|union]] of two (or more) [[Definition:Separated Sets|separated sets]].
+So $H$ can have only one point in it.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Totally Separated Space is Completely Hausdorff and Urysohn}
+Tags: Totally Separated Spaces, Completely Hausdorff Spaces, Urysohn Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Totally Separated Space|totally separated]].
+Then $T$ is [[Definition:Completely Hausdorff Space|completely Hausdorff]] and [[Definition:Urysohn Space|Urysohn]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Totally Separated Space|totally separated space]].
+Then for every $x, y \in S: x \ne y$ there exists a [[Definition:Separation (Topology)|separation]] $U \mid V$ of $T$ such that $x \in U, y \in V$.
+Consider a [[Definition:Mapping|mapping]] $f: S \to \closedint 0 1$ such that:
+:$\Img {f {\restriction_U} } = \set 0$
+:$\Img {f {\restriction_V} } = \set 1$
+where:
+:$f {\restriction_U}$ and $f {\restriction_V}$ denote the [[Definition:Restriction of Mapping|restrictions]] of $f$ to $U$ and $V$ respectively
+:$\Img g$ denotes the [[Definition:Image of Subset under Mapping|image set]] of a [[Definition:Mapping|mapping]] $g$.
+From [[Constant Mapping is Continuous]], both $f {\restriction_U}$ and $f {\restriction_V}$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] on $U$ and $V$ respectively.
+From [[Continuous Mapping of Separation]], it follows that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] on $S$ itself.
+By definition, $f$ is then an [[Definition:Urysohn Function|Urysohn function]] on $x$ and $y$.
+Hence $T$ is [[Definition:Urysohn Space|Urysohn]].
+Having proved that $T$ is an [[Definition:Urysohn Space|Urysohn space]], it follows from [[Urysohn Space is Completely Hausdorff Space]] that $T$ is also [[Definition:Completely Hausdorff Space|completely Hausdorff]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Morphism is Unique}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Category|category]], and $X$ and [[Definition:Object|object]] of $\mathbf C$.
+Then the [[Definition:Identity Morphism|identity morphism]] $\operatorname{id}_X : X \to X$ is unique.
+\end{theorem}
+
+\begin{proof}
+{{MissingLinks}}
+Let $\operatorname{id}_X^1$, $\operatorname{id}_X^2$ be two identity morphisms for $X$.
+By definition, for any morphism $f : Y \to X$, we have $\operatorname{id}_X^1 \circ f = f$.
+In particular, taking $Y = X$ and $f = \operatorname{id}_X^2$, we have $\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^2$.
+Similarly for any morphism $g : X \to Y$ we have $g \circ \operatorname{id}_X^2 = g$.
+So taking $Y = X$ and $g = \operatorname{id}_X^1$ we have $\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^1$.
+Putting this together we have:
+:$\operatorname{id}_X^2 = \operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^1$
+as required.
+{{qed}}
+[[Category:Category Theory]]
+97fsz6fhmtxyo5cohrtntajdnwm5695
+\end{proof}<|endoftext|>
+\section{Product Category is Category}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ and $\mathbf D$ be [[Definition:Metacategory|metacategories]].
+Then the [[Definition:Product Category|product category]] $\mathbf C \times \mathbf D$ is a [[Definition:Metacategory|metacategory]].
+\end{theorem}
+
+\begin{proof}
+Let $\left({X,Y}\right), \left({X',Y'}\right) \in \mathbf C \times \mathbf D$.
+Let $\left({f,g}\right) : \left({X,Y}\right) \to \left({X',Y'}\right)$ and $\left({h,k}\right) : \left({X',Y'}\right) \to \left({X,Y}\right)$ be [[Definition:Morphism (Category Theory)|morphisms]].
+Let $\operatorname{id}_X$, $\operatorname{id}_Y$ be the [[Definition:Identity Morphism|identity morphisms]] for the [[Definition:Object|objects]] $X$ and $Y$ respectively.
+Then:
+{{begin-eqn}}
+{{eqn|l=\left({f, g}\right) \circ \left({\operatorname{id}_X, \operatorname{id}_Y}\right)
+     |r=\left({f \circ \operatorname{id}_X, g \circ \operatorname{id}_Y}\right)
+     |c=By the definition of [[Definition:Composition of Morphisms|composition]] in the product category.
+}}
+{{eqn|l=
+     |r=\left({f, g}\right)
+     |c=By the definition of the [[Definition:Identity Morphism|identity morphisms]]
+}}
+{{end-eqn}}
+Similarly,
+{{begin-eqn}}
+{{eqn|l=\left({\operatorname{id}_X, \operatorname{id}_Y}\right) \circ \left({h, k}\right)
+     |r=\left({\operatorname{id}_X \circ h, \operatorname{id}_Y \circ k}\right)
+     |c=By the definition of [[Definition:Composition of Morphisms|composition]] in the product category.
+}}
+{{eqn|r=\left({h, k}\right)
+     |c=By the definition of the [[Definition:Identity Morphism|identity morphisms]]
+}}
+{{end-eqn}}
+Therefore $\left({\operatorname{id}_X, \operatorname{id}_Y}\right)$ satisfies the property of an identity morphism.
+Now let $\left({f, g}\right)$, $\left({h, k}\right)$ and $\left({\ell, m}\right)$ be [[Definition:Composable Morphisms|composable morphisms]] of $\mathbf C \times \mathbf D$. We have:
+{{begin-eqn}}
+{{eqn|l=\left({ \left({f, g}\right) \circ \left({h, k}\right) }\right) \circ \left({\ell, m}\right)
+     |r= \left({f \circ h, g \circ k}\right) \circ \left({\ell, m}\right)
+     |c=By the definition of [[Definition:Composition of Morphisms|composition]] in the product category.
+}}
+{{eqn|r=\left({ \left({f \circ h}\right) \circ \ell, \left({g \circ k}\right) \circ m}\right)
+     |c=
+}}
+{{eqn|r=\left({f \circ \left({h \circ \ell}\right), g \circ \left({k \circ m}\right) }\right)
+     |c=By [[Definition:Morphism (Category Theory)|associativity]] of morphisms of $\mathbf C$ and $\mathbf D$.
+}}
+{{eqn|l=
+     |r=\left({f, g}\right) \circ \left({ \left({h, k}\right) \circ \left({\ell, m}\right) }\right)
+     |c=
+}}
+{{end-eqn}}
+Therefore composition of morphisms in $\mathbf C \times \mathbf D$ is also associative.
+{{qed}}
+[[Category:Category Theory]]
+dhcnr0or9t8cmmt0o8r8c1lf4voy08r
+\end{proof}<|endoftext|>
+\section{Category of Sets is Category}
+Tags: Examples of Categories, Category of Sets
+
+\begin{theorem}
+Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
+Then $\mathbf{Set}$ is a [[Definition:Metacategory|metacategory]].
+\end{theorem}
+
+\begin{proof}
+Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a [[Definition:Metacategory|metacategory]].
+For any two mappings their [[Definition:Composition of Mappings|composition]] (in the usual [[Definition:Set Theory|set theoretic]] sense) is again a mapping by [[Composite Mapping is Mapping]].
+For any set $X$, we have the [[Definition:Identity Mapping|identity mapping]] $\operatorname{id}_X$.
+By [[Identity Mapping is Left Identity]] and [[Identity Mapping is Right Identity]], this is the [[Definition:Identity Morphism|identity morphism]] for $X$.
+Finally by [[Composition of Mappings is Associative]], the associative property is satisfied.
+Hence $\mathbf{Set}$ is a [[Definition:Metacategory|metacategory]].
+{{qed}}
+[[Category:Examples of Categories]]
+[[Category:Category of Sets]]
+rpd0zesyfpvfk4le8e9ow3t0xrsnjyl
+\end{proof}<|endoftext|>
+\section{Projection from Product Category}
+Tags: Category Theory, Projections
+
+\begin{theorem}
+Let $\mathcal C$ and $\mathcal D$ be [[Definition:Category|categories]].
+Let $\mathcal C \times \mathcal D$ be the [[Definition:Product Category|product category]].
+The ''projections'':
+:$\pi_{\mathcal C} : \mathcal C \times \mathcal D \to \mathcal C : (f,g) \mapsto f$
+:$\pi_{\mathcal D} : \mathcal C \times \mathcal D \to \mathcal D : (f,g) \mapsto g$
+where $(f,g) \in \operatorname{ob}(\mathcal C \times \mathcal D)$ or $(f,g)\in \operatorname{mor}(\mathcal C \times \mathcal D)$ are [[Definition:Functor|functors]].
+Moreover, $\pi_{\mathcal C}$ and $\pi_{\mathcal D}$ satisfy the following universal property:
+:''For any category $\mathcal E$ and any [[Definition:Functor|functors]] $F : \mathcal E \to \mathcal C$, $G : \mathcal E \to \mathcal D$, there exists a unique functor $H : \mathcal E \to \mathcal C \times \mathcal D$ such that $F = \pi_{\mathcal C}H$ and $G = \pi_{\mathcal D}H$''
+That is, the following diagram commutes:
+:[[File:Product_Category_Universal_Property.png|380px]]
+{{improve|XyJax?}}
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Category Theory]]
+[[Category:Projections]]
+87j70os0sequzprz1zt8rxavkud0mdf
+\end{proof}<|endoftext|>
+\section{Extremally Disconnected Space is Totally Separated}
+Tags: Totally Separated Spaces, Extremally Disconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be an [[Definition:Extremally Disconnected Space|extremally disconnected topological space]].
+Then $T$ is [[Definition:Totally Separated Space|totally separated]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be an [[Definition:Extremally Disconnected Space|extremally disconnected topological space]].
+Then by definition $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]] such that the [[Definition:Closure (Topology)|closure]] of every [[Definition:Open Set (Topology)|open set]] of $T$ is [[Definition:Open Set (Topology)|open]].
+Let $x, y \in S: x \ne y$.
+As $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]], there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $x$ and $y$ respectively.
+Let $U^-$ denote the [[Definition:Closure (Topology)|closure]] of $U$.
+From [[Topological Closure is Closed]], $U^-$ is [[Definition:Closed Set (Topology)|closed]].
+But $U$ is [[Definition:Open Set (Topology)|open]], so from the definition of [[Definition:Extremally Disconnected Space|extremally disconnected space]], then $U^-$ is also [[Definition:Open Set (Topology)|open]].
+So $U^-$ is [[Definition:Clopen Set|clopen]].
