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+\section{Infinite Set in Compact Space has Omega-Accumulation Point}
+Tags: Compact Spaces, Omega-Accumulation Points
+
+\begin{theorem}
+Let $\struct {X, \tau}$ be a [[Definition:Compact Topological Space|compact topological space]].
+Let $A \subseteq X$ be [[Definition:Infinite Set|infinite]].
+Then $A$ has an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $A$ has no [[Definition:Omega-Accumulation Point|$\omega$-accumulation points]] in $X$.
+Then for any $x \in X$, there exists some [[Definition:Open Set (Topology)|open set]] $U_x$ such that $U_x$ only contains a [[Definition:Finite|finite]] number of points in $A$.
+The collection of all such [[Definition:Open Set (Topology)|open sets]] $\CC = \set {U_x: x \in X}$ is an [[Definition:Open Cover|open cover]] for $X$.
+Since $X$ is [[Definition:Compact Topological Space|compact]], $\CC$ has a [[Definition:Finite Subcover|finite subcover]] $\CC'$.
+$\CC'$ contains a [[Definition:Finite|finite]] number of sets in $\CC$, and each of those sets contain a [[Definition:Finite|finite]] number of points in $A$.
+Therefore $\bigcup \CC'$ contains a [[Definition:Finite|finite]] number of points in $A$.
+However by definition of a [[Definition:Subcover|subcover]], $A \subseteq X \subseteq \bigcup \CC'$.
+Thus $\bigcup \CC'$ contains all points in $A$, of which there are [[Definition:Infinite|infinite]].
+This is a [[Definition:Contradiction|contradiction]].
+Therefore $A$ must have an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$.
+{{qed}}
+[[Category:Compact Spaces]]
+[[Category:Omega-Accumulation Points]]
+kg2bxqe4nthappgc1mzu9tkh6hfzjkl
+\end{proof}<|endoftext|>
+\section{Half-Open Real Interval is not Open Set}
+Tags: Real Intervals, Open Sets
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
+Let $\hointr a b \subset \R$ be a [[Definition:Half-Open Real Interval|half-open interval]] of $\R$.
+Then $\hointr a b$ is not an [[Definition:Open Set (Metric Space)|open set]] of $\R$.
+Similarly, the [[Definition:Half-Open Real Interval|half-open interval]] $\hointl a b \subset \R$ is not an [[Definition:Open Set (Metric Space)|open set]] of $\R$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+Let $\map {B_\epsilon} a$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$.
+We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\hointr a b$.
+Thus $\hointr a b$ is not a [[Definition:Neighborhood (Metric Space)|neighborhood]] $a$.
+It follows that $\hointr a b$ is not an [[Definition:Open Set (Metric Space)|open set]] of $\R$.
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], the argument also shows that $\hointl a b \subset \R$ is not an [[Definition:Open Set (Metric Space)|open set]] of $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Open in Metric Space}
+Tags: Metric Spaces, Empty Set, Open Sets
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then the [[Definition:Empty Set|empty set]] $\varnothing$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Open Set (Metric Space)|open set]] $S \subseteq A$ is one where every point inside it is an element of an [[Definition:Open Ball|open ball]] contained entirely within that set.
+That is, there are no points in $S$ which have an [[Definition:Open Ball|open ball]] some of whose elements are not in $S$.
+As there are no elements in $\varnothing$, the result follows [[Definition:Vacuous Truth|vacuously]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Open in Itself}
+Tags: Metric Spaces, Open Sets
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then the [[Definition:Set|set]] $A$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Open Set (Metric Space)|open set]] $S \subseteq A$ is one where every point inside it is an element of an [[Definition:Open Ball|open ball]] contained entirely within that set.
+Let $x \in A$.
+An [[Definition:Open Ball|open ball]] of $x$ in $M$ is by definition a [[Definition:Subset|subset]] of $A$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Standard Discrete Metric Space is Open}
+Tags: Discrete Metrics
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Standard Discrete Metric|standard discrete metric space]].
+Let $S \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Then $S$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Standard Discrete Metric|standard discrete metric]]:
+:$\forall x, y \in A: d \left({x, y}\right) = \begin{cases}
+0 & : x = y \\
+1 & : x \ne y
+\end{cases}$
+Let $\epsilon \in \R_{>0}$ be such that $0 < \epsilon \le 1$.
+Let $x \in S$.
+Let $B_\epsilon \left({x}\right)$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$.
+Then by definition of $\epsilon$ and $d$:
+:$B_\epsilon \left({x}\right) = \left\{{x}\right\}$
+Thus:
+: $\forall x \in S: B_\epsilon \left({x}\right) \subseteq S$
+Hence the result by definition of [[Definition:Open Set (Metric Space)|open set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Open Set under Continuous Mapping in Metric Space may not be Open}
+Tags: Open Sets, Continuity
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Continuous Mapping (Metric Spaces)|$\left({d_1, d_2}\right)$-continuous mapping]] from $A_1$ to $A_2$.
+Let $U \subseteq A_1$ be an [[Definition:Open Set (Metric Space)|open set]] of $M_1$.
+Then it is not necessarily the case that $f \left({U}\right)$ is an [[Definition:Open Set (Metric Space)|open set]] of $M_2$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Constant Mapping|constant mapping]] $f_0: \R^2 \to \R$ defined as:
+:$\forall \left({x, y}\right) \in \R^2: f \left({x, y}\right) = 0$
+Then by [[Constant Mapping is Continuous]] $f$ is a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+But consider any [[Definition:Open Set (Metric Space)|open set]] $U \subseteq A_1$ of $M_1$ such that $U \ne \varnothing$.
+Then $f \left({U}\right) = \left\{{0}\right\} = \left[{0 \,.\,.\, 0}\right]$ which is a [[Definition:Closed Real Interval|closed interval]] of $\R$.
+The result follows from [[Closed Real Interval is not Open Set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Intersection of Open Sets of Metric Space may not be Open}
+Tags: Open Sets, Set Intersection
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $U_1, U_2, U_3, \ldots$ be an [[Definition:Infinite Set|infinite set]] of [[Definition:Open Set (Metric Space)|open sets]] of $M$.
+Then it is not necessarily the case that $\displaystyle \bigcap_{n \mathop \in \N} U_n$ is itself an [[Definition:Open Set (Metric Space)|open set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Open Real Interval|open real interval]] $\openint {-\dfrac 1 n} {\dfrac 1 n} \subseteq \R$.
+From [[Open Real Interval is Open Set]], $\openint {-\dfrac 1 n} {\dfrac 1 n}$  is [[Definition:Open Set (Metric Space)|open]] in $\R$ for all $n \in \N$.
+But:
+:$\displaystyle \bigcap_{n \mathop \in \N} \openint {-\dfrac 1 n} {\dfrac 1 n} = \set 0 = \closedint 0 0$
+which is a [[Definition:Closed Real Interval|closed interval]] of $\R$.
+The result follows from [[Closed Real Interval is not Open Set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Adherent Point}
+Tags: Set Closures, Equivalence of Definitions of Adherent Point
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$.
+{{TFAE|def = Adherent Point|view = adherent point of $A$}}
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 2|Definition from Open Neighborhood is equivalent to Definition from Closure]] ===
+{{:Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 2}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 3| Definition from Open Neighborhood is equivalent to Definition from Neighborhood]] ===
+{{:Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 3}}{{qed}}
+[[Category:Set Closures]]
+[[Category:Equivalence of Definitions of Adherent Point]]
+82ykqg71wrlnlh3p49zzvwysi1pyw8m
+\end{proof}<|endoftext|>
+\section{Metric Space is Compact iff Countably Compact}
+Tags: Metric Spaces, Compact Spaces, Countably Compact Spaces
+
+\begin{theorem}
+A [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]] {{iff}} it is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Compact Space is Countably Compact]]
+:[[Countably Compact Metric Space is Compact]]
+{{qed}}
+{{ACC|Countably Compact Metric Space is Compact}}
+\end{proof}<|endoftext|>
+\section{Countably Compact Metric Space is Sequentially Compact}
+Tags: Metric Spaces, Countably Compact Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+Let $M$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:Metric Space|metric space]].
+Then $M$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from the results:
+:[[Metric Space is First-Countable]]
+:[[Countably Compact First-Countable Space is Sequentially Compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point}
+Tags: Countably Compact Spaces, Infinite Sets, Countable Sets
+
+\begin{theorem}
+Let $\struct {X, \tau}$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:Topological Space|topological space]].
+Let $A \subseteq X$ be [[Definition:Countable Set|countably infinite]].
+Then $A$ has an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Contradiction]].
+{{AimForCont}} that $A$ does not have an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] in $X$.
+Let $\SS \subseteq \powerset A$ be the [[Definition:Set|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$.
+By [[Set of Finite Subsets of Countable Set is Countable]], we have that $\SS$ is [[Definition:Countable Set|countable]].
+For all ([[Definition:Finite Set|finite]]) $F \in \SS$, define:
+:$U_F = \paren {F \cup \paren {X \setminus A} }^\circ$
+where ${}^\circ$ denotes the [[Definition:Interior (Topology)|interior]].
+By [[Image of Countable Set under Mapping is Countable]], we have that $\CC = \set {U_F: F \in \SS}$ is [[Definition:Countable Set|countable]].
+By the definition of an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]], it follows that:
+:$\forall x \in X: \exists U \in \tau: x \in U: \exists F \in \SS: U \cap A = F$
+By [[Set Difference Union Intersection]], we have:
+:$U = \paren {U \cap A} \cup \paren {U \setminus A} \subseteq F \cup \paren {X \setminus A}$
+By [[Set Interior is Largest Open Set]], it follows that $\CC$ is a [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] for $X$.
+Hence, by the definition of a [[Definition:Countably Compact Space|countably compact space]], there exists a [[Definition:Finite Subcover|finite subcover]] $\CC'$ of $\CC$ for $X$.
+By the [[Principle of Finite Choice]], there exists an [[Definition:Indexed Family of Subsets|indexed family]] $\family {G_V}_{V \mathop \in \CC'}$ of [[Definition:Element|elements]] of $\SS$ such that:
+:$\forall V \in \CC': V = U_{G_V} = \paren {G_V \cup \paren {X \setminus A} }^\circ$
+Define:
+:$\displaystyle B = \bigcup_{V \mathop \in \CC'} G_V \subseteq A$
+By [[Union of Finite Sets is Finite]], it follows that $B$ is [[Definition:Finite Set|finite]].
+We can use the definition of a [[Definition:Cover of Set|cover]] to conclude that:
+:$X \subseteq B \cup \paren {X \setminus A}$
+By [[Relative Complement inverts Subsets]], it follows that:
+:$\paren {X \setminus B} \cap A = \O$
+Hence, $A \subseteq B$.
+But then by [[Subset of Finite Set is Finite]], it follows that $A$ is [[Definition:Finite Set|finite]], a [[Definition:Contradiction|contradiction]].
+\end{proof}<|endoftext|>
+\section{First-Countable Space is Sequentially Compact iff Countably Compact}
+Tags: Sequentially Compact Spaces, Countably Compact Spaces, First-Countable Spaces
+
+\begin{theorem}
+A [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]] is [[Definition:Sequentially Compact Space|sequentially compact]] {{iff}} it is [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Sequentially Compact Space is Countably Compact]]
+:[[Countably Compact First-Countable Space is Sequentially Compact]]
+{{qed}}
+{{ACC|Sequentially Compact Space is Countably Compact}}
+\end{proof}<|endoftext|>
+\section{Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit}
+Tags: First-Countable Spaces, Sequences
+
+\begin{theorem}
+Let $\struct {X, \tau}$ be a [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
+Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $X$.
+Let $x$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$.
+Then $x$ is a [[Definition:Subsequential Limit|subsequential limit]] of $\sequence {x_n}$.
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:First-Countable Space|first-countable space]], there exists a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] $\BB$ at $x$.
+By [[Surjection from Natural Numbers iff Countable]], there exists a [[Definition:Surjection|surjection]] $\phi: \N \to \BB$.
+For all $n \in \N$, define the [[Definition:Set|set]]:
+:$\displaystyle U_n = \bigcap_{k \mathop = 0}^n \map \phi k$
+By [[General Intersection Property of Topological Space]], it follows that $U_n$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+{{explain|What is the exact connection with PoRD?}}
+Using the [[Principle of Recursive Definition]], we construct a [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] $\sequence {n_k}_{k \mathop \in \N}$ in $\N$.
+By the definition of an [[Definition:Accumulation Point of Sequence|accumulation point]], we can choose $n_0 \in \N$ such that $x_{n_0} \in U_0$.
+For all $k \in \N$, let $n_{k + 1} > n_k$ be the ([[Smallest Element is Unique|unique]]) [[Definition:Smallest Element|smallest]] [[Definition:Natural Numbers|natural number]] such that $x_{n_{k + 1} } \in U_{k + 1}$.
+Such an $n_{k + 1}$ exists by the definition of an [[Definition:Accumulation Point of Sequence|accumulation point]], and by the [[Well-Ordering Principle|well-ordering principle]].
+We now show that $x$ is a [[Definition:Limit Point of Sequence|limit point]] of $\sequence {x_{n_k} }$.
+Let $U$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+By the definition of a [[Definition:Local Basis|local basis]], there exists an $H \in \BB$ such that $H \subseteq U$.
+By the definition of a [[Definition:Surjection|surjection]], there exists a [[Definition:Natural Numbers|natural number]] $m$ such that $H = \map \phi m$.
+By construction, we have:
+:$\forall k \in \N: k > m \implies x_{n_k} \in U_k \subseteq \map \phi m = H$
+Hence the result, by the definition of a [[Definition:Limit Point of Sequence|limit point]].
+{{qed}}
+[[Category:First-Countable Spaces]]
+[[Category:Sequences]]
+ldlam0wvs63p4v87p2v8zkb910vd0b1
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Space is Totally Bounded}
+Tags: Metric Spaces, Sequentially Compact Spaces, Totally Bounded Metric Spaces, Sequentially Compact Metric Space is Totally Bounded
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $M$ be [[Definition:Sequentially Compact Space|sequentially 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 $\epsilon \in \R_{>0}$ 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{Separable Metric Space is Second-Countable}
+Tags: Metric Spaces, Second-Countable Spaces, Separable Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $M$ be [[Definition:Separable Space|separable]].
+Then $M$ is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Separable Space|separability]], we can choose a [[Definition:Subset|subset]] $S \subseteq X$ that is [[Definition:Countable Set|countable]] and [[Definition:Everywhere Dense|everywhere dense]].
+Define:
+:$\BB = \set {\map {B_{1/n} } x: x \in S, \, n \in \N_{>0} }$
+where $\map {B_\epsilon } x$ denotes the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball of $x$ in $M$]].
+We have that [[Cartesian Product of Countable Sets is Countable]].
+Hence, by [[Image of Countable Set under Mapping is Countable]], it follows that $\BB$ is [[Definition:Countable Set|countable]].
+Let $\tau$ denote the [[Definition:Topology Induced by Metric|topology on $X$ induced by the metric $d$]].
+It suffices to show that $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
+From [[Open Ball of Metric Space is Open Set]], we have that $\BB \subseteq \tau$.
+We use [[Equivalence of Definitions of Analytic Basis]].
+Let $y \in U \in \tau$.
+By the definition of an [[Definition:Open Set of Metric Space|open set]], there exists a [[Definition:Strictly Positive Real Number|strictly positive real number]] $\epsilon$ such that $\map {B_\epsilon} y \subseteq U$.
+By the [[Archimedean Principle]], there exists a [[Definition:Natural Number|natural number]] $n > \dfrac 2 \epsilon$.
+That is:
+:$\dfrac 2 n < \epsilon$
+and so:
+:$\map {B_{2/n} } y \subseteq \map {B_\epsilon} y$.
+From [[Subset Relation is Transitive]], we have $\map {B_{2/n} } y \subseteq U$.
+By the definition of [[Definition:Everywhere Dense|everywhere denseness]], and by [[Equivalence of Definitions of Adherent Point]], there exists an $x \in S \cap \map {B_{1/n} } y$.
+By [[Definition:Metric Space Axioms|axiom $\left({M3}\right)$ for a metric]], it follows that $y \in \map {B_{1/n} } x$.
+For all $z \in \map {B_{1/n} } x$, we have:
+{{begin-eqn}}
+{{eqn | l = \map d {z, y}
+      | o = \le
+      | r = \map d {z, x} + \map d {x, y}
+      | c = [[Definition:Metric Space Axioms|Metric Space Axiom $(\text M 2)$]]
+}}
+{{eqn | r = \map d {z, x} + \map d {y, x}
+      | c = [[Definition:Metric Space Axioms|Metric Space Axiom $(\text M 3)$]]
+}}
+{{eqn | o = <
+      | r = \frac 2 n
+}}
+{{end-eqn}}
+That is:
+:$\map {B_{1/n} } x \subseteq \map {B_{2/n} } y$
+From [[Subset Relation is Transitive]], we have:
+:$y \in \map {B_{1/n} } x \subseteq U$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lipschitz Equivalence is Equivalence Relation}
+Tags: Lipschitz Equivalence, Equivalence Relations
+
+\begin{theorem}
+Let $A$ be a [[Definition:Set|set]].
+Let $\mathcal D$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
+Let $\sim$ be the [[Definition:Relation|relation]] on $\mathcal D$ defined as:
+:$\forall d_1, d_2 \in \mathcal D: d_1 \sim d_2 \iff d_1$ is [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]] to $d_2$
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Let $A$ be a [[Definition:Set|set]] and let $\mathcal D$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
+In the following, let $d_1, d_2, d_3 \subseteq \mathcal D$ be arbitrary.
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+Let $d_1$ be a [[Definition:Metric|metric]] on $A$.
+Then trivially:
+:$\forall x, y \in A: 1 \times d_1 \left({x, y}\right) \le d_1 \left({x, y}\right) \le 1 \times d_1 \left({x, y}\right)$
+That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $d_1 \sim d_2$.
+That is, let $d_1, d_2$ be [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent metrics]] on $A$.
+Then by definition:
+: $\forall x, y \in A: h d_1 \left({x, y}\right) \le d_2 \left({x, y}\right) \le k d_1 \left({x, y}\right)$
+for some $h, k \in \R_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = h d_1 \left({x, y}\right)
+      | o = \le
+      | r = d_2 \left({x, y}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_1 \left({x, y}\right)
+      | o = \le
+      | r = \frac 1 h d_2 \left({x, y}\right)
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = d_2 \left({x, y}\right)
+      | o = \le
+      | r = k d_1 \left({x, y}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac 1 k d_2 \left({x, y}\right)
+      | o = \le
+      | r = d_1 \left({x, y}\right)
+      | c = 
+}}
+{{end-eqn}}
+That is:
+: $\forall x, y \in A: \dfrac 1 h d_2 \left({x, y}\right) \le d_1 \left({x, y}\right) \le \dfrac 1 k d_2 \left({x, y}\right)$
+for some $\dfrac 1 h, \dfrac 1 k \in \R_{>0}$.
+That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+Note that in the definition of [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent metrics]], in the expression $d_1 \sim d_2$ it is not explicitly specified which of $d_1$ and $d_2$ goes in the middle of the defining statement.
+This result demonstrates that it does not actually matter.
+{{qed|lemma}}
+=== Transitivity ===
+Let $d_1 \sim d_2$ and $d_2 \sim d_3$.
+Then by definition:
+: $\forall x, y \in A: h_1 d_1 \left({x, y}\right) \le d_2 \left({x, y}\right) \le k_1 d_1 \left({x, y}\right)$
+: $\forall x, y \in A: h_2 d_2 \left({x, y}\right) \le d_3 \left({x, y}\right) \le k_2 d_2 \left({x, y}\right)$
+for some $h_1, k_1, h_2, k_2 \in \R_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = h_1 d_1 \left({x, y}\right)
+      | o = \le
+      | r = d_2 \left({x, y}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = h_1 h_2 d_1 \left({x, y}\right)
+      | o = \le
+      | r = h_2 d_2 \left({x, y}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = h_1 h_2 d_1 \left({x, y}\right)
+      | o = \le
+      | r = d_3 \left({x, y}\right)
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = d_2 \left({x, y}\right)
+      | o = \le
+      | r = k_1 d_1 \left({x, y}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = k_2 d_2 \left({x, y}\right)
+      | o = \le
+      | r = k_1 k_2 d_1 \left({x, y}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_3 \left({x, y}\right)
+      | o = \le
+      | r = k_1 k_2 d_1 \left({x, y}\right)
+      | c = 
+}}
+{{end-eqn}}
+So:
+: $\forall x, y \in A: h_1 h_2 d_1 \left({x, y}\right) \le d_3 \left({x, y}\right) \le k_1 k_2 d_1 \left({x, y}\right)$
+That is, $d_1 \sim d_3$ and 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{Local Membership of Equalizer}
+Tags: Subobjects, Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $e: E \to C$ be the [[Definition:Equalizer|equalizer]] of $f,g : C \to D$.
+Then a [[Definition:Variable Element|variable element]] $z: Z \to C$ is a [[Definition:Local Membership Relation|local member]] of $e$ [[Definition:Iff|iff]] $f \circ z = g \circ z$:
+:$z \in_C e \iff f \circ z = g \circ z$
+\end{theorem}
+
+\begin{proof}
+Firstly, note that by [[Equalizer is Monomorphism]], [[Definition:Local Membership Relation|local membership]] of $e$ is defined.
+=== Necessary Condition ===
+Suppose that $z \in_C e$.
+By definition of [[Definition:Local Membership Relation|local membership]], there is an $h: Z \to E$ such that $z = e \circ h$.
+Then since $e$ is the [[Definition:Equalizer|equalizer]] of $f$ and $g$:
+:$f \circ e = g \circ e$
+from which we deduce:
+:$f \circ z = f \circ e \circ h = g \circ e \circ h = g \circ z$
+by using [[Definition:Metacategory|metacategory]] axiom $(C3)$, i.e. associativity of $\circ$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $f \circ z = g \circ z$.
+Since $e$ is the [[Definition:Equalizer|equalizer]] of $f$ and $g$, we find $h: Z \to E$ such that:
+:$z = e \circ h$
+Hence $z \in_C e$, by definition of [[Definition:Local Membership Relation|local membership]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Space is Separable}
+Tags: Separable Spaces, Metric Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from:
+* [[Sequentially Compact Metric Space is Totally Bounded]]
+* [[Totally Bounded Metric Space is Separable]].
+{{qed}}
+{{ACC|Totally Bounded Metric Space is Separable}}
+[[Category:Separable Spaces]]
+[[Category:Metric Spaces]]
+[[Category:Sequentially Compact Spaces]]
+5ngsevn0ih6kfj87bntzl2pno8t3wip
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Space is Second-Countable}
+Tags: Second-Countable Spaces, Metric Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from:
+: [[Sequentially Compact Metric Space is Separable]]
+: [[Separable Metric Space is Second-Countable]].
+{{qed}}
+{{ACC|Sequentially Compact Metric Space is Separable}}
+[[Category:Second-Countable Spaces]]
+[[Category:Metric Spaces]]
+[[Category:Sequentially Compact Spaces]]
+au7mekk5yhl715pl6nvwpjyl5tw6cpe
+\end{proof}<|endoftext|>
+\section{Topological Equivalence is Equivalence Relation}
+Tags: Topologically Equivalent Metrics, Equivalence Relations
+
+\begin{theorem}
+Let $A$ be a [[Definition:Set|set]].
+Let $\mathcal D$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
+Let $\sim$ be the [[Definition:Relation|relation]] on $\mathcal D$ defined as:
+:$\forall d_1, d_2 \in \mathcal D: d_1 \sim d_2 \iff d_1$ is [[Definition:Topologically Equivalent Metrics|topologically equivalent]] to $d_2$
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Let $A$ be a [[Definition:Set|set]] and let $\mathcal D$ be the [[Definition:Set|set]] of all [[Definition:Metric|metrics]] on $A$.
+In the following, let $d_1, d_2, d_3 \subseteq \mathcal D$ be arbitrary.
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+Let $d_1$ be a [[Definition:Metric|metric]] on $A$.
+Then trivially:
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]] $\iff$ $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+That is, $d_1 \sim d_1$ and so $\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $d_1 \sim d_2$.
+That is, let $d_1, d_2$ be [[Definition:Topologically Equivalent Metrics|topologically equivalent metrics]] on $A$.
+Then by definition:
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]] $\iff$ $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+It follows from [[Biconditional is Commutative]] that:
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]] $\iff$ $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+That is, $d_2 \sim d_1$ and so $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $d_1 \sim d_2$ and $d_2 \sim d_3$.
+Then by definition:
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]] $\iff$ $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]] $\iff$ $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_3$-open]].
+Then by [[Biconditional is Transitive]]:
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]] $\iff$ $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_3$-open]].
+That is, $d_1 \sim d_3$ and 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{Lipschitz Equivalent Metrics are Topologically Equivalent}
+Tags: Lipschitz Equivalence, Topologically Equivalent Metrics, Lipschitz Equivalent Metrics are Topologically Equivalent
+
+\begin{theorem}
+Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
+Let $d_1$ and $d_2$ be [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
+Then $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Identity Mapping|identity mapping]]:
+: $f: A \to A: \forall x \in A: f \left({x}\right) = x$
+Then $f: \left({A, d_1}\right) \to \left({A, d_2}\right)$ can be considered as a [[Definition:Lipschitz Equivalent Metric Spaces|Lipschitz equivalence]].
+The result then follows from [[Lipschitz Equivalent Metric Spaces are Homeomorphic]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalence]]:
+: $\forall x, y \in A: h \map {d_2} {x, y} \le \map {d_1} {x, y} \le k \map {d_2} {x, y}$
+for some $h, k \in \R_{>0}$.
+Let $x \in A$.
+Let $\epsilon \in \R_{>0}$.
+Let $\map {B_{h \epsilon} } {x; d_1}$ denote the [[Definition:Open Ball|open $h \epsilon$-ball]] with respect to $d_1$ of $x \in A$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = \map {B_{h \epsilon} } {x; d_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_1} {x, y}
+      | o = <
+      | r = h \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_2} {x, y}
+      | o = \le
+      | r = \frac {\map {d_1} {x, y} } h
+      | c = 
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = \map {B_\epsilon} {x; d_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {B_{h \epsilon} } {x; d_1}
+      | o = \subseteq
+      | r = \map {B_\epsilon} {x; d_2}
+      | c = 
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = \map {B_{\epsilon / k} } {x; d_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_2} {x, y}
+      | o = <
+      | r = \frac \epsilon k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_1} {x, y}
+      | o = \le
+      | r = k \frac {\map {d_2} {x, y} } h
+      | c = 
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = \map {B_\epsilon} {x; d_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {B_{\epsilon / k} } {x; d_2}
+      | o = \subseteq
+      | r = \map {B_\epsilon} {x; d_1}
+      | c = 
+}}
+{{end-eqn}}
+Now suppose $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+Let $x \in U$.
+Then:
+:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {x; d_1} \subseteq U$
+Thus:
+:$\map {B_{\epsilon / k} } {x; d_2} \subseteq \map {B_\epsilon} {x; d_1} \subseteq U$
+and so $U$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], if $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]], it follows that $U$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalence]]:
+$\exists K_1, K_2 \in \R_{>0}$ such that:
+: $(1): \quad \forall x, y \in A: d_2 \left({x, y}\right) \le K_1 d_1 \left({x, y}\right)$
+: $(2): \quad \forall x, y \in A: d_1 \left({x, y}\right) \le K_2 d_2 \left({x, y}\right)$
+By [[Identity Mapping between Metrics separated by Scale Factor is Continuous]]:
+: the [[Definition:Identity Mapping|identity mapping]] from $M_1$ to $M_2$ is [[Definition:Continuous on Metric Space|continuous]]
+: the [[Definition:Identity Mapping|identity mapping]] from $M_2$ to $M_1$ is [[Definition:Continuous on Metric Space|continuous]].
+Hence the result by definition of [[Definition:Topologically Equivalent Metrics|topological equivalence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Metric Space is Complete}
+Tags: Complete Metric Spaces, Compact Spaces
+
+\begin{theorem}
+A [[Definition:Compact Metric Space|compact metric space]] is [[Definition:Complete Metric Space|complete]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Compact Space is Countably Compact]]
+:[[Countably Compact Metric Space is Sequentially Compact]]
+:[[Sequentially Compact Metric Space is Complete]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pullback as Equalizer}
+Tags: Pullbacks
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]] [[Definition:Category with Binary Products|with binary products]].
+Suppose that the following diagram [[Definition:Commutative Diagram|commutes]] (possibly except for the square):
+::$\begin{xy}\xymatrix@+1em{
+ P
+  \ar@/^/[rrd]^*+{p_1}
+  \ar@/_/[ddr]_*+{p_2}
+  \ar[rd]^*+{e}
+\\
+&
+ A \times B
+  \ar[r]_*+{\pi_1}
+  \ar[d]^*+{\pi_2}
+&
+ A
+  \ar[d]^*+{f}
+\\
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+where $A \times B$ is the [[Definition:Binary Product (Category Theory)|binary product]] of $A$ and $B$, and $\pi_1, \pi_2$ are the associated projections.