+From [[Disjoint Open Sets remain Disjoint with one Closure]], $U^- \cap V = \O$, so $y \notin U^-$.
+Thus $\set {U^- \mid S \setminus U^-}$ is a [[Definition:Partition (Topology)|partition]] where $x \in U^-$ and $y \in X \setminus U^-$.
+Thus, by definition, $T$ is [[Definition:Totally Separated Space|totally separated]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Dimensional Space is T3}
+Tags: T3 Spaces, Zero Dimensional Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional topological space]].
+Then $T$ is a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional space]].
+Let $F \subseteq S$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
+Let also $y \notin F$.
+Then by definition of [[Definition:Closed Set (Topology)|closed]], $\relcomp S F$ is [[Definition:Open Set (Topology)|open]] in $T$, where $\relcomp S F$ is the [[Definition:Relative Complement|complement of $F$ in $S$]].
+As $T$ is [[Definition:Zero Dimensional Space|zero dimensional]], it has a [[Definition:Basis (Topology)|basis]] $\BB$ which consists entirely of [[Definition:Clopen Set|clopen sets]].
+As $\BB$ is a [[Definition:Basis (Topology)|basis]] for $T$, it follows that:
+:$\displaystyle \exists \UU \subseteq \BB: \relcomp S F = \bigcup \UU$
+that is, $\relcomp S F$ is the [[Definition:Set Union|union]] of a [[Definition:Subset|subset]] of [[Definition:Element|elements]] of $\BB$.
+Thus, there is a [[Definition:Set|set]] $U \in \UU$ such that $y \in U$ and $F \subseteq \relcomp S U$.
+But the [[Definition:Element|elements]] of $\UU$ are [[Definition:Clopen Set|clopen sets]], so $U$ is itself [[Definition:Clopen Set|clopen]].
+Thus, by definition, $\relcomp S U$  is also [[Definition:Clopen Set|clopen]].
+So we have that $U$ and $\relcomp S U$ are [[Definition:Open Set (Topology)|open sets]] in $T$ such that:
+:$\exists W, R \in \tau: F \subseteq W, y \in R: R \cap W = \O$
+by setting $R = U$ and $W = \relcomp S U$.
+That is, $T$ is a [[Definition:T3 Space|$T_3$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Dimensional T0 Space is Totally Separated}
+Tags: T0 Spaces, Zero Dimensional Spaces, Totally Separated Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional topological space]] which is also a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Then $T$ is [[Definition:Totally Separated Space|totally separated]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Zero Dimensional Space|zero dimensional space]] which is also a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+As $T$ is [[Definition:Zero Dimensional Space|zero dimensional]], there exists a [[Definition:Basis (Topology)|basis]] $\BB$ whose sets are all [[Definition:Clopen Set|clopen]].
+Let $x, y \in S$.
+As $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]]:
+:$\exists U \in \tau: x \in U, y \notin U$
+or:
+:$\exists U \in \tau: y \in U, x \notin U$
+{{WLOG}}, suppose that $\exists U \in \tau: x \in U, y \notin U$.
+Then:
+:$\exists V \in \BB: x \in V$ and $V \subseteq U$
+by definition of [[Definition:Basis (Topology)|basis]].
+The set $V$ is [[Definition:Clopen Set|clopen]] by the definition of $\BB$.
+But then $x \in V$ which is [[Definition:Open Set (Topology)|open]] and $y \in S \setminus V$ which is also [[Definition:Open Set (Topology)|open]].
+$\set {V \mid S \setminus V}$ is a [[Definition:Partition (Topology)|partition]] and hence $T$ is [[Definition:Totally Separated Space|totally separated]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Scattered T1 Space is Totally Disconnected}
+Tags: T1 Spaces, Scattered Spaces, Totally Disconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Scattered Space|scattered topological space]] which is also a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Then $T$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Scattered Space|scattered space]] which is also a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+We have that every [[Non-Trivial Connected Set in T1 Space is Dense-in-itself|Non-Trivial Connected Set in $T_1$ Space is Dense-in-itself]].
+As $T$ is [[Definition:Scattered Space|scattered]], every $H \subseteq S$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+So $H$ is not [[Definition:Dense-in-itself|dense-in-itself]] and so if $H$ has more than one [[Definition:Element|element]] it can not be [[Definition:Connected Set (Topology)|connected]].
+As $H$ is arbitrary, it follows that $T$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Connected Set in T1 Space is Dense-in-itself}
+Tags: T1 Spaces, Denseness, Connected Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) topological space]].
+Let $H \subseteq S$ be [[Definition:Connected Set (Topology)|connected]] in $T$.
+If $H$ has more than one element, then $H$ is [[Definition:Dense-in-itself|dense-in-itself]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $H$ is not [[Definition:Dense-in-itself|dense-in-itself]].
+Then $\exists x \in H$ such that $x$ is [[Definition:Isolated Point of Subset|isolated]] in $H$.
+That is, $\exists U \in \tau: U \cap H = \set x$.
+Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$.
+Since $T$ is [[Definition:Fréchet Space (Topology)|$T_1$]]:
+:$\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y \in V_y, x \notin V_y}$
+Define $V = \displaystyle \bigcup_{y \in H\\ y \ne x} V_y$.
+Then $\forall z \in H$:
+:$z = x \implies z \in U$
+:$z \ne x \implies z \in V_z \subseteq V$
+Showing that $H \subseteq U \cup V$.
+This also shows that both $U \cap H$ and $V \cap H$ are non-empty.
+Since $x \notin V_y$ for any $y \in H$, 
+we have $x \notin V$. 
+This implies $\set x \cap V = \O$
+So $H \cap U \cap V = \O$.
+Thus we have:
+:$H \subseteq U \cup V$
+:$H \cap U \cap V = \O$
+:$U \cap H \ne \O$
+:$V \cap H \ne \O$
+Showing that $H$ is [[Definition:Disconnected Set (Topology)|disconnected]].
+This is a contradiction.
+Hence we must have $H$ is [[Definition:Dense-in-itself|dense-in-itself]]. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Disconnectedness Properties}
+Tags: Disconnected Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Disconnected Space|disconnectedness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Topological Space|topological space]] $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | [[Definition:Regular Space|Regular]]||
+| align="center" | $\impliedby$ ||
+| align="center" | [[Definition:Zero Dimensional Space|Zero Dimensional]] and [[Definition:Kolmogorov Space|$T_0$]]||
+| align="center" | $\impliedby$ ||
+| align="center" | [[Definition:Discrete Space|Discrete Space]] ||
+| align="center" | $\implies$ ||
+| align="center" | $\implies$ ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Scattered Space|Scattered]] and [[Definition:Fréchet Space (Topology)|$T_1$]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | [[Definition:Extremally Disconnected Space|Extremally Disconnected]] ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\implies$ ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Totally Separated Space|Totally Separated]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Urysohn Space|Urysohn]] ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Fréchet Space (Topology)|$T_1$]] ||
+| align="center" | $\impliedby$ ||
+| align="center" | [[Definition:Totally Disconnected Space|Totally Disconnected]] ||
+| align="center" | $\impliedby$ ||
+| align="center" | $\impliedby$ ||
+| align="center" | $\impliedby$ ||
+| align="center" | $\impliedby$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Totally Pathwise Disconnected Space|Totally Pathwise Disconnected]] ||
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+* [[Discrete Space satisfies all Separation Properties|Discrete Space is $T_0$]] and [[Discrete Space is Zero Dimensional]].
+* [[Discrete Space satisfies all Separation Properties|Discrete Space is $T_1$]] and [[Discrete Space is Scattered|Discrete Space is Scattered]].
+* [[Zero Dimensional Space is T3]], and by definition of a [[Definition:Regular Space|regular space]] as being both [[Definition:T3 Space|$T_3$]] and [[Definition:Kolmogorov Space|$T_0$]].
+* [[Zero Dimensional T0 Space is Totally Separated|Zero Dimensional $T_0$ Space is Totally Separated]].
+* [[Discrete Space is Extremally Disconnected]].
+* [[Extremally Disconnected Space is Totally Separated]].
+* [[Totally Separated Space is Completely Hausdorff and Urysohn]].
+* [[Totally Separated Space is Totally Disconnected]].
+* [[Scattered T1 Space is Totally Disconnected|Scattered $T_1$ Space is Totally Disconnected]].
+* [[Totally Disconnected Space is Totally Pathwise Disconnected]].
+* [[Totally Disconnected Space is T1|Totally Disconnected Space is $T_1$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Scattered Space is T0}
+Tags: T0 Spaces, Scattered Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Scattered Space|scattered topological space]].
+Then $T$ is also a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Suppose $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+From [[Equivalence of Definitions of T0 Space|Equivalence of Definitions of $T_0$ Space]], there exist $x, y \in S$ such that $x$ and $y$ are both [[Definition:Limit Point of Point|limit points]] of each other.
+So by definition of [[Definition:Isolated Point (Topology)/Subset/Definition 2|isolated point]], neither $x$ nor $y$ are [[Definition:Isolated Point of Subset|isolated in $\set {x, y}$]].
+Thus we have found a [[Definition:Subset|subset]] $\set {x, y} \subseteq T$ such that $\set {x, y}$ is by definition [[Definition:Dense-in-itself|dense-in-itself]].
+So $T$ is not [[Definition:Scattered Space|scattered]].
+Hence the result by [[Rule of Transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set with Dispersion Point is Biconnected}
+Tags: Biconnected Sets, Dispersion Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$ be a [[Definition:Connected Set (Topology)|connected set]] in $T$.
+Let $p \in H$ be a [[Definition:Dispersion Point|dispersion point]] of $H$.
+Then $H$ is [[Definition:Biconnected Set|biconnected]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $H$ is not [[Definition:Biconnected Set|biconnected]].
+Then by definition there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Non-Degenerate Connected Set|non-degenerate connected sets]] $U, V$ such that $H = U \cup V$.
+{{WLOG}}, let $p \in U$.
+Then $V \subset H \setminus \set p$.
+As $p$ is a [[Definition:Dispersion Point|dispersion point]] of $H$, $H \setminus \set p$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+Thus, by definition, $H \setminus \set p$ contains no [[Definition:Non-Degenerate Connected Set|non-degenerate connected sets]].
+But this contradicts our definition of $V$, as $V \subset H \setminus \set p$.
+It follows by [[Proof by Contradiction]] that $H$ is [[Definition:Biconnected Set|biconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Totally Disconnected Space is Punctiform}
+Tags: Totally Disconnected Spaces, Punctiform Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Totally Disconnected Space|totally disconnected]].
+Then $T$ is [[Definition:Punctiform Space|punctiform]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be [[Definition:Totally Disconnected Space|totally disconnected]].
+Then by definition its [[Definition:Component (Topology)|components]] are [[Definition:Singleton|singletons]].
+Thus by definition each of its [[Definition:Connected Set (Topology)|connected sets]] are [[Definition:Degenerate Connected Set|degenerate]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance in Pseudometric is Non-Negative}
+Tags: Pseudometric Spaces
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]] on which a [[Definition:Pseudometric|pseudometric]] $d: X \times X \to \R$ has been imposed.
+Then:
+:$\forall x, y \in X: \map d {x, y} \ge 0$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Pseudometric|pseudometric]], we have that:
+{{begin-axiom}}
+{{axiom | n = \text M 1
+        | q = \forall x \in A
+        | m = \map d {x, x} = 0
+}}
+{{axiom | n = \text M 2
+        | q = \forall x, y, z \in A
+        | m = \map d {x, y} + \map d {y, z} \ge \map d {x, z}
+}}
+{{axiom | n = \text M 3
+        | q = \forall x, y \in A
+        | m = \map d {x, y} = \map d {y, x}
+}}
+{{end-axiom}}
+Hence:
+{{begin-eqn}}
+{{eqn | lo= \forall x, y \in X:
+      | l = \map d {x, y} + \map d {y, x}
+      | o = \ge
+      | r = \map d {x, x}
+      | c = $\text M 2$
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \map d {x, y}
+      | o = \ge
+      | r = 0
+      | c = $\text M 1$ and $\text M 3$
+}}
+{{eqn | ll= \leadsto
+      | l = \map d {x, y}
+      | o = \ge
+      | r = 0
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Pseudometric Spaces]]
+hhxteeliri2ilwxojg011vsuv8hloiz
+\end{proof}<|endoftext|>
+\section{Metric Defines Norm iff it Preserves Linear Structure}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $\struct {k, \norm{\,\cdot\,}_k}$ be a [[Definition:Valued Field|valued field]].
+Let $V$ be a [[Definition:Vector Space|vector space]] over the [[Definition:Valued Field|valued field]] $\struct {k, \norm{\,\cdot\,}_k}$.
+Let $d : V \times V \to k$ be a [[Definition:Metric|metric]] on $V$.
+Then the function $\norm{ v } := d(v,0)$ is a [[Definition:Norm on Vector Space|norm]] on $V$ if and only if for all $x,y,z \in V$, $\lambda \in k$:
+:$(1): \quad d(x+z,y+z) = d(x,y)$ ([[Definition:Homogeneous (Metric Spaces)|homogeneity]] or [[Definition:Translation|translation]] [[Definition:Invariance|invariance]])
+:$(2): \quad d(\lambda x, \lambda y) = \norm{\lambda}_k d(x,y)$ (the [[Definition:Enlargement Property|enlargement property]])
+\end{theorem}
+
+\begin{proof}
+Suppose first that $d$ satisfies the [[Definition:Hypothesis|hypotheses]] 1. and 2..
+Since $d(u,v) \geq 0$ for all $u,v \in V$, $\| u \| = d(u,0) \geq 0$ for all $u \in V$.
+Moreover if $\| u \| = 0$ then $d(u,0) = 0$, so $u = 0$.
+Now let $\lambda \in K$, $u \in V$.
+Then, using the enlargement property of $d$:
+{{begin-eqn}}
+{{eqn|l=\norm{\lambda v}
+     |r=d(\lambda v, 0 )
+     |c=By the definition of $\norm{\,\cdot\,}$
+}}
+{{eqn|l=
+     |r=d(\lambda v, \lambda \cdot 0)
+     |c=
+}}
+{{eqn|l=
+     |r=\norm{\lambda}_k d(v,0)
+     |c=By the enlargement property
+}}
+{{eqn|l=
+     |r=\norm{\lambda}_k \norm{v}
+     |c=By the definition of $\norm{\,\cdot\,}$
+}}
+{{end-eqn}}
+Finally if $u,v \in V$, then we have
+{{begin-eqn}}
+{{eqn|l= \Vert u + v \Vert
+     |r= d(u + v, 0)
+     |c=By the definition of $\Vert \cdot \Vert$
+}}
+{{eqn|l=
+     |r= d(u, -v)
+     |c=By translation invariance of $d$
+}}
+{{eqn|l=
+     |r=d(u, 0) + d(0,-v)
+     |o=\leq
+     |c=By the triangle inequality for metrics.
+}}
+{{eqn|l=
+     |r=d(u, 0) + d(v,0)
+     |c=By the enlargement property
+}}
+{{eqn|l=
+     |r=\Vert u \Vert + \Vert v \Vert
+     |c=By the definition of $\Vert \cdot \Vert$
+}}
+{{end-eqn}}
+{{stub|converse to add}}
+{{qed}}
+[[Category:Metric Spaces]]
+18ys98ewi06rum6rtjtl9lio10uxcsm
+\end{proof}<|endoftext|>
+\section{Set of all Sets}
+Tags: Antinomies
+
+\begin{theorem}
+Forming the set $\SS$ of all [[Definition:Set|sets]] leads to a [[Definition:Contradiction|contradiction]].
+\end{theorem}
+
+\begin{proof}
+Let $\SS$ be the set of all sets.
+Then $\SS$ must be an [[Definition:Element|element]] of itself, in symbols, $\SS \owns \SS$.
+Thus we have an infinite descending sequence of membership:
+:$\SS \owns \SS \owns \SS \owns \cdots$
+But by the [[Axiom:Axiom of Foundation|axiom of foundation]], no such sequence exists, a [[Definition:Contradiction|contradiction]].
+{{qed}}
+{{questionable|This is circular: We need to demonstrate the contradiction and then demonstrate why ZF (specifically the AoF) prevents it happening.}}
+[[Category:Antinomies]]
+87mh65z8bqgp2d2r28fe681ukov2elw
+\end{proof}<|endoftext|>
+\section{Real Numbers form Perfect Set}
+Tags: Real Numbers, Perfect Sets
+
+\begin{theorem}
+Consider the [[Definition:Real Number|set of real numbers]] $\R$ as a [[Real Number Line is Complete Metric Space|(complete) metric space]] with the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]].
+Then $\R$ forms a [[Definition:Perfect Set|perfect set]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Perfect Set|perfect set]] is a set which equals its set of [[Definition:Limit Point of Set|limit points]].
+Let $x \in \R$.
+Consider the [[Definition:Sequence|sequence]]:
+:$\left \langle{y_k}\right \rangle = x + \dfrac 1 k$
+Then as $\left \langle{z_k}\right \rangle = \dfrac 1 k$ is a [[Definition:Basic Null Sequences|basic null sequence]] it follows that:
+:$\displaystyle \lim_{n \to \infty} \left \langle{y_k}\right \rangle = x$
+Thus we see that $x$ is a [[Definition:Limit Point of Set|limit point]] of $S$.
+{{qed}}
+[[Category:Real Numbers]]
+[[Category:Perfect Sets]]
+p7sv1yxcmud2uc77hsudekcha3vuuoi
+\end{proof}<|endoftext|>
+\section{Closed Interval in Reals is Uncountable}
+Tags: Uncountable Sets
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Extended Real Number Line|extended real numbers]] such that $a < b$.
+Then the [[Definition:Closed Real Interval|closed interval]] $\set {x \in \R : a \le x \le b} \subseteq \R$ is [[Definition:Uncountable Set|uncountable]].
+\end{theorem}
+
+\begin{proof}
+First suppose that $a, b \in \R$.
+We have that the [[Real Numbers are Uncountable/Cantor's Diagonal Argument|unit interval is uncountable]].
+{{MissingLinks|[[Unit Interval is Uncountable]] (page does not exist at time of editing)}}
+Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b - a} x$.
+Then if $\map f {x_1} = \map f {x_2}$, we have:
+:$a + \paren {b - a} x_1 = a + \paren {b - a} x_2$
+Because $b > a$:
+:$b - a > 0$
+so this implies:
+:$x_1 = x_2$
+Therefore $f$ is [[Definition:Injective|injective]].
+Now if $\closedint a b$ were [[Definition:Countable Set|countable]], there would be an [[Definition:Injective|injective function]] $g : \closedint a b \to \N$.
+By [[Composite of Injections is Injection]], this implies that $g \circ f$ is an injection, so $\closedint 0 1$ is [[Definition:Countable Set|countable]], a contradiction.
+If one or both of $a, b$ is (are) not real, then we can pick a [[Definition:Closed Real Interval|closed interval]] $\closedint c d \subseteq \closedint a b$ with $c, d \in \R, c < d$.
+We know by the above that $\closedint c d$ is [[Definition:Uncountable Set|uncountable]].
+Let $S = \set {x \in \R : a \le x \le b}$. 
+The identity function $I_\R : \closedint c d \to S$ is trivially [[Definition:Injection|injective]].
+If $S$ were [[Definition:Uncountable Set|uncountable]], we would have an [[Definition:Injection|injective function]] $g : S \to \N$.
+Then by [[Composite of Injections is Injection]] we would have an [[Definition:Injection|injection]] $\closedint c d \to \N$ given by $g \circ I_\R$.
+This was shown above to be false, so $S$ is [[Definition:Uncountable Set|uncountable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is T5}
+Tags: Metric Spaces, T5 Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $S, T \subseteq A$ such that $S$ and $T$ are [[Definition:Separated Sets|separated]] in $A$.
+Then:
+:each point $x \in S$ has an [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] $\map {B_{\epsilon_x} } x$ which is [[Definition:Disjoint Sets|disjoint]] from $T$
+:each point $y \in T$ has an [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] $\map {B_{\epsilon_y} } y$ which is [[Definition:Disjoint Sets|disjoint]] from $S$.
+Then:
+:$U_S = \displaystyle \bigcup_{x \mathop \in S} \map {B_{\epsilon_x / 2} } x$
+:$U_T = \displaystyle \bigcup_{y \mathop \in T} \map {B_{\epsilon_y / 2} } y$
+are [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Neighborhood|open neighborhoods]] of $S$ and $T$ respectively.
+{{finish|that's cutting the corner}}
+Hence the result by the definition of [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Perfectly T4}
+Tags: Metric Spaces, Perfectly T4 Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Metric Space is T4|metric space is $T_4$]].
+We also have that every [[Closed Set in Metric Space is G-Delta|closed set in a metric space is a $G_\delta$ set]].