+Then the border of the diagram, i.e.:
+::$\begin{xy}\xymatrix{
+ P
+  \ar[r]^*+{p_1}
+  \ar[d]_*+{p_2}
+&
+ A
+  \ar[d]^*+{f}
+\\
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+is a [[Definition:Pullback (Category Theory)|pullback]] of $f$ and $g$ [[Definition:Iff|iff]] $e: P \to A \times B$ is the [[Definition:Equalizer|equalizer]] of $f \circ \pi_1$ and $g \circ \pi_2$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Still need to look for a nice formulation}}
+\end{proof}<|endoftext|>
+\section{Minkowski's Inequality for Sums/Index 2}
+Tags: Algebra, Real Analysis
+
+\begin{theorem}
+Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \ge 0$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
+Then:
+:$\displaystyle \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^2}^{1/2} \le \paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2
+      | r = \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2}
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2
+}}
+{{eqn | o = \le
+      | r = \sum_{k \mathop = 1}^n a_k^2 + 2 \paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} + \sum_{k \mathop = 1}^n b_k^2
+      | c = [[Cauchy's Inequality]]
+}}
+{{eqn | r = \paren {\paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} }^2
+}}
+{{end-eqn}}
+The result follows from [[Order is Preserved on Positive Reals by Squaring]].
+\end{proof}<|endoftext|>
+\section{Banach-Steinhaus Theorem}
+Tags: Banach Spaces
+
+\begin{theorem}
+Let $X$ be a [[Definition:Banach Space|Banach space]].
+Let $Y$ be a [[Definition:Normed Vector Space|normed vector space]] with norm $\norm {\,\cdot\,}_Y$.
+Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an [[Definition:Indexed Family|$A$-indexed family]] of [[Definition:Bounded Linear Transformation|bounded linear transformations]] from $X$ to $Y$.
+Suppose that:
+:$\displaystyle \forall x \in X: \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y < \infty$
+Then:
+:$\displaystyle \sup_{\alpha \mathop \in A} \norm {T_\alpha} < \infty$
+where $\norm {T_\alpha}$ denotes the [[Definition:Norm on Bounded Linear Transformation|norm of the linear transformation $T_\alpha$]].
+{{explain|We may need a definitive explanation for the intuitively understood $< \infty$.}}
+\end{theorem}<|endoftext|>
+\section{Minkowski's Inequality for Sums/Index Greater than 1}
+Tags: Algebra, Analysis
+
+\begin{theorem}
+Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
+Let $p \in \R$ be a [[Definition:Real Number|real number]] such that $p > 1$.
+Then:
+:$\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/p} \le \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p}$
+\end{theorem}
+
+\begin{proof}
+Without loss of generality, assume that:
+:$\displaystyle \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p \ne 0$
+Define:
+:$q = \dfrac p {p-1}$
+Then:
+:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p-1} p = 1$
+It follows that:
+{{begin-eqn}}
+{{eqn|l = \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p
+     |r = \sum_{k \mathop = 1}^n a_k \left({a_k + b_k}\right)^{p-1} + \sum_{k \mathop = 1}^n b_k \left({a_k + b_k}\right)^{p-1}
+}}
+{{eqn|o = \le
+     |r = \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q} 
+     |c = by [[Hölder's Inequality for Sums]], and because $\left({p - 1}\right) q = p$, [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn|o = 
+     |ro= +
+     |r = \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}
+}}
+{{eqn|r = \left({\left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} }\right) \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}
+}}
+{{end-eqn}}
+{{explain|Too many steps are being taken at the crucial point where it needs to be explained carefully, at "by [[Hölder's Inequality for Sums]], and because $\left({p - 1}\right) q {{=}} p$, [[Definition:By Hypothesis|by hypothesis]]. Remedy needed: (a) separate out the manipulation of the indices according to the relation between $p$ and $q$ as one step, (b) use an instantiation of [[Hölder's Inequality for Sums]] which explains it in terms of summations (at the moment it is expressed in the compact and gnomic notation of $\left\Vert{x}\right\Vert$, which requires one to go to a separate page again for the explanation of this notation) then (c) use as many steps of [[Hölder's Inequality for Sums]] (I'm not sure, but I believe there are two separate applications of it) as is necessary to provide the solution.}}
+The result follows by dividing both sides of the above inequality by $\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}$, and using the equation $\displaystyle 1 - \frac 1 q = \frac 1 p$.
+\end{proof}<|endoftext|>
+\section{Minkowski's Inequality for Sums/Index Less than 1}
+Tags: Algebra, Analysis
+
+\begin{theorem}
+Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
+Let $p \in \R$ be a [[Definition:Real Number|real number]].
+If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be [[Definition:Strictly Positive/Real Number|strictly positive]].
+If $p < 1$, $p \ne 0$, then:
+:$\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/p} \ge \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p}$
+\end{theorem}
+
+\begin{proof}
+In this case, $p$ and $q$ have opposite sign.
+The proof then follows the same lines as [[Minkowski's Inequality for Sums/Index Greater than 1|the proof for $p > 1$]], except that the [[Reverse Hölder's Inequality for Sums]] is applied instead.
+\end{proof}<|endoftext|>
+\section{Monotone Real Function is Darboux Integrable}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]], where $a < b$.
+Let $f: \closedint a b \to \R$ be a [[Definition:Monotone Real Function|monotone real function]].
+Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] over $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+We consider the case where $f$ is [[Definition:Increasing Real Function|increasing]]; the case where $f$ is [[Definition:Decreasing Real Function|decreasing]] is handled similarly.
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+By the [[Archimedean Principle]], there exists a [[Definition:Natural Numbers|natural number]] $n$ such that:
+:$n > \dfrac {\paren {b - a} \paren {\map f b - \map f a} } {\epsilon}$
+For $k \in \set {0, 1, 2, \ldots, n}$, define:
+:$x_k = a + k \dfrac {b - a} n$
+Then:
+:$a = x_0 < x_1 < \cdots < x_n = b$
+and hence $P = \set {x_0, x_1, \ldots, x_n}$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$.
+The [[Definition:Lower Sum|lower sum]] and [[Definition:Upper Sum|upper sum]] of $f$ belonging to the [[Definition:Subdivision (Real Analysis)|subdivision]] $P$ are, respectively:
+:$\displaystyle \map L P = \sum_{k \mathop = 1}^n \map f {x_{k - 1} } \paren {x_k - x_{k - 1} } = \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_{k - 1} }$
+:$\displaystyle \map U P = \sum_{k \mathop = 1}^n \map f {x_k} \paren {x_k - x_{k - 1} } = \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_k}$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \le
+      | r = \overline {\int_a^b} \map f x \rd x - \underline {\int_a^b} \map f x \rd x
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \map U P - \map L P
+      | c = 
+}}
+{{eqn | r = \frac {\paren {b - a} \paren {\map f b - \map f a} } n
+      | c = 
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+Since $\epsilon \in \R_{>0}$ is arbitrary, the result follows.
+{{qed}}
+{{link wanted|proof of $\displaystyle \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x$}}
+\end{proof}<|endoftext|>
+\section{Topologies Not Always Comparable by Coarseness}
+Tags: Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] with at least $2$ [[Definition:Element|elements]].
+Let $\mathbb T$ be the [[Definition:Set|set]] of all [[Definition:Topology|topologies]] on $S$.
+For two [[Definition:Topology|topologies]] $\tau_a, \tau_b \in \mathbb T$, let $\tau_a \le \tau_b$ denote that $\tau_a$ is [[Definition:Coarser Topology|coarser]] than $\tau_b$.
+Then there exist $\tau_1, \tau_2 \in \mathbb T$ such that neither:
+:$\tau_1 \le \tau_2$
+nor:
+:$\tau_2 \le \tau_1$
+That is, there are always [[Definition:Topology|topologies]] on $S$ which are [[Definition:Comparable Topologies|non-comparable]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$.
+Let:
+: $\tau_a$ be the [[Definition:Particular Point Topology|particular point topology]] with respect to $a$ on $S$
+: $\tau_b$ be the [[Definition:Particular Point Topology|particular point topology]] with respect to $b$ on $S$
+From [[Homeomorphic Non-Comparable Particular Point Topologies]]:
+:neither $\tau_a \le \tau_b$ nor $\tau_b \le \tau_a$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Plane is Abstract Geometry}
+Tags: Geometry
+
+\begin{theorem}
+The [[Definition:Euclidean Plane|Euclidean plane]] $\left({\R^2, L_E}\right)$ is an [[Definition:Abstract Geometry|abstract geometry]].
+\end{theorem}
+
+\begin{proof}
+We will show that the axioms for an [[Definition:Abstract Geometry|abstract geometry]] hold.
+=== Axiom 1 ===
+Let $P = \left({x_1, y_1}\right)$ and $Q = \left({x_2, y_2}\right)$ be two distinct [[Definition:Point (Abstract Geometry)|points]] in $\left({\R^2, L_E}\right)$.
+If $x_1 = x_2 = a$ then $P, Q \in L_a$.
+If $x_1 \ne x_2$ then let:
+:$m = \dfrac {y_2 - y_1} {x_2 - x_1}$
+:$b = y_2 - m x_2$
+Then $P, Q \in L_{m,b}$, since:
+:$b + m x_1 = y_2 - m \left({x_2 - x_1}\right) = y_2 - \left({y_2 - y_1}\right) = y_1$
+:$b + m x_2 = y_2 - m \left({x_2 - x_2}\right) = y_2$
+So any two [[Definition:Point (Abstract Geometry)|points]] in $\R^2$ lie on a [[Definition:Line (Abstract Geometry)|line]] in $L_E$.
+{{qed|lemma}}
+=== Axiom 2 ===
+For any $a \in \R$:
+:$\left({a, 0}\right), \left({a, 1}\right) \in L_a$
+Also for any $m, b \in \R$:
+:$\left({0, b}\right),\left({1, m + b}\right) \in L_{m,b}$
+So every [[Definition:Line (Abstract Geometry)|line]] in $L_E$ has at least two distinct [[Definition:Point (Abstract Geometry)|points]].
+{{qed|lemma}}
+Hence $\left({\R^2, L_E}\right)$ is an [[Definition:Abstract Geometry|abstract geometry]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pullback of Subset Inclusion}
+Tags: Pullbacks, Category of Sets, Subsets, Inclusion Mappings
+
+\begin{theorem}
+Denote with $\mathbf{Set}$ the [[Definition:Category of Sets|category of sets]].
+Let $A, B$ be [[Definition:Set|sets]], and let $f: A \to B$ be a [[Definition:Mapping|mapping]].
+Let $V \subseteq B$ be a [[Definition:Subset|subset]] of $B$.
+Denote with $i: V \to B$ the [[Definition:Inclusion Mapping|inclusion mapping]].
+Let $f^{-1} \left({V}\right) \subseteq A$ be the [[Definition:Preimage of Subset under Mapping|preimage]] of $V$ under $f$.
+Denote with $j: f^{-1} \left({V}\right) \to A$ the [[Definition:Inclusion Mapping|inclusion mapping]].
+Denote with $\bar f = f \restriction_{f^{-1} \left({V}\right)}$ the [[Definition:Restriction of Mapping|restriction]] of $f$ to $f^{-1} \left({V}\right)$.
+Then:
+::$\begin{xy}\xymatrix{
+ f^{-1} \left({V}\right)
+  \ar[r]^*+{\bar f}
+  \ar[d]_*+{j}
+&
+ V
+  \ar[d]^*+{i}
+\\
+ A
+  \ar[r]_*+{f}
+&
+ B
+}\end{xy}$
+is a [[Definition:Pullback (Category Theory)|pullback diagram]] in $\mathbf{Set}$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Pullback (Category Theory)|pullback]], given a [[Definition:Commutative Diagram|commutative diagram]]:
+::$\begin{xy}\xymatrix{
+ Q
+  \ar[r]^*+{q_1}
+  \ar[d]_*+{q_2}
+&
+ V
+  \ar[d]^*+{i}
+\\
+ A
+  \ar[r]_*+{f}
+&
+ B
+}\end{xy}$
+we need to find a [[Definition:Mapping|mapping]] $u: Q \to f^{-1} \left({V}\right)$ such that $\bar f \circ u = q_1$ and $j \circ u = q_2$.
+Let $x \in Q$; then $q_1 \left({x}\right) \in V \subseteq B$.
+Note that the commutativity of the diagram implies that $q_1 \left({x}\right) = f \left({q_2 \left({x}\right)}\right)$ since $i$ is an [[Definition:Inclusion Mapping|inclusion]].
+Hence also $f \left({q_2 \left({x}\right)}\right) \in V$.
+It follows by definition of [[Definition:Preimage of Subset under Mapping|preimage]] that $q_2 \left({x}\right) \in f^{-1} \left({V}\right)$.
+Thus define $u: Q \to f^{-1} \left({V}\right)$ by [[Definition:Restriction of Mapping|restriction]] of $q_2$:
+:$u \left({x}\right) := q_2 \restriction_{Q \times f^{-1} \left({V}\right)} \left({x}\right)$
+Then we have:
+:$j \circ u \left({x}\right) = u \left({x}\right) = q_2 \left({x}\right)$
+since $j$ is an [[Definition:Inclusion Mapping|inclusion mapping]].
+Hence $q_2 = j \circ u$ by [[Equality of Mappings]].
+Furthermore:
+:$\bar f \circ u \left({x}\right) = \bar f \circ q_2 \left({x}\right) = f \left({q_2 \left({x}\right)}\right) = q_1 \left({x}\right)$
+since $\bar f$ and $u$ are [[Definition:Restriction of Mapping|restrictions]] of $f$ and $q_2$, respectively.
+Hence $\bar f \circ u = q_1$ by [[Equality of Mappings]].
+Observe that any $u': Q \to f^{-1} \left({V}\right)$ must satisfy:
+:$j \circ u' = q_2 = j \circ u$
+By [[Inclusion Mapping is Injection]], $j$ is an [[Definition:Injection|injection]].
+From [[Injection iff Monomorphism in Category of Sets]], we conclude $j$ is a [[Definition:Monomorphism (Category Theory)|monomorphism]], and so $u = u'$.
+This establishes that $u: Q \to f^{-1} \left({V}\right)$ is [[Definition:Unique|unique]].
+Therefore, we have established that:
+::$\begin{xy}\xymatrix{
+ f^{-1} \left({V}\right)
+  \ar[r]^*+{\bar f}
+  \ar[d]_*+{j}
+&
+ V
+  \ar[d]^*+{i}
+\\
+ A
+  \ar[r]_*+{f}
+&
+ B
+}\end{xy}$
+is a [[Definition:Pullback (Category Theory)|pullback diagram]] in $\mathbf{Set}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Poincaré Plane is Abstract Geometry}
+Tags: Geometry
+
+\begin{theorem}
+The [[Definition:Poincaré Plane|Poincaré plane]] $\struct {\H, L_H}$ is an [[Definition:Abstract Geometry|abstract geometry]].
+\end{theorem}
+
+\begin{proof}
+We will show that the axioms for an [[Definition:Abstract Geometry|abstract geometry]] hold.
+=== Axiom 1 ===
+Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be two [[Definition:Point (Abstract Geometry)|points]] in $\struct {\H, L_H}$.
+If $x_1 = x_2 = a$ then $P, Q \in {}_a L$.
+If $x_1 \ne x_2$ then let:
+:$c = \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {2 \paren {x_2 - x_1} }$
+:$r = \sqrt{\paren{x_1 - c}^2 + y_1^2}$
+From definition of ${}_c L_r$:
+:${}_c L_r := \set {\tuple {x, y} \in \H : \paren {x - c}^2 + y^2 = r^2}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren{x_1 - c}^2 + y_1^2 
+      | r = r^2
+}}
+{{eqn | l = \paren{x_2 - c}^2 + y_2^2 
+      | r = y_2^2 + x_2^2 - 2 x_2 c + c^2
+}}
+{{eqn | r = y_2^2 + x_2^2 - 2 x_2 \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {2 \paren {x_2 - x_1} } + c^2
+}}
+{{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {y_2^2 x_2 - y_2^2 x_1 + x_2^3 - x_2^2 x_1 - x_2 y_2^2 + x_2 y_1^2 - x_2^3 + x_2 x_1^2}
+}}
+{{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {- y_2^2 x_1 - x_2^2 x_1 + x_2 y_1^2 + x_2 x_1^2}
+}}
+{{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {- y_2^2 x_1 - x_2^2 x_1 + x_2 y_1^2 + x_2 x_1^2 + y_1^2 x_1 - y_1^2 x_1 + x_1^3 - x_1^3}
+}}
+{{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {x_2 y_1^2 - y_1^2 x_1 + x_2 x_1^2 - x_1^3 - y_2^2 x_1 + y_1^2 x_1 - x_2^2 x_1 + x_1^3}
+}}
+{{eqn | r = y_1^2 + x_1^2 - x_1 \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {x_2 - x_1} + c^2
+}}
+{{eqn | r = y_1^2 + \paren {x_1 - c}^2
+}}
+{{eqn | r = r^2
+}}
+{{end-eqn}}
+Thus $P, Q \in {}_c L_r$.
+So any two points in $\H$ lie on a [[Definition:Line (Abstract Geometry)|line]] in $L_H$.
+{{qed|lemmma}}
+=== Axiom 2 ===
+For any $a \in \R$:
+:$\tuple {a, 1}, \tuple {a, 2} \in {}_a L$
+Also for any $c \in \R$ and $r \in \R_{>0}$:
+:$\tuple {c + \dfrac 1 2 r, \dfrac {\sqrt 3} 2 r}, \tuple {c - \dfrac 1 2 r, \dfrac {\sqrt 3} 2 r} \in {}_c L_r$
+So every line in $L_H$ has at least two points.
+{{qed|lemma}}
+Hence $\struct {\H, L_H}$ is an [[Definition:Abstract Geometry|abstract geometry]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pullback Lemma}
+Tags: Named Theorems, Pullbacks, Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Suppose that the following is a [[Definition:Commutative Diagram|commutative diagram]] in $\mathbf C$:
+::$\begin{xy}\xymatrix@+0.5em{
+ F
+  \ar[d]_*+{h''}
+  \ar[r]^*+{f'}
+&
+ E
+  \ar[d]^*+{h'}
+  \ar[r]^*+{g'}
+&
+ D
+  \ar[d]^*+{h}
+\\
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+Suppose furthermore that the right square is a [[Definition:Pullback (Category Theory)|pullback]].
+Then the left square is a [[Definition:Pullback (Category Theory)|pullback]] [[Definition:Iff|iff]] the outer rectangle is.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that the left square is a [[Definition:Pullback (Category Theory)|pullback]].
+To verify the outer rectangle is a [[Definition:Pullback (Category Theory)|pullback]], suppose we are given the following [[Definition:Commutative Diagram|commutative diagram]]:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@/^1em/[rrrd]^*+{p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+\\ & & &
+ D
+  \ar[d]^*+{h}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+The right square being a [[Definition:Pullback (Category Theory)|pullback]] implies there is a [[Definition:Unique|unique]] $v: P \to E$ making the following commute:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@/^1em/[rrrd]^*+{p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+  \ar@{-->}@/^/[rrd]_*+{v}
+\\ & &
+ E
+  \ar[d]^*+{h'}
+  \ar[r]^*+{g'}
+&
+ D
+  \ar[d]^*+{h}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+Since the left square is also a [[Definition:Pullback (Category Theory)|pullback]], we subsequently find a [[Definition:Unique|unique]] $u: P \to F$ making the following commute:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@/^1em/[rrrd]^*+{p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+  \ar@/^/[rrd]_*+{v}
+  \ar@{-->}[rd]_*+{u}
+\\ &
+ F
+  \ar[d]_*+{h''}
+  \ar[r]^*+{f'}
+&
+ E
+  \ar[d]^*+{h'}
+  \ar[r]^*+{g'}
+&
+ D
+  \ar[d]^*+{h}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+In conclusion, the outer rectangle is a [[Definition:Pullback (Category Theory)|pullback]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that the outer rectangle is a [[Definition:Pullback (Category Theory)|pullback]].
+To verify the left square is a [[Definition:Pullback (Category Theory)|pullback]], suppose we are given the following [[Definition:Commutative Diagram|commutative diagram]]:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@/^/[rrd]^*+{p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+\\ & &
+ E
+  \ar[d]^*+{h'}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+}\end{xy}$
+Evidently, we can amend this diagram as follows:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@/^/[rrd]_*{p_1}
+  \ar@/^1em/[rrrd]^*+{g' \circ p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+\\ & &
+ E
+  \ar[d]^*+{h'}
+  \ar[r]^*+{g'}
+&
+ D
+  \ar[d]^*+{h}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+Because the outer rectangle is a [[Definition:Pullback (Category Theory)|pullback]], we thus obtain a [[Definition:Unique|unique]] $u: P \to F$ making the following commute:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@/^1em/[rrrd]^*+{g' \circ p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+  \ar@{-->}[rd]_*+{u}
+\\ &
+ F
+  \ar[d]_*+{h''}
+  \ar[r]^*+{f'}
+&
+ E
+  \ar[d]^*+{h'}
+  \ar[r]^*+{g'}
+&
+ D
+  \ar[d]^*+{h}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+If we now can prove that $f' \circ u = p_1$, then we're done.
+To this end, remark that they both complete the following [[Definition:Commutative Diagram|commutative diagram]]:
+::$\begin{xy}\xymatrix@+0.5em{
+ P
+  \ar@{-->}[rrd]
+  \ar@/^1em/[rrrd]^*+{g' \circ p_1}
+  \ar@/_1em/[ddr]_*+{p_2}
+\\ & &
+ E
+  \ar[d]^*+{h'}
+  \ar[r]^*+{g'}
+&
+ D
+  \ar[d]^*+{h}
+\\ &
+ A
+  \ar[r]_*+{f}
+&
+ B
+  \ar[r]_*+{g}
+&
+ C
+}\end{xy}$
+Since the right square is a [[Definition:Pullback (Category Theory)|pullback]], there is only one such [[Definition:Morphism (Category Theory)|morphism]].
+Therefore, we conclude $f' \circ u = p_1$.
+Hence the left square is also a [[Definition:Pullback (Category Theory)|pullback]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strong Separation Theorem}
+Tags: Convex Analysis, Named Theorems
+
+\begin{theorem}
+Let $C \subset \R^\ell$ be closed and convex.
+Let $D = \set {\mathbf v} \subset C^c$.
+Then $C$ and $D$ can be strongly separated.
+\end{theorem}
+
+\begin{proof}
+=== Lemma 1 ===
+For any $\mathbf y \ne 0$ and any $\mathbf z \in \map {H_{\mathbf y}^<} r$, $r = \mathbf y \mathbf y$, the line segment joining $\mathbf y$ and $\mathbf z$ contains points with lengths strictly less than $\mathbf y$.
+{{explain|Because of the loose use of commas to separate clauses, it is unclear what the above is supposed to mean. Is $r {{=}} \mathbf y \mathbf y$ part of the premise or conclusion? That is, should it be taken to mean "the line segment joining $\mathbf y$ and $\mathbf z$? (If so, then perhaps it ought to be $r {{=}} \mathbf y \mathbf z$.) Or is it one of the conditions on $\mathbf y$ and $\mathbf z$ which needs to be fulfilled in order for the conclusion to follow? (In that case, its meaning needs to be explained.}}
+{{explain|What is the domain of $\mathbf y$? Presumably it is a vector of some kind, in which case the nature of $0$ also needs to be defined.}}
+{{explain|$\map {H_{\mathbf y}^<} r$}}
+=== Proof of Lemma 1 ===
+Since $\mathbf z \in \map {H_\mathbf y} r^<$:
+:$\mathbf z = \mathbf x - s \mathbf y$
+for some $\mathbf x \perp \mathbf y$ and $s > 0$.
+{{explain|$\mathbf x \perp \mathbf y$}}
+Let:
+:$\map f \gamma = \norm {\gamma \mathbf y + \paren {1 - \gamma} \mathbf z}^2$
+for $\gamma \in \closedint 0 1$.
+We show that:
+:$\map f \gamma < \norm {\mathbf y}^2$
+as $\gamma \to 1$.
+Expand $\map f \gamma$:
+{{begin-eqn}}
+{{eqn | l = \map f \gamma
+      | r = \norm {\gamma \, \mathbf y + \paren {1 - \gamma} \mathbf z}^2
+      | c = Definition of $\map f \gamma$ 
+}}
+{{eqn | r = \paren {\gamma \, \mathbf y + \paren {1 - \gamma} \mathbf z} \cdot \paren {\gamma \mathbf y + \paren {1 - \gamma} \mathbf z}
+      | c = Definition of norm on vector space $\R^\ell$
+}}
+{{eqn | r = \paren {\gamma \, \mathbf y + \paren {1 - \gamma} \paren {\mathbf x - s \, \mathbf y} } \cdot \paren {\gamma \, \mathbf y + \paren {1 - \gamma} \paren {\mathbf x - s \, \mathbf y} }
+      | c = as $\mathbf z = \mathbf x - s \, \mathbf y$
+}}
+{{eqn | r = \gamma^2 \mathbf y \mathbf y + 2 \gamma \paren {1 - \gamma} \mathbf y \paren {\mathbf x - s \mathbf y} + \paren {1 - \gamma}^2 \paren {\mathbf x - s \, \mathbf y} \paren {\mathbf x - s \, \mathbf y}
+      | c = Distributive property of dot product
+}}
+{{eqn | r = \gamma^2 \mathbf y \mathbf y - 2 \gamma \paren {1 - \gamma} s \, \mathbf y \mathbf y + \paren {1 - \gamma}^2 s^2 \mathbf y \mathbf y
+      | c = Distributive property of dot product
+}}
+{{eqn | o = 
+      | ro= +
+      | r = 2 \gamma \paren {1 - \gamma} \mathbf x \mathbf y - 2 \paren {1 - \gamma}^2 s \, \mathbf x \mathbf y + \paren {1 - \gamma}^2 \mathbf x \mathbf x 
+      | c = 
+}}
+{{eqn | r = \mathbf y \mathbf y \paren {\gamma^2 - 2 \gamma \paren {1 - \gamma} s + \paren {1 - \gamma}^2 s^2} + \paren {1 - \gamma}^2 \mathbf x \mathbf x 
+      | c = Definition of orthogonal vectors 
+}}
+{{eqn | r = \mathbf y \mathbf y \paren {\gamma - s \paren {1 - \gamma} }^2 + \paren {1 - \gamma}^2 \mathbf x \mathbf x 
+      | c = [[Binomial Theorem]]
+}}
+{{end-eqn}}
+Since $\map {f'} 1 = 2 \paren {1 + s} > 0$:
+:$\map f \gamma < \map f 1$
+as $\gamma \to 1$.
+Since $\map f 1 = \mathbf y \mathbf y = \norm {\mathbf y}^2$:
+:$\map f \gamma < \norm {\mathbf y}^2$
+as $\gamma \to 1$.
+{{qed|lemma}}
+=== Proof of Theorem ===
+Translating $C$ and $D = \set {\mathbf v}$ by $-\mathbf v$:
+We show that the theorem holds for $\mathbf v = 0$.
+Pick $n > 0$ such that:
+:$K = \map \cl {\map {B_n} {\mathbf v} } \cap C \ne \O$
+Since $C$ is closed, $K$ is closed.
+Since $\map {B_n} {\mathbf v}$ is bounded, $K$ is bounded.
+Since $K$ is closed and bounded, $K$ is compact.
+Let $\map f {\mathbf x} = \norm {\mathbf x}$.
+Since $\norm {\mathbf x}^2$ and $\sqrt x: \R_+ \to \R_+$ are both continuous, $\map f {\mathbf x}$ is continuous.
+Since $K$ is compact and $\map f {\mathbf x}$ is continuous, $\map f {\mathbf x}$ achieves its minimum at $\mathbf y \in C$ by the Weierstrass Theorem.
+Let $\map L {\mathbf x} = \mathbf y \mathbf x$ and $r = \mathbf y \mathbf y$.
+We show that:
+:$C \subset \map {L^{-1} } {\hointr r \to}$
+Suppose $C$ is not a subset of $\map {L^{-1} } {\hointr r \to}$.
+There exists $\mathbf z \in C \cap \map {L^{-1} } {\closedint {-\infty} r}$
+So:
+:$\mathbf z \in \map {H_\mathbf y} r$
+As $\gamma \to 1$, there exists a $0 < \gamma < 1$ such that:
+:$\norm {\mathbf y \, \gamma \, \mathbf z} < \norm {\mathbf y}$
+by Lemma 1.
+Since $C$ is convex, $\mathbf y \in C$ and $\mathbf z \in C$, convex combination $\mathbf y \, \gamma \, \mathbf z \in C$.
+{{explain|It is not clear which parts of the above compound "sentence" are in the premise and which in the conclusion.}}
+Since $\mathbf y $ is a minimmizer of $\norm {\mathbf x}$:
+:$\norm {\mathbf y} \le \norm {\mathbf y \, \gamma \, \mathbf z}$
+which is a contradiction.
+{{explain|Make it plain what it contradicts, and what the conclusion is of that [[Proof by Contradiction]].}}
+For strong separation, let $\epsilon < r / 2$.
+We have:
+:$C \subset \map {L^{-1} } {\hointr r \to} \subset \map {L^{-1} } {\hointr {r/2 + \epsilon} \infty}$
+Since:
+:$\map L {\mathbf 0} = \mathbf y \mathbf 0 = 0$
+and:
+:$r/2 - \epsilon > 0$
+it follows that:
+:$\mathbf v = \mathbf 0 \subset \map {L^{-1} } {\hointl {-\infty} {r/2 - \epsilon} }$
+{{explain|Explain how the conclusion of the proof follows from that.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Carathéodory's Theorem (Convex Analysis)}
+Tags: Convex Analysis
+
+\begin{theorem}
+Let $E \subset \R^\ell$.
+Let $\mathbb x \in \map {\operatorname{co} } E$.