+Hence the result, by definition of a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space fulfils all Separation Axioms}
+Tags: Metric Spaces, Separation Axioms
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then [[Metric Induces Topology|$M$, considered as a topological space]], fulfils all [[Definition:Separation Axioms|separation axioms]]:
+:$M$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]]
+:$M$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]]
+:$M$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]]
+:$M$ is a [[Definition:Semiregular Space|semiregular space]]
+:$M$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]] 
+:$M$ is a [[Definition:T3 Space|$T_3$ space]] 
+:$M$ is a [[Definition:Regular Space|regular space]]
+:$M$ is an [[Definition:Urysohn Space|Urysohn space]]
+:$M$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]
+:$M$ is a [[Definition:Tychonoff Space|Tychonoff (completely regular) space]]
+:$M$ is a [[Definition:T4 Space|$T_4$ space]]
+:$M$ is a [[Definition:Normal Space|normal space]]
+:$M$ is a [[Definition:T5 Space|$T_5$ space]] 
+:$M$ is a [[Definition:Completely Normal Space|completely normal space]]
+:$M$ is a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]
+:$M$ is a [[Definition:Perfectly Normal Space|perfectly normal space]]
+\end{theorem}
+
+\begin{proof}
+We have that:
+:A [[Metric Space is Hausdorff|metric space is a $T_2$ (Hausdorff) space]].
+:A [[Metric Space is Perfectly T4|metric space is a perfectly $T_4$ space]].
+:A [[Metric Space is T5|metric space is a $T_5$ space]].
+From [[Sequence of Implications of Separation Axioms]] we then have:
+:[[T5 Space is T4 Space|$T_5$ space is $T_4$ space]].
+:[[T2 Space is T1 Space|$T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space]].
+:[[T1 Space is T0 Space|$T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space]].
+By definition, a [[Definition:Perfectly Normal Space|perfectly normal space]] is:
+:a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]
+:a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]]
+So $M$ is a [[Definition:Perfectly Normal Space|perfectly normal space]].
+By definition, a [[Definition:Completely Normal Space|completely normal space]] is:
+:a [[Definition:T5 Space|$T_5$ space]]
+:a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]]
+So $M$ is a [[Definition:Completely Normal Space|completely normal space]].
+Then from [[Sequence of Implications of Separation Axioms]] we can complete the chain:
+:[[Completely Normal Space is Normal Space]].
+:[[Normal Space is Tychonoff Space|Normal Space is Tychonoff (Completely Regular) Space]].
+:[[Definition:Completely Regular Space|Completely Regular (Tychonoff)]] implies [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$]] by definition.
+:[[Tychonoff Space is Regular, T2 and T1|Completely Regular (Tychonoff) Space is Regular Space]].
+:[[Tychonoff Space is Urysohn Space|Completely Regular (Tychonoff) Space is Urysohn Space]].
+:[[Definition:Regular Space|Regular]] implies [[Definition:T3 Space|$T_3$]] by definition.
+:[[Regular Space is Completely Hausdorff Space]].
+:[[Regular Space is Semiregular Space]].
+{{qed}}
+There are other chains of implications which can be used, but the above are sufficient to prove the hypothesis.
+\end{proof}<|endoftext|>
+\section{Metric Space is Fully T4}
+Tags: Metric Spaces, Fully T4 Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is a [[Definition:Fully T4 Space|fully $T_4$ space]].
+\end{theorem}<|endoftext|>
+\section{Metric Space is Fully Normal}
+Tags: Metric Spaces, Fully Normal Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is a [[Definition:Fully Normal Space|fully normal space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Metric Space is Fully T4|metric space is fully $T_4$]].
+We also have that a [[Metric Space fulfils all Separation Axioms|metric space is a $T_1$ (Fréchet) space]].
+Hence the result, by definition of a [[Definition:Fully Normal Space|fully normal space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Paracompact}
+Tags: Metric Spaces, Paracompact Spaces, Metric Space is Paracompact
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is a [[Definition:Paracompact Space|paracompact space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Metric Space is Fully Normal]].
+Then we have that a [[Fully Normal Space is Paracompact]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\sequence {C_\alpha}$ be an [[Definition:Open Cover|open cover]] of $M$ [[Definition:Indexed Set|indexed]] by [[Definition:Ordinal|ordinals]].
+Let $\map {B_r} x$ be the [[Definition:Open Ball|open $r$-ball]] in $M$ around $x$.
+For $n \in \N$, let $D_{\alpha n}$ denote the [[Definition:Set Union|union]] of all the [[Definition:Open Ball|open $r$-balls]] $\map {B_{2^{-n} } } x$ such that:
+:$(1): \quad$ $\alpha$ is the smallest [[Definition:Ordinal|ordinal]] such that $x \in C_\alpha$
+:$(2): \quad$ $x \notin D_{\beta j}$ for $j < n$
+:$(3): \quad$ $\map {B_{3 \times 2^{-n} } } x \subseteq C_\alpha$
+It is to be shown that $\sequence {D_{\alpha n} }$ is a [[Definition:Locally Finite Cover|locally finite]] [[Definition:Open Refinement|open refinement]] of $\sequence {C_\alpha}$ which [[Definition:Cover|covers]] $M$.
+By construction, $D_{\alpha n}$ is a [[Definition:Refinement of Cover|refinement]] of $\sequence {C_\alpha}$.
+Let $x \in A$.
+Then there exists a smallest [[Definition:Ordinal|ordinal]] $\alpha$ such that $x \in C_\alpha$.
+Then there exists $n$ [[Definition:Sufficiently Large|sufficiently large]] such that $(3)$ holds.
+Then by $(2)$:
+:$x \in D_{\beta j}$
+for some $j \le n$.
+Thus $\sequence {D_{\alpha n} }$ [[Definition:Cover|covers]] $M$.
+Now assume $x \in A$.
+Let $\alpha$ be the smallest [[Definition:Ordinal|ordinal]] such that $x \in D_{\alpha n}$ for some $n$.
+Choose $j$ such that $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$.
+Let $i \ge n$.
+As $i > n$, every one of the [[Definition:Open Ball|open balls]] of [[Definition:Radius of Open Ball|radius]] $2^{-i}$ used to define $D_{\beta i}$ has its [[Definition:Center of Open Ball|center]] $y$ outside $D_{\alpha n}$
+And because $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$, we have that $\map d {x, y} \ge 2^{-j}$.
+But $i \ge j + 1$ and $n + j \ge j + 1$.
+So:
+:$\map {B_{2^{-n - j} } } x \cap \map {B_{2^{-j} } } y = \varnothing$
+Thus if $i \ge n$, then $\map {B_{2^{-n - j} } } x$ [[Definition:Set Intersection|intersects]] no $D_{\beta i}$.
+Let $i < n$.
+Let us choose:
+:$p \in D_{\beta i}$
+:$q \in D_{\gamma i}$
+:$\beta < \gamma$
+There exist points $y$ and $z$ such that:
+:$p \in \map {B_{2^{-j} } } y \subseteq D_{\beta i}$
+:$q \in \map {B_{2^{-j} } } z \subseteq D_{\gamma i}$
+By $(3)$:
+:$\map {B_{3 \times 2^{-n} } } y \subseteq C_\beta$
+But by $(2)$:
+:$z \notin C_\beta$
+So:
+:$\map d {y, z} \ge 3 \times 2^{-i}$
+and:
+:$\map d {p, q} > 2^{-i} \ge 2^{-n - j + 1}$
+Thus if $i < n$, then $\map {B_{2^{-n - j} } } x$ [[Definition:Set Intersection|intersects]] $D_{\beta i}$ for at most one $\beta$.
+Thus $D_{\alpha n}$ is [[Definition:Locally Finite Cover|locally finite]].
+Then $D_{\alpha n}$ is a [[Definition:Locally Finite Cover|locally finite]] [[Definition:Open Refinement|open refinement]] of $\sequence {C_\alpha}$ which [[Definition:Cover|covers]] $M$.
+Hence by definition $M$ is [[Definition:Paracompact Space|paracompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is First-Countable}
+Tags: Metric Spaces, First-Countable Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in A$.
+Let:
+:$\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$
+where $\map {B_\epsilon} x$ denotes the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball of $x$ in $M$]].
+By [[Surjection from Natural Numbers iff Countable]], we have that $\BB$ is [[Definition:Countable Set|countable]].
+By the definition of a [[Definition:First-Countable Space|first-countable space]], it suffices to show that $\BB$ is a [[Definition:Local Basis|local basis]] at $x$.
+By [[Open Ball of Metric Space is Open Set]], every [[Definition:Element|element]] of $\BB$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+By the definition of an [[Definition:Open Set (Metric Space)|open set]], there exists a [[Definition:Strictly Positive Real Number|strictly positive real number]] $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.
+By the [[Archimedean Principle]], there exists a [[Definition:Natural Number|natural number]] $n > \dfrac 1 \epsilon$.
+That is, $\dfrac 1 n < \epsilon$, and so:
+:$\map {B_{1/n} } x \subseteq \map {B_\epsilon} x \subseteq U$
+The result follows from [[Subset Relation is Transitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Separable iff Second-Countable}
+Tags: Metric Spaces, Second-Countable Spaces, Separable Spaces
+
+\begin{theorem}
+A [[Definition:Metric Space|metric space]] is [[Definition:Separable Space|separable]] {{iff}} it is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Separable Metric Space is Second-Countable]]
+:[[Second-Countable Space is Separable]]
+{{qed}}
+{{ACC|Second-Countable Space is Separable}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Lindelöf iff Second-Countable}
+Tags: Metric Spaces, Lindelöf Spaces, Second-Countable Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is [[Definition:Lindelöf Space|Lindelöf]] {{iff}} $M$ is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+We have from [[Second-Countable Space is Lindelöf]] that [[Definition:Second-Countable Space|second-countability]] implies [[Definition:Lindelöf Space|Lindelöf]] in all [[Definition:Topological Space|topological spaces]], regardless of whether they are [[Definition:Metric Space|metric spaces]] or not.
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose $M = \struct {A, d}$ is [[Definition:Lindelöf Space|Lindelöf]].
+Let us define the [[Definition:Open Cover|open covers]] on $A$:
+:$\CC_k = \set {\map {N_{1/k} } x: x \in S}$
+for all $k \in \N_{>0}$.
+As $M$ is [[Definition:Lindelöf Space|Lindelöf]], each one of these has a [[Definition:Countable Subcover|countable subcover]].