+{{explain|Definition of $\map {\operatorname{co} } E$}}
+Then $\mathbf x$ is a convex combination of affinely independent points of $E$. 
+In particular, $\mathbf x $ is a convex combination of at most $\ell + 1$ points of $E$.
+\end{theorem}
+
+\begin{proof}
+Since $\mathbb x \in \map {\operatorname{co} } E$, $\mathbf x$ is a convex combination of points in $E$.
+From the definition of convex combination:
+:$\displaystyle \mathbf x = \sum_{i \mathop = 1}^k \gamma_i \mathbf y_i$
+with:
+:$\gamma_i \ge 0$
+:$\displaystyle \sum_{i \mathop = 1}^k \gamma_i = 1$
+and:
+:$\mathbf y_i \in E$
+Let $K \subset \N$ be the set of all possible $k$ such that $\mathbf x$ is a convex combination of $k$ elements of $E$.
+By the [[Well-Ordering Principle]], $K$ is [[Definition:Well-Ordering|well-ordered]]. 
+$K$ has a smallest element $k_s$ by the definition of [[Definition:Well-Ordering|well-ordered sets]].
+Suppose that the set $\set {\mathbf y_i: i \le k_s}$ is affinely dependent.
+From the condition for affinely dependent set, there exists a set $\set {\alpha_i: i \le k_s}$ such that for some $\alpha_i > 0$:
+:$\displaystyle \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i = 0$
+and:
+:$\displaystyle \sum_{i \mathop = 1}^{k_s} \alpha_i = 0$
+Pick the smallest $\dfrac {\gamma_j} {\alpha_j} > 0$.
+Then:
+:$\displaystyle \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} + 0 = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} - \paren {\frac {\gamma_j} {\alpha_j} \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i}$ 
+Rearrange to get:
+:$\displaystyle \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \paren {\gamma_i - \frac {\gamma_j} {\alpha_j} \alpha_i} \mathbf y_i}$
+Since $\gamma_j > 0$ is made the smallest and $\alpha_j > 0$ is made the greatest:
+:$\dfrac {\alpha_i} {\alpha_j} \gamma_j \le \gamma_j \le \gamma_i$
+for all $i \le k_s$.
+Thus:
+:$\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} \ge 0$
+For $i = j$:
+:$\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} = 0$
+For $i \ne j$, let:
+:$\gamma'_{i'} \equiv \gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i$
+We can express:
+:$\displaystyle x = \sum_{i' \mathop = 1}^{k_s-1} \gamma'_{i'} \mathbf y_i$
+which [[Proof by Contradiction|contradicts]] the assumption that $k_s = \min K$.
+{{qed}}
+{{Namedfor|Constantin Carathéodory|cat = Carathéodory}}
+\end{proof}<|endoftext|>
+\section{Limit of Sine of X over X/Corollary}
+Tags: Differential Calculus, Sine Function, Limit of Sine of X over X
+
+\begin{theorem}
+:$\displaystyle \lim_{x \mathop \to 0} \frac x {\sin x} = 1$
+\end{theorem}
+
+\begin{proof}
+We have the inequality:
+:$1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$
+for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.
+{{explain|Find a link for this}}
+Taking the limit of the leftmost term and the rightmost term:
+:$\displaystyle \lim_{\theta \mathop \to 0} \ 1 = 1$
+:$\displaystyle \lim_{\theta \mathop \to 0} \frac 1 {\cos\theta} = 1$
+So by the [[Squeeze Theorem]]:
+:$\displaystyle \lim_{\theta \mathop \to 0} \frac \theta {\sin\theta} = 1$
+{{qed}}
+[[Category:Differential Calculus]]
+[[Category:Sine Function]]
+[[Category:Limit of Sine of X over X]]
+917kwt3kc1unm5swydrnipque9ae2rk
+\end{proof}<|endoftext|>
+\section{Dicyclic Group is Group/Corollary}
+Tags: Dicyclic Groups
+
+\begin{theorem}
+The [[Definition:Quaternion Group|quaternion group]] $Q_4$ is a [[Definition:Abelian Group|non-abelian]] [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+As the [[Definition:Quaternion Group|quaternion group]] $Q_4$ is an example of the [[Definition:Dicyclic Group|dicyclic group]] with $4$ elements, is also a [[Definition:Abelian Group|non-abelian]] [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Constant/Definite}
+Tags: Definite Integrals
+
+\begin{theorem}
+:$\displaystyle \int_a^b c \ \ \mathrm dx = c \left({b-a}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $f_c: \R \to \R$ be the [[Definition:Constant Mapping|constant function]].
+By definition:
+: $\forall x \in \R: f_c \left({x}\right) = c$
+Thus:
+: $\sup \left({f_c}\right) = \inf \left({f_c}\right) = c$
+So from [[Upper and Lower Bounds of Integral]], we have:
+: $\displaystyle c \left({b-a}\right) \le \int_a^b c \ \ \mathrm dx \le c \left({b-a}\right)$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Constant}
+Tags: Primitives
+
+\begin{theorem}
+:$\ds \int c \rd x = c x + C$ where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\ds \map F x = \int c \rd x$
+From the definition of [[Definition:Primitive (Calculus)|primitive]]:
+:$\map {F'} x = c$
+From [[Derivative of Function of Constant Multiple]]:
+:$\map {\dfrac \d {\d x} } {c x} = c$
+From [[Primitives which Differ by Constant]]:
+:$\map {\dfrac \d {\d x} } {c x + C} = c$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Derivative of Constant Multiple]]:
+:$\map {\dfrac \d {\d x} } {c \map f x} = c \map {\dfrac \d {\d x} } {\map f x}$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+\end{proof}<|endoftext|>
+\section{Unbounded Set of Real Numbers is not Compact}
+Tags: Compact Spaces, Analysis, Unbounded Set of Real Numbers is not Compact
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number|set of real numbers]] considered as a [[Definition:Euclidean Space|Euclidean space]].
+Let $S \subseteq \R$ be [[Definition:Unbounded Metric Space|unbounded]] in $\R$.
+Then $S$ is not a [[Definition:Compact Subspace|compact subspace]] of $\R$.
+\end{theorem}
+
+\begin{proof}
+By the [[Rule of Transposition|rule of transposition]], it suffices to show that if $S$ is a [[Definition:Compact Subspace|compact subspace]] of $\R$, then $S$ is [[Definition:Bounded Metric Space|bounded]].
+Let $\mathcal C$ be the [[Definition:Set|set]] of all [[Definition:Open Ball|open $\epsilon$-balls]] of $0$ in $\R$:
+:$\mathcal C = \set {\map {B_\epsilon} 0: \epsilon \in \R_{>0}}$
+We have that:
+:$\displaystyle \bigcup \mathcal C = \R \supseteq S$
+From [[Open Ball of Metric Space is Open Set]], it follows that $\mathcal C$ is an [[Definition:Open Cover|open cover]] for $S$.
+Let $\mathcal F$ be a [[Definition:Finite Subcover|finite subcover]] of $\mathcal C$ for $S$.
+Then $\displaystyle \bigcup \mathcal F$ is the largest [[Definition:Open Ball|open $\epsilon$-ball]] in $\mathcal F$.
+Thus:
+:$S \subseteq \displaystyle \bigcup \mathcal F \in \mathcal C$
+Hence, $S$ is [[Definition:Bounded Metric Space|bounded]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From:
+* [[Real Number Line is Metric Space]]
+* [[Compact Metric Space is Totally Bounded]]
+* [[Totally Bounded Metric Space is Bounded]]
+the result follows by the [[Rule of Transposition|rule of transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Mapping is Sequentially Continuous}
+Tags: Continuous Mappings
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
+Let $x \in X$.
+Let $f: X \to Y$ be a [[Definition:Mapping|mapping]] that is [[Definition:Continuous Mapping at Point (Topology)|continuous at $x$]].
+Then $f$ is [[Definition:Sequential Continuity|sequentially continuous at $x$]].
+\end{theorem}
+
+\begin{proof}
+Let $\left\langle{x_n}\right\rangle_{n \ge 1}$ be a [[Definition:Sequence|sequence]] in $X$ [[Definition:Convergent Sequence (Topology)|converging]] to $x$.
+Let $V$ be a [[Definition:Neighborhood of Point|neighborhood]] of $f \left({x}\right)$ in $Y$.
+We are required to show that there exists $N \in \N$ such that $f \left({x_n}\right) \in V$ for all $n \ge N$.
+By [[Definition:Everywhere Continuous Mapping (Topology)|continuity]] of $f$, choose a [[Definition:Neighborhood of Point|neighborhood]] $U$ of $x$ in $X$ such that $f \left({U}\right) \subseteq V$.
+Since $\left\langle{x_n}\right\rangle_{n \ge 1}$ converges, there exists $N \in \N$ such that $x_n \in U$ for all $n \ge N$.
+Therefore we must have that $f \left({x_n}\right) \in f \left({U}\right) \subseteq V$ for all $n \ge N$, as required.
+{{qed}}
+[[Category:Continuous Mappings]]
+niicbib9rld1rf8s68zdpaatvlg3sns
+\end{proof}<|endoftext|>
+\section{Sequential Continuity is Equivalent to Continuity in Metric Space}
+Tags: Metric Spaces, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $\left({X, d}\right)$ and $\left({Y, e}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
+Let $x \in X$.
+Then $f$ is [[Definition:Continuous at Point of Metric Space|continuous at $x$]] {{iff}} $f$ is [[Definition:Sequential Continuity|sequentially continuous at $x$]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Continuous Mapping is Sequentially Continuous]].
+To prove the [[Definition:Converse Statement|converse]], by the [[Rule of Transposition]] we may prove the [[Definition:Contrapositive Statement|contrapositive]]:
+:If $f$ is not [[Definition:Continuous at Point of Metric Space|continuous at $x$]], then $f$ is not [[Definition:Sequential Continuity|sequentially continuous at $x$]].
+We suppose therefore that there exists $\epsilon_0 > 0$  such that for all $\delta > 0$ there exists $y \in X$ such that $d \left({x, y}\right) < \delta$ and $e \left({f \left({x}\right), f \left({y}\right)}\right) \ge \epsilon_0$.
+For $n \ge 1$, define $\delta_n = \dfrac 1 n$.
+For $n \ge 1$, we may choose $y_n \in X$ such that $d \left({x, y_n}\right) < \delta_n$ and $e \left({f \left({x}\right), f \left({y_n}\right)}\right) \ge \epsilon_0$.
+Therefore, by definition the [[Definition:Sequence|sequence]] $\left\langle{y_n}\right\rangle_{n \ge 1}$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$.
+However, by definition the [[Definition:Sequence|sequence]] $\left\langle{f \left({y_n}\right)}\right\rangle_{n \ge 1}$ does not [[Definition:Convergent Sequence (Metric Space)|converge]] to $f \left({x}\right)$.
+That is, $f$ is not [[Definition:Sequential Continuity|sequentially continuous]] at $x$.
+{{Qed}}
+[[Category:Metric Spaces]]
+[[Category:Continuous Mappings on Metric Spaces]]
+p90ycd9zttj9bebjtd673xiqu3exi30
+\end{proof}<|endoftext|>
+\section{Pullback of Commutative Triangle}
+Tags: Pullbacks
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Suppose that the following is a [[Definition:Commutative Diagram|commutative diagram]] in $\mathbf C$:
+::$\begin{xy}\xymatrix@+1em@L+2px{
+ A'
+  \ar[rr]^*{h_\alpha}
+  \ar[dd]_*{\alpha'}
+& &
+ A
+  \ar[rd]^*{\gamma}
+  \ar[dd]^(.4)*{\alpha}
+\\ &
+ B'
+  \ar[ld]^*{\beta'}
+  \ar[rr] |{\hole} ^(.3)*{h_\beta}
+& &
+ B
+  \ar[ld]^*{\beta}
+\\
+ C'
+  \ar[rr]_*{h}
+& &
+ C
+}\end{xy}$
+and that the two squares in it are [[Definition:Pullback Diagram|pullback diagrams]].
+Then there is a [[Definition:Unique|unique]] [[Definition:Morphism (Category Theory)|morphism]] $\gamma': A' \to B'$ making the following commute:
+::$\begin{xy}\xymatrix@+1em@L+2px{
+ A'
+  \ar[rr]^*{h_\alpha}
+  \ar[dd]_*{\alpha'}
+  \ar@{-->}[rd]^*{\gamma'}
+& &
+ A
+  \ar[rd]^*{\gamma}
+  \ar[dd]^(.4)*{\alpha}
+\\ &
+ B'
+  \ar[ld]^*{\beta'}
+  \ar[rr] |{\hole} ^(.3)*{h_\beta}
+& &
+ B
+  \ar[ld]^*{\beta}
+\\
+ C'
+  \ar[rr]_*{h}
+& &
+ C
+}\end{xy}$
+Furthermore, $\gamma'$ makes the following a [[Definition:Pullback (Category Theory)|pullback]]:
+::$\begin{xy}\xymatrix@+.5em@L+2px{
+ A'
+  \ar[r]^*{h_\alpha}
+  \ar[d]_*{\gamma'}
+&
+ A
+  \ar[d]^*{\gamma}
+\\
+ B'
+  \ar[r]_*{h_\beta}
+&
+ B
+}\end{xy}$
+\end{theorem}
+
+\begin{proof}
+The first diagram above can be distorted into:
+::$\begin{xy}\xymatrix@+.5em@L+2px{
+ A'
+  \ar@/^/[rrd]^*{\gamma \circ h_\alpha}
+  \ar@/_/[ddr]_*{\alpha'}
+\\ &
+ B'
+  \ar[r]^*{h_\beta}
+  \ar[d]^*{\beta'}
+&
+ B
+  \ar[d]^*{\beta}
+\\ &
+ C'
+  \ar[r]_*{h}
+&
+ C
+}\end{xy}$
+and since the square is a [[Definition:Pullback (Category Theory)|pullback]], there is a [[Definition:Unique|unique]] $\gamma': A' \to B'$ as desired.
+That:
+::$\begin{xy}\xymatrix@+.5em@L+2px{
+ A'
+  \ar[r]^*{h_\alpha}
+  \ar[d]_*{\gamma'}
+&
+ A
+  \ar[d]^*{\gamma}
+\\
+ B'
+  \ar[r]_*{h_\beta}
+&
+ B
+}\end{xy}$
+is a [[Definition:Pullback (Category Theory)|pullback]] follows immediately from the [[Pullback Lemma]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Associative Commutative Idempotent Operation is Distributive over Itself}
+Tags: Abstract Algebra, Commutativity, Associativity, Idempotence, Distributive Operations
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]], such that:
+:$(1): \quad \circ$ is [[Definition:Associative|associative]]
+:$(2): \quad \circ$ is [[Definition:Commutative Operation|commutative]]
+:$(3): \quad \circ$ is [[Definition:Idempotent Operation|idempotent]].
+Then $\circ$ is [[Definition:Distributive Operation|distributive]] over itself.
+That is:
+:$\forall a, b, c \in S: \left({a \circ b}\right) \circ \left({a \circ c}\right) = a \circ b \circ c = \left({a \circ c}\right) \circ \left({b \circ c}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left({a \circ b}\right) \circ \left({a \circ c}\right)
+      | r = a \circ \left({b \circ a}\right) \circ c
+      | c = $\circ$ is [[Definition:Associative|associative]]
+}}
+{{eqn | r = a \circ \left({a \circ b}\right) \circ c
+      | c = $\circ$ is [[Definition:Commutative Operation|commutative]]
+}}
+{{eqn | r = \left({a \circ a}\right) \circ b \circ c
+      | c = $\circ$ is [[Definition:Associative|associative]]
+}}
+{{eqn | r = a \circ b \circ c
+      | c = $\circ$ is [[Definition:Idempotent Operation|idempotent]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \left({a \circ c}\right) \circ \left({b \circ c}\right)
+      | r = a \circ \left({c \circ b}\right) \circ c
+      | c = $\circ$ is [[Definition:Associative|associative]]
+}}
+{{eqn | r = a \circ \left({b \circ c}\right) \circ c
+      | c = $\circ$ is [[Definition:Commutative Operation|commutative]]
+}}
+{{eqn | r = a \circ b \circ \left({c \circ c}\right)
+      | c = $\circ$ is [[Definition:Associative|associative]]
+}}
+{{eqn | r = a \circ b \circ c
+      | c = $\circ$ is [[Definition:Idempotent Operation|idempotent]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Abstract Algebra]]
+[[Category:Commutativity]]
+[[Category:Associativity]]
+[[Category:Idempotence]]
+[[Category:Distributive Operations]]
+8kmodbsjdxvs9f00hcmjpy84c56ct7e
+\end{proof}<|endoftext|>
+\section{Intersection Distributes over Intersection}
+Tags: Set Intersection, Distributive Operations, Intersection Distributes over Intersection
+
+\begin{theorem}
+[[Definition:Set Intersection|Set intersection]] is [[Definition:Distributive Operation|distributive]] over itself:
+:$\forall A, B, C: \paren {A \cap B} \cap \paren {A \cap C} = A \cap B \cap C = \paren {A \cap C} \cap \paren {B \cap C}$
+where $A, B, C$ are [[Definition:Set|sets]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Intersection is Associative]]
+:[[Intersection is Commutative]]
+:[[Intersection is Idempotent]]
+The result follows from [[Associative Commutative Idempotent Operation is Distributive over Itself]].
+{{qed}}
+[[Category:Set Intersection]]
+[[Category:Distributive Operations]]
+[[Category:Intersection Distributes over Intersection]]
+sm1iv7tccv98f1b00fj1mhvpqppzw5x
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping is Restriction of Identity}
+Tags: Inclusion Mappings, Identity Mappings
+
+\begin{theorem}
+Let $T$ be a [[Definition:Set|set]].
+Let $S \subseteq T$ be a [[Definition:Subset|subset]] of $T$.
+Let $i_S: S \to T$ be the [[Definition:Inclusion Mapping|inclusion mapping]] on $S$.
+Then $i_S$ is the [[Definition:Restriction of Mapping|restriction]] of the [[Definition:Identity Mapping|identity mapping]] $I_T: T \to T$ on $T$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Inclusion Mapping|inclusion mapping]]:
+:$i_S: S \to T: \forall x \in S: \map {i_S} x = x$
+By definition of [[Definition:Identity Mapping|identity mapping]]:
+:$I_T: T \to T: \forall x \in T: \map {I_T} x = x$
+The result follows by definition of [[Definition:Restriction of Mapping|restriction of mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Subset under Inclusion Mapping}
+Tags: Inclusion Mappings, Subsets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Let $i_H: H \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] on $H$.
+Let $T \subseteq S$.
+Then:
+:$i_H^{-1} \left({T}\right) = T \cap H$
+where $i_H^{-1} \left({T}\right)$ is the [[Definition:Preimage of Subset under Mapping|preimage]] of $T$ under $i_H$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in i_H^{-1} \left({T}\right)$.
+By the definition of [[Definition:Inclusion Mapping|inclusion mapping]]:
+:$i_H \left({x}\right) = x$
+By definition of [[Definition:Preimage of Subset under Mapping|preimage]] of $T$:
+:$i_H \left({x}\right) \in T$
+Thus $x \in T$.
+By definition of [[Definition:Inclusion Mapping|inclusion mapping]]:
+:$x \in \operatorname{Dom} \left({i_H}\right) = H$
+So $x \in T$ and $x \in H$ and so by definition of [[Definition:Set Intersection|set intersection]]:
+:$x \in T \cap H$
+Thus by definition of [[Definition:Subset|subset]]:
+:$i_H^{-1} \left({T}\right) \subseteq T \cap H$
+Now let $x \in T \cap H$.
+As $H = \operatorname{Dom} \left({i_H}\right)$ it follows that:
+: $x \in \operatorname{Dom} \left({i_H}\right)$
+As $x \in T \cap H$ it follows by definition of [[Definition:Set Intersection|set intersection]] that $x \in T$.
+Thus:
+: $i_H \left({x}\right) \in T$
+and so by definition of [[Definition:Preimage of Subset under Mapping|preimage]] of $T$:
+: $x \in i_H^{-1} \left({T}\right)$
+Thus by definition of [[Definition:Subset|subset]]:
+:$T \cap H \subseteq i_H^{-1} \left({T}\right)$
+The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+[[Category:Inclusion Mappings]]
+[[Category:Subsets]]
+5qor5qq5mjgdiuhzmypk9lb04k7ew86
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping is Continuous}
+Tags: Continuous Mappings, Inclusion Mappings
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T_H = \struct {H, \tau_H}$ be a [[Definition:Topological Subspace|topological subspace]] of $T$ where $H \subseteq S$.
+Let $i_H: H \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] on $H$.
+Then $i_H$ is a [[Definition:Everywhere Continuous Mapping (Topology)|$\struct {\tau_H, \tau}$-continuous mapping]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$.
+Then from [[Preimage of Subset under Inclusion Mapping]]:
+:$\map {i_H^{-1} } U = U \cap H$
+From the definition of the [[Definition:Topological Subspace|subspace topology]]:
+:$U \cap H \in \tau_H$
+Hence the result by definition of a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+{{qed}}
+[[Category:Continuous Mappings]]
+[[Category:Inclusion Mappings]]
+qr62gddavqaem41gk9q2zvk9fjbyzex
+\end{proof}<|endoftext|>
+\section{Continuity of Composite with Inclusion/Mapping on Inclusion}
+Tags: Topological Subspaces, Continuous Mappings, Inclusion Mappings, Composite Mappings
+
+\begin{theorem}
+If $f$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau, \tau'}$-continuous]], then $f \circ i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau'}$-continuous]]
+\end{theorem}
+
+\begin{proof}
+From [[Inclusion Mapping is Continuous]], $i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau}$-continuous]].
+It follows from [[Continuity of Composite Mapping]] that $f \circ i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau'}$-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Composite with Inclusion/Inclusion on Mapping}
+Tags: Topological Subspaces, Continuous Mappings, Inclusion Mappings, Composite Mappings
+
+\begin{theorem}
+$g$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau', \tau_H}\right)$-continuous]] [[Definition:Iff|iff]] $i \circ g$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau', \tau}\right)$-continuous]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $g$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau', \tau_H}\right)$-continuous]].
+From [[Inclusion Mapping is Continuous]], $i$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau_H, \tau}\right)$-continuous]].
+It follows from [[Continuity of Composite Mapping]] that $i \circ g$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau', \tau}\right)$-continuous]].
+=== Sufficient Condition ===
+Suppose $i \circ g$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau', \tau}\right)$-continuous]].
+Let $V \in \tau_H$.
+Then from the definition of [[Definition:Topological Subspace|topological subspace]]:
+: $V = U \cap H$ for some $U \in \tau$
+and by [[Preimage of Subset under Inclusion Mapping]]:
+: $U \cap H = i^{-1} \left({U}\right)$
+So by [[Preimage of Subset under Composite Mapping]]:
+: $g^{-1} \left({V}\right) = g^{-1} \left({i^{-1} \left({U}\right)}\right) = \left({i \circ g}\right)^{-1} \left({U}\right)$
+Thus:
+: $g^{-1} \left({V}\right) \in \tau'$
+by continuity of $i \circ g$.
+Hence $g$ is [[Definition:Continuous Mapping (Topology)|$\left({\tau', \tau_H}\right)$-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Composite with Inclusion/Uniqueness of Induced Topology}
+Tags: Topological Subspaces, Continuous Mappings, Inclusion Mappings, Composite Mappings
+
+\begin{theorem}
+The [[Definition:Topological Subspace|induced topology]] $\tau_H$ is the ''only'' [[Definition:Topology|topology]] on $H$ satisfying [[Continuity of Composite with Inclusion/Inclusion on Mapping|Continuity of Composite with Inclusion: Inclusion on Mapping]] for all possible $g$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\tau''$ is a [[Definition:Topology|topology]] on $H$ such that:
+:$(1) \quad$ For any [[Definition:Topological Space|topological space]] $T' = \struct {A', \tau'}$, and
+:$(2) \quad$ For any [[Definition:Mapping|mapping]] $g: A' \to H$:
+$g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau', \tau''}$-continuous]] {{iff}} $i \circ g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau', \tau}$-continuous]].
+It needs to be shown that $\tau''$ must be the same as $\tau_H$.
+Let $A' = H$ and $\tau' = \tau''$.
+Let $g$ be the [[Definition:Identity Mapping|identity mapping]] on $H$.
+From [[Identity Mapping is Continuous]], $g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau'', \tau''}$-continuous]]
+Thus from [[Continuity of Composite Mapping]], $i \circ g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau'', \tau}$-continuous]].
+Hence for any $U \in \tau$:
+:$\map {\paren {i \circ g}^{-1} } U \in \tau''$
+But:
+:$\map {\paren {i \circ g}^{-1} } U = \map {i^{-1} } U = U \cap H$
+Hence $\tau_H \subseteq \tau''$.
+Next, take take $A' = H$ and $\tau' = \tau_H$.
+Let $g$ be the [[Definition:Identity Mapping|identity mapping]] on $H$.
+We have that $i \circ g = i$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau}$-continuous]],
+From [[Continuity of Composite with Inclusion/Inclusion on Mapping|Continuity of Composite with Inclusion: Inclusion on Mapping]], it follows that $g$ is [[Definition:Continuous Mapping (Topology)|$\tuple {\tau_H, \tau''}$-continuous]].
+But by definition of [[Definition:Continuous Mapping (Topology)|continuity]], this is the same as saying $\tau'' \subseteq \tau_H$.
+So $\tau'' = \tau_H$, as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Products of Open Sets form Local Basis in Product Space}
+Tags: Product Spaces
+
+\begin{theorem}
+Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $\struct{T, \tau} = T_1 \times T_2$ be the [[Definition:Product Space (Topology)|product space]] of $T_1$ and $T_2$.
+Let $\tuple{x, y} \in A_1 \times A_2$.
+Let $W \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$ such that $\tuple{x, y} \in W$.
+Then:
+:$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple{x, y} \in U_1 \times U_2 \subseteq W$
+That is, [[Definition:Cartesian Product|products]] of [[Definition:Open Set (Topology)|open sets]] from $T_1$ and $T_2$ form a [[Definition:Local Basis|local basis]] at $\tuple{x, y}$.
+\end{theorem}
+
+\begin{proof}
+Let $W \in \tau$ such that $\tuple{x, y} \in W$.
+From [[Natural Basis of Tychonoff Topology of Finite Product]], $\tau$ is the [[Definition:Topology|topology]] with [[Definition:Basis (Topology)|basis]]:
+: $\BB = \set{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}$
+Thus $W$ is the [[Definition:Set Union|union]] of sets of the form $U_1 \times U_2$ where $U_1 \in \tau_1$ and $U_2 \in \tau_2$.
+That is:
+:$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple{x, y} \in U_1 \times U_2 \subseteq W$
+which is what was to be proved.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Real Intervals are Homeomorphic}
+Tags: Homeomorphic Metric Spaces
+
+\begin{theorem}
+Consider the [[Definition:Real Number|real numbers]] $\R$ as a [[Real Number Line is Metric Space|metric space]] under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]].
+Let $I_1 := \left({a \,.\,.\, b}\right)$ and $I_2 := \left({c \,.\,.\, d}\right)$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Real Interval|open real intervals]].
+Then $I_1$ and $I_2$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Open Real Interval|open real interval]], for $I_1$ and $I_2$ to be [[Definition:Non-Empty Set|non-empty]] it must be the case that $a < b$ and $c < d$.
+In particular it is noted that $a \ne b$ and $c \ne d$.
+Thus $a - b \ne 0$ and $c - d \ne 0$.
+Consider the [[Definition:Real Function|real function]] $f: I_1 \to I_2$ defined as:
+:$\forall x \in I_1: f \left({x}\right) = c + \dfrac {\left({d - c}\right) \left({x - a}\right)} {b - a}$
+Then after some algebra:
+:$\forall x \in I_2: f^{-1} \left({x}\right) = a + \dfrac {\left({b - a}\right) \left({x - c}\right)} {d - c}$
+Both of these are defined as $a - b \ne 0$ and $c - d \ne 0$.
+By the [[Combination Theorem for Continuous Functions]], both $f$ and $f^{-1}$ are [[Definition:Continuous Mapping (Metric Spaces)|continuous]] on the [[Definition:Open Real Interval|open real intervals]] on which they are defined.
+Hence the result by definition of [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Half-Open Real Interval is not Closed in Real Number Line}
+Tags: Real Intervals, Closed Sets
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
+Let $\left[{a \,.\,.\, b}\right) \subset \R$ be a [[Definition:Half-Open Real Interval|half-open interval]] of $\R$.
+Then $\left[{a \,.\,.\, b}\right)$ is not a [[Definition:Closed Set (Metric Space)|closed set]] of $\R$.
+Similarly, the [[Definition:Half-Open Real Interval|half-open interval]] $\left({a \,.\,.\, b}\right] \subset \R$ is not a [[Definition:Closed Set (Metric Space)|closed set]] of $\R$.
+\end{theorem}
+
+\begin{proof}
+Consider:
+: $A := \R \setminus \left[{a \,.\,.\, b}\right) = \left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$
+Let $A:= \left({-\infty \,.\,.\, a}\right)$ and let $B := \left[{b \,.\,.\, \infty}\right)$.
+Let $B_\epsilon \left({b}\right)$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $b$.
+We have that $b - \epsilon < b$ and so $B_\epsilon \left({b}\right) = \left({b - \epsilon \,.\,.\, b + \epsilon}\right)$ does not lie entirely in $\left[{b \,.\,.\, \infty}\right)$.
+Now $b - \epsilon$ may itself lie in $A$.
+However:
+: $\left[{a \,.\,.\, b}\right) \cap B_\epsilon \left({b}\right) \ne \varnothing$
+and so there are elements of $B_\epsilon \left({b}\right)$ which are not in $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$.
+So $\left({-\infty \,.\,.\, a}\right) \cup \left[{b \,.\,.\, \infty}\right)$ is not an [[Definition:Open Set (Metric Space)|open set]] of $\R$.
+Thus, by definition, $\left[{a \,.\,.\, b}\right)$ is not a [[Definition:Closed Set (Metric Space)|closed set]] of $\R$.
+Hence the result.
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is neither an [[Definition:Open Set (Metric Space)|open set]] nor a [[Definition:Closed Set (Metric Space)|closed set]] of $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Half-Open Real Interval is Closed in some Open Intervals}
+Tags: Real Intervals, Closed Sets
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as an [[Definition:Euclidean Space|Euclidean space]].
+Let $\left[{a \,.\,.\, b}\right) \subset \R$ be a [[Definition:Half-Open Real Interval|half-open interval]] of $\R$.
+Let $c < a$.
+Then $\left[{a \,.\,.\, b}\right)$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({c \,.\,.\, b}\right)$.
+Similarly, let $d > b$.
+Then the [[Definition:Half-Open Real Interval|half-open interval]] $\left({a \,.\,.\, b}\right]$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({a \,.\,.\, d}\right)$.
+\end{theorem}
+
+\begin{proof}
+Consider:
+: $A := \left({c \,.\,.\, b}\right) \setminus \left[{a \,.\,.\, b}\right) = \left({c \,.\,.\, a}\right)$
+Then $A$ is an [[Definition:Open Real Interval|open interval]].
+By [[Open Real Interval is Open Set]], $A$ is [[Definition:Open Set (Metric Space)|open]] in $\R$.
+By the definition of the [[Definition:Topological Subspace|subspace topology]] it follows that $A$ is [[Definition:Open Set (Metric Space)|open]] in $\left({c \,.\,.\, b}\right)$.
+Thus by the definition of [[Definition:Closed Set (Metric Space)|closed set]], $\left({c \,.\,.\, b}\right) \setminus A = \left[{a \,.\,.\, b}\right)$ is [[Definition:Closed Set (Metric Space)|closed]] in $\left({c \,.\,.\, b}\right)$.
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], the argument also shows that $\left({a \,.\,.\, b}\right] \subset \R$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({a \,.\,.\, d}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Underlying Set of Topological Space is Clopen}
+Tags: Clopen Sets
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Then the [[Definition:Underlying Set of Topological Space|underlying set]] $S$ of $T$ is [[Definition:Clopen Set|both open and closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Topology|topology]], $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Empty Set is Element of Topology]], $\varnothing$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Underlying Set of Topological Space is Closed]] $S$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Closed in Topological Space}
+Tags: Empty Set, Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $\O$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Closed Set (Topology)|closed set]], $U$ is [[Definition:Open Set (Topology)|open]] in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Underlying Set of Topological Space is Clopen]], $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Set Difference with Self is Empty Set]], we have $S \setminus S = \O$, so $\O$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis induces Local Basis}
+Tags: Topology
+
+\begin{theorem}
+Let $\left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $\mathcal B$ be an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
+Let $x \in S$, and denote with $\mathcal B_x$ the set $\left\{{B \in \mathcal B: x \in B}\right\}$.
+Then $\mathcal B_x$ is a [[Definition:Local Basis|local basis]] at $x$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in U$, with $U$ an [[Definition:Open Set (Topology)|open set]] of $\tau$.
+Since $\mathcal B$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$, we have:
+:$U = \displaystyle \bigcup \mathcal A$
+for some $\mathcal A \subseteq \mathcal B$.
+Since $x \in \displaystyle \bigcup \mathcal A$, there is a $B \in \mathcal A$ such that $x \in B$, by definition of [[Definition:Union of Set of Sets|set union]].
+Hence, by definition of $\mathcal B_x$, $B \in \mathcal B_x$.
+From [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]], we also have:
+:$B \subseteq U$
+Hence $\mathcal B_x$ is a [[Definition:Local Basis|local basis]] at $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is T0}
+Tags: T0 Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau_p}\right)$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite|infinite set]] $S$.
+Then $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+: [[Fort Space is T1|Fort Space is $T_1$]]
+: [[T1 Space is T0 Space|$T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space]]
+{{qed}}
+[[Category:T0 Spaces]]
+[[Category:Fort Spaces]]
+3yo8x75pff8wnyiw5wr8rl4jbiwiao4
+\end{proof}<|endoftext|>
+\section{Pullback Functor is Functor}
+Tags: Pullbacks
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]] having all [[Definition:Pullback (Category Theory)|pullbacks]].
+Let $f: C \to D$ be a [[Definition:Morphism (Category Theory)|morphism]] of $\mathbf C$.
+Let $\mathbf C \mathbin / C$ and $\mathbf C \mathbin / D$ be the [[Definition:Slice Category|slice categories]] over $C$ and $D$, respectively.
+Let $f^* : \mathbf C \mathbin / D \to \mathbf C \mathbin / C$ be the [[Definition:Pullback Functor|pullback functor]] defined by $f$.
+Then $f^*$ is a [[Definition:Covariant Functor|functor]].
+\end{theorem}
+
+\begin{proof}
+Let $\alpha: A \to D$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C \mathbin / D$.
+Then the [[Definition:Identity Morphism|identity morphism]] $\operatorname{id}_\alpha: \alpha \to \alpha$ is by definition $\operatorname{id}_A: A \to A$.
+Thus $f^* \operatorname{id}_\alpha$ is by definition the unique [[Definition:Morphism (Category Theory)|morphism]] fitting:
+::$\begin{xy}\xymatrix@+1em{
+ A'
+  \ar[rr]^*{f_\alpha}
+  \ar[dd]_*{f^* \alpha}
+  \ar@{-->}[rd]^*{f^* \mathrm{id}_\alpha}
+& &
+ A
+  \ar[rd]^*{\mathrm{id}_\alpha}
+  \ar[dd]^(.4)*{\alpha}
+\\ &
+ A'
+  \ar[ld]^*{f^* \alpha}
+  \ar[rr] |{\hole} ^(.3)*{f_\alpha}
+& &
+ A
+  \ar[ld]^*{\alpha}
+\\
+ C
+  \ar[rr]_*{f}
+& &
+ D
+}\end{xy}$
+It is apparent that $\operatorname{id}_{A'}$ satisfies this condition.
+Since by definition, $\operatorname{id}_{f^* \alpha} = \operatorname{id}_{A'}$, this shows that:
+:$f^* \operatorname{id}_\alpha = \operatorname{id}_{f^* \alpha}$
+Now let $\gamma_1: \alpha_2 \to \alpha_1$ and $\gamma_2: \alpha_3 \to \alpha_2$ be [[Definition:Morphism (Category Theory)|morphisms]] of $\mathbf C \mathbin / D$.
+Then $f^* \left({\gamma_1 \circ \gamma_2}\right)$ is the unique morphism fitting the dashed arrow in:
+::$\begin{xy}\xymatrix@+1em{
+ A_3'
+  \ar[rr]^*{f_{\alpha_3}}
+  \ar[dd]_*{f^* \alpha_3}
+  \ar@{-->}[rd]
+& &
+ A_3
+  \ar[rd]^*{\gamma_1 \circ \gamma_2}
+  \ar[dd]^(.4)*{\alpha_3}
+\\ &
+ A_1'
+  \ar[ld]^*{f^* \alpha_1}
+  \ar[rr] |{\hole} ^(.3)*{f_{\alpha_1}}
+& &
+ A_1
+  \ar[ld]^*{\alpha_1}
+\\
+ C
+  \ar[rr]_*{f}
+& &
+ D
+}\end{xy}$
+Unfolding the composition $\gamma_1 \circ \gamma_2$ yields the expanded diagram:
+::$\begin{xy}\xymatrix@+1em{
+ A_3'
+  \ar[rr]^*{f_{\alpha_3}}
+  \ar[ddd]_*{f^* \alpha_3}
+  \ar[rd]^*{f^* \gamma_2}
+  \ar@{-->}@/_/[rdd]
+& &
+ A_3
+  \ar[rd]^*{\gamma_2}
+  \ar[ddd]^*{\alpha_3}
+\\ &
+ A_2'
+  \ar[d]^*{f^* \gamma_1}
+  \ar[rr] |{\hole} ^(.3)*{f_{\alpha_2}}
+& &
+ A_2
+  \ar[d]^*{\gamma_1}
+\\ &
+ A_1'
+  \ar[ld]^*{f^* \alpha_1}
+  \ar[rr] |{\hole} ^(.3)*{f_{\alpha_1}}
+& &
+ A_1
+  \ar[ld]^*{\alpha_1}
+\\
+ C
+  \ar[rr]_*{f}
+& &
+ D
+}\end{xy}$
+in which $f^* \left({\gamma_1 \circ \gamma_2}\right)$ again fits the dashed arrow.
+But this means, in particular, that we must have:
+:$f^* \left({\gamma_1 \circ \gamma_2}\right) = f^* \gamma_1 \circ f^* \gamma_2$
+Therefore, $f^*$ is a [[Definition:Covariant Functor|functor]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Bounded Above Set of Reals is in Closure}
+Tags: Set Closures, Real Analysis
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]].
+Let $H \subseteq \R$ be a [[Definition:Bounded Above Set|bounded above subset]] of $\R$ such that $H \ne \varnothing$.
+Let $u = \map \sup H$ be the [[Definition:Supremum of Set|supremum]] of $H$.
+Then:
+:$u \in \map \cl H$
+where $\map \cl H$ denotes the [[Definition:Closure (Topology)|closure]] of $H$ in $\R$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $\map {B_\epsilon} u$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $u$ in $\R$.
+From [[Distance from Subset of Real Numbers]]:
+:$\map d {u, H} = 0$
+Thus by definition of [[Definition:Distance from Subset|distance from subset]]:
+:$\exists x \in H: \map d {u, x} < \epsilon$
+Thus $x \in \map {B_\epsilon} u$.
+As $x \in H$ and $x \in \map {B_\epsilon} u$, from the definition of [[Definition:Set Intersection|intersection]]:
+:$x \in H \cap \map {B_\epsilon} u$
+The result follows from [[Condition for Point being in Closure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Space is Open Neighborhood of Subset}
+Tags: Topology, Neighborhoods
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Then $S$ is an [[Definition:Open Neighborhood|open neighborhood]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From [[Underlying Set of Topological Space is Clopen]], $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
+[[Definition:By Hypothesis|By hypothesis]], $H \subseteq S$.
+The result follows from [[Open Superset is Open Neighborhood]].
+{{qed}}
+[[Category:Topology]]
+[[Category:Neighborhoods]]
+23fxtpbd2ihzx2sxohd59o0o3q2xfuf
+\end{proof}<|endoftext|>
+\section{Open Superset is Open Neighborhood}
+Tags: Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+Let $H \subseteq U$.
+Then $U$ is an [[Definition:Open Neighborhood|open neighborhood]] of $H$.
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Also [[Definition:By Hypothesis|by hypothesis]], $H \subseteq U$.
+The result follows by definition of [[Definition:Open Neighborhood|open neighborhood]].
+{{qed}}
+[[Category:Topology]]
+aqh7jnykp72r6rr7jvroil8njg00al0
+\end{proof}<|endoftext|>
+\section{Category has Products and Equalizers iff Pullbacks and Terminal Object}
+Tags: Pullbacks, Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Then the following are equivalent:
+:$(1): \mathbf C$ has all [[Definition:Finite Product (Category Theory)|finite products]] and [[Definition:Equalizer|equalizers]].
+:$(2): \mathbf C$ has all [[Definition:Pullback (Category Theory)|pullbacks]] and a [[Definition:Terminal Object|terminal object]].
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Suppose $\mathbf C$ has all [[Definition:Product (Category Theory)|finite products]] and [[Definition:Equalizer|equalizers]].
+That $\mathbf C$ has [[Definition:Pullback (Category Theory)|pullbacks]] follows from [[Pullback as Equalizer]].
+That $\mathbf C$ has a [[Definition:Terminal Object|terminal object]] follows by [[Empty Product is Terminal Object]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Suppose $\mathbf C$ has all [[Definition:Pullback (Category Theory)|pullbacks]] and a [[Definition:Terminal Object|terminal object]].
+From [[Category has Finite Products iff Terminal Object and Binary Products]], it suffices to check that $\mathbf C$ has [[Definition:Binary Product (Category Theory)|binary products]].
+That this is the case follows from [[Product as Pullback]].
+That $\mathbf C$ has [[Definition:Equalizer|equalizers]] follows from [[Equalizer as Pullback]], as it is established that $\mathbf C$ has [[Definition:Finite Product (Category Theory)|finite products]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Bounded Below Set of Reals is in Closure}
+Tags: Set Closures, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line with Euclidean Metric|real number line with the  usual (Euclidean} metric]].
+Let $H \subseteq \R$ be a [[Definition:Bounded Below Set|bounded below subset]] of $\R$ such that $H \ne \O$.
+Let $l = \map \inf H$ be the [[Definition:Infimum of Set|infimum]] of $H$.
+Then:
+:$l \in \map \cl H$
+where $\map \cl H$ denotes the [[Definition:Closure (Topology)|closure]] of $H$ in $\R$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $\map {B_\epsilon} l$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $l$ in $\R$.
+From [[Distance from Subset of Real Numbers]]:
+:$\map d {l, H} = 0$
+Thus by definition of [[Definition:Distance from Subset|distance from subset]]:
+:$\exists x \in H: \map d {l, x} < \epsilon$
+Thus $x \in \map {B_\epsilon} l$.
+As $x \in H$ and $x \in \map {B_\epsilon} l$, from the definition of [[Definition:Set Intersection|intersection]]:
+:$x \in H \cap \map {B_\epsilon} l$
+The result follows from [[Condition for Point being in Closure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Infinite Union may not equal Union of Closures}
+Tags: Set Union, Set Closures, Closure of Infinite Union may not equal Union of Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $I$ be an [[Definition:Infinite Set|infinite]] [[Definition:Indexing Set|indexing set]].
+Let $\family {H_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family of subsets]] of a set $S$.
+Let $\displaystyle H = \bigcup_{i \mathop \in I} H_i$ be the [[Definition:Union of Family|union]] of $\family {H_i}_{i \mathop \in I}$.
+Then it is not always the case that:
+:$\displaystyle \bigcup_{i \mathop \in I} \map \cl {H_i} = \map \cl {\bigcup_{i \mathop \in I} H_i}$
+where $\map \cl {H_i}$ denotes the [[Definition:Closure (Topology)|closure]] of $H_i$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Consider the [[Definition:Real Number Line with Euclidean Topology|real number line $\struct {\R, \tau_d}$ with the usual (Euclidean) topology]] $\tau_d$.
+Let:
+:$H_n \subseteq \R: H_n = \closedint {\dfrac 1 n} 1$ for $n \ge 2$
+where $\closedint {\dfrac 1 n} 1$ denotes the [[Definition:Closed Real Interval|closed real interval]] from $\dfrac 1 n$ to $1$.
+From [[Closed Real Interval is Closed Set]], $\closedint {\dfrac 1 n} 1$ is a [[Definition:Closed Set (Topology)|closed set]] of $\struct {\R, \tau_d}$.
+Then from [[Set is Closed iff Equals Topological Closure]]:
+:$\map \cl {H_n} = H_n$
+Also:
+:$\displaystyle \bigcup_{n \mathop \ge 2} \map \cl {H_n} = \bigcup_{n \mathop \ge 2} H_n = \hointl 0 1$
+However:
+:$\displaystyle \map \cl {\bigcup_{n \mathop \ge 2} H_n} = \closedint 0 1$
+So:
+:$\displaystyle \bigcup_{n \mathop \ge 2} \map \cl {H_n} \ne \map \cl {\bigcup_{n \mathop \ge 2} H_n}$
+\end{proof}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $\struct {\R, \tau_d}$ denote the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]] $\tau_d$.
+Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]], also under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]] $\tau_d$.
+For a [[Definition:Rational Number|rational number]] $\alpha \in \Q$, let $B_\alpha$ denote the [[Definition:Singleton|singleton]] containing $\alpha$.
+From [[Union of Closures of Singleton Rationals is Rational Space]]:
+:$\displaystyle \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \Q$
+From [[Closure of Union of Singleton Rationals is Real Number Line]]:
+:$\displaystyle \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha} = \R$
+So it is seen that:
+:$\displaystyle \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} \subset \displaystyle \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha}$
+but it is not the case that:
+:$\displaystyle \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \displaystyle \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha}$
+and the result is apparent.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Reciprocals of Positive Integers is Nowhere Dense in Reals}
+Tags: Denseness, Real Number Line with Euclidean Metric, Integer Reciprocal Space
+
+\begin{theorem}
+Let $N$ be the [[Definition:Set|set]] defined as:
+:$N := \set {\dfrac 1 n: n \in \Z_{>0} }$
+where $\Z_{>0}$ is the [[Definition:Set|set]] of [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $\R$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Then $N$ is [[Definition:Nowhere Dense|nowhere dense]] in $\R$.
+\end{theorem}
+
+\begin{proof}
+Consider the set $\R \setminus N$, the [[Definition:Relative Complement|relative complement]] of $N$.
+By definition:
+:$\displaystyle \R \setminus N = \openint 1 \to \cup \paren {\bigcup_{n \mathop \in \N} \openint {\frac 1 {n + 1} } {\frac 1 n} }$
+{{mistake|You forgot $\hointl \gets 0$ for a start}}
+That is, $\R \setminus N$ is a [[Definition:Countable Union|(countable) union]] of [[Definition:Open Real Interval|open real intervals]].
+From [[Open Real Interval is Open Set]] and its [[Open Real Interval is Open Set/Corollary|corollary]], each one of these is an [[Definition:Open Set (Topology)|open set]] of $\R$.
+{{mistake|... except $\hointl \gets 0$ is not -- it's closed.}}
+From [[Definition:Open Set Axioms|open set axiom $\text O 1$]], it follows that $\R \setminus N$ is itself an [[Definition:Open Set (Topology)|open set]] of $\R$.
+So by definition, $N$ is [[Definition:Closed Set (Topology)|closed in $\R$]].
+From [[Closed Set Equals its Closure]]:
+:$N = N^-$
+where $N^-$ is the [[Definition:Closure (Topology)|closure]] of $N$ in $\R$.
+{{mistake|Not quite right -- from [[Closure of Integer Reciprocal Space]] we have that $N^- {{=}} N \cup \set 0$}}
+But $N$ contains no [[Definition:Open Real Interval|open real intervals]] and so has no [[Definition:Subset|subset]] which is [[Definition:Open Set (Topology)|open]] in $\R$.
+Thus by definition $N$ is [[Definition:Nowhere Dense|nowhere dense]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary}
+Tags: Denseness, Set Closures, Closed Sets
+
+\begin{theorem}
+Let $H$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+Then $H$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$ {{iff}} $S \setminus H$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Closed Set equals its Closure]], $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}}:
+: $H = H^-$
+where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$.
+The result then follows directly from [[Nowhere Dense iff Complement of Closure is Everywhere Dense]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Topology Generated by Synthetic Basis}
+Tags: Topological Bases
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
+{{TFAE|def = Topology Generated by Synthetic Basis}}
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 2|Definition 1 iff Definition 2]] ===
+{{:Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 2}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 3|Definition 1 iff Definition 3]] ===
+{{:Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 3}}{{qed}}
+[[Category:Topological Bases]]
+56z39n330e0jrnu19q4hfi6yq956j5y
+\end{proof}<|endoftext|>
+\section{Union from Synthetic Basis is Topology}
+Tags: Topological Bases, Union from Synthetic Basis is Topology
+
+\begin{theorem}
+Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] on a [[Definition:Set|set]] $X$.
+Let $\displaystyle \tau = \set {\bigcup \AA: \AA \subseteq \BB}$.
+Then $\tau$ is a [[Definition:Topology|topology]] on $X$.
+\end{theorem}
+
+\begin{proof}
+We proceed to verify the [[Definition:Open Set Axioms|open set axioms]] for $\tau$ to be a [[Definition:Topology|topology]] on $S$.
+=== $(\text O 1):$ Union of Open Sets ===
+Let $\AA \subseteq \tau$.
+It is to be shown that:
+:$\displaystyle \bigcup \AA \in \tau$
+Define:
+:$\displaystyle \AA' = \bigcup_{U \mathop \in \AA} \set {B \in \BB: B \subseteq U}$
+By [[Union is Smallest Superset/Family of Sets|Union is Smallest Superset: Family of Sets]], it follows that $\AA' \subseteq \BB$.
+Hence, by [[Equivalence of Definitions of Topology Generated by Synthetic Basis]] and [[General Distributivity of Set Union]]:
+:$\displaystyle \bigcup \AA = \bigcup_{U \mathop \in \AA} \bigcup \set {B \in \BB: B \subseteq U} = \bigcup \AA' \in \tau$
+{{qed|lemma}}
+=== $(\text O 2):$ Pairwise Intersection of Open Sets ===
+Let $U, V \in \tau$.
+It is to be shown that:
+:$U \cap V \in \tau$
+Define:
+:$\OO = \set {A \cap B: A, B \in \BB, \, A \subseteq U, \, B \subseteq V}$
+By the definition of a [[Definition:Synthetic Basis|synthetic basis]]:
+:$\forall A, B \in \BB: A \cap B \in \tau$
+Hence, by the definition of a [[Definition:Subset|subset]], it follows that $\OO \subseteq \tau$.
+By [[Definition:Open Set Axioms|open set axiom]] $(1)$, which has been verified above for $\tau$, we have:
+:$\displaystyle \bigcup \OO \in \tau$
+Since [[Set Intersection Preserves Subsets|set intersection preserves subsets]], we have:
+:$\displaystyle \forall W \in \OO: \exists A, B \in \BB: W = A \cap B \subseteq U \cap V$
+By [[Union is Smallest Superset/General Result|Union is Smallest Superset: General Result]], it follows that:
+:$\displaystyle \bigcup \OO \subseteq U \cap V$
+By the definition of $\tau$, it follows from [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]] that:
+:$\displaystyle \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: x \in A \cap B \subseteq \bigcup \OO$
+That is, by the definition of a [[Definition:Subset|subset]]:
+:$\displaystyle U \cap V \subseteq \bigcup \OO$
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\displaystyle U \cap V = \bigcup \OO \in \tau$
+{{qed|lemma}}
+=== $(\text O 3):$ Set Itself ===
+By the definition of a [[Definition:Synthetic Basis|synthetic basis]], $\BB$ is a [[Definition:Cover of Set|cover]] for $S$.
+By [[Equivalent Conditions for Cover by Collection of Subsets]], it follows that:
+:$\displaystyle S = \bigcup \BB \in \tau$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We use [[Equivalence of Definitions of Topology Generated by Synthetic Basis]].
+We proceed to verify the [[Definition:Open Set Axioms|open set axioms]] for $\tau$ to be a [[Definition:Topology|topology]] on $X$.
+=== $(\text O 1):$ Union of Open Sets ===
+Let $\AA \subseteq \tau$.
+It is to be shown that:
+:$\displaystyle \bigcup \AA \in \tau$
+By the definition of $\tau$, it follows from [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]] that:
+:$\displaystyle \forall x \in \bigcup \AA: \exists U \in \AA: \exists B \in \BB: x \in B \subseteq U \subseteq \bigcup \AA$
+By the [[Subset Relation is Transitive|transitivity of $\subseteq$]], the result follows.
+{{qed|lemma}}
+=== $(\text O 2):$ Pairwise Intersection of Open Sets ===
+Let $U, V \in \tau$.
+It is to be shown that:
+:$U \cap V \in \tau$
+By the definition of a [[Definition:Synthetic Basis|synthetic basis]], we have that:
+:$\forall A, B \in \BB: A \cap B \in \tau$
+Therefore, it follows from [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]] and [[Set Intersection Preserves Subsets]] that:
+:$\displaystyle \forall x \in U \cap V: \exists A, B \in \BB: A \subseteq U, \, B \subseteq V: \exists C \in \BB: x \in C \subseteq A \cap B \subseteq U \cap V$
+By the [[Subset Relation is Transitive|transitivity of $\subseteq$]], the result follows.
+{{qed|lemma}}
+=== $(\text O 3):$ Set Itself ===
+By the definition of a [[Definition:Synthetic Basis|synthetic basis]], we have that:
+:$\forall x \in X: \exists B \in \BB: x \in B \subseteq X$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product as Limit}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $C_1, C_2$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$.
+Let their [[Definition:Binary Product (Category Theory)|binary product]] $C_1 \times C_2$ exist in $\mathbf C$.
+Then $C_1 \times C_2$ is the [[Definition:Limit (Category Theory)|limit]] of the [[Definition:Diagram (Category Theory)|diagram]] $D: 2 \to \mathbf C$ defined by:
+:$D_0 := C_1, D_1 := C_2$
+where $2$ is the [[Definition:Discrete Category|discrete category]] with two [[Definition:Object (Category Theory)|objects]] $0, 1$.
+\end{theorem}
+
+\begin{proof}
+Since there are no non-[[Definition:Identity Morphism|identity morphisms]], a [[Definition:Cone (Category Theory)|cone]] to $D$ is simply a pair:
+::$\begin{xy}\xymatrix@+1em@L+3px{
+ C_1
+&
+ C
+  \ar[l]_*+{f_1}
+  \ar[r]^*+{f_2}
+&
+ C_2
+}\end{xy}$
+of [[Definition:Morphism (Category Theory)|morphisms]] with common [[Definition:Domain (Category Theory)|domain]] $C$.
+By the [[Definition:UMP|UMP]] of the [[Definition:Binary Product (Category Theory)|binary product]] $C_1 \times C_2$, for such a [[Definition:Cone (Category Theory)|cone]] to $D$ there is a [[Definition:Unique|unique]] $u: C \to C_1 \times C_2$ making:
+::$\begin{xy}\xymatrix@L+3px@+1em{
+&
+ C
+  \ar[dl]_*+{f_1}
+  \ar@{-->}[d]^*{u}
+  \ar[dr]^*+{f_2}
+\\
+ C_1
+&
+ C_1 \times C_2
+  \ar[l]_*+{p_1}
+  \ar[r]^*+{p_2}
+&
+ C_2
+}\end{xy}$
+a [[Definition:Commutative Diagram|commutative diagram]].
+The conditions $p_1 \circ u = f_1$ and $p_2 \circ u = f_2$ precisely mean that $u: C \to C_1 \times C_2$ is a [[Definition:Morphism of Cones|morphism of cones]].
+Thus for every [[Definition:Cone (Category Theory)|cone]] $\left({C, f_1, f_2}\right)$ to $D$, there is a [[Definition:Unique|unique]] [[Definition:Morphism of Cones|morphism of cones]] to the [[Definition:Cone (Category Theory)|cone]] $\left({C_1 \times C_2, p_1, p_2}\right)$.
+That is, $\left({C_1 \times C_2, p_1, p_2}\right)$ is a [[Definition:Terminal Object|terminal object]] in $\mathbf{Cone} \left({D}\right)$, the [[Definition:Category of Cones|category of cones]] to $D$.
+Hence the result, by definition of [[Definition:Limit (Category Theory)|limit]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equalizer as Limit}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $f_1, f_2: C_1 \to C_2$ be [[Definition:Morphism (Category Theory)|morphisms]] of $\mathbf C$.
+Let their [[Definition:Equalizer|equalizer]] $e: E \to C_1$ exist in $\mathbf C$.
+Then $\left({E, e}\right)$ is the [[Definition:Limit (Category Theory)|limit]] of the [[Definition:Diagram (Category Theory)|diagram]] $D: \mathbf J \to \mathbf C$ defined by:
+::$\begin{xy}\xymatrix@+1em@L+3px{
+ \mathbf{J}:
+&
+ \ast
+  \ar[r]<2pt>
+  \ar[r]<-2pt>
+&
+ \star
+}\end{xy}$
+::$\begin{xy}\xymatrix@+1em@L+3px{
+ D:
+&
+ C_1
+  \ar[r]<2pt>^*+{f_1}
+  \ar[r]<-2pt>_*+{f_2}
+&
+ C_2
+}\end{xy}$
+\end{theorem}<|endoftext|>
+\section{Terminal Object as Limit}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $\mathbf C$ have a [[Definition:Terminal Object|terminal object]] $1$.
+Then $1$ is the [[Definition:Limit (Category Theory)|limit]] of the [[Definition:Unique|unique]] [[Definition:Diagram (Category Theory)|diagram]] $D: \mathbf 0 \to \mathbf C$, where $\mathbf 0$ is the [[Definition:Zero (Category)|zero category]].
+\end{theorem}<|endoftext|>
+\section{Pullback as Limit}
+Tags: Category Theory, Pullbacks
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $f_1: A \to C$ and $f_2: B \to C$ be [[Definition:Morphism (Category Theory)|morphisms]] of $\mathbf C$.
+Let their [[Definition:Pullback (Category Theory)|pullback]]:
+::$\begin{xy}\xymatrix@+1em@L+3px{
+ P
+  \ar[r]^*+{p_2}
+  \ar[d]_*+{p_1}
+&
+ A
+  \ar[d]^*+{f_1}
+\\
+ B
+  \ar[r]_*+{f_2}
+&
+ C
+}\end{xy}$
+exist in $\mathbf C$.