+The [[Definition:Set Union|union]] of all these [[Definition:Countable Subcover|subcovers]] is a [[Definition:Countable Basis|countable basis]] for the [[Metric Induces Topology|topology on $A$]].
+Hence the result, by definition of [[Definition:Second-Countable Space|second-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Countably Compact iff Sequentially Compact}
+Tags: Metric Spaces, Countably Compact Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+Let $M$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is [[Definition:Countably Compact Space|countably compact]] {{iff}} $M$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from the results:
+:[[Countably Compact Metric Space is Sequentially Compact]]
+:[[Sequentially Compact Space is Countably Compact]]
+{{qed}}
+{{ACC|Sequentially Compact Space is Countably Compact}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Weakly Countably Compact iff Countably Compact}
+Tags: Metric Spaces, Countably Compact Spaces, Weakly Countably Compact Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} $M$ is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Metric Space fulfils all Separation Axioms]], a [[Definition:Metric Space|metric space]] is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+The result follows from [[T1 Space is Weakly Countably Compact iff Countably Compact|$T_1$ Space is Weakly Countably Compact iff Countably Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Compact Metric Space is Compact}
+Tags: Metric Spaces, Countably Compact Spaces, Compact Spaces, Countably Compact Metric Space is Compact
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $M$ be [[Definition:Countably Compact Space|countably compact]].
+Then $M$ is [[Definition:Compact Metric Space|compact]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from:
+: [[Metric Space is First-Countable]]
+: [[Countably Compact First-Countable Space is Sequentially Compact]]
+: [[Sequentially Compact Metric Space is Second-Countable]]
+: [[Second-Countable Space is Compact iff Countably Compact]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that a [[Metric Space is Countably Compact iff Sequentially Compact]].
+Then we have that a [[Sequentially Compact Metric Space is Separable]].
+For each $n$, a [[Definition:Metric Space|metric space]] which is [[Definition:Countably Compact|countably compact]] can be covered by [[Definition:Finite Set|finitely]] many [[Definition:Open Ball|open $\paren {1/n}$-balls]]: $\map {B_{1/n} } {x_i}$.
+{{explain|why?}}
+So $\set {x_i}$ is a [[Definition:Everywhere Dense|dense subset]] of $A$ which is [[Definition:Countable|countable]].
+So if a [[Definition:Metric Space|metric space]] is [[Definition:Countably Compact|countably compact]] it is by definition [[Definition:Second-Countable Space|second-countable]].
+{{explain|by definition? See unresolved issues in Talk page.}}
+The result follows from [[Second-Countable Space is Compact iff Countably Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Weakly Locally Compact iff Strongly Locally Compact}
+Tags: Metric Spaces, Weakly Locally Compact Spaces, Strongly Locally Compact Spaces, Locally Compact Hausdorff Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]] {{iff}} $M$ is [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+\end{theorem}
+
+\begin{proof}
+We have from [[Strongly Locally Compact Space is Weakly Locally Compact]] that [[Definition:Strongly Locally Compact Space|strong local compactness]] implies [[Definition:Weakly Locally Compact Space|weak local compactness]] in all [[Definition:Topological Space|topological spaces]], regardless of whether they are [[Definition:Metric Space|metric spaces]] or not.
+So all we need to do is demonstrate that if $M$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]] then it is [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+We have that a [[Metric Space fulfils all Separation Axioms|metric space is a $T_2$ (Hausdorff) space]].
+The result follows from [[Weakly Locally Compact Hausdorff Space is Strongly Locally Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Implications of Metric Space Compactness Properties}
+Tags: Metric Spaces, Compact Spaces
+
+\begin{theorem}
+Let $P_1$ and $P_2$ be [[Definition:Compact Topological Space|compactness properties]] and let:
+:$P_1 \implies P_2$
+mean:
+:If a [[Definition:Metric Space|metric space]] $M$ satsifies property $P_1$, then $M$ also satisfies property $P_2$.
+Then the following sequence of implications holds:
+{|
+|-
+| align="center" | [[Definition:Sequentially Compact Space|Sequentially Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Weakly Sigma-Locally Compact Space|Weakly $\sigma$-Locally Compact]] ||
+| align="center" | $\implies$ ||
+| align="center" | [[Definition:Weakly Locally Compact Space|Weakly Locally Compact]] ||
+|-
+| align="center" | $\Big\Updownarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Updownarrow$ ||
+|-
+| align="center" | [[Definition:Countably Compact Space|Countably Compact]] ||
+| align="center" | ||
+| align="center" | [[Definition:Sigma-Compact Space|$\sigma$-Compact]] ||
+| align="center" | ||
+| align="center" | [[Definition:Strongly Locally Compact Space|Strongly Locally Compact]] ||
+|-
+| align="center" | $\Big\Updownarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Downarrow$ ||
+|-
+| align="center" | [[Definition:Compact Topological Space|Compact]] ||
+| align="center" | ||
+| align="center" | [[Definition:Separable Space|Separable]] ||
+|-
+| align="center" | $\Big\Updownarrow$ ||
+| align="center" | ||
+| align="center" | $\Big\Updownarrow$ ||
+|-
+| align="center" | [[Definition:Weakly Countably Compact Space|Weakly Countably Compact]] ||
+| align="center" | ||
+| align="center" | [[Definition:Lindelöf Space|Lindelöf]] ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | $\Big\Updownarrow$ ||
+|-
+| align="center" | ||
+| align="center" | ||
+| align="center" | [[Definition:Second-Countable Space|Second-Countable]] ||
+|}
+\end{theorem}
+
+\begin{proof}
+The relevant justifications are listed as follows:
+:[[Metric Space is Compact iff Countably Compact]].
+:[[Metric Space is Countably Compact iff Sequentially Compact]].
+:[[Metric Space is Weakly Countably Compact iff Countably Compact]].
+:[[Compact Space is Weakly Sigma-Locally Compact|Compact Space is Weakly $\sigma$-Locally Compact]].
+:By definition, a [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space]] is both [[Definition:Weakly Locally Compact Space|weakly locally compact]] and [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+:[[Sigma-Compact Space is Lindelöf|$\sigma$-Compact Space is Lindelöf]].
+:[[Metric Space is Lindelöf iff Second-Countable]].
+:[[Metric Space is Separable iff Second-Countable]].
+:[[Metric Space is Weakly Locally Compact iff Strongly Locally Compact]].
+{{qed}}
+{{ACC}}
+{{explain|Indicate which of the above theorems uses ACC in its proof}}
+\end{proof}<|endoftext|>
+\section{Extremally Disconnected Metric Space is Discrete}
+Tags: Metric Spaces, Extremally Disconnected Spaces, Discrete Topology
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Extremally Disconnected Space|extremally disconnected]].
+Then $M$ is the [[Definition:Discrete Topology|discrete topology]].
+\end{theorem}
+
+\begin{proof}
+Let $M = \struct {A, d}$ be [[Definition:Extremally Disconnected Space|extremally disconnected]].
+Let $p \in A$.
+As $M$ is a [[Definition:Metric Space|metric space]], $\set p$ can be expressed as:
+:$\set p = \displaystyle \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$
+where:
+:$\map {B_{1 / n} } p$ denotes the [[Definition:Open Ball of Metric Space|open $1 / n$-ball]] of $p$
+:$\paren {\map {B_{1 / n} } p}^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\map {B_{1 / n} } p$
+That is, as the [[Definition:Set Intersection|intersection]] of the [[Definition:Closure (Topology)|closures]] of the [[Definition:Open Ball of Metric Space|open $1 / n$-ball]] of $p$ for all [[Definition:Natural Numbers|non-zero natural numbers]].
+Now let:
+:$\displaystyle U = \bigcup_{n \mathop \in \N_{>0} } \map {B_{1 / 2 n} } p \setminus \paren {\map {B_{1 / \paren {2 n + 1} } } p}^-$
+Then either $U$ or the [[Definition:Set Complement|complementary set of annuli]] is an [[Definition:Open Set (Topology)|open set]] which has $p$ as a non-[[Definition:Interior (Topology)|interior]] [[Definition:Limit Point of Set|limit point]] provided $\set p$ is not [[Definition:Open Set (Topology)|open]].
+{{explain|Take another run at that last sentence, it needs amplification.}}
+So if $M$ is not the [[Definition:Discrete Topology|discrete topology]], it cannot be [[Definition:Extremally Disconnected Space|extremally disconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Range of Characters}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] of [[Definition:Order of Structure|order]] $m$.
+Let $\chi: G \to \C^\times$ be a [[Definition:Character (Number Theory)|character]] on $G$.
+Then for any $g \in G$, $\map \chi g$ is an [[Definition:Complex Roots of Unity|$m$th root of unity]].
+If $e$ is the [[Definition:Identity Element|identity]] of $G$ then $\map \chi g = 1$.
+\end{theorem}
+
+\begin{proof}
+The claim that $\map \chi e = 1$ is shown by [[Group Homomorphism Preserves Identity]].
+Let $g \in G$ be arbitrary.
+Let $k$ be the [[Definition:Order of Group Element|order]] of $g$.
+By [[Order of Element Divides Order of Finite Group]]:
+:$\exists l \in \Z: m = k l$
+Therefore:
+:$g^m = \paren {g^k}^l = e^l = e$
+By the [[Definition:Group Homomorphism|homomorphism property]]:
+:$1 = \map \chi e = \map \chi {g^m} = \map \chi g^m$
+Hence the result.
+{{Qed}}
+[[Category:Analytic Number Theory]]
+0ze9r5yuv8ntst78141gfauk9wn6afm
+\end{proof}<|endoftext|>
+\section{Indicator is Well-Defined}
+Tags: Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]], and $a \in G$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then the [[Definition:Indicator of Group Element|indicator]] of $a$ in $H$ is well defined.
+\end{theorem}
+
+\begin{proof}
+If $a \in H$, then for $n = 1$, $a^n \in H$.
+Suppose that $a \notin H$. By [[Identity of Subgroup]], we have that the [[Definition:Identity Element|identity]] $e$ of $G$ is in $H$.
+Moreover, by [[Element of Finite Group is of Finite Order]], there is $n \in \N$ such that $a^n = e \in H$.
+Therefore, for any $g \in G$, there is ''some'' $n \in \N$ such that $a^n \in H$.
+That is, the set of [[Definition:Positive Integer|positive integers]] $n$ such that $a^n \in H$ is [[Definition:Non-Empty Set|non-empty]].
+So by the [[Well-Ordering Principle]], this set has a least element, and the [[Definition:Indicator of Group Element|indicator]] is well defined.
+{{qed}}
+[[Category:Subgroups]]
+9ellxqergqaoxuoyn896eccdklzha7o
+\end{proof}<|endoftext|>
+\section{Compact Metric Space is Totally Bounded}
+Tags: Totally Bounded Metric Spaces, Compact Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Compact Metric Space|compact]].
+Then $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+By definition, $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]] only if there exists a [[Definition:Finite Net|finite $\epsilon$-net]] for $M$.
+We use a [[Proof by Contraposition]].
+To that end, suppose that there exists no [[Definition:Finite Net|finite $\epsilon$-net]] for $M$.
+The aim is to construct an [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n} _{n \in \N}$ in $A$ that has no [[Definition:Convergent Sequence (Metric Space)|convergent]] [[Definition:Subsequence|subsequence]].
+Suppose that $x_0, x_1, \ldots, x_r \in A$ have been defined such that:
+:$\forall m, n \in \set {0, 1, \ldots, r} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
+That is, any two [[Definition:Distinct Elements|distinct elements]] of $\set {x_0, x_1, \ldots, x_r}$ are at least $\epsilon$ apart.
+[[Definition:By Hypothesis|By hypothesis]], there exists no [[Definition:Finite Net|finite $\epsilon$-net]] for $M$.
+Specifically, $\set {x_0, x_1, \ldots, x_r}$ is therefore not an [[Definition:Net (Metric Space)|$\epsilon$-net]] for $M$.
+So, by definition of a [[Definition:Net (Metric Space)|$\epsilon$-net]]:
+:$\displaystyle A \nsubseteq \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$
+where $\map {B_\epsilon} {x_i}$ denotes the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball of $x_i$ in $M$]].
+Thus there must exist $x_{r+1} \in A$ such that:
+:$\displaystyle x_{r+1} \notin \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$
+That is:
+:$\exists x_{r+1} \in A: \forall m, n \in \set {0, 1, \ldots, r, r + 1} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
+Thus, by [[Principle of Mathematical Induction|induction]], the [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n}$ in $A$ has been constructed such that:
+:$\forall m, n \in \N: m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
+Thus $\sequence {x_n}$ has no [[Definition:Cauchy Sequence (Metric Space)|Cauchy]] [[Definition:Subsequence|subsequence]].
+Since a [[Convergent Sequence in Metric Space is Cauchy Sequence|convergent sequence is Cauchy]], it has no [[Definition:Convergent Sequence (Metric Space)|convergent]] [[Definition:Subsequence|subsequence]] either. 
+Thus, by definition, $M$ is not [[Definition:Sequentially Compact Space|sequentially compact]].
+From [[Rule of Transposition]], if $M$ is sequentially compact, then there exists no [[Definition:Finite Net|finite $\epsilon$-net]] for $M$.
+So if $M$ is sequentially compact, it is certainly not totally bounded.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $M = \struct {A, d}$ be [[Definition:Compact Metric Space|compact]].
+Let $\epsilon > 0$.
+Then the family $\set {\map {B_\epsilon} x: x \in A}$ of [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] forms an [[Definition:Open Cover|open cover]] of $A$.
+By the definition of [[Definition:Compact Metric Space|compact]], there exists a [[Definition:Finite Subcover|finite subcover]].
+That is, there are points $x_0, \ldots, x_n$ such that:
+:$\displaystyle A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$
+as required. 
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+By definition, $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]] only if there exists a [[Definition:Finite Net|finite $\epsilon$-net]] for $M$.
+We use a [[Proof by Contradiction]].
+That is, suppose that there exists no [[Definition:Finite Net|finite $\epsilon$-net]] for $M$.
+The aim is to construct an [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n}_{n \ge 1}$ in $A$ that has no [[Definition:Convergent Sequence (Metric Space)|convergent]] [[Definition:Subsequence|subsequence]].
+For all [[Definition:Natural Numbers|natural numbers]] $n \ge 1$, define the [[Definition:Set|set]]:
+:$\mathcal S_n = \set{F \subseteq A: \size F = n: \forall x, y \in F: x \ne y \implies \map d {x, y} \ge \epsilon}$
+where $\size F$ denotes the [[Definition:Cardinality|cardinality]] of $F$.
+We use the [[Principle of Mathematical Induction]] to prove that $\mathcal S_n$ is [[Definition:Non-Empty Set|non-empty]].
+It is [[Definition:Vacuous Truth|vacuously true]] that any [[Definition:Singleton|singleton]] $\set x \subseteq A$ is an [[Definition:Element|element]] of $\mathcal S_1$.
+Since $A$ is [[Definition:Non-Empty Set|non-empty]] by the definition of a [[Definition:Metric Space|metric space]], it follows from [[Existence of Singleton Set]] that $\mathcal S_1$ is [[Definition:Non-Empty Set|non-empty]].
+Let $F \in \mathcal S_n$.
+By definition, $F$ is [[Definition:Finite Set|finite]].
+So $F$ is not an [[Definition:Net (Metric Space)|$\epsilon$-net]] for $M$, [[Definition:By Hypothesis|by hypothesis]].
+Hence, there exists an $x \in A$ such that:
+:$\forall y \in F: \map d {x, y} \ge \epsilon$
+Note that, by [[Definition:Metric Space Axioms|axiom $\left({M1}\right)$ for a metric]]:
+:$x \notin F$
+Consider the [[Definition:Set|set]]:
+:$F' := F \cup \set x$.
+Then:
+:$\left\vert{F'}\right\vert = n + 1$
+and by [[Definition:Metric Space Axioms|axiom $\paren{M3}$ for a metric]]:
+:$F' \in \mathcal S_{n+1}$
+Thus, we have proven that $\mathcal S_n$ is [[Definition:Non-Empty Set|non-empty]] for all [[Definition:Natural Numbers|natural numbers]] $n \ge 1$.
+Therefore, using the [[Axiom:Axiom of Countable Choice|axiom of countable choice]], we can obtain an [[Definition:Infinite Sequence|infinite sequence]] $\sequence {F_n}_{n \ge 1}$ such that:
+: $\forall n \in \N_{\ge 1}: F_n \in \mathcal S_n$
+From [[Countable Union of Countable Sets is Countable]], there exists an [[Definition:Injection|injection]]:
+:$\displaystyle \phi: \bigcup_{n \mathop \ge 1} F_n \to \N$
+We now construct an [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n}_{n \ge 1}$ in $A$.
+{{explain|This seems a colloquial sense of PoRD, which needs to be formalised in some way}}
+To do this, we use the [[Principle of Recursive Definition]] to define the [[Definition:Sequence|sequence]] $\sequence {\tuple {x_1, x_2, \ldots, x_n} }_{n \ge 1}$ of [[Definition:Ordered Tuple|ordered $n$-tuples]].
+Let $x_1 \in F_1$.
+Suppose that $x_1, x_2, \ldots, x_n$ have been defined, and let:
+:$T_n = \set {x_1, x_2, \ldots, x_n}$
+Define:
+:$D_n = \set {x \in F_{n+1}: \forall y \in T_n: \map d {x, y} \ge \dfrac \epsilon 2}$
+Using a [[Proof by Contradiction]], we show that $D_n$ is [[Definition:Non-Empty Set|non-empty]].
+For all $x \in F_{n+1}$, define:
+:$\map {C_n} x = \set {y \in T_n: \map d {x, y} < \dfrac \epsilon 2}$
+Let $x, x' \in F_{n+1}$ be [[Definition:Distinct Elements|distinct]].
+Let $y \in \map {C_n} x$.
+Then it follows from:
+:the definition of $F_{n+1}$
+:[[Definition:Metric Space Axioms|axioms $\paren {M2}$ and $\paren {M3}$ for a metric]]
+that:
+:$\map d {x', y} \ge \map d {x, x'} - \map d {x, y} > \dfrac \epsilon 2$
+Hence, $y \notin \map {C_n} {x'}$.
+That is, the [[Definition:Indexed Family of Sets|indexed family of sets]]:
+:$\sequence {\map {C_n} x}_{x \in F_{n + 1}}$
+is [[Definition:Pairwise Disjoint Family|pairwise disjoint]].
+Suppose that $D_n$ is [[Definition:Empty Set|empty]].
+That is:
+:$\forall x \in F_{n+1}: \map {C_n} x$ is [[Definition:Non-Empty Set|non-empty]]
+Then, from [[Cardinality is Additive Function]], [[Finite Union of Sets in Additive Function]], and [[Cardinality of Subset of Finite Set]], we have:
+:$\displaystyle \size {F_{n+1}} \le \sum_{x \mathop \in F_{n+1}} \size {\map {C_n} x} \le \size {T_n} < \size {F_{n+1}}$
+which is a [[Definition:Contradiction|contradiction]].
+{{explain|"Therefore ..." and explain what assumption was made so as to give rise to this contradiction.}}
+From the [[Well-Ordering Principle|well-ordering principle]], we have that $\struct {\N, \le}$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+Let $\le_{\phi}$ be the [[Definition:Ordering Induced by Injection|ordering induced by $\phi$]].
+Then [[Injection Induces Well-Ordering|$\le_{\phi}$ is a well-ordering]].
+We define $x_{n+1}$ as the ([[Smallest Element is Unique|unique]]) [[Definition:Smallest Element|$\le_{\phi}$-smallest element]] of $D_n$.
+By construction:
+:$\forall m, n \in \N_{>0}: m \le n \implies \map d {x_{n+1}, x_m} \ge \dfrac \epsilon 2$
+Hence, by [[Principle of Mathematical Induction|induction]], it follows from [[Definition:Metric Space Axioms|axiom $\paren {M3}$ for a metric]] that:
+:$\forall m, n \in \N_{>0}: m \ne n \implies \map d {x_m, x_n} \ge \dfrac \epsilon 2$
+Therefore, the [[Definition:Sequence|sequence]] $\sequence {x_n}$ has no [[Definition:Cauchy Sequence (Metric Space)|Cauchy]] [[Definition:Subsequence|subsequence]].
+From [[Convergent Sequence in Metric Space is Cauchy Sequence]], $\sequence {x_n}$ has no [[Definition:Convergent Sequence (Metric Space)|convergent]] [[Definition:Subsequence|subsequence]] either.
+Thus, by definition, $M$ is not [[Definition:Sequentially Compact Space|sequentially compact]].
+But this [[Definition:Contradiction|contradicts]] the original assumption that $M$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+Thus the assumption that there exists no [[Definition:Finite Net|finite $\epsilon$-net]] for $M$ was false.