+Then $\left({P, p_1, p_2}\right)$ is the [[Definition:Limit (Category Theory)|limit]] of the [[Definition:Diagram (Category Theory)|diagram]] $D: \mathbf J \to \mathbf C$ defined by:
+::$\begin{xy}\xymatrix@+1em@L+3px{
+ \save[]+<0em,-2em>*{\mathbf{J}:} \restore
+& &
+ \cdot
+  \ar[d]
+\\ &
+ \cdot
+  \ar[r]
+&
+ \cdot
+}\end{xy}$
+::$\begin{xy}\xymatrix@+1em@L+3px{
+ \save[]+<0em,-2em>*{D:} \restore
+& &
+ A
+  \ar[d]^*+{f_1}
+\\ &
+ B
+  \ar[r]_*+{f_2}
+&
+ C
+}\end{xy}$
+\end{theorem}<|endoftext|>
+\section{Identification Topology is Topology}
+Tags: Identification Topology
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ be a [[Definition:Topological Space|topological space]].
+Let $S_2$ be a [[Definition:Set|set]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $\tau_2$ be the [[Definition:Identification Topology|identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$]].
+Then $\tau_2$ is a [[Definition:Topology|topology]] on $S_2$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$
+We examine each of the [[Definition:Open Set Axioms|open set axioms]] in turn:
+=== $(\text O 1)$: Union of Open Sets ===
+Let $\family {U_i}_{i \matho \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Element|elements]] of $\tau_2$.
+Let $\ds V = \bigcup_{i \mathop \in I} U_i$ be the [[Definition:Union of Family|union]] of $\family {U_i}_{i \mathop \in I}$.
+From [[Preimage of Union under Mapping/Family of Sets|Preimage of Union under Mapping: Family of Sets]]:
+:$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in I} U_i} = \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i}$
+[[Definition:By Hypothesis|By hypothesis]] $f^{-1} \sqbrk {U_i} \in \tau_1$.
+As $\tau_1$ is a [[Definition:Topology|topology]]:
+:$\ds \bigcup_{i \mathop \in I} f^{-1} \sqbrk {U_i} \in \tau_1$
+So $f^{-1} \sqbrk V \in \tau_1$.
+Thus $V \in \tau_2$ and so $V$ is [[Definition:Open Set (Topology)|open]] by definition.
+{{qed|lemma}}
+=== $(\text O 2)$: Intersection of Open Sets ===
+Let $U$ and $V$ be [[Definition:Element|elements]] of $\tau_2$.
+From [[Preimage of Intersection under Mapping]]:
+:$f^{-1} \sqbrk {U \cap V} = f^{-1} \sqbrk U \cap f^{-1} \sqbrk V$
+As $\tau_1$ is a [[Definition:Topology|topology]]:
+:$f^{-1} \sqbrk U \cap f^{-1} \sqbrk V \in \tau_1$
+Hence $U \cap V$ is [[Definition:Open Set (Topology)|open]] by definition.
+{{qed|lemma}}
+=== $(\text O 3)$: Set Itself ===
+As $f$ is a [[Definition:Mapping|mapping]] its [[Definition:Domain of Mapping|domain]] is $S_1$.
+That is:
+:$f^{-1} \sqbrk {S_2} = S_1$
+By definition of a [[Definition:Topology|topology]], $S_1$ is [[Definition:Open Set (Topology)|open]] in $T_1$.
+Thus by definition $S_2 \in \tau_2$.
+{{qed|lemma}}
+All the [[Definition:Open Set Axioms|open set axioms]] are fulfilled, and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identification Mapping is Continuous}
+Tags: Identification Topology
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ be a [[Definition:Topological Space|topological space]].
+Let $S_2$ be a [[Definition:Set|set]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $\tau_2$ be the [[Definition:Identification Topology|identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$]].
+Then the [[Definition:Identification Mapping|identification mapping]] $f$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Identification Topology|identification topology]]:
+:$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$
+That is:
+:$V \in \tau_2 \implies f^{-1} \sqbrk V \in \tau_1$
+This is precisely the definition of a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
+{{qed}}
+[[Category:Identification Topology]]
+p1p49egjbhzpe3wej3gre4srnw99xg3
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Identification Topology}
+Tags: Identification Topology
+
+\begin{theorem}
+Let $T_1 = \left({S_1, \tau_1}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $S_2$ be a [[Definition:Set|set]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $\tau_2$ be the [[Definition:Identification Topology|identification topology on $S_2$ with respect to $f$ and $\left({S_1, \tau_1}\right)$]].
+Then the [[Definition:Identification Topology|identification topology]] on $S_2$ with respect to $f$ and $\left({S_1, \tau_1}\right)$ always exists and is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let $\tau_2$ be the [[Definition:Identification Topology|identification topology]] on $S_2$ with respect to $f$ and $\left({S_1, \tau_1}\right)$.
+By definition:
+:$\tau_2 := \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\} \subseteq \mathcal P \left({S_2}\right)$
+where $\mathcal P \left({S_2}\right)$ is the [[Definition:Power Set|power set]] of $S_2$.
+Let $V \subseteq S_2$.
+Then either:
+: $f^{-1} \left({V}\right) \in \tau_1 \implies V \in \tau_2$
+or:
+: $f^{-1} \left({V}\right) \notin \tau_1 \implies V \notin \tau_2$
+In particular:
+: $f^{-1} \left({S_2}\right)  = S_1 \in \tau_1 \implies S_2 \in \tau_2$
+Thus $\tau_2$ exists.
+By the same coin, given any $V \subseteq S_2$, either $V \in \tau_2$ or $V \notin \tau_2$, and so the set:
+:$\tau_2 := \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\}$
+consists of a uniquely defined set of elements.
+{{qed}}
+[[Category:Identification Topology]]
+r7e2utcv5y2icdnbuoqigd7zekj6z1o
+\end{proof}<|endoftext|>
+\section{Category has Finite Limits iff Finite Products and Equalizers}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Then:
+:$\mathbf C$ has all [[Definition:Finite Limit (Category Theory)|finite limits]]
+{{iff}}:
+:$\mathbf C$ has all [[Definition:Product (Category Theory)|finite products]] and [[Definition:Equalizer|equalizers]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+By definition, [[Definition:Finite Product (Category Theory)|finite products]] are instances of [[Definition:Finite Limit (Category Theory)|finite limits]].
+So are [[Definition:Equalizer|equalizers]], by [[Equalizer as Limit]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $C: \mathbf J \to \mathbf C$ be a [[Definition:Diagram (Category Theory)|diagram]], with $\mathbf J$ [[Definition:Finite Category|finite]].
+The objective is to construct the [[Definition:Limit (Category Theory)|limit]] $D = \varprojlim_j C_j$, that is:
+:an [[Definition:Object|object]] $D$
+:[[Definition:Morphism (Category Theory)|morphisms]] $p_j: D \to C_j$
+such that:
+:$C_\alpha p_i = p_j$ for each [[Definition:Morphism (Category Theory)|$\mathbf J$-morphism]] $\alpha: i \to j$
+:for each $E$ and $q_j: E \to C_j$ such that $C_\alpha q_i = q_j$ for each $\alpha$, there is a [[Definition:Unique|unique]] $f: E \to D$ such that $q_j = p_j f$ for each $j$
+To this end, consider the [[Definition:Finite Product (Category Theory)|finite products]]:
+:$\displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ and $\displaystyle \prod_{\alpha \mathop \in \operatorname{mor} \mathbf J} C_{\operatorname{cod} \alpha}$
+of the [[Definition:Object (Category Theory)|objects]] of the [[Definition:Diagram (Category Theory)|diagram]] and the [[Definition:Codomain (Category Theory)|codomains]] of its [[Definition:Morphism (Category Theory)|morphisms]].
+Next, define the [[Definition:Morphism (Category Theory)|morphisms]] $\pi, \varepsilon: \displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j \to \prod_{\alpha \mathop \in \operatorname{mor} \mathbf J} C_{\operatorname{cod} \alpha}$ by:
+:$\pi_\alpha = \pr_{C_{\operatorname{cod} \alpha}}$
+:$\varepsilon_\alpha = C_\alpha \pr_{C_{\operatorname{dom} \alpha}}$
+Now let $e: D \to \displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ be the [[Definition:Equalizer|equalizer]] of $\pi$ and $\varepsilon$.
+Also, define $p_j: D \to C_j$ by $p_j = \pr_{C_j} e$ for each $j$.
+Now for every $\alpha: i \to j$:
+{{begin-eqn}}
+{{eqn | l = C_\alpha p_i
+      | r = C_\alpha \pr_{C_j} e
+      | c = Definition of $p_i$
+}}
+{{eqn | r = \varepsilon_\alpha e
+      | c = Definition of $\varepsilon_\alpha$
+}}
+{{eqn | r = \pi_\alpha e
+      | c = $e$ [[Definition:Equalizer|equalizes]] $\pi$ and $\varepsilon$
+}}
+{{eqn | r = \pr_{C_j} e
+      | c = Definition of $\pi_\alpha$
+}}
+{{eqn | r = p_j
+      | c = Definition of $p_j$
+}}
+{{end-eqn}}
+Lastly, suppose that $E$ and $q_j$ have the properties stated.
+By definition of the [[Definition:Product (Category Theory)/General Definition|product]] $\displaystyle \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$, there is a [[Definition:Morphism (Category Theory)|morphism]]:
+:$\displaystyle q: E \to \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$
+Now, for every $\alpha: i \to j$:
+{{begin-eqn}}
+{{eqn | l = \pi_\alpha q
+      | r = \pr_j q
+      | c = Definition of $\pi_\alpha$
+}}
+{{eqn | r = q_j
+      | c = Definition of $q$
+}}
+{{eqn | r = C_\alpha q_i
+      | c = Assumption on the $q_i$
+}}
+{{eqn | r = C_\alpha \pr_i q
+      | c = Definition of $q$
+}}
+{{eqn | r = \varepsilon_\alpha q
+      | c = Definition of $\varepsilon_\alpha$
+}}
+{{end-eqn}}
+Hence, since $e$ [[Definition:Equalizer|equalizes]] $\pi$ and $\varepsilon$, it follows that there exists a [[Definition:Unique|unique]] $f: E \to D$ such that:
+:$q = ef$
+Thus, for each $j$:
+:$\pr_j q = \pr_j e f$
+which, by the definitions of $q_j$ and $p_j$, amounts to:
+:$q_j = p_j f$
+Hence $D$ and the $q_j$ form the desired [[Definition:Limit (Category Theory)|limit]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identification Topology is Finest Topology for Mapping to be Continuous}
+Tags: Identification Topology
+
+\begin{theorem}
+Let $T_1 := \struct {S_1, \tau_1}$ be a [[Definition:Topological Space|topological space]].
+Let $S_2$ be a [[Definition:Set|set]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $\tau_2$ be the [[Definition:Identification Topology|identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$]].
+Let $T_2 := \struct {S_2, \tau_2}$ be the resulting [[Definition:Topological Space|topological space]].
+Then $\tau_2$ is the [[Definition:Finer Topology|finest topology]] on $S_2$ such that $f: T_1 \to T_2$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+It is established in [[Identification Mapping is Continuous]] that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Let $\tau \subseteq \powerset {S_2}$ be a [[Definition:Topology|topology]] on $S_2$ for which $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|$\struct {\tau_1, \tau}$-continuous]].
+Let $U \in \tau$.
+Then by definition of [[Definition:Everywhere Continuous Mapping (Topology)|$\struct {\tau_1, \tau}$-continuous]]:
+:$f^{-1} \sqbrk U \in \tau_1$
+By definition of $\tau_2$ it then follows that $U \in \tau_2$.
+By definition of [[Definition:Subset|subset]] it follows that $\tau \subseteq \tau_2$.
+Thus it has been shown that $\tau_2$ is [[Definition:Finer Topology|finer]] than any [[Definition:Topology|topology]] on $S_2$ for which $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|$\struct {\tau_1, \tau}$-continuous]].
+Hence the result, by definition of [[Definition:Finer Topology|finest topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identification Topology equals Quotient Topology on Induced Equivalence}
+Tags: Identification Topology
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ be a [[Definition:Topological Space|topological space]].
+Let $S_2$ be a [[Definition:Set|set]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $\tau_2$ be the [[Definition:Identification Topology|identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$]]:
+:$\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$
+Let $\RR_f \subseteq S_1 \times S_1$ be the [[Definition:Equivalence Relation Induced by Mapping|equivalence on $S_1$ induced by $f$]]:
+:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
+Let $\tau_{\RR_f}$ be the [[Definition:Quotient Topology|quotient topology on $S / \RR_f$ by $q_{\RR_f}$]]:
+:$\tau_{\RR_f} := \set {U \subseteq S / \RR_f: q_{\RR_f}^{-1} \sqbrk U \in \tau_1}$
+Then:
+:$\tau_2$ is [[Definition:Homeomorphism|homeomorphic]] to $\tau_{\RR_f}$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Identification Topology]]
+qmjbugl9b3dquhx5a120fjnyyf82gbu
+\end{proof}<|endoftext|>
+\section{Continuity of Composite with Identification Mapping}
+Tags: Identification Topology
+
+\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 $f_1: T_1 \to T_2$ be a [[Definition:Mapping|mapping]].
+Let $S_3$ be a [[Definition:Set|set]].
+Let $p: S_1 \to S_3$ be a [[Definition:Mapping|mapping]].
+Let $\tau_3$ be the [[Definition:Identification Topology|identification topology on $S_3$ with respect to $p$ and $T_1$]].
+Let $T_3 = \struct {S_3, \tau_3}$ be the resulting [[Definition:Topological Space|topological space]].
+Let $f_2: S_3 \to S_2$ be a [[Definition:Mapping|mapping]] such that:
+:$f_1 = f_2 \circ p$
+where $f_2 \circ p$ is the [[Definition:Composition of Mappings|composition]] of $f_2$ with $p$.
+Then $f_1$ is a [[Definition:Continuous Mapping|continuous mapping]] {{iff}} $f_2$ is a [[Definition:Continuous Mapping|continuous mapping]].
+\end{theorem}<|endoftext|>
+\section{Young's Inequality for Increasing Functions}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $a_0$ and $b_0$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a [[Definition:Strictly Increasing Real Function|strictly increasing]] [[Definition:Bijection|bijection]].
+Let $a$ and $b$ be [[Definition:Real Number|real numbers]] such that $0 \le a \le a_0$ and $0 \le b \le b_0$.
+Then:
+:$\displaystyle ab \le \int_0^a \map f u \rd u + \int_0^b \map {f^{-1}} v \rd v$
+where $\displaystyle \int$ denotes the [[Definition:Definite Integral|definite integral]].
+\end{theorem}
+
+\begin{proof}
+[[File:Young's_Ineq_for_Increasing_Functions.png|200px|thumb|right|The blue colored region corresponds to $\displaystyle \int_0^a \map f u \rd u$ and the red colored region to $\displaystyle \int_0^b\map {f^{-1}} v \rd v$.]]
+{{ProofWanted}}
+{{Namedfor|William Henry Young|cat = Young}}
+\end{proof}<|endoftext|>
+\section{Set is Closed iff Equals Topological Closure}
+Tags: Closed Sets, Set Closures, Set is Closed iff Equals Topological Closure
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Then $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}}:
+:$H = \map \cl H$
+\end{theorem}
+
+\begin{proof}
+Let $H^{\complement}$ denote the [[Definition:Relative Complement|relative complement]] of $H$ in $T$.
+By definition, we have that $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}} $H^{\complement}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+By [[Set is Open iff Neighborhood of all its Points]], this is [[Definition:Logical Equivalence|equivalent]] to:
+:$\forall x \in H^{\complement}: \exists U \in \tau: x \in U \subseteq H^{\complement}$
+By [[Empty Intersection iff Subset of Complement]], we have that:
+:$U \subseteq H^{\complement} \iff U \cap H = \O$
+By [[Condition for Point being in Closure]], it follows that $H^{\complement}$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}}:
+:$\forall x \in H^{\complement}: x \notin \map \cl H$
+By the [[Rule of Transposition]], this is [[Definition:Logical Equivalence|equivalent]] to $\map \cl H \subseteq H$.
+From [[Set is Subset of its Topological Closure]], we have that $H \subseteq \map \cl H$.
+By definition of [[Definition:Set Equality/Definition 2|set equality]], this is [[Definition:Logical Equivalence|equivalent]] to:
+:$H = \map \cl H$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $H'$ denote the [[Definition:Derived Set|derived set]] of $H$.
+By [[Closed Set iff Contains all its Limit Points]], $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}} $H' \subseteq H$.
+By [[Union with Superset is Superset]], $H' \subseteq H$ {{iff}} $H = H \cup H'$.
+The result follows from the definition of [[Definition:Closure (Topology)/Definition 1|closure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Compact Topological Subspace}
+Tags: Compact Spaces
+
+\begin{theorem}
+{{TFAE|def = Compact Subspace}}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $T_H = \left({H, \tau_H}\right)$ be a [[Definition:Topological Subspace|topological subspace]] of $T$, where $H \subseteq S$.
+\end{theorem}
+
+\begin{proof}
+=== 1 implies 2 ===
+Suppose $T_H$ is compact in the sense of [[Definition:Compact Space/Topology/Subspace/Definition 1|Definition 1]].
+Let $\mathcal C$ be a [[Definition:Cover of Set|cover]] of $H$ by [[Definition:Open Set (Topology)|open sets]] of $T$.
+Then for each $U \in \mathcal C$, $U \cap H$ is [[Definition:Open Set (Topology)|open]] in $T_H$ by definition of the [[Definition:Subspace Topology|subspace topology]].
+Since $\mathcal C$ is a [[Definition:Cover of Set|cover]] of $H$ it follows that $\mathcal C' = \left\{{U \cap H: U \in \mathcal C}\right\}$ is also a [[Definition:Cover of Set|cover]] of $H$.
+[[Definition:By Hypothesis|By hypothesis]], there exists a [[Definition:Finite Subcover|finite subcover]] $\mathcal F'$ of $\mathcal C'$ for $H$.
+Thus let $\mathcal F' = \left\{{U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}\right\}$ where $U_1, U_2, \ldots, U_n \in \mathcal C$.
+It follows that $\mathcal F = \left\{{U_1, U_2, \ldots, U_n}\right\}$ is a [[Definition:Finite Subcover|finite subcover]] of $H$.
+That is, $T_H$ is compact in the sense of [[Definition:Compact Space/Topology/Subspace/Definition 2|Definition 2]].
+{{qed|lemma}}
+=== 2 implies 1 ===
+Suppose $T_H$ is compact in the sense of [[Definition:Compact Space/Topology/Subspace/Definition 2|Definition 2]].
+Let $\mathcal C$ be a [[Definition:Cover of Set|cover]] of $H$ by [[Definition:Open Set (Topology)|open sets]] of $H$.
+By definition of the [[Definition:Subspace Topology|subspace topology]], each $V \in \mathcal C$ is of the form $U \cap H$ for some $U \in \tau$.
+It follows that the collection of all such $U$ is a [[Definition:Cover of Set|cover]] of $H$ by [[Definition:Open Set (Topology)|open sets]] of $T$.
+[[Definition:By Hypothesis|By hypothesis]], there exists a [[Definition:Finite Subcover|finite subcover]] $\mathcal F = \left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal C$ for $H$.
+Thus $\mathcal F' = \left\{{U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}\right\}$ is also a [[Definition:Finite Subcover|finite subcover]] of $H$
+That is, $T_H$ is compact in the sense of [[Definition:Compact Space/Topology/Subspace/Definition 1|Definition 1]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Covariant Hom Functor is Continuous}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
+Let $\mathbf C$ be a [[Definition:Locally Small Category|locally small category]].
+Let $C$ be an [[Definition:Object|object]] of $\mathbf C$, and let $\hom \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the [[Definition:Covariant Hom Functor|covariant hom functor]] based at $C$.
+Then $\hom \paren {C, \cdot}$ is a [[Definition:Continuous Functor|continuous functor]].
+\end{theorem}<|endoftext|>
+\section{Contravariant Hom Functor maps Colimits to Limits}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
+Let $\mathbf C$ be a [[Definition:Locally Small Category|locally small category]].
+Let $C$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
+Let $\map \hom {\cdot, C}: \mathbf C \to \mathbf{Set}$ be the [[Definition:Contravariant Hom Functor|contravariant hom functor]] based at $C$.
+Then $\map \hom {\cdot, C}$ maps every [[Definition:Colimit|colimit]] to a [[Definition:Limit|limit]], in that:
+:$\map \hom { {\varinjlim \,}_j \,D_j, C} \cong {\varprojlim \,}_j \, \map \hom {D_j, C}$
+for every [[Definition:Diagram (Category Theory)|diagram]] $D: \mathbf J \to \mathbf C$.
+\end{theorem}<|endoftext|>
+\section{Composite Mapping is Mapping}
+Tags: Composite Mappings
+
+\begin{theorem}
+Let $S_1, S_2, S_3$ be [[Definition:Set|sets]].
+Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Mapping|mappings]].
+Then the [[Definition:Composite Mapping|composite mapping]] $g \circ f$ is also a [[Definition:Mapping|mapping]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Composite Mapping|composite of $f$ and $g$]] is defined as:
+:$g \circ f := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f \land \tuple {y, z} \in g}$
+It is necessary to show that $g \circ f$ is both [[Definition:Left-Total Relation|left-total]] and [[Definition:Many-to-One Relation|many-to-one]].
+=== Left-total ===
+From the definition of a [[Definition:Mapping|mapping]]:
+{{begin-eqn}}
+{{eqn | lo= \forall x \in S_1: \exists y \in S_2:
+      | l = \map f x
+      | r = y
+      | c = {{Defof|Mapping}}
+}}
+{{eqn | lo= \forall y \in S_2: \exists z \in S_2:
+      | l = \map g y
+      | r = z
+      | c = {{Defof|Mapping}}
+}}
+{{eqn | ll=\leadsto
+      | lo= \forall x \in S_1: \exists z \in S_3:
+      | l = \map g {\map f x}
+      | r = z
+      | c = 
+}}
+{{eqn | ll=\leadsto
+      | lo= \forall x \in S_1: \exists z \in S_3:
+      | l = \map {g \circ f} x
+      | r = z
+      | c = {{Defof|Composite Mapping}}
+}}
+{{end-eqn}}
+That is, $g \circ f$ is [[Definition:Left-Total Relation|left-total]].
+{{qed|lemma}}
+=== Many-to-one ===
+Suppose $x_1, x_2 \in S_1: x_1 = x_2$.
+Then as $f$ is a [[Definition:Mapping|mapping]] and so by definition [[Definition:Many-to-One Relation|many-to-one]]:
+:$\map f {x_1} = \map f {x_2}$
+and so:
+:$\map g {\map f {x_1} } = \map g {\map f {x_2} }$
+demonstrating that $g \circ f$ is similarly [[Definition:Many-to-One Relation|many-to-one]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Space is Regular Space}
+Tags: Regular Spaces, Normal Spaces
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+Then $\struct {S, \tau}$ is also a [[Definition:Regular Space|regular space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
+From [[Normal Space is T3 Space|Normal Space is $T_3$ Space]], we have that $T$ is a [[Definition:T3 Space|$T_3$ space]].
+We also have by definition of [[Definition:Normal Space|normal space]] that $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+From [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]] we have that $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]]
+So $T$ is both a [[Definition:T3 Space|$T_3$ space]] and a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Hence $T$ is a [[Definition:Regular Space|regular space]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Hausdorff Factor Spaces is Hausdorff/General Result}
+Tags: Hausdorff Spaces, Product Spaces
+
+\begin{theorem}
+Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for $\alpha$ in some [[Definition:Indexing Set|indexing set]] $I$.
+Let $\displaystyle T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\SS$.
+Let each of $\struct {S_\alpha, \tau_\alpha}$ for $\alpha \in I$ be [[Definition:Hausdorff Space|$T_2$ (Hausdorff) spaces]].
+Then $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S : x \ne y$.
+Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$.
+Since $\struct {S_\alpha, \tau_\alpha}$ is [[Definition:Hausdorff Space|Hausdorff]] then:
+:$\exists U, V \in \tau_\alpha: x_\alpha \in U, y_\alpha \in V : U \cap V = \O$
+Let $\pr_\alpha: S \to S_\alpha$ be the [[Definition:Projection (Cartesian Product)|projection]] of $S$ to $S_\alpha$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\pr_\alpha^\gets} U \cap \map {\pr_\alpha^\gets} V
+      | r = \map {\pr_\alpha^\gets} {U \cap V}
+      | c = [[Preimage of Intersection under Mapping]]
+}}
+{{eqn | r = \map {\pr_\alpha^\gets} \O
+}}
+{{eqn | r = \O
+}}
+{{end-eqn}} 
+By definition of the [[Definition:Projection (Cartesian Product)|projection]] $\pr_\alpha$:
+:$\map {\pr_\alpha} x = x_\alpha \in U$
+By definition of the [[Definition:Preimage under Mapping|preimage]] under $\pr_\alpha$:
+:$x \in \map {\pr_\alpha^\gets} U$
+Similarly:
+:$y \in \map {\pr_\alpha^\gets} V$
+By definition of the [[Definition:Tychonoff Topology|Tychonoff topology]] $\tau$:
+:$\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$
+Since $x, y \in S$ were arbitrary, it follows that $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+{{qed}}
+[[Category:Hausdorff Spaces]]
+[[Category:Product Spaces]]
+ia1v6mckgd7rylx9ukwbewey9ye5e33
+\end{proof}<|endoftext|>
+\section{Product of Hausdorff Factor Spaces is Hausdorff}
+Tags: Hausdorff Spaces, Product Spaces
+
+\begin{theorem}
+Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be [[Definition:Topological Space|topological spaces]].
+Let $T = T_\alpha \times T_\beta$ be the [[Definition:Product Space (Topology)|product space]] of $T_\alpha$ and $T_\beta$
+Let $T_\alpha$ and $T_\beta$ both be [[Definition:Hausdorff Space|$T_2$ (Hausdorff) spaces]].
+Then $T$ is also a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T_\alpha$ and $T_\beta$ be [[Definition:Hausdorff Space|Hausdorff spaces]].
+Let $\tuple {a, b}$ and $\tuple {c, d}$ be two [[Definition:Distinct|distinct]] points of the [[Definition:Product Space (Topology)|product space]] $T$.
+Let $a = c$.
+Then as $\tuple {a, b} \ne \tuple {c, d}$ it follows that $b \ne d$.
+As $T_\beta$ is [[Definition:Hausdorff Space|Hausdorff]], there exists two [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \subseteq T_\beta$ such that $b \in U, d \in V$.
+Then $T_\alpha \times U$ and $T_\alpha \times V$ are two [[Definition:Open Set (Topology)|open]] [[Definition:Disjoint Sets|disjoint sets]] in the [[Definition:Product Space (Topology)|product space]] $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$.
+Let $a \ne c$.
+As $T_\alpha$ is [[Definition:Hausdorff Space|Hausdorff]], there exists two [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \subseteq T_\alpha$ such that $a \in U, c \in V$.
+Then $U \times T_\beta$ and $V \times T_\beta$ are two [[Definition:Open Set (Topology)|open]] [[Definition:Disjoint Sets|disjoint sets]] in the [[Definition:Product Space (Topology)|product space]] $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Continuous Injection to Hausdorff Space is Hausdorff}
+Tags: Hausdorff Spaces, Injections, Continuous Mappings
+
+\begin{theorem}
+Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be [[Definition:Topological Space|topological spaces]].
+Let $f: S_\alpha \to S_\beta$ be a [[Definition:Continuous Mapping (Topology)|continuous mapping]] which is an [[Definition:Injection|injection]].
+If $T_\beta$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]], then $T_\alpha$ is also a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S_\alpha$ be [[Definition:Distinct Elements|distinct]] [[Definition:Point|points]].
+We want to find [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau_\alpha$ containing $x$ and $y$ respectively.
+Since $f$ is [[Definition:Injection|injective]] the [[Definition:Point|points]] $\map f x, \map f y \in S_\beta$ are [[Definition:Distinct Elements|distinct]].
+By assumption $T_\beta$ is [[Definition:Hausdorff Space|Hausdorff]].
+Therefore we can choose [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U', V'$ of $T_\beta$ such that $\map f x \in U'$ and $\map f y \in V'$.
+Since $f$ is [[Definition:Continuous Mapping (Topology)|continuous]], the [[Definition:Set|sets]] $U = f^{-1} \sqbrk {U'}$ and $V = f^{-1} \sqbrk {V'}$ are [[Definition:Open Set (Topology)|open]].
+Moreover we have $x \in U$ and $y \in V$.
+Finally we show that the [[Definition:Set|sets]] $U$ and $V$ are [[Definition:Disjoint Sets|disjoint]].
+Suppose $z \in U \cap V$.
+Then by the definition of $U$, $V$ we have $\map f z \in U'$ and $\map f z \in V'$.
+This is a [[Proof by Contradiction|contradiction]], since $U' \cap V' = \O$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Homeomorphic Topological Spaces}
+Tags: Homeomorphisms
+
+\begin{theorem}
+Let $T_\alpha = \left({S_\alpha, \tau_\alpha}\right)$ and $T_\beta = \left({S_\beta, \tau_\beta}\right)$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_\alpha \to T_\beta$ be a [[Definition:Bijection|bijection]].
+{{TFAE|def = Homeomorphism (Topological Spaces)|view = homeomorphism}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 iff Definition 2 ===
+Let $f$ and $f^{-1}$ both be [[Definition:Continuous Mapping (Topology)|continuous]].