+Therefore, by definition, $M$ is [[Definition:Totally Bounded Metric Space|totally bounded]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup Generated by Subgroup and Element}
+Tags: Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
+Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
+Let $a \in G \setminus H$.
+Let $n$ be the [[Definition:Indicator of Group Element|indicator]] of $a$ in $H$.
+Then:
+:$K = \left\{{x a^k: x \in H, \ 0 \le k < n}\right\}$
+is a [[Definition:Subgroup|subgroup]] of $G$ such that $H \subseteq K$, and each [[Definition:Element|element]] of $K$ has a [[Definition:Unique|unique]] representation in this form.
+Moreover:
+:$K = \left\langle{H, a}\right\rangle$
+where $\left\langle{H, a}\right\rangle$ denotes the [[Definition:Generator of Subgroup|subgroup generated by $\left\{{H, a}\right\}$]], and:
+:$\left\vert{K}\right\vert = n \left\vert{H}\right\vert$
+where $\left\vert{K}\right\vert$ denotes the [[Definition:Order of Structure|order of $K$]].
+\end{theorem}
+
+\begin{proof}
+=== $K$ is [[Definition:Subgroup|Subgroup]] of $G$ ===
+We first show that $K$ is a [[Definition:Subgroup|subgroup]] of $G$ using the [[Two-Step Subgroup Test]].
+$(1): \quad K \ne \varnothing$:
+From [[Identity of Subgroup]]:
+:$e \in H$
+From [[Indicator is Well-Defined]], $n > 0$ and so $n - 1 \ge 0$.
+Thus $k = 0$ fulfils the condition that $0 \le k < n$, and so:
+:$e = e a^0 \in K$
+Thus $K \ne \varnothing$.
+$(2): \quad K$ is [[Definition:Closure (Abstract Algebra)|closed]]:
+Let $r = x a^k \in K$ and $s = y a^l \in K$ where $x, y \in H$ and $0 \le k < n, 0 \le l < n$.
+$G$ is [[Definition:Abelian Group|abelian]], so $r$, $s$ and $a$ [[Definition:Commute|commute]].
+So:
+:$r s = x y a^{k + l}$.
+Since $H$ is a [[Definition:Group|group]] and $x, y \in H$ then also $x y \in H$.
+Let $k + l \le n - 1$.
+Then $r s \in K$ by definition of $K$.
+Let $n \le k + l \le 2 n$.
+Then:
+:$r s = x y a^n a^{k + l - n}$
+By definition of [[Definition:Indicator of Group Element|indicator]], $a^n \in H$.
+Thus as $x, y, a^n \in H$ it follows that $x y a^n \in H$.
+As $0 \le k + l - n \le n - 1$ it follows that $\left({x y a^n}\right) a^{k + l - n} \in K$ by definition of $K$. 
+So again $r s \in K$.
+Thus $K$ is [[Definition:Closure (Abstract Algebra)|closed]].
+$(3): \quad K$ is [[Definition:Closure (Abstract Algebra)|closed]] under [[Definition:Inverse Element|inversion]]:
+Let $r = x a^k \in K$ where $x \in H$ and $0 \le k < n$.
+Then $x^{-1} \in H$.
+Let $k = 0$.
+Then 
+:$r^{-1} = x^{-1} e = x^{-1} a^0 \in K$
+Let $k > 0$.
+Then:
+:$r^{-1} = x^{-1} a^{-k} = x^{-1} a^{-n} a^{n - k}$
+By definition of [[Definition:Indicator of Group Element|indicator]], $a^n \in H$.
+We have that $x^{-1}, a^n \in H$.
+As $G$ is [[Definition:Abelian Group|abelian]], $x$ and $a$ [[Definition:Commute|commute]].
+Thus from [[Inverse of Commuting Pair]]:
+: $x^{-1} a^{-n} = \left({x a^n}\right)^{-1} \in H$
+Also:
+:$0 \le n - k < n -1$
+so:
+:$r^{-1} = \left({x a^n}\right)^{-1} a^{n - k} \in K$
+Therefore by the [[Two-Step Subgroup Test]] we have shown that $K$ is a [[Definition:Subgroup|subgroup]] of $G$.
+=== $H$ is [[Definition:Subgroup|Subgroup]] of $K$ ===
+As:
+:$\forall x \in H: x = x a^0 \in K$
+and so $H \subseteq K$.
+=== [[Definition:Order of Structure|Order]] of $K$ ===
+It remains to be shown that:
+:$\left\vert{K}\right\vert = n q$
+where $q = \left\vert{H}\right\vert$.
+Let $x a^k = y a^l \in K$.
+[[Definition:WLOG|WLOG]], let $k \ge l$ (if not, just relabel the two).
+We have:
+: $a^{k - l} = x^{-1} y$
+Since $H$ is a subgroup of $G$:
+:$x^{-1} y \in H$
+and therefore:
+:$a^{k - l} \in H$
+By definition of [[Definition:Indicator of Group Element|indicator]], $n$ is the least [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $a^n \in H$
+Hence because $n > k \ge k - l$:
+:$k - l = 0$
+Therefore $k = l$, and $x = y$.
+Therefore the elements $x a^k$ with $x \in H$ and $0 \le k < n$ are [[Definition:Distinct|distinct]].
+So for each $x \in H$ there are $n$ elements of the form $x a^k$ in $K$.
+We have that there are $q$ elements in $H$.
+Thus it follows that there are $n q$ such [[Definition:Element|elements]].
+By definition of $K$, these form all of $K$.
+Thus:
+:$\left\vert{K}\right\vert = n q$
+as required
+{{qed}}
+[[Category:Subgroups]]
+ccxml4gy512g7balk0j604fvssw1m8y
+\end{proof}<|endoftext|>
+\section{Number of Characters on Finite Abelian Group}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
+Then the number of [[Definition:Character (Number Theory)|characters]] $G \to \C^\times$ is $\left\vert{G}\right\vert$.
+\end{theorem}
+
+\begin{proof}
+=== Lemma ===
+Let $H \le G$ be a [[Definition:Subgroup|subgroup]].
+Let $\chi: H \to \C^\times$ be a [[Definition:Character (Number Theory)|character]] on $G$.
+Let $a \in G \mathrel \backslash H$ where $\backslash$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then:
+:$\chi$ [[Definition:Extension of Mapping|extends]] to $\left[{G : H}\right]$ distinct [[Definition:Character (Number Theory)|characters]] on $G$
+where $\left[{G : H}\right]$ denotes the [[Definition:Index of Subgroup|index]] of $H$ in $G$.
+=== Proof ===
+The proof proceeds by [[Second Principle of Mathematical Induction|stroing induction on $\left[{G : H}\right]$]].
+If $\left[{G : H}\right] = 1$, then $H = G$ by [[Lagrange's Theorem (Group Theory)|Langrange's theorem]].
+Hence (trivially) the result.
+Suppose that $\left[{G : H}\right] > 1$, and the result holds for all subgroups of smaller index in $G$.
+Let $a \notin H$ and let $n$ be the [[Definition:Indicator of Group Element|indicator]] of $a$ in $H$.
+Let $K = \left\langle{H, a}\right \rangle$ denote the [[Definition:Generator of Subgroup|subgroup generated]] by $H$ and $a$.
+From [[Subgroup Generated by Subgroup and Element]], each element of $K$ has a unique representation in the form $x a^k$ with:
+:$x \in H$ and $0 \le k \le n-1$
+and:
+:$\left|{K}\right| = n \left|{H}\right|$
+Let $\tilde \chi$ be a character [[Definition:Extension of Mapping|extending]] $\chi$.
+Then:
+:$(1): \quad \tilde \chi \left({x a^n}\right) = \tilde \chi \left({x}\right) \tilde \chi \left({a}\right)^n$
+and:
+:$(2): \quad \tilde \chi \left({x a^n}\right) = \tilde \chi \left({x}\right) \tilde \chi \left({a^n}\right)$
+Since $x, a^n \in H$, $\tilde \chi$ and $\chi$ agree on these values, so equating $(1)$ and $(2)$ we obtain:
+:$\tilde \chi \left({a}\right)^n = \tilde \chi \left({a^n}\right)$
+That is:
+:$\tilde \chi \left({a}\right)$ is an $n$th root of $\tilde \chi \left({a^n}\right)$.
+So by [[nth roots of a complex number are distinct]] there are $n$ distinct possibilities.
+We choose one of these possibilities and define:
+: $\tilde \chi \left({x a^k}\right) = \chi \left({x}\right) \tilde \chi \left({a}\right)^k$
+for $x a^k \in K$.
+We check that $\tilde \chi$ is [[Definition:Multiplicative Arithmetic Function|multiplicative]]:
+For $x,y\in H$,
+{{begin-eqn}}
+{{eqn | l = \tilde \chi \left({x a^k y a^l}\right)
+      | r = \tilde \chi \left({x y a^{k + l} }\right)
+}}
+{{eqn | r = \chi \left({x y}\right) \tilde \chi \left({a}\right)^{k+l}
+}}
+{{eqn | r = \left[{\chi \left({x}\right) \tilde \chi \left({a}\right)^k}\right] \left[{\chi \left({y}\right) \tilde \chi \left({a}\right)^l}\right]
+}}
+{{eqn | r = \tilde \chi \left({x a^k}\right) \tilde \chi \left({y a^l}\right)
+}}
+{{end-eqn}}
+By [[Lagrange's Theorem (Group Theory)|Lagrange's theorem]]:
+: $n = \dfrac {\left|{K}\right|} {\left|{H}\right|} = \left[{K : H}\right]$
+Also by Lagrange's Theorem:
+:$\left[{G : K}\right] < \left[{G : H}\right]$
+since $H \subsetneq K$.
+So by the induction hypothesis, each $\tilde \chi$ extends to $\left[{G : K}\right]$ characters on $G$.
+Therefore by the [[Tower Law for Subgroups]], $\chi$ extends to $\left[{K : H}\right] \left[{G : K}\right] = \left[{G : H}\right]$ characters on $G$.
+{{qed|lemma}}
+Since a character is a [[Definition:Group Homomorphism|homomorphism]], it must [[Group Homomorphism Preserves Identity|preserve the identity]].
+Therefore there is only one character $\chi : \left\{{e}\right\} \to \C^\times : e \mapsto 1$ on the [[Definition:Trivial Subgroup|trivial subgroup]].