+As $f$ is [[Definition:Continuous Mapping (Topology)|continuous]], then by definition:
+: $V \in \tau_\beta \implies f^{-1} \left({V}\right) \in \tau_\alpha$
+and as $f^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]], then by definition:
+: $U \in \tau_\alpha \implies \left({f^{-1}}\right)^{-1} \left({U}\right) = f \left({U}\right) \in \tau_\beta$
+That is:
+: $\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \left({U}\right) \in \tau_\beta$
+[[Definition:Converse Statement|Conversely]], suppose that:
+: $\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \left({U}\right) \in \tau_\beta$
+By definition:
+: $U \in \tau_\alpha \implies f \left({U}\right) \in \tau_\beta$ [[Definition:Iff|iff]] $f^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Also by definition:
+: $f \left({U}\right) \in \tau_\beta \implies U = f^{-1} \left({f \left({U}\right)}\right) \in \tau_\alpha$ [[Definition:Iff|iff]] $f$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+That is, both $f$ and $f^{-1}$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed|lemma}}
+=== Definition 1 iff Definition 3 ===
+Let $f$ and $f^{-1}$ both be [[Definition:Continuous Mapping (Topology)|continuous]].
+Then by [[Bijection is Open iff Inverse is Continuous]], $f$ is [[Definition:Open Mapping|open]].
+Thus $f$ is both [[Definition:Open Mapping|open]] and [[Definition:Continuous Mapping (Topology)|continuous]].
+[[Definition:Converse Statement|Conversely]], let $f$ be both [[Definition:Open Mapping|open]] and [[Definition:Continuous Mapping (Topology)|continuous]].
+Then also by [[Bijection is Open iff Inverse is Continuous]], $f^{-1}$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Thus both $f$ and $f^{-1}$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed|lemma}}
+=== Definition 3 iff Definition 4 ===
+Let $f$ be both [[Definition:Open Mapping|open]] and [[Definition:Continuous Mapping (Topology)|continuous]].
+Then by [[Bijection is Open iff Closed]] it follows that $f$ is both [[Definition:Closed Mapping|closed]] and [[Definition:Continuous Mapping (Topology)|continuous]].
+[[Definition:Converse Statement|Conversely]], let $f$ be both both [[Definition:Closed Mapping|closed]] and [[Definition:Continuous Mapping (Topology)|continuous]].
+Then also by [[Bijection is Open iff Closed]] it follows that $f$ is both [[Definition:Open Mapping|open]] and [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed}}
+[[Category:Homeomorphisms]]
+tpsbuo6gn0rvywf6lq00ny7v7gw4urr
+\end{proof}<|endoftext|>
+\section{Quotient Space of Hausdorff Space is not necessarily Hausdorff}
+Tags: Hausdorff Spaces, Quotient Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
+Let $\RR \subseteq S \times S$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
+Let $T_\RR := \struct {S / \RR, \tau_\RR}$ be the [[Definition:Quotient Space (Topology)|quotient space]] of $S$ by $\RR$.
+Then $T_\RR$ is not necessarily also a [[Definition:Hausdorff Space|Hausdorff space]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Real Number Line with Euclidean Topology|real number line with the Euclidean topology]] $\struct {\R, \tau}$.
+By [[Real Number Line satisfies all Separation Axioms]], $\struct {\R, \tau}$ is a [[Definition:Hausdorff Space|Hausdorff space]].
+By [[Quotient Space of Real Line may not be Kolmogorov]], there is a relation $\RR$ on $\R$ such that the [[Definition:Quotient Space (Topology)|quotient space]] $\struct {\R / \RR, \tau_\RR}$ is not a [[Definition:Kolmogorov Space|Kolmogorov space]].
+Thus the theorem holds by [[Sequence of Implications of Separation Axioms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Direct Limit of Sequence of Groups}
+Tags: Group Theory, Existence and Uniqueness of Direct Limit of Sequence of Groups
+
+\begin{theorem}
+Let $\sequence {G_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Group|groups]].
+Let $\sequence {g_n}_{n \mathop \in \N}: g_n: G_n \to G_{n + 1}$ be a [[Definition:Sequence|sequence]] of [[Definition:Group Homomorphism|group homomorphisms]].
+Then their [[Definition:Direct Limit of Sequence of Groups/Definition 1|direct limit]] $G_\infty$ exists and is [[Definition:Unique|unique]] up to [[Definition:Unique|unique]] [[Definition:Group Isomorphism|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+{{refactor|Transclude the lemmata}}
+=== Existence ===
+Define $\widehat G_\infty$ by:
+:$\widehat G_\infty := \displaystyle \bigsqcup_{n \mathop \in \N} G_n$
+where $\displaystyle \bigsqcup$ signifies a [[Definition:Disjoint Union (Set Theory)|disjoint union]].
+For $m \ge n$, denote with $g_{n, m}$ the [[Definition:Composition|composition]] $g_{m - 1} \circ \cdots \circ g_n: G_n \to G_m$.
+In particular, $g_{n, n}: G_n \to G_n$ is taken to be the [[Definition:Identity Mapping|identity mapping]] on $G_n$.
+Now define an [[Definition:Equivalence Relation|equivalence relation]] $\sim$ on $\hat G_\infty$ by putting:
+:$\tuple {x_n, n} \sim \tuple {y_m, m} \iff \exists k \ge n, m: \map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$
+which is established to be indeed an equivalence relation in [[Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 1|Lemma 1]].
+Now define $G_\infty$ as the [[Definition:Quotient Set|quotient set]]:
+:$G_\infty := \widehat G_\infty / \sim$
+We can equip $G_\infty$ with a group structure as follows:
+Let $\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$, where $\eqclass {\tuple {x_n, n} } {}$ denotes the [[Definition:Equivalence Class|equivalence class]] of $\tuple{x_n,n}$.
+Let $l:= \max \set {m, n}$.
+We define a group operation on $G_\infty$ via 
+:$\tuple {\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} } \longmapsto \eqclass {\tuple {\map {g_{n l} } {x_n} \map {g_{m l} } {y_m}, l} } {}$.
+We shall also refer to this operation as multiplication.
+A proof that this is indeed a well-defined group can be found in [[Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 2|Lemma 2]].
+Now define $u_n: G_n \to G_\infty$ by:
+:$\map {u_n} {x_n} := \eqclass {\tuple {x_n, n} } {}$
+That this is a group homomorphism follows immediately from the definition of the group structure.
+Now let $H$ be a [[Definition:Group|group]] and $h_n: G_n \to H$ be [[Definition:Group Homomorphism|group homomorphisms]] such that $h_{n + 1} \circ g_n = h_n$.
+Define $h_\infty: G_\infty \to H$ by:
+:$\map {h_\infty} {\eqclass {\tuple {x_n, n} } {} } = \map {h_n} {x_n}$
+Under this definition:
+:$\forall x_n \in G_n: \map {h_\infty \circ u_n} {x_n} = \map {h_n} {x_n}$
+By [[Equality of Mappings]], it follows that:
+:$h_\infty \circ u_n = h_n$
+It remains to be shown that $h_\infty$ is [[Definition:Well-Defined Mapping|well-defined]] and a [[Definition:Group Homomorphism|group homomorphism]].
+This is done in [[Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 3|Lemma 3]].
+{{qed|lemma}}
+=== Uniqueness ===
+A direct proof of uniqueness can be formulated as follows:
+{{tidy}}
+Let $G_\infty'$ be a second direct limit group  with homomorphisms $u_n': G_n \rightarrow G_\infty'$. 
+By the above universal property we can construct homomorphisms $u'_{\infty} : G_{\infty} \rightarrow G_{\infty}'$ (taking $H = G_{\infty}'$ and $h'_n = u'_n$) and $u_{\infty} : G'_{\infty} \rightarrow G_{\infty}$ (taking $H = G_{\infty}$ and $h_n = u_n$). 
+We shall prove that $u'_\infty$ is a [[Definition:Inverse (Abstract Algebra)/Left Inverse|left inverse]] of $u_{\infty}$. Showing that it is also a [[Definition:Inverse (Abstract Algebra)/Right Inverse|right inverse]] follows analogously by exchanging $u_{\infty}$ and $u'_{\infty}$. 
+From 
+:$u'_{\infty} \circ u_n = u_n' = \mathop{Id}_{G'_\infty} \circ u'_n$
+we find that the map $h_\infty : G'_\infty \to G'_\infty$ corresponding to $H = G'_\infty$ and $h_n = u'_{\infty} \circ u_n$ must be $\mathop{Id}_{G'_\infty}$ by uniqueness.
+We also have
+:$u'_{\infty} \circ u_n  = u'_{\infty} \circ \mathop{Id}_{G_\infty}  \circ u_n $
+thus by the same argument $h_\infty$ must also equal $u_\infty'\circ u_\infty$. 
+We conclude that $u_\infty'$ is a right inverse of $u_\infty$.
+{{qed}}
+{{refactor|Extract the below and implement it as a second proof on its own page.}}
+For a more categorical point of view:
+From [[Equivalence of Definitions of Direct Limit of Sequence of Groups]], $G_\infty$ can be described as a [[Definition:Colimit|colimit]] in the [[Definition:Category of Groups|category of groups]] $\mathbf{Grp}$.
+By [[Colimit is Unique]], it therefore is [[Definition:Unique|unique]] up to [[Definition:Unique|unique]] [[Definition:Isomorphism (Category Theory)|$\mathbf{Grp}$-isomorphism]].
+By [[Isomorphism in Category of Groups iff Group Isomorphism]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Product of Compact Spaces/Finite Product}
+Tags: Topology
+
+\begin{theorem}
+Let $T_1, T_2, \ldots, T_n$ be [[Definition:Topological Space|topological spaces]].
+Let $\displaystyle \prod_{i \mathop = 1}^n T_i$ be the [[Definition:Product Space (Topology)|product space]] of $T_1, T_2, \ldots, T_n$.
+Then $\displaystyle \prod_{i \mathop = 1}^n T_i$ is [[Definition:Compact Topological Space|compact]] {{iff}} all of $T_1, T_2, \ldots, T_n$ are [[Definition:Compact Topological Space|compact]].
+\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]]:
+:$\displaystyle \prod_{i \mathop = 1}^n T_i$ is [[Definition:Compact Topological Space|compact]] {{iff}} all of $T_1, T_2, \ldots, T_n$ are [[Definition:Compact Topological Space|compact]]
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$T_1$ is [[Definition:Compact Topological Space|compact]] {{iff}} $T_1$ is [[Definition:Compact Topological Space|compact]]
+which is trivially true.
+Thus $\map P 1$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \prod_{i \mathop = 1}^k T_i$ is [[Definition:Compact Topological Space|compact]] {{iff}} all of $T_1, T_2, \ldots, T_k$ are [[Definition:Compact Topological Space|compact]]
+from which it is to be shown that:
+:$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i$ is [[Definition:Compact Topological Space|compact]] {{iff}} all of $T_1, T_2, \ldots, T_{k + 1}$ are [[Definition:Compact Topological Space|compact]]
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+We have:
+:$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \prod_{i \mathop = 1}^{k + 1} T_i
+      | o = \text {is}
+      | r = \text {compact}
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = \prod_{i \mathop = 1}^k T_i, T_{k + 1}
+      | o = \text {are}
+      | r = \text {compact}
+      | c = [[Topological Product of Compact Spaces]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = T_1, T_2, \ldots, T_k, T_{k + 1}
+      | o = \text {are}
+      | r = \text {compact}
+      | c = [[Topological Product of Compact Spaces/Finite Product#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and thus it follows by the [[Principle of Mathematical Induction]] that for any $n \ge 1$:
+:$\displaystyle \prod_{i \mathop = 1}^n T_i$ is [[Definition:Compact Topological Space|compact]] {{iff}} all of $T_1, T_2, \ldots, T_n$ are [[Definition:Compact Topological Space|compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uniform Continuity on Metric Space does not imply Compactness}
+Tags: Metric Spaces, Compact Spaces, Uniformly Continuous Mappings
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous mapping]] on $A_1$.
+Then $M_1$ does not necessarily have to be a [[Definition:Compact Metric Space|compact metric space]].
+\end{theorem}
+
+\begin{proof}
+Let $M_1 = \left({A_1, d_1}\right)$ be any [[Definition:Metric Space|metric space]] which is not [[Definition:Compact Metric Space|compact]].
+Let $I_{M_1}: M_1 \to M_1$ be the [[Definition:Identity Mapping|identity mapping]].
+From [[Identity Mapping is Uniformly Continuous]], $I_{M_1}$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous]] on $M_1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uniformly Continuous Function is Continuous/Metric Space}
+Tags: Metric Spaces, Continuous Mappings on Metric Spaces, Uniformly Continuous Mappings
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_1, d_1}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let the [[Definition:Mapping|mapping]] $f: M_1 \to M_2$ be [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous on $M_1$]].
+Then $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous on $M_1$]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous on $M_1$]].
+Let $x \in M_1$.
+Let $\epsilon > 0$.
+As $f$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous]], $\exists \delta > 0$ such that:
+:$\forall y \in M_1: d_1 \left({x, y}\right) < \delta: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$
+Thus by definition $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]] at $x$.
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:Continuous Mappings on Metric Spaces]]
+[[Category:Uniformly Continuous Mappings]]
+qdpnhtcjgrywzny4gfvtlykqmhh0bj9
+\end{proof}<|endoftext|>
+\section{Uniformly Continuous Function is Continuous/Real Function}
+Tags: Real Analysis, Continuous Real Functions, Uniformly Continuous Real Functions, Uniformly Continuous Real Function is Continuous
+
+\begin{theorem}
+Let $I$ be an [[Definition:Real Interval|interval]] of $\R$.
+Let $f: I \to \R$ be a [[Definition:Uniformly Continuous Real Function|uniformly continuous real function]] on $I$.
+Then $f$ is [[Definition:Continuous on Interval|continuous]] on $I$.
+\end{theorem}<|endoftext|>
+\section{Young's Inequality for Increasing Functions/Equality}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $a_0$ and $b_0$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a [[Definition:Strictly Increasing Real Function|strictly increasing]] [[Definition:Bijection|bijection]].
+Let $a$ and $b$ be [[Definition:Real Number|real numbers]] such that $0 \le a \le a_0$ and $0 \le b \le b_0$.
+Then $b = \map f a$ {{iff}}:
+:$\displaystyle a b = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
+where $\displaystyle \int$ denotes the [[Definition:Darboux Integral|Darboux integral]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+{{ProofWanted}}
+=== Sufficient Condition ===
+By [[Monotone Real Function is Darboux Integrable]], $f$ and $f^{-1}$ are [[Definition:Darboux Integrable Function|Darboux integrable]].
+Let $b = \map f a$.
+Define:
+:$\displaystyle A = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
+Consider any [[Definition:Subdivision (Real Analysis)|subdivision]] $P = \set {x_0, x_1, \ldots, x_n}$ of the [[Definition:Closed Real Interval|closed real interval]] $\closedint 0 a$.
+Then:
+:$f P = \set {\map f {x_0}, \map f {x_1}, \ldots, \map f {x_n} }$
+is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint 0 b$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map U {P, f} + \map L {f P, f^{-1} }
+      | r = \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} } \map f {x_k} + \sum_{k \mathop = 1}^n \paren {\map f {x_k} - \map f {x_{k - 1} } } \map {f^{-1} } {\map f {x_{k - 1} } }
+      | c = because $f$ and $f^{-1}$ are [[Definition:Increasing Real Function|increasing]]
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {x_k \map f {x_k} - x_{k - 1} \map f {x_{k - 1} } }
+      | c = by [[Summation is Linear]]
+}}
+{{eqn | r = a b
+}}
+{{end-eqn}}
+Similarly:
+:$\map L {P, f} + \map U {f P, f^{-1} } = a b$
+Let $\epsilon \in \R_{>0}$ be an arbitrary [[Definition:Strictly Positive Real Number|strictly positive real number]].
+By the definition of the [[Definition:Darboux Integral|Darboux integral]], there exist [[Definition:Subdivision (Real Analysis)|subdivisions]] $P_+$ and $P_-$ of $\closedint 0 a$ such that:
+:$a b = \map U {P_+, f} + \map L {f P_+, f^{-1} } < A + \epsilon$
+:$a b = \map L {P_-, f} + \map U {f P_-, f^{-1} } > A - \epsilon$
+The result follows from [[Real Plus Epsilon]].
+{{qed}}
+{{Namedfor|William Henry Young}}
+\end{proof}<|endoftext|>
+\section{Category of Finite Sets is Cartesian Closed}
+Tags: Category of Finite Sets
+
+\begin{theorem}
+Let $\mathbf{Finset}$ be the [[Definition:Category of Finite Sets|category of finite sets]].
+Then $\mathbf{Finset}$ is [[Definition:Cartesian Closed Category|Cartesian closed]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|will be tedious until some theorems on subcategories are available}}
+\end{proof}<|endoftext|>
+\section{Components of Separation are Clopen}
+Tags: Disconnected Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A \mid B$ be a [[Definition:Separation (Topology)|separation]] of $T$.
+Then both $A$ and $B$ are [[Definition:Clopen Set|clopen]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Set with Relative Complement forms Partition]]:
+:$A = \complement_S \left({B}\right)$
+and:
+:$B = \complement_S \left({A}\right)$
+where $\complement_S$ denotes the [[Definition:Relative Complement|complement relative to $S$]].
+As $A$ and $B$ are both [[Definition:Open Set (Topology)|open]], it follows by definition that $A$ and $B$ are also both [[Definition:Closed Set (Topology)|closed]].
+That is, by definition, they are [[Definition:Clopen Set|clopen]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets}
+Tags: Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+{{TFAE|def = Connected Topological Space}}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+From [[Biconditional Equivalent to Biconditional of Negations/Formulation 1|Biconditional Equivalent to Biconditional of Negations]] it follows that the statement can be expressed as:
+:$T$ admits a [[Definition:Separation (Topology)|separation]]
+{{iff}}:
+:there exist two [[Definition:Closed Set (Topology)|closed sets]] of $T$ which form a [[Definition:Partition (Set Theory)|(set) partition]] of $S$.
+By definition, a [[Definition:Separation (Topology)|separation]] of $T$ is a [[Definition:Partition (Set Theory)|(set) partition]] of $S$ by $A, B$ which are [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Complements of Components of Two-Component Partition form Partition]]:
+:$A \mid B$ is a [[Definition:Partition (Set Theory)|(set) partition]] of $S$ {{iff}} $\complement_S \left({A}\right) \mid \complement_S \left({B}\right)$ is a [[Definition:Partition (Set Theory)|(set) partition]] of $S$.
+Hence the result by definition of [[Definition:Closed Set (Topology)|closed set]].
+\end{proof}<|endoftext|>
+\section{Empty Set Satisfies Topology Axioms}
+Tags: Topology, Empty Set
+
+\begin{theorem}
+Let $T = \struct {\O, \set \O}$ where $\O$ denotes the [[Definition:Empty Set|empty set]].
+Then $T$ satisfies the [[Definition:Open Set Axioms|open set axioms]] for a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+We proceed to verify the [[Definition:Open Set Axioms|open set axioms]] for $\set \O$ to be a [[Definition:Topology|topology]] on $\O$.
+Let $\tau = \set \O$.
+=== {{OpenSetAxiom|1}} ===
+By [[Union of Empty Set]]:
+:$\displaystyle \bigcup \tau = \O \in \tau$
+Thus [[Definition:Open Set Axioms|open set axiom $\paren {\text O 1}$]] is satisfied.
+{{qed|lemma}}
+=== {{OpenSetAxiom|2}} ===
+From [[Intersection with Empty Set]]:
+:$\O \cap \O = \O \in \tau$
+and so [[Definition:Open Set Axioms|open set axiom $\paren {\text O 2}$]] is satisfied.
+{{qed|lemma}}
+=== {{OpenSetAxiom|3}} ===
+By definition $\O \in \tau$ and so [[Definition:Open Set Axioms|open set axiom $\paren {\text O 3}$]] is satisfied.
+{{qed}}
+[[Category:Topology]]
+[[Category:Empty Set]]
+gqv71ragp7zb18f9xx2eolvz3ta8dr1
+\end{proof}<|endoftext|>
+\section{Complements of Components of Two-Component Partition form Partition}
+Tags: Relative Complement
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] with at least two [[Definition:Element|elements]].
+Let $A, B \subseteq S$.
+Let $\complement_S$ denote the [[Definition:Relative Complement|complement relative to $S$]].
+$A \mid B$ is a [[Definition:Set Partition|partition]] of $S$ {{iff}} $\relcomp S A \mid \relcomp S B$ is a [[Definition:Set Partition|partition]] of $S$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $A \mid B$ be a [[Definition:Set Partition|partition]] of $S$.
+That is, by definition:
+:$(1): \quad A \cap B = \O$
+:$(2): \quad A \cup B = S$
+:$(3): \quad A, B \ne \O$
+Thus:
+{{begin-eqn}}
+{{eqn | l = A \cap B 
+      | r = \O
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S {A \cap B}
+      | r = S
+      | c = [[Relative Complement of Empty Set]]
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S A \cup \relcomp S B
+      | r = S
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = A \cup B 
+      | r = S
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S {A \cup B}
+      | r = \O
+      | c = [[Relative Complement with Self is Empty Set]]
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S A \cap \relcomp S B
+      | r = \O
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union|De Morgan's Laws: Complement of Union]]
+}}
+{{end-eqn}}
+{{WLOG}}, now suppose that $\relcomp S A = \O$.
+{{begin-eqn}}
+{{eqn | l = \relcomp S A
+      | r = \O
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = S
+      | c = [[Relative Complement of Empty Set]]
+}}
+{{eqn | ll= \leadsto
+      | l = B
+      | r = \O
+      | c = as $A \cap B = \O$
+}}
+{{end-eqn}}
+Thus by [[Rule of Transposition]]:
+:$B \ne \O \implies \relcomp S A \ne \O$
+[[Definition:Mutatis Mutandis|Mutatis mutandis]]:
+:$A \ne \O \implies \relcomp S B \ne \O$
+[[Definition:By Hypothesis|by hypothesis]]:
+:$A, B \ne \O$
+and so:
+:$\relcomp S A, \relcomp S B \ne \O$
+Thus all three conditions are satisfied for $\relcomp S A \mid \relcomp S B$ to be a [[Definition:Set Partition|partition]] of $S$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\relcomp S A \mid \relcomp S B$ be a [[Definition:Set Partition|partition]] of $S$.
+By the [[Complements of Components of Two-Component Partition form Partition#Necessary Condition|necessary condition]], it follows that $\relcomp S {\relcomp S A} \mid \relcomp S {\relcomp S B}$ is a [[Definition:Set Partition|partition]] of $S$.
+From [[Relative Complement of Relative Complement]]:
+:$\relcomp S {\relcomp S A} = A$ and $\relcomp S {\relcomp S B} = B$
+and so $A \mid B$ is a [[Definition:Set Partition|partition]] of $S$.
+{{qed}}
+[[Category:Relative Complement]]
+nqeqddcvfi8knul9tva7cbh8tv0x4yp
+\end{proof}<|endoftext|>
+\section{Sigma-Algebra Closed under Finite Intersection}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $A_1, \ldots, A_n \in \Sigma$.
+Then $\displaystyle \bigcap_{k \mathop = 1}^n A_k \in \Sigma$.
+\end{theorem}
+
+\begin{proof}
+Define for $k \in \N, k > n: A_k = X$.
+By axiom $(1)$ of a [[Definition:Sigma-Algebra|$\sigma$-algebra]], it follows that $\forall k \in \N, k > n: A_k \in \Sigma$.
+From [[Sigma-Algebra Closed under Countable Intersection]], it follows that $\displaystyle \bigcap_{k \mathop \in \N} A_k = \bigcap_{k \mathop = 1}^n A_k \in \Sigma$.
+{{qed}}
+[[Category:Sigma-Algebras]]
+4cocq4ef5t089luotyuid8duajhg9f6
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary}
+Tags: Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $T$ have no two [[Definition:Disjoint Sets|disjoint]] [[Definition:Non-Empty Set|non-empty]] [[Definition:Closed Set (Topology)|closed sets]] whose [[Definition:Set Union|union]] is $S$.
+Then the only [[Definition:Subset|subsets]] of $S$ whose [[Definition:Boundary (Topology)|boundary]] is [[Definition:Empty Set|empty]] are $S$ and $\varnothing$.
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] whose [[Definition:Boundary (Topology)|boundary]] $\partial H$ is [[Definition:Empty Set|empty]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = \partial H
+      | r = \varnothing
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = H^- \cap \left({S \setminus H}\right)^-
+      | r = \varnothing
+      | c = [[Boundary is Intersection of Closure with Closure of Complement]]
+}}
+{{end-eqn}}
+From [[Topological Closure is Closed]], both $H^-$ and $\left({S \setminus H}\right)^-$ are [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+From [[Union of Closure with Closure of Complement is Whole Space]]:
+:$H^- \cup \left({S \setminus H}\right)^- = S$
+Thus $H^-$ and $\left({S \setminus H}\right)^-$ are two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] of $T$ whose [[Definition:Set Union|union]] is $S$.
+Hence, [[Definition:By Hypothesis|by hypothesis]], one of them must be [[Definition:Empty Set|empty]].
+Suppose $H$ is not [[Definition:Empty Set|empty]].
+It must therefore follow that:
+: $S \setminus H = \varnothing$
+Therefore $H = S$.
+Thus the only [[Definition:Subset|subsets]] of $S$ whose [[Definition:Boundary (Topology)|boundary]] is [[Definition:Empty Set|empty]] are $S$ and $\varnothing$.
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Subsets with Empty Boundary implies No Clopen Sets}
+Tags: Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $T$ be such that the only [[Definition:Subset|subsets]] of $S$ whose [[Definition:Boundary (Topology)|boundary]] is [[Definition:Empty Set|empty]] are $S$ and $\varnothing$.
+Then the only [[Definition:Clopen Set|clopen sets]] of $T$ are $S$ and $\varnothing$.
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$ be a [[Definition:Clopen Set|clopen set]] of $T$.
+From [[Set is Clopen iff Boundary is Empty]], $H$ has an [[Definition:Empty Set|empty]] [[Definition:Boundary (Topology)|boundary]].
+[[Definition:By Hypothesis|By hypothesis]], $H = S$ or $H = \varnothing$.
+That is, the only [[Definition:Clopen Set|clopen sets]] of $T$ are $S$ and $\varnothing$.
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets}
+Tags: Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let the only [[Definition:Clopen Set|clopen sets]] of $T$ be $S$ and $\O$.
+Then there are no two [[Definition:Non-Empty Set|non-empty]] [[Definition:Separated Sets|separated sets]] of $T$ whose [[Definition:Set Union|union]] is $S$.
+\end{theorem}
+
+\begin{proof}
+Suppose $A$ and $B$ are [[Definition:Separated Sets|separated subsets]] of $T$ such that $A \cup B = S$.
+By definition of [[Definition:Separated Sets|separated sets]]:
+:$A \cap B^- = \O$
+Then:
+{{begin-eqn}}
+{{eqn | l = S
+      | r = A \cup B
+      | c = 
+}}
+{{eqn | o = \subseteq
+      | r = A \cup B^-
+      | c = [[Set is Subset of its Topological Closure]]
+}}
+{{eqn | o = \subseteq
+      | r = S
+      | c = by definition of $S$
+}}
+{{end-eqn}}
+Hence $A = S \setminus B^-$.
+From [[Topological Closure is Closed]], $B^-$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Thus $A$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Also by definition of [[Definition:Separated Sets|separated sets]]:
+:$A^- \cap B = \O$
+Hence, by the same reasoning, $B$ must also be [[Definition:Open Set (Topology)|open]].
+But:
+:$A \cap B \subseteq A \cap B^- = \O$
+and $A \cup B = S$, by assumption.
+So:
+:$A = S \setminus B$ and $B = S \setminus A$
+and we conclude that both $A$ and $B$ are [[Definition:Clopen Set|clopen]].
+Therefore, [[Definition:By Hypothesis|by hypothesis]], one of them must be $S$ and the other must be $\O$.
+That is, there are no two [[Definition:Non-Empty Set|non-empty]] [[Definition:Separated Sets|separated sets]] of $T$ whose [[Definition:Set Union|union]] is $S$.
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space}
+Tags: Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T$ be such that there are no two [[Definition:Non-Empty Set|non-empty]] [[Definition:Separated Sets|separated sets]] whose [[Definition:Set Union|union]] is $S$.
+Then there exists no [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]] from $T$ onto a [[Definition:Discrete Topology|discrete two-point space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] such that there are no two [[Definition:Non-Empty Set|non-empty]] [[Definition:Separated Sets|separated sets]] whose [[Definition:Set Union|union]] is $S$.
+Let $D = \struct {\set {0, 1}, \tau}$ be the [[Definition:Discrete Space|discrete two-point space]] on $\set {0, 1}$.
+{{AimForCont}} $f: T \to \set {0, 1}$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
+By definition of [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]]:
+:$\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are [[Definition:Open Set (Topology)|open sets]] of $T$.
+From the definition of a [[Definition:Mapping|mapping]]:
+:$\map {f^{-1} } 0 \cup \map {f^{-1} } 1 = S$
+and
+:$\map {f^{-1} } 0 \cap \map {f^{-1} } 1 = \O$
+Then:
+:$\map {f^{-1} } 0 = S \setminus \map {f^{-1} } 1$
+and:
+:$\map {f^{-1} } 1 = T \setminus \map {f^{-1} } 0$
+are [[Definition:Clopen Set|clopen]].
+From [[Closed Set equals its Closure]] they are their respective [[Definition:Closure (Topology)|closures]].
+It follows from the definition that $\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are [[Definition:Separated Sets|separated subsets]] of $T$ whose [[Definition:Set Union|union]] is $S$.
+Hence, [[Definition:By Hypothesis|by hypothesis]], one of them must be [[Definition:Empty Set|empty]], and the other one must be $S$.
+Therefore $f$ is [[Definition:Constant Mapping|constant]], and so is not a [[Definition:Surjection|surjection]].
+This [[Definition:Contradiction|contradicts]] the original hypothesis.
+That is, there exists no [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]] from $T$ onto a [[Definition:Discrete Topology|discrete two-point space]].
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation}
+Tags: Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $T$ be such that there exists no [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]] from $T$ onto a [[Definition:Discrete Topology|discrete two-point space]].
+Then there exist no [[Definition:Open Set (Topology)|open sets]] $A, B \in \tau$ such that $A, B \ne \varnothing$, $A \cup B = S$ and $A \cap B = \varnothing$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]] such that there exists no [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]] from $T$ onto a [[Definition:Discrete Topology|discrete two-point space]].
+Let $D = \left({\left\{{0, 1}\right\}, \tau}\right)$ be the [[Definition:Discrete Space|discrete two-point space]] on $\left\{{0, 1}\right\}$.
+Let $A$ and $B$ be [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] of $T$ such that $A \cup B = S$.
+The aim is to show that one of them is [[Definition:Empty Set|empty]].
+Let us define the [[Definition:Mapping|mapping]] $f: S \to \left\{{0, 1}\right\}$ by:
+:$f \left({x}\right) = \begin{cases}
+0 & : x \in A \\
+1 & : x \in B
+\end{cases}$
+There are only four open sets in $\left\{{0, 1}\right\}$, namely:
+: $\varnothing$
+: $\left\{{0}\right\}$
+: $\left\{{1}\right\}$
+: $\left\{{0, 1}\right\}$
+We have that:
+:$f^{-1} \left[{\varnothing}\right] = \varnothing$
+:$f^{-1} \left[{\left\{{0}\right\}}\right] = A$
+:$f^{-1} \left[{\left\{{1}\right\}}\right] = B$
+:$f^{-1} \left[{\left\{{0, 1}\right\}}\right] = S$
+where $f^{-1} \left[{X}\right]$ denotes the [[Definition:Preimage of Subset under Mapping|preimage]] of the [[Definition:Set|set]] $X$.
+But [[Definition:By Hypothesis|by hypothesis]] all of $\varnothing, A, B, S$ are [[Definition:Open Set (Topology)|open sets]] of $T$.
+So by definition $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Also [[Definition:By Hypothesis|by hypothesis]], $f$ cannot be [[Definition:Surjection|surjective]].
+It follows that $f$ must be [[Definition:Constant Mapping|constant]].
+So either $A$ or $B$ must be [[Definition:Empty Set|empty]], and the other one must be $S$.
+Hence the result.
+\end{proof}<|endoftext|>
+\section{Existence of Vector Space Bases implies Axiom of Choice}
+Tags: Bases of Vector Spaces, Axiom of Choice
+
+\begin{theorem}
+The supposition that every [[Definition:Vector Space|vector space]] has a [[Definition:Basis of Vector Space|basis]], along with the [[Definition:Zermelo-Fraenkel Axioms|Zermelo-Fraenkel axioms]], implies that the [[Axiom:Axiom of Choice|axiom of choice]] holds.
+\end{theorem}
+
+\begin{proof}
+[http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf Blass, 1984]
+{{proof wanted}}
+\end{proof}<|endoftext|>
+\section{Deterministic Time Hierarchy Theorem}
+Tags: Complexity Theory, Computer Science, Named Theorems
+
+\begin{theorem}
+Let $\map f n$ be a [[Definition:Time-Constructible Function|time-constructible function]].
+Then there exists a [[Definition:Decision Problem|decision problem]] which:
+:can be solved in [[Definition:Worst-Case Deterministic Time|worst-case deterministic time]] $\map f {2 n + 1}^3$
+but:
+:cannot be solved in [[Definition:Worst-Case Deterministic Time|worst-case deterministic time]] $\map f n$.
+In other words, the [[Definition:Complexity Class|complexity class]] $\map {\mathsf {DTIME} } {\map f n} \subsetneq \map {\mathsf {DTIME} } {\map f {2 n + 1}^3}$.
+\end{theorem}
+
+\begin{proof}
+Let $H_f$ be a [[Definition:Set|set]] defined as follows:
+:$H_f = \set {\tuple {\sqbrk M, x}: \text {$M$ accepts $x$ in $\map f {\size x}$ steps} }$
+where:
+:$M$ is a [[Definition:Turing Machine|(deterministic) Turing machine]]
+:$x$ is its input (the initial contents of its tape)
+:$\sqbrk M$ denotes an input that encodes the Turing machine $M$
+Let $m$ be the size of $\tuple {\sqbrk M, x}$.
+We know that we can decide membership of $H_f$ by way of a [[Definition:Turing Machine|(deterministic) Turing machine]] that:
+:$(1): \quad$ calculates $f \left({\size x}\right)$
+:$(2): \quad$ writes out a row of $0$s of that length
+:$(3): \quad$ uses this row of $0$s as a counter to simulate $M$ for at most that many steps.
+At each step, the simulating [[Definition:Turing Machine|machine]] needs to look through the definition of $M$ to decide what the next action would be.
+It is safe to say that this takes at most $\map f m^3$ operations, so:
+:$ H_f \in \map {\mathsf{DTIME} } {\map f m^3}$
+{{handwaving|"We know that ...", "It is safe to say ..."}}
+{{AimForCont}}:
+:$H_f \in \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$
+Then we can construct some [[Definition:Turing Machine|machine]] $K$ which:
+:given some [[Definition:Turing Machine|machine]] description $\sqbrk {M_K}$ and input $x$
+:decides within $\map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$ whether $\tuple {\sqbrk {M_K}, x} \in H_f$.
+Construct another [[Definition:Turing Machine|machine]] $N$ which:
+:takes a machine description $\sqbrk {M_N}$
+:runs $K$ on $\left({ \sqbrk {M_N}, \sqbrk {M_N} }\right)$
+:accepts only if $K$ rejects, and rejects if $K$ accepts.
+Let $m_n$ be the length of $\sqbrk {M_N}$.
+Then $m$ (the length of the input to $K$) is twice $m_n$ plus some delimiter symbol, so:
+:$m = 2m_n + 1$
+$N$'s running time is thus:
+{{begin-eqn}}
+{{eqn | l = \map {\mathsf{DTIME} } {\map f {\floor {\frac m 2} } }
+      | r = \map {\mathsf{DTIME} } {\map f {\floor {\frac {2 m_n + 1} 2} } }
+      | c = 
+}}
+{{eqn | r = \map {\mathsf{DTIME} } {\map f {m_n} }
+      | c = 
+}}
+{{end-eqn}}
+Now consider the case $M_N = N$.
+That is we feed $\sqbrk N$ as input into $N$ itself).
+In this case $m_n$ is the length of $\sqbrk N$.
+* If $N$ '''accepts''' $\sqbrk N$ (which we know it does in at most $\map f {m_n}$ operations):
+** By the definition of $N$, $K$ '''rejects''' $\tuple {\sqbrk N, \sqbrk N}$
+** Therefore, by the definition of $K$, $\tuple {\sqbrk N, \sqbrk N} \notin H_f$
+** Therefore, by the definition of $H_f$, $N$ does not accept $\sqbrk N$ in $\map f {m_n}$ steps -- a [[Definition:Contradiction|contradiction]].
+* If $N$ '''rejects''' $\sqbrk N$ (which we know it does in at most $\map f {m_n}$ operations):
+** By the definition of $N$, $K$ '''accepts''' $\tuple {\sqbrk N, \sqbrk N}$
+** Therefore, by the definition of $K$, $\tuple {\sqbrk N, \sqbrk N} \in H_f$
+** Therefore, by the definition of $H_f$, $N$ '''does''' accept $\sqbrk N$ in $\map f {m_n}$ steps -- a [[Definition:Contradiction|contradiction]].
+Therefore, $K$ does not exist.
+So, by [[Proof by Contradiction]]:
+:$H_f \notin \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$
+Substituting $2 n + 1$ for $m$, we get:
+:$H_f \notin \map {\mathsf{DTIME} } {\map f n}$
+and, from the earlier result:
+:$H_f \in \map {\mathsf{DTIME} } {\map f {2 n + 1}^3}$
+{{qed}}
+[[Category:Complexity Theory]]
+[[Category:Computer Science]]
+[[Category:Named Theorems]]
+23hffzr83e7hv17hpn8s4dsqf0rrlhl
+\end{proof}<|endoftext|>
+\section{Category of Posets is Cartesian Closed}
+Tags: Category of Ordered Sets
+
+\begin{theorem}
+Let $\mathbf{OrdSet}$ be the [[Definition:Category of Ordered Sets|category of ordered sets]].
+Then $\mathbf{OrdSet}$ is [[Definition:Cartesian Closed Category|Cartesian closed]].
+\end{theorem}<|endoftext|>
+\section{Definition:Exponentiation Functor}
+Tags: Definitions: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Cartesian Closed Category|Cartesian closed metacategory]].
+Let $A$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
+Then '''exponentiation by $A$''', denoted $\left({-}\right)^A: \mathbf C \to \mathbf C$, is the [[Definition:Covariant Functor|functor]] defined by:
+{{begin-axiom}}
+{{axiom|lc= [[Definition:Object Functor|Object functor]]:
+       |m = C^A := C^A
+       |rc= $C^A$ is the [[Definition:Exponential (Category Theory)|exponential]] of $C$ by $A$
+}}
+{{axiom|lc= [[Definition:Morphism Functor|Morphism functor]]:
+       |m = f^A := \widetilde{\left({f \circ \epsilon}\right)}
+       |rc= $f: B \to C$ is a [[Definition:Morphism (Category Theory)|morphism]] of $\mathbf C$
+}}
+{{end-axiom}}
+Here $\epsilon: B^A \times A \to B$ denotes the [[Definition:Evaluation Morphism|evaluation morphism]], and $\widetilde{\left({f \circ \epsilon}\right)}: B^A \to C^A$ is the [[Definition:Exponential Transpose|exponential transpose]] of $f \circ \epsilon$.
+That it is in fact a [[Definition:Covariant Functor|functor]] is shown on [[Exponentiation Functor is Functor]].
+\end{theorem}<|endoftext|>
+\section{Exponentiation Functor is Functor}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Cartesian Closed Category|Cartesian closed metacategory]].
+Let $A$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
+Let $\left({-}\right)^A: \mathbf C \to \mathbf C$ be the [[Definition:Exponentiation Functor|exponentiation functor]].
+Then $\left({-}\right)^A$ is a [[Definition:Covariant Functor|functor]].
+\end{theorem}
+
+\begin{proof}
+Let $B$ be an [[Definition:Object (Category Theory)|object]] of $\mathbf C$.
+Let $\epsilon_B: B^A \times A \to B$ be the [[Definition:Evaluation Morphism|evaluation morphism]] at $B$.
+Then, since:
+{{begin-eqn}}
+{{eqn|l = \operatorname{id}_B \epsilon_B
+     |r = \epsilon_B
+}}
+{{eqn|r = \epsilon_B \operatorname{id}_{B^A \times A}
+}}
+{{eqn|r = \epsilon_B \left({\operatorname{id}_{B^A} \times \operatorname{id}_A}\right)
+}}
+{{end-eqn}}
+it follows that $\left({\operatorname{id}_B}\right)^A = \operatorname{id}_{B^A}$.
+Next, let $f: B \to C, g: C \to D$ be [[Definition:Composable Morphisms|composable morphisms]].
+Then:
+{{begin-eqn}}
+{{eqn|l = g f \epsilon_B
+     |r = g \epsilon_C \left({f^A \times \operatorname{id}_A}\right)
+     |c = Definition of $f^A$
+}}
+{{eqn|r = \epsilon_D \left({g^A \times \operatorname{id}_A}\right) \left({f^A \times \operatorname{id}_A}\right)
+     |c = Definition of $g^A$
+}}
+{{eqn|r = \epsilon_D \left({g^A f^A \times \operatorname{id}_A}\right)
+}}
+{{end-eqn}}
+Thus, it is seen that $g^A f^A$ is the [[Definition:Morphism (Category Theory)|morphism]] satisfying the [[Definition:UMP|UMP]] for $\left({g f}\right)^A$.
+That is to say:
+:$\left({g f}\right)^A = g^A f^A$
+Hence $\left({-}\right)^A$ is a [[Definition:Covariant Functor|functor]].
+{{qed}}
+{{expand|diagram}}
+\end{proof}<|endoftext|>
+\section{Cosine Exponential Formulation}
+Tags: Cosine Function, Cosine Exponential Formulation
+
+\begin{theorem}
+:$\cos z = \dfrac {\map \exp {i z} + \map \exp {-i z} } 2$
+\end{theorem}
+
+\begin{proof}
+Recall the definition of the [[Definition:Complex Cosine Function|cosine function]]:
+{{begin-eqn}}
+{{eqn | l = \cos z
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}
+      | c = 
+}}
+{{eqn | r = 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \frac {z^6} {6!} + \cdots + \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Recall the definition of the [[Definition:Exponential Function/Complex/Sum of Series|exponential as a power series]]:
+{{begin-eqn}}
+{{eqn | l = \exp z
+      | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}
+      | c = 
+}}
+{{eqn | r = 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Then, starting from the {{RHS}}:
+{{begin-eqn}}
+{{eqn | l = \frac {\exp \paren {i z} + \exp \paren {-i z} } 2
+      | r = \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }
+}}
+{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n + \paren {-i z}^n} {n!} }
+      | c = [[Cosine Function is Absolutely Convergent]]
+}}
+{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} + \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }
+      | c = split into [[Definition:Even Integer|even]] and [[Definition:Odd Integer|odd]] $n$
+}}
+{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!}
+      | c = $\paren {-i z}^{2 n + 1} = -\paren {i z}^{2 n + 1}$
+}}
+{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n} } {\paren {2 n}!}
+      | c = $\left({ -1 }\right)^{2n} = 1$
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} } {\paren {2 n}!}
+      | c = cancel $2$
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}
+      | c = $i^{2 n} = \paren {-1}^n$
+}}
+{{eqn | r = \cos z
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Recall [[Euler's Formula]]:
+:$\exp \paren {i z} = \cos z + i \sin z$
+Then, starting from the {{RHS}}:
+{{begin-eqn}}
+{{eqn | l = \frac {\exp \paren {i z} + \exp \paren {-i z} } 2
+      | r = \frac {\cos z + i \sin z + \cos \paren {-z} + i \sin \paren {-z} } 2
+}}
+{{eqn | r = \frac {\cos z + \cos \paren {-z} } 2
+      | c = [[Sine Function is Odd]]
+}}
+{{eqn | r = \frac {2 \cos z} 2
+      | c = [[Cosine Function is Even]]
+}}
+{{eqn | r = \cos z
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \exp \paren {i z}
+      | r = \cos z + i \sin z
+      | c = [[Euler's Formula]]
+}}
+{{eqn | n = 2
+      | l = \exp \paren {-i z}
+      | r = \cos z - i \sin z
+      | c = [[Euler's Formula/Corollary|Euler's Formula: Corollary]]
+}}
+{{eqn | ll= \leadsto
+      | l = \exp \paren {i z} + \exp \paren {-i z}
+      | r = \paren {\cos z + i \sin z} + \paren {\cos z - i \sin z}
+      | c = $(1) + (2)$
+}}
+{{eqn | r = 2 \cos z
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\exp \paren {i z} + \exp \paren {-i z} } 2
+      | r = \cos z
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reciprocal of i}
+Tags: Complex Numbers, Examples of Reciprocals
+
+\begin{theorem}
+Let $i$ be the [[Definition:Imaginary Unit|imaginary unit]] such that $i^2 = -1$.
+Then $\dfrac 1 i = -i$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = i^2
+      | r = -1
+      | c = {{Defof|Imaginary Unit}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {i^2} i
+      | r = \frac {-1} i
+}}
+{{eqn | ll= \leadsto
+      | l = i
+      | r = \frac {-1} i
+}}
+{{eqn | ll= \leadsto
+      | l = -i
+      | r = \frac 1 i
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Complex Numbers]]
+[[Category:Examples of Reciprocals]]
+03drb2g0e9qp2h96h6eix9w996f6zwc
+\end{proof}<|endoftext|>
+\section{Squeeze Theorem/Functions/Proof 3}
+Tags: Squeeze Theorem
+
+\begin{theorem}
+Let $a$ be a point on an [[Definition:Open Real Interval|open real interval]] $I$.
+Also let $f$, $g$ and $h$ be [[Definition:Real Function|real functions]] defined at all points of $I$ except for possibly at point $a$.
+Suppose that:
+: $\forall x \ne a \in I: g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
+: $\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right) = L$
+Then:
+: $\displaystyle \lim_{x \mathop \to a} \ f \left({x}\right) = L$
+\end{theorem}
+
+\begin{proof}
+By the definition of the [[Definition:Limit of Real Function|limit of a real function]], we have to prove that:
+: $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{f \left({x}\right) - L}\right| < \epsilon}\right)$
+Let $\epsilon \in \R_{>0}$ be given.
+We have:
+: $\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right)$
+Hence by [[Sum Rule for Limits of Functions]]:
+:$\displaystyle \lim_{x \mathop \to a} \ h \left({x}\right) - g \left({x}\right) = 0$
+By the definition of the [[Definition:Limit of Real Function|limit of a real function]]:
+: $(1): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - L}\right| < \epsilon'}\right)$
+: $(2): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{g \left({x}\right) - L}\right| < \epsilon'}\right)$
+: $(3): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - g \left({x}\right)}\right| < \epsilon'}\right)$
+Take $\epsilon' = \dfrac {\epsilon} 3$ in $(1)$, $(2)$, $(3)$.
+Then there exists $\delta_1, \delta_2, \delta_3$ that satisfies $(1)$, $(2)$, $(3)$ with $\epsilon' = \dfrac \epsilon 3$. 
+Take $\delta = \min\{\delta_1, \delta_2, \delta_3\}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left\vert{x - a}\right\vert
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{h \left({x}\right) - L}\right\vert
+      | o = <
+      | r = \frac {\epsilon} 3
+      | c =
+}}
+{{eqn | lo= \land
+      | l = \left\vert{g \left({x}\right) - L}\right\vert
+      | o = <
+      | r = \frac {\epsilon} 3
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert
+      | o = <
+      | r = \frac {\epsilon} 3
+      | c = 
+}}
+{{end-eqn}}
+So, if $\left\vert{x - a}\right\vert<\delta$:
+{{begin-eqn}}
+{{eqn | l = \left\vert{f \left({x}\right) - L}\right\vert
+      | r = \left\vert{f \left({x}\right) - g \left({x}\right) + h \left({x}\right) - L + g \left({x}\right) - h \left({x}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert + \left\vert{h \left({x}\right) - L}\right\vert + \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert + \left\vert{h \left({x}\right) - L}\right\vert + \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac {\epsilon} 3 + \frac {\epsilon} 3 + \frac {\epsilon} 3
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers are not Connected}
+Tags: Connected Spaces, Rational Numbers
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ is not a [[Definition:Connected Topological Space|connected topological space]].
+\end{theorem}
+
+\begin{proof}
+Let $\alpha \in \R$ be an [[Definition:Irrational Number|irrational number]].
+By definition, $\alpha \notin \Q$.
+Consider the [[Definition:Set|sets]]:
+:$S := \Q \cap \left({-\infty \,.\,.\, \alpha}\right)$
+:$T := \Q \cap \left({\alpha \,.\,.\, \infty}\right)$
+Let $x \in S$.
+Let $B_\epsilon \left({x}\right)$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$ in $\Q$.
+Then:
+:$\forall x \in S: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq S$
+by setting $\epsilon = \alpha - x$.
+Similarly:
+:$\forall x \in T: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq T$
+by setting $\epsilon = x - \alpha$.
+Thus $S$ and $T$ are [[Definition:Open Set (Metric Space)|open sets]] of $\Q$.
+Then:
+:$S \cup T = \Q$
+:$S \cap T = \O$
+:$S, T \ne \O$
+So $S$ and $T$ fulfil the conditions for $S \mid T$ to be a [[Definition:Separation (Topology)|separation]] of $\Q$.
+Hence the result by definition of [[Definition:Connected Topological Space|connected topological space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Connected Set is Connected}
+Tags: Set Closures, Connected Sets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H$ be a [[Definition:Connected Set (Topology)|connected set]] of $T$.
+Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $T$. 
+Then $H^-$ is [[Definition:Connected Set (Topology)|connected]] in $T$.
+\end{theorem}
+
+\begin{proof}
+By [[Set is Subset of Itself]], the result follows by setting $K = H^-$ in [[Set between Connected Set and Closure is Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Topologist's Sine Curve is Connected}
+Tags: Topology, Sine Function, Reciprocals, Connected Sets
+
+\begin{theorem}
+Let $G$ be the [[Definition:Graph of Mapping|graph]] of the [[Definition:Real Function|function]] $y = \map \sin {\dfrac 1 x}$ for $x > 0$.
+Let $J$ be the [[Definition:Line Segment|line segment]] joining the points $\tuple {0, -1}$ and $\tuple {0, 1}$ in $\R^2$.
+Then $G \cup J$ is [[Definition:Connected Set (Topology)|connected]].
+\end{theorem}
+
+\begin{proof}
+Since the [[Definition:Open Real Interval|open interval]] $\openint 0 \infty$ is [[Definition:Connected (Topology)|connected]], then so is $G$ by [[Continuous Image of Connected Space is Connected]].
+It is enough, from [[Set between Connected Set and Closure is Connected]], to show that $J \subseteq \map \cl G$.
+Let $p \in J$, say, $\tuple {0, y}$ where $-1 \le y \le 1$.
+We need to show that:
+:$\forall \epsilon > 0: \map {N_\epsilon} p \cap G \ne \O$
+where $\map {N_\epsilon} p$ is the [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhood]] of $p$.
+Let us choose $n \in \N: \dfrac 1 {2 n \pi} < \epsilon$.
+From [[Sine of Half-Integer Multiple of Pi]]:
+:$\map \sin {\dfrac {\paren {4 n + 1} \pi} 2} = 1$
+and:
+:$\map \sin {\dfrac {\paren {4 n + 3} \pi} 2} = -1$
+So by the [[Intermediate Value Theorem]], $\map \sin {\dfrac 1 x}$ takes every value between $-1$ and $1$ in the [[Definition:Closed Real Interval|closed interval]] $\closedint {\dfrac 2 {\paren {4 n + 3} \pi} } {\dfrac 2 {\paren {4 n + 1} \pi} }$.
+In particular, $\map \sin {\dfrac 1 {x_0} } = y$ for some $x_0$ in this interval.
+The distance between the points $\tuple {0, y}$ and $\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } = \tuple {x_0, y}$ is $x_0 < \epsilon$.
+So:
+:$\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } \in \map {N_\epsilon} p \cap G$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separated Sets are Clopen in Union}
+Tags: Separated Sets
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A$ and $B$ be [[Definition:Separated Sets|separated sets]] in $T$.
+Let $H = A \cup B$ be given the [[Definition:Subspace Topology|subspace topology]].
+Then $A$ and $B$ are each [[Definition:Clopen Set|both open and closed]] in $H$.
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]], $A$ and $B$ are [[Definition:Separated Sets|separated]]:
+:$A \cap B^- = A^- \cap B = \O$
+Then:
+{{begin-eqn}}
+{{eqn | l = H \cap B^-
+      | r = \paren {A \cup B} \cap B^-
+}}
+{{eqn | r = \paren {A \cap B^-} \cup \paren {B \cap B^-}
+      | c = [[Intersection Absorbs Union]]
+}}
+{{eqn | r = \O \cup B
+      | c = [[Set is Subset of its Topological Closure]] and [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = B
+      | c = [[Union with Empty Set]]
+}}
+{{end-eqn}}
+Since the [[Definition:Set Intersection|intersection]] of a [[Definition:Closed Set (Topology)|closed set]] with a [[Definition:Subspace Topology|subspace]] is [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Subspace Topology|subspace]], $B$ is [[Definition:Closed Set (Topology)|closed]] in $H$.
+{{explain|Link to the above result}}
+Since $A = H \setminus B$ and $B$ is [[Definition:Closed Set (Topology)|closed]] in $H$, $A$ is [[Definition:Open Set (Topology)|open]] in $H$.
+{{explain|$A {{=}} H \setminus B$: it may be trivial but all statements are linked to or explained.}}
+By the same argument with the roles of $A$ and $B$ reversed, $A$ is [[Definition:Closed Set (Topology)|closed]] in $H$ and $B$ is [[Definition:Open Set (Topology)|open]] in $H$.
+Hence the result.
+{{qed}}
+[[Category:Separated Sets]]
+m1f02h1s1lcidd77jo0wfdkarkqzuml
+\end{proof}<|endoftext|>
+\section{Compatibility of Atlases is Equivalence Relation}
+Tags: Manifolds
+
+\begin{theorem}
+Let $M$ be a [[Definition:Topological Space|topological space]].
+Let $d$ and $k$ be [[Definition:Natural Number|natural numbers]].
+Let $\mathcal A$ denote the [[Definition:Set|set]] of all $d$-[[Definition:Dimension of Atlas|dimensional]] [[Definition:Atlas|atlases]] of [[Definition:Class of Atlas|class]] $\mathcal C^k$ on $M$.
+Define a [[Definition:Relation|relation]] $\sim$ on $\mathcal A$ by putting, for any two [[Definition:Atlas|$\mathcal C^k$-atlases]] $\mathcal F$ and $\mathcal G$:
+:$\mathcal F \sim \mathcal G$ {{iff}} $\mathcal F$ and $\mathcal G$ are $C^k$-[[Definition:Compatible Atlases|compatible]].
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]] on $\mathcal A$.
+\end{theorem}
+
+\begin{proof}
+It is to be shown that $\sim$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+=== Reflexive ===
+Let $\mathcal F \in \mathcal A$ be a [[Definition:Atlas|$C^k$-atlas]].
+That $\mathcal F$ is [[Definition:Compatible Atlases|compatible]] with itself is precisely condition $(2)$ of the definition of [[Definition:Atlas|$C^k$-atlas]].
+{{qed|lemma}}
+=== Symmetric ===
+Let $\mathcal F$ and $\mathcal G$ be [[Definition:Atlas|$C^k$-atlases]] and suppose that $\mathcal F \sim \mathcal G$.
+Let $\left({U, \phi}\right) \in \mathcal F$ and let $\left({V, \psi}\right) \in \mathcal G$ be [[Definition:Chart|charts]].
+Then by hypothesis:
+:$\phi \circ \psi^{-1}: \psi \left({U \cap V}\right) \to \phi \left({U \cap V}\right)$
+is a [[Definition:Differentiability Class|$C^k$ mapping]].
+Because $\phi$ and $\psi$ are [[Definition:Homeomorphism|homeomorphisms]], we have that:
+:$\psi \circ \phi^{-1}: \phi \left({U \cap V}\right) \to \psi \left({U \cap V}\right)$
+is also a [[Definition:Homeomorphism|homeomorphism]], and in particular [[Definition:Continuous Mapping (Topology)|continuous]].
+By the [[Inverse Function Theorem]], $\psi \circ \phi^{-1}$ is also a [[Definition:Differentiability Class|$C^k$ mapping]].
+Since the [[Definition:Chart|charts]] were arbitrary, we conclude that $\mathcal G \sim \mathcal F$.
+{{qed|lemma}}
+=== Transitive ===
+Let $\mathcal F$, $\mathcal G$ and $\mathcal H$ be [[Definition:Atlas|$C^k$-atlases]], and suppose that $\mathcal F \sim \mathcal G$ and $\mathcal G \sim \mathcal H$.
+{{finish|That's messy; probably easiest to use partition of unity compatible with $\mathcal G$, or via common refinement}}
+{{explain|Clarification is needed as to why this result should be categorised in [[:Category:Manifolds]].}}
+[[Category:Manifolds]]
+3mb3byl8pdxtdl55kkak3fht2cf2sff
+\end{proof}<|endoftext|>
+\section{Addition of Real and Imaginary Parts}
+Tags: Complex Addition
+
+\begin{theorem}
+Let $z_0, z_1 \in \C$ be two [[Definition:Complex Number|complex numbers]].
+Then:
+:$\map \Re {z_0 + z_1} = \map \Re {z_0} + \map \Re {z_1}$
+and:
+:$\map \Im {z_0 + z_1} = \map \Im {z_0} + \map \Im {z_1}$
+Here, $\map \Re {z_0}$ denotes the [[Definition:Real Part|real part]] of $z_0$, and $\map \Im {z_0}$ denotes the [[Definition:Imaginary Part|imaginary part]] of $z_0$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = z_0 + z_1
+      | r = \paren {\map \Re {z_0} + i \, \map \Im {z_0} } + \paren {\map \Re {z_1} + i \, \map \Im {z_1} }
+      | c = {{Defof|Complex Number}}
+}}
+{{eqn | r = \paren {\map \Re {z_0} + \map \Re {z_1} } + i \paren {\map \Im {z_0} + \map \Im {z_1} }
+      | c = {{Defof|Complex Addition}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map \Re {z_0 + z_1}
+      | r = \map \Re {z_0} + \map \Re {z_1}
+      | c = {{Defof|Real Part}}
+}}
+{{eqn | l = \map \Im {z_0 + z_1}
+      | r = \map \Im {z_0} + \map \Im {z_1}
+      | c = {{Defof|Imaginary Part}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Complex Addition]]
+ao3h6u72s4zm8bjt7jh7d7re2gufozy
+\end{proof}<|endoftext|>
+\section{Multiplication of Real and Imaginary Parts}
+Tags: Complex Multiplication
+
+\begin{theorem}
+Let $w, z \in \C$ be [[Definition:Complex Number|complex numbers]].