+Moreover, by [[Lagrange's Theorem (Group Theory)|Langrange's theorem]], the [[Definition:Trivial Subgroup|trivial subgroup]] has index $\left[{G : \left\{{e}\right\}}\right] = \left|{G}\right|$.
+So by the lemma there are $\left|{G}\right|$ distinct characters on $G$.
+{{qed}}
+[[Category:Analytic Number Theory]]
+bo9cl4unpag8bnu29nzn572d5idkni7
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Space is Complete}
+Tags: Sequentially Compact Spaces, Complete Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]] which is [[Definition:Sequentially Compact Space|sequentially compact]].
+Then $M$ is [[Definition:Complete Metric Space|complete]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$.
+As $M$ is [[Definition:Sequentially Compact Space|sequentially compact]], $\sequence {x_n}$ has a [[Definition:Convergent Sequence (Metric Space)|convergent]] [[Definition:Subsequence|subsequence]].
+By [[Convergent Subsequence of Cauchy Sequence|Convergent Subsequence of Cauchy Sequence in Metric Space]], this implies that the entire [[Definition:Sequence|sequence]] $\sequence {x_n}$ is [[Definition:Convergent Sequence (Metric Space)|convergent]].
+Hence, $M$ is [[Definition:Complete Metric Space|complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Completeness is Topological Property}
+Tags: Topologically Complete Spaces
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]] which are [[Definition:Homeomorphic Topological Spaces|homeomorphic]].
+Let $T_1$ be [[Definition:Topologically Complete Space|topologically complete]].
+Then $T_2$ is also [[Definition:Topologically Complete Space|topologically complete]].
+\end{theorem}<|endoftext|>
+\section{Topological Completeness is Weakly Hereditary}
+Tags: Topologically Complete Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Topologically Complete Space|topologically complete]].
+Let $V \subseteq S$ be a [[Definition:Closed Set (Topology)|closed]] [[Definition:Topological Subspace|subspace]] of $T$.
+Then $V$ is also [[Definition:Topologically Complete Space|topologically complete]].
+That is, [[Definition:Topologically Complete Space|topological completeness]] is [[Definition:Weakly Hereditary Property|weakly hereditary]].
+\end{theorem}
+
+\begin{proof}
+{{finish|Use [[Subspace of Complete Metric Space is Closed iff Complete]]}}
+\end{proof}<|endoftext|>
+\section{Metric Space Compact iff Complete in All Equivalent Metrics}
+Tags: Complete Metric Spaces, Compact Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A, d_1}$ be a [[Definition:Metric Space|metric space]].
+Then $M_1$ is [[Definition:Compact Metric Space|compact]] {{iff}} $M_2 = \struct {A, d_2}$ is a [[Definition:Complete Metric Space|complete metric space]] whenever $d_2$ is [[Definition:Equivalent Metrics|equivalent]] to $d_1$.
+\end{theorem}<|endoftext|>
+\section{Metric Space Completeness is Preserved by Isometry}
+Tags: Metric Space Completeness is Preserved by Isometry, Complete Metric Spaces, Isometries
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi: M_1 \to M_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
+If $M_1$ is [[Definition:Complete Metric Space|complete]] then so is $M_2$.
+\end{theorem}
+
+\begin{proof}
+Let $\tau_1$ be the [[Definition:Topology Induced by Metric|topology on $A_1$ induced by $d_1$]].
+Let $\tau_2$ be the [[Definition:Topology Induced by Metric|topology on $A_2$ induced by $d_2$]].
+Let $\left\langle{x_n}\right\rangle$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A_2$.
+From [[Inverse of Isometry of Metric Spaces is Isometry]], $\phi^{-1}$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+Since [[Isometric Image of Cauchy Sequence is Cauchy Sequence]], $\left\langle{\phi^{-1} \left({x_n}\right)}\right\rangle$ is a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]].
+Since $M_1$ is a [[Definition:Complete Metric Space|complete metric space]], $\left\langle{\phi^{-1} \left({x_n}\right)}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]].
+Since [[Isometry Preserves Sequence Convergence]], $\left\langle{\phi \left({\phi^{-1} \left({x_n}\right)}\right)}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]].
+But:
+: $\left\langle{\phi \left({\phi^{-1} \left({x_n}\right)}\right)}\right\rangle = \left\langle{x_n}\right\rangle$
+so $\left\langle{x_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]].
+Thus each [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $M_2$ [[Definition:Convergent Sequence (Metric Space)|converges]].
+It follows by definition that so $M_2$ is a [[Definition:Complete Metric Space|complete metric space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+Let $\left\langle{b_n}\right\rangle$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A_2$.
+Thus:
+: $\exists N_1 \in \N: d_2 \left({b_n, b_m}\right) < \epsilon$
+whenever $n, m \ge N_1$ and $b_n, b_m \in A_2$.
+We have that $M_1$ is [[Definition:Isometry (Metric Spaces)|isometric]] to $M_2$.
+[[Isometry is Equivalence Relation]] and so in particular [[Definition:Symmetric Relation|symmetric]].
+Hence $M_2$ is [[Definition:Isometry (Metric Spaces)|isometric]] to $M_1$, via $\phi^{-1}$.
+Thus:
+: $d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b_m}\right)}\right) = d_2 \left({b_n, b_m}\right) < \epsilon$
+whenever $n, m \ge N_1$ and $\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b_m}\right) \in A$.
+So $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ is [[Definition:Cauchy Sequence (Metric Space)|Cauchy]] in $A_1$.
+Since $A_1$ is [[Definition:Complete Metric Space|complete]], $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] in $A_1$ to, say, $a$.
+By definition of [[Definition:Isometry (Metric Spaces)|isometry]], $\phi^{-1}$ is a [[Definition:Bijection|bijection]], and in particular [[Definition:Surjection|surjective]]
+Thus there exists some $b \in A_2$ such that $\phi^{-1} \left({b}\right) = a$.
+Since $\left\langle{\phi^{-1} \left({b_n}\right)}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $\phi^{-1} \left({b}\right)$, there exists some $N_2 \in \N$ such that:
+:$d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right)}\right) < \epsilon$
+whenever $n \ge N_2$ and $\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right) \in A_1$.
+Since $M_2$ is  [[Definition:Isometry (Metric Spaces)|isometric]] to $M_1$, we have:
+:$d_1 \left({\phi^{-1} \left({b_n}\right), \phi^{-1} \left({b}\right)}\right) = d_2 \left({b_n, b}\right)$
+and so:
+:$d_2 \left({b_n, b}\right) < \epsilon$
+whenever $n \ge N_2$ and $b_n, b \in A_2$.
+Thus $\left\langle{b_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] in $A_2$.
+Since $\left\langle{b_n}\right\rangle$ was an arbitrary [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]], we have that $M_2$ is [[Definition:Complete Metric Space|complete]], as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Urysohn's Metrization Theorem}
+Tags: Metrizable Topologies, Regular Spaces, Second-Countable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Regular Space|regular]] and [[Definition:Second-Countable Space|second-countable]].
+Then $T$ is [[Definition:Metrizable Topology|metrizable]].
+\end{theorem}<|endoftext|>
+\section{Metrizable iff Regular and has Sigma-Locally Finite Basis}
+Tags: Metrizable Topologies, Regular Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $T$ is [[Definition:Metrizable Topology|metrizable]] {{iff}} $T$ is [[Definition:Regular Space|regular]] and has a [[Definition:Sigma-Locally Finite Basis|$\sigma$-locally finite basis]].
+\end{theorem}<|endoftext|>
+\section{Uniformity iff Quasiuniformity has Symmetric Basis}
+Tags: Uniformities
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\UU$ be a [[Definition:Quasiuniformity|quasiuniformity]] on $S$.
+Then $\UU$ is a [[Definition:Uniformity|uniformity]] {{iff}} $\UU$ has a [[Definition:Symmetric Filter Basis|symmetric filter basis]].
+\end{theorem}
+
+\begin{proof}
+Let $\UU$ be a [[Definition:Quasiuniformity|quasiuniformity]] on $S$ which has a [[Definition:Symmetric Filter Basis|symmetric filter basis]] $\BB$.
+From the definition of [[Definition:Filter Basis|filter basis]], all the elements of $\UU$ can be formed from [[Definition:Set Intersection|intersections]] of [[Definition:Element|elements]] of $\BB$.
+But from [[Intersection of Symmetric Relations is Symmetric]], it follows that all [[Definition:Element|elements]] of $\UU$ are [[Definition:Symmetric Entourage|symmetric]].
+Now suppose $\UU$ is a [[Definition:Uniformity|uniformity]].
+If $\BB$ is a [[Definition:Filter Basis|filter basis]] of $\UU$ then all the [[Definition:Element|elements]] of $\BB$ are also [[Definition:Element|elements]] of $\UU$.
+Hence $\BB$ is a [[Definition:Symmetric Filter Basis|symmetric filter basis]] of $\UU$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quasiuniformity Induces Topology}
+Tags: Uniformities
+
+\begin{theorem}
+Let $\UU$ be a [[Definition:Quasiuniformity|quasiuniformity]] on a [[Definition:Set|set]] $S$.
+Let $\tau \subseteq \powerset S$ be a [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] of $S$, created from $\UU$ by:
+:$\tau := \set {\map u x: u \in \UU, x \in S}$
+where:
+:$\forall x \in S: \map u x = \set {y: \tuple {x, y} \in u}$
+That is, where $\map u x$ is the [[Definition:Image of Element under Relation|image]] of $x$ under $u$, where $u$ is considered as a [[Definition:Relation|relation]] on $S$.
+Then $\tau$ is a [[Definition:Topology|topology]] on $S$.
+That is, the [[Definition:Quasiuniform Space|quasiuniform space]] $\struct {\struct {S, \UU}, \tau}$ is also the [[Definition:Topological Space|topological space]] $\struct {S, \tau}$.
+\end{theorem}<|endoftext|>
+\section{Topological Space is Quasiuniformizable}
+Tags: Uniformities
+
+\begin{theorem}
+Every [[Definition:Topological Space|topological space]] is [[Definition:Quasiuniformizable Space|quasiuniformizable]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\BB$ be defined as:
+:$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$
+Then $\BB$ is a [[Definition:Filter Sub-Basis|filter sub-basis]] for a [[Definition:Quasiuniformity|quasiuniformity]] on $S$ such that $\struct {\struct {S, \UU}, \tau}$ is a [[Definition:Quasiuniform Space|quasiuniform space]].
+{{finish|Prove that above assertion.}}
+\end{proof}
\ No newline at end of file