+$(1)$ If $w$ is [[Definition:Wholly Real|wholly real]], then:
+:$\map \Re {w z} = w \, \map \Re z$
+and:
+:$\map \Im {w z} = w \, \map \Im z$
+$(2)$ If $w$ is [[Definition:Wholly Imaginary|wholly imaginary]], then:
+:$\map \Re {w z} = -\map \Im w \, \map \Im z$
+and:
+:$\map \Im {w z} = \map \Im w \, \map \Re z$
+Here, $\map \Re z$ denotes the [[Definition:Real Part|real part]] of $z$, and $\map \Im z$ denotes the [[Definition:Imaginary Part|imaginary part]] of $z$.
+\end{theorem}
+
+\begin{proof}
+Assume that $w$ is [[Definition:Wholly Real|wholly real]].
+Then:
+{{begin-eqn}}
+{{eqn | l = w z
+      | r = \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}
+      | c = {{Defof|Complex Multiplication}}
+}}
+{{eqn | r = w \, \map \Re z + i w \, \map \Im z 
+      | c = as $\map \Re w = w$ and $\map \Im w = 0$
+}}
+{{end-eqn}}
+This equation shows that $\map \Re {w z} = w \, \map \Re z$, and $\map \Im {w z} = w \, \map \Im z$.
+This proves $(1)$.
+Now, assume that $w$ is [[Definition:Wholly Imaginary|wholly imaginary]].
+Then:
+{{begin-eqn}}
+{{eqn | l = w z
+      | r = \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}
+}}
+{{eqn | r = -\map \Im w \, \map \Im z) + i \, \map \Im w \, \map \Re z
+      | c = as $\map \Re w = 0$
+}}
+{{end-eqn}}
+This equation shows that $\map \Re {w z} = -\map \Im w \, \map \Im z$, and $\map \Im {w c} = \map \Im w \, \map \Re z$.
+This proves $(2)$.
+{{qed}}
+[[Category:Complex Multiplication]]
+9aoaf7jbh49ghocleog40chvkz0qv08
+\end{proof}<|endoftext|>
+\section{Rational Numbers are Totally Disconnected}
+Tags: Rational Numbers, Totally Disconnected Spaces, Rational Numbers are Totally Disconnected
+
+\begin{theorem}
+Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]] under the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]] $\tau_d$.
+Then $\struct {\Q, \tau_d}$ is a [[Definition:Totally Disconnected Space|totally disconnected space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \Q$ such that $x \ne y$.
+From [[Between two Rational Numbers exists Irrational Number]], there exists $\alpha \in \R \setminus \Q$ such that $x < \alpha < y$.
+From [[Rational Numbers are not Connected]], it follows that $x$ and $y$ belong to different [[Definition:Component (Topology)|components]] of $\Q$.
+As $x$ and $y$ are arbitrary, it follows that no [[Definition:Rational Number|rational number]] is in the same [[Definition:Component (Topology)|component]] as any other [[Definition:Rational Number|rational number]].
+That is, the [[Definition:Component (Topology)|components]] of $\Q$ are [[Definition:Singleton|singetons]].
+Hence the result, by definition of a [[Definition:Totally Disconnected Space|totally disconnected space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows from:
+: [[Rational Number Space is Totally Separated]]
+: [[Totally Separated Space is Totally Disconnected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Totally Disconnected}
+Tags: Totally Disconnected Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Then $T$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+\end{theorem}
+
+\begin{proof}
+Follows from:
+* [[Discrete Space is Extremally Disconnected]]
+* [[Extremally Disconnected Space is Totally Separated]]
+* [[Totally Separated Space is Totally Disconnected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers are not Discrete Space}
+Tags: Rational Numbers, Totally Disconnected Spaces
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ does not form a [[Definition:Discrete Space|discrete space]].
+\end{theorem}
+
+\begin{proof}
+Follows from:
+:[[Rational Numbers are not Extremally Disconnected]]
+:[[Discrete Space is Extremally Disconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers are not Extremally Disconnected}
+Tags: Rational Numbers, Extremally Disconnected Spaces
+
+\begin{theorem}
+Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]] under the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]] $\tau_d$.
+Then $\struct {\Q, \tau_d}$ does not form an [[Definition:Extremally Disconnected Space|extremally disconnected space]].
+\end{theorem}
+
+\begin{proof}
+The definition of [[Definition:Extremally Disconnected Space|extremally disconnected space]] can be stated as:
+:A [[Definition:Hausdorff Space|$T_2$ (Hausdorff) topological space]] $T = \struct {S, \tau}$ is '''extremally disconnected''' {{iff}} the [[Definition:Closure (Topology)|closure]] of every [[Definition:Open Set (Topology)|open set]] of $T$ is [[Definition:Open Set (Topology)|open]].
+Hence the existence will be demonstrated of an [[Definition:Open Set (Topology)|open set]] in $\struct {\Q, \tau_d}$ whose [[Definition:Closure (Topology)|closure]] is not [[Definition:Open Set (Topology)|open]].
+First we establish that $\struct {\Q, \tau_d}$ is indeed a [[Definition:Hausdorff Space|Hausdorff space]].
+Indeed, we have:
+:[[Rational Numbers form Metric Space]]
+:[[Metric Space is Hausdorff]]
+We have that $\openint 0 1$ is [[Definition:Open Set (Topology)|open]] in $\struct {\R, \tau_d}$.
+Thus $\openint 0 1 \cap \Q$ is [[Definition:Open Set (Topology)|open]] in $\struct {\Q, \tau_d}$.
+We show that $0$ and $1$ are [[Definition:Limit Point of Set|limit points]] of $\openint 0 1 \cap \Q$.
+For any $\epsilon \in \R{>0}$:
+:$\openint {-\epsilon} \epsilon \cap \Q \cap \openint 0 1 = \openint 0 \epsilon \cap \Q$
+From [[Between two Real Numbers exists Rational Number]]:
+:$\openint 0 \epsilon \cap \Q \ne \O$
+Similarly:
+:$\openint {1 - \epsilon} 1 \cap \Q \ne \O$
+Thus $0$ and $1$ are [[Definition:Limit Point of Set|limit points]] of $\openint 0 1 \cap \Q$.
+Hence:
+:$0, 1 \in \paren {\openint 0 1 \cap \Q}'$
+Thus:
+:$\closedint 0 1 \cap \Q \subseteq \paren {\openint 0 1 \cap \Q}^-$
+where $\paren {\openint 0 1 \cap \Q}^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\openint 0 1 \cap \Q$.
+Now we show that for any $x \notin \closedint 0 1$, $x$ is not a [[Definition:Limit Point of Set|limit point]] of $\openint 0 1 \cap \Q$.
+Suppose $x < 0$.
+Then $\dfrac x 2 < 0$.
+Thus:
+:$\openint {x + \dfrac x 2} {x - \dfrac x 2} \cap \Q \cap \openint 0 1 = \O$
+Similarly for $y > 1$:
+:$\openint {y - \dfrac {y - 1} 2} {y + \dfrac {y - 1} 2} \cap \Q \cap \openint 0 1 = \O$
+Thus:
+:$\map \complement {\closedint 0 1 \cap \Q} \subseteq \map \complement {\paren {\openint 0 1 \cap \Q}^-}$
+Hence we have:
+:$\closedint 0 1 \cap \Q = \paren {\openint 0 1 \cap \Q}^-$
+From [[Between two Real Numbers exists Rational Number]]:
+:$\openint {-\epsilon} \epsilon \cap \Q \cap \map \complement {\closedint 0 1} = \openint {-\epsilon} 0 \cap \Q \ne \O$
+A [[Definition:Neighborhood (Topology)|neighborhood]] of $0$ must [[Definition:Set Intersection|intersect]] $\map \complement {\closedint 0 1}$.
+So $\closedint 0 1 \cap \Q$ is not [[Definition:Open Set (Topology)|open]] in $\struct {\Q, \tau_d}$.
+Hence $\openint 0 1$ is that [[Definition:Open Set (Topology)|open set]] in $\struct {\Q, \tau}$ whose [[Definition:Closure (Topology)|closure]] is not [[Definition:Open Set (Topology)|open]] of which we were to demonstrate the existence.
+Thus the result follows from definition of [[Definition:Extremally Disconnected Space|extremally disconnected space]].
+{{qed}}
+[[Category:Rational Numbers]]
+[[Category:Extremally Disconnected Spaces]]
+pj3e3go1uulkw6i5pzacjqxou6r42t6
+\end{proof}<|endoftext|>
+\section{Points in Product Spaces are Near Open Sets}
+Tags: Product Spaces
+
+\begin{theorem}
+Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]], where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $X = \displaystyle \prod_{i \mathop \in I} X_i$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\family {X_i}_{i \mathop \in I}$.
+Let $U$ be [[Definition:Non-Empty Set|nonempty]] [[Definition:Open Set|open subset]] of $X$.
+Let $x$ be a [[Definition:Element|point]] in $X$.
+For each [[Definition:Element|point]] $y$ in $X$, let $\map K y = \set {i \in I : y_i \ne x_i}$.
+Then there exists a [[Definition:Element|point]] $u$ in $U$ such that $\map K u$ is [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+Let $q$ be any [[Definition:Element|point]] in $U$.
+The [[Definition:Topology|topology]] on the [[Definition:Product Space of Topological Spaces|product space]] $X$ is the [[Definition:Tychonoff Topology|Tychonoff topology]] which has the [[Definition:Natural Basis|natural basis]] as a [[Definition:Synthetic Basis|(synthetic) basis]].
+Then:
+:$\exists Q \in \BB : q \in Q \subseteq U$
+where $\BB$ is the [[Definition:Natural Basis|natural basis]] for [[Definition:Tychonoff Topology|Tychonoff topology]] on $X$.
+From [[Natural Basis of Tychonoff Topology]]:
+:$\displaystyle Q = \prod_{i \mathop \in I} Q_i$ where:
+::for all $i \in I : Q_i$ is [[Definition:Open Set (Topology)|open]] in $X_i$
+::$J = \set {i \in I : Q_i \ne X_i}$ is [[Definition:Finite Set|finite]]
+Let $u$ be the [[Definition:Element|point]] defined by:
+:$u_i = \begin {cases} q_i & i \in J \\ x_i & i \notin J \end {cases}$
+By definition of $u$:
+:$\forall i \in I : u_i \in Q_i$
+By definition of the [[Definition:Cartesian Product of Family|Cartesian product]]:
+:$u \in Q \subseteq U$
+By definition of $u$:
+:$\forall i \in I: i \notin J \implies u_i = x_i$
+So by [[Rule of Transposition]]:
+:$\forall i \in I: u_i \ne x_i \implies i \in J$
+Thus:
+:$\map K u \subseteq J$ 
+From [[Subset of Finite Set is Finite]], and because $J$ is [[Definition:Finite Set|finite]], $\map K u$ is also [[Definition:Finite Set|finite]].
+{{qed}}
+[[Category:Product Spaces]]
+2b0la43c8y2wvbel7lntglyjkor0rw5
+\end{proof}<|endoftext|>
+\section{Real and Imaginary Part Projections are Continuous}
+Tags: Complex Numbers, Continuous Functions
+
+\begin{theorem}
+Define the [[Definition:Real-Valued Function|real-valued functions]] $x, y: \C \to \R$ by:
+:$\forall z \in \C: \map x z = \map \Re z$
+:$\forall z \in \C: \map y z = \map \Im z$
+Equip $\R$ with the usual [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]].
+Equip $\C$ with the usual [[Complex Plane is Metric Space|Euclidean metric]].
+{{refactor|Reconsider the above link so that it goes to an actual definition page.}}
+Then both $x$ and $y$ are [[Definition:Continuous on Metric Space|continuous functions]].
+\end{theorem}
+
+\begin{proof}
+Let $z \in \C$, and let $\epsilon \in \R_{>0}$.
+Put $\delta = \epsilon$.
+For all $w \in \C$ with $\cmod {w - z} < \delta$:
+{{begin-eqn}}
+{{eqn | l = \cmod {\map \Re w - \map \Re z} + \cmod {\map \Im w - \map \Im z}
+      | r = \cmod {\map \Re w - \map \Re z} + \cmod {i \paren {\map \Im w - \map \Im z} }
+      | c = as $\cmod i = 1$
+}}
+{{eqn | o = \le
+      | r = \cmod {\map \Re w - \map \Re z + i \map \Im w - i \map \Re z}
+      | c = [[Triangle Inequality for Complex Numbers]]
+}}
+{{eqn | r = \cmod {w - z}
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+{{Mistake|$\cmod {\map \Re w - \map \Re z} + \cmod {i \paren {\map \Im w - \map \Im z} } \le \cmod {\map \Re w - \map \Re z + i \map \Im w - i \map \Im z}$ is wrong -- triangle inequality works the other way}}
+This equation shows that $\cmod {\map x w - \map x z} < \epsilon$, and $\cmod {\map y w - \map y z} < \epsilon$.
+It follows by definition that $x$ and $y$ are both [[Definition:Continuous on Metric Space|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Complex Function is Complex Riemann Integrable}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $f: \closedint a b \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex function]].
+Then $f$ is [[Definition:Complex Riemann Integral|complex Riemann integrable]] over $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+Define the [[Definition:Real Function|real function]] $x: \closedint a b \to \R$ by:
+:$\forall t \in \closedint a b : \map x t = \map \Re {\map f t}$
+Define the real function $y: \closedint a b \to \R$ by:
+:$\forall t \in \closedint a b : \map y t = \map \Im {\map f t}$
+where:
+:$\map \Re {\map f t}$ denotes the [[Definition:Real Part|real part]] of the [[Definition:Complex Number|complex number]] $\map f t$
+:$\map \Im {\map f t}$ denotes the [[Definition:Imaginary Part|imaginary part]] of $\map f t$.
+From [[Real and Imaginary Part Projections are Continuous]], it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are [[Definition:Continuous Mapping (Metric Spaces)|continuous functions]].
+From [[Continuity of Composite Mapping]], it follows that $x$ and $y$ are continuous.
+From [[Continuous Real Function is Darboux Integrable]], it follows that $x$ and $y$ are [[Definition:Darboux Integral|Darboux integrable]] over $\closedint a b$.
+By definition, it follows that $f$ is [[Definition:Complex Riemann Integral|complex Riemann integrable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Subspace of Metric Space is Sequentially Compact in Itself}
+Tags: Compact Spaces, Sequentially Compact Spaces, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $C \subseteq M$ be a [[Definition:Metric Subspace|subspace]] of $M$ such that $C$ is [[Definition:Compact Metric Space|compact]].
+Then $C$ is [[Definition:Sequentially Compact In Itself|sequentially compact in itself]].
+\end{theorem}
+
+\begin{proof}
+Let $C \subseteq M$ be [[Definition:Compact Metric Space|compact]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $C$]].
+Let $S$ be the [[Definition:Range of Sequence|range of $\sequence {x_n}$]].
+Thus $S \subseteq C$ and $S$ may be either [[Definition:Finite Set|finite]] or [[Definition:Infinite Set|infinite]].
+=== Finite Range ===
+Let $S$ be [[Definition:Finite Set|finite]].
+Then at least one $x \in S$ must be repeated infinitely often in $\sequence {x_n}$.
+{{MissingLinks|Infinite Pigeonhole Principle}}
+Its occurrences form a [[Definition:Subsequence|subsequence]] converging to $x$.
+{{qed|lemma}}
+=== Infinite Range ===
+Let $S$ be [[Definition:Infinite Set|infinite]].
+From [[Subsequence of Sequence in Metric Space with Limit]], it is enough to show that $S$ has a [[Definition:Limit Point (Metric Space)|limit point]] in $C$.
+{{AimForCont}} $S$ has no [[Definition:Limit Point (Metric Space)|limit point]] in $C$.
+Then for each $x \in C$, there exists $\epsilon \in \R_{>0}$ such that the [[Definition:Open Ball|open $\epsilon$-ball]] $\map {B_\epsilon} x$ contains no $y \in S: y \ne x$.
+That is:
+:$\map {B_\epsilon} x \cap S = \set x$
+or
+:$\map {B_\epsilon} x \cap S = \O$
+Because $C$ is [[Definition:Compact Metric Space|compact]], the [[Definition:Open Cover|open cover]] $\mathcal B := \set {\map {B_\epsilon} x: x \in C}$ has a [[Definition:Finite Subcover|finite subcover]].
+But each $\map {B_\epsilon} x \in \mathcal B$ contains at most one point of $S$.
+Therefore the [[Definition:Set Union|union]] of any [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\mathcal B$ contains only finitely many points of $S$.
+So no [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\mathcal B$ [[Definition:Open Cover|covers]] $S$, let alone $C$.
+Thus $S$ must have a [[Definition:Limit Point (Metric Space)|limit point]] in $C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Products of Products are Homeomorphic to Collapsed Products}
+Tags: Product Spaces
+
+\begin{theorem}
+Let $I$ be an index set, and for each $i \in I$ let $J_i$ be an index set.
+Suppose that the sets $J_i$ are pairwise disjoint.
+Let $\displaystyle J = \bigcup_{i \mathop \in I} J_i$
+For each $j \in J$, let $X_j$ be a topological space.
+Then $\displaystyle \prod_{j \mathop \in J} X_j$ is homeomorphic to $\displaystyle \prod_{i \mathop \in I} \prod_{j \mathop \in J_i} X_j$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Product Spaces]]
+f2toq82p2egm620n0lem44e8oq1wnqf
+\end{proof}<|endoftext|>
+\section{Sum of Complex Integrals on Adjacent Intervals}
+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 complex function]].
+Let $c \in \left[{a \,.\,.\, b}\right]$.
+Then:
+:$\displaystyle \int_a^c f \left({t}\right) \ \mathrm dt + \int_c^b f \left({t}\right) \ \mathrm dt = \int_a^b f \left({t}\right) \ \mathrm dt$
+\end{theorem}
+
+\begin{proof}
+From [[Continuous Complex Function is Complex Riemann Integrable]], it follows that all three [[Definition:Complex Riemann Integral|complex Riemann integrals]] are well defined.
+From [[Real and Imaginary Part Projections are Continuous]], it follows that $\operatorname{Re}: \C \to \R$ and $\operatorname{Im}: \C \to \R$ are [[Definition:Continuous Mapping (Metric Spaces)|continuous functions]].
+{{explain|Revisit the above link -- see if there is a more appropriate one to use so as not to invoke the concept of metric spaces}}
+From [[Continuity of Composite Mapping]], it follows that $\operatorname{Re} \circ f: \R \to \R$ and $\operatorname{Im} \circ f: \R \to \R$ are [[Definition:Continuous Real Function|continuous real functions]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_a^b f \left({t}\right) \, \mathrm d t
+      | r = \int_a^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_a^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t
+      | c = Definition of [[Definition:Complex Riemann Integral|Complex Riemann Integral]]
+}}
+{{eqn | r = \int_a^c \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \left({\int_a^c \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t}\right)
+      | c = [[Sum of Integrals on Adjacent Intervals for Continuous Functions]]
+}}
+{{eqn | r = \int_a^c \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_a^c \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_c^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t
+}}
+{{eqn | r = \int_a^c f \left({t}\right) \, \mathrm d t + \int_c^b f \left({t}\right) \, \mathrm d t
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Complex Analysis]]
+m40944pv7962zynma06z15sdjap8lli
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Metric Space Continuity at Point}
+Tags: Continuity
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$.
+Let $a \in A_1$ be a point in $A_1$.
+{{TFAE|def = Continuous at Point of Metric Space|continuity at a point}}
+\end{theorem}
+
+\begin{proof}
+=== $\epsilon$-$\delta$ Definition iff Definition by Limits ===
+This is proved in [[Metric Space Continuity by Epsilon-Delta]].
+{{qed|lemma}}
+=== $\epsilon$-$\delta$ Definition iff $\epsilon$-Ball Definition ===
+This is proved in [[Metric Space Continuity by Open Ball]].
+{{qed}}
+=== $\epsilon$-Ball Definition iff Definition by Neighborhood ===
+This is proved in [[Metric Space Continuity by Neighborhood]].
+{{qed}}
+[[Category:Continuity]]
+nq08azkwsibqfx0z1ch9d3jk75i17qu
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Limit of Function in Metric Space}
+Tags: Connected Spaces
+
+\begin{theorem}
+{{TFAE|def = Limit of Function (Metric Space)|view = Limit of Function|context = Metric Space|contextview = Metric Spaces}}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $c$ be a [[Definition:Limit Point (Metric Space)|limit point]] of $M_1$.
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$ defined everywhere on $A_1$ ''except possibly'' at $c$.
+Let $L \in M_2$.
+\end{theorem}
+
+\begin{proof}
+=== $\epsilon$-$\delta$ Condition implies $\epsilon$-Ball Condition ===
+Suppose that $f$ satisfies the [[Definition:Limit of Function (Metric Space)/Epsilon-Delta Condition|$\epsilon$-$\delta$ condition]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < d_1 \left({x, c}\right) < \delta \implies d_2 \left({f \left({x}\right), L}\right) < \epsilon$
+Let $y \in f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right)$.
+By definition of [[Definition:Open Ball|open $\epsilon$-ball]], this means:
+:$\exists x \in A_1: 0 < d_1 \left({x, c}\right) < \delta: y = f \left({x}\right)$
+[[Definition:By Hypothesis|By hypothesis]], it follows that $d_2 \left({y, L}\right) < \epsilon$
+That is, that $y \in B_\epsilon \left({L; d_2}\right)$.
+By definition of [[Definition:Subset|subset]]:
+:$f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$
+Thus it follows that $f$ satisfies the [[Definition:Limit of Function (Metric Space)/Epsilon-Ball Condition|$\epsilon$-ball condition]].
+{{qed|lemma}}
+=== $\epsilon$-Ball Condition implies $\epsilon$-$\delta$ Condition ===
+Suppose that $f$ satisfies the [[Definition:Limit of Function (Metric Space)/Epsilon-Ball Condition|$\epsilon$-ball condition]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$.
+Let $0 < d_1 \left({x, c}\right) < \delta$.
+Then by definition of [[Definition:Open Ball|open $\epsilon$-ball]], this means:
+:$x \in B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}$
+[[Definition:By Hypothesis|By hypothesis]], it follows that:
+:$f \left({x}\right) \in B_\epsilon \left({L; d_2}\right)$
+Thus by definition of [[Definition:Open Ball|open $\epsilon$-ball]], this means:
+:$d_2 \left({f \left({x}\right), L}\right) < \epsilon$
+That is, $f$ satisfies the [[Definition:Limit of Function (Metric Space)/Epsilon-Delta Condition|$\epsilon$-$\delta$ condition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering Principle}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then there exists a [[Definition:Total Ordering|total ordering]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
+From its definition, $\struct {S, \circ, \preceq}$ is [[Definition:Well-Ordered Set|well-ordered]] by $\preceq$.
+The result follows.
+As $\N_{\ne 0} = \N \setminus \set 0$, by [[Set Difference is Subset]] $\N_{\ne 0} \subseteq \N$.
+As $\N$ is [[Definition:Well-Ordered Set|well-ordered]], by definition, every subset of $\N$ has a [[Definition:Smallest Element|smallest element]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|set of natural numbers]] $\N$.
+We take as [[Definition:Axiom|axiomatic]] that $\N$ is itself a [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers]] $\R$.
+Thus $S \subseteq \R$.
+By definition:
+:$\forall n \in \N: n \ge 0$
+and so:
+:$\forall n \in S: n \ge 0$
+Hence $0$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] of $S$.
+This establishes the fact that $S$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]].
+By the [[Continuum Property]], we have that $S$ admits an [[Definition:Infimum of Subset of Real Numbers|infimum]].
+Hence let $b = \inf S$.
+Because $b$ is the [[Definition:Infimum of Subset of Real Numbers|infimum]] of $S$, it follows that $b + 1$ is not a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] of $S$.
+So, for some $n \in S$:
+:$n < b + 1$
+{{AimForCont}} $n$ is not the [[Definition:Smallest Element|smallest element]] of $S$.
+Then there exists $m \in S$ such that:
+:$b \le m < n < b + 1$
+from which it follows that:
+:$0 < n - m < 1$
+But there exist no [[Definition:Natural Number|natural numbers]] $k$ such that $0 < k < 1$.
+Hence $n = b$ is the [[Definition:Smallest Element|smallest element]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number Smaller than Lebesgue Number is also Lebesgue Number}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Lebesgue Number|Lebesgue number]] for $M$.
+Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.
+Then $\epsilon'$ is also a [[Definition:Lebesgue Number|Lebesgue number]] for $M$.
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]], let $\epsilon \in \R_{>0}$ be a [[Definition:Lebesgue Number|Lebesgue number]] for $M$.
+Then by definition:
+:$\forall x \in A: \exists U \left({x}\right) \in \mathcal U: B_\epsilon \left({x}\right) \subseteq U \left({x}\right)$
+where $B_\epsilon \left({x}\right)$ is the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] of $x$ in $M$.
+Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.
+Let $y \in B_{\epsilon'} \left({x}\right)$.
+{{begin-eqn}}
+{{eqn | l = d \left({y, x}\right)
+      | o = <
+      | r = \epsilon'
+      | c = {{Defof|Open Ball of Metric Space}}
+}}
+{{eqn | ll= \implies
+      | l = d \left({y, x}\right)
+      | o = <
+      | r = \epsilon
+      | c = as $\epsilon' < \epsilon$
+}}
+{{eqn | ll= \implies
+      | l = y
+      | o = \in
+      | r = B_\epsilon \left({x}\right)
+      | c = {{Defof|Open Ball of Metric Space}}
+}}
+{{eqn | ll= \implies
+      | l = y
+      | o = \in
+      | r = U \left({x}\right)
+      | c = {{Defof|Subset}}: $B_\epsilon \left({x}\right) \subseteq U \left({x}\right)$
+}}
+{{end-eqn}}
+That is:
+:$B_{\epsilon'} \left({x}\right) \subseteq U \left({x}\right)$
+The result follows by definition of [[Definition:Lebesgue Number|Lebesgue number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Cover may not have Lebesgue Number}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\mathcal C$ be an [[Definition:Open Cover|open cover]] of $M$.
+Then it may not necessarily be the case that $\mathcal C$ has a [[Definition:Lebesgue Number|Lebesgue number]].
+\end{theorem}
+
+\begin{proof}
+Let $M := \left({0 \,.\,.\, 1}\right) \subseteq \R$ under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]].
+Let $\mathcal C := \left\{{\left({\dfrac 1 n \,.\,.\, 1}\right): n \ge 2}\right\}$.
+For any $\epsilon \in \R_{>0}$, take $n > \dfrac 1 \epsilon$ and $x = \dfrac 1 n$.
+There is no $\left({\dfrac 1 m \,.\,.\, 1}\right)$ such that $B_\epsilon \left({x}\right) \subseteq \left({\dfrac 1 m \,.\,.\, 1}\right)$, since $B_\epsilon \left({x}\right) = B_\epsilon \left({\dfrac 1 n}\right) = \left({0 \,.\,.\, \dfrac 1 n + \epsilon}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Burnside's Theorem}
+Tags: Solvable Groups, Finite Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let the [[Definition:Order of Structure|order]] of $G$ be $p^m q^n$ where:
+: $p, q$ are [[Definition:Prime Number|prime]]
+: $m, n \in \N$
+Then $G$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|William Burnside|cat = Burnside}}
+[[Category:Solvable Groups]]
+[[Category:Finite Groups]]
+0u0lkofp75qtf49p8lyq9sbe9s0g5fn
+\end{proof}<|endoftext|>
+\section{Constant Function is Uniformly Continuous/Metric Space}
+Tags: Constant Mappings, Metric Spaces, Uniformly Continuous Mappings
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] from $A_1$ to $A_2$:
+:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
+That is, every [[Definition:Element|point]] in $A_1$ maps to the same [[Definition:Element|point]] $c$ in $A_2$.
+Then $f_c$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+\end{theorem}
+
+\begin{proof}
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] between two [[Definition:Metric Space|metric spaces]] $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
+Let $\epsilon \in \R: \epsilon > 0$.
+Pick any $\delta \in \R$ such that $\delta > 0$.
+Let $x, y \in A_1$ such that $d_1 \left({x, y}\right) < \delta$.
+Now we have:
+:$f_c \left({x}\right) = c = f_c \left({y}\right)$
+Hence:
+:$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$
+By definition of a [[Definition:Metric|metric]]:
+:$d_2 \left({c, c}\right) = 0 < \epsilon$.
+Thus we have that:
+:$\forall \epsilon > 0: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$
+Hence by definition of [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniform continuity on a metric space]], $f_c$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+{{qed}}
+[[Category:Constant Mappings]]
+[[Category:Metric Spaces]]
+[[Category:Uniformly Continuous Mappings]]
+dwll4poxn4ev0ljeemh7zgf578t7aky
+\end{proof}<|endoftext|>
+\section{Constant Function is Uniformly Continuous/Real Function}
+Tags: Constant Mappings, Real Analysis, Uniformly Continuous Real Functions
+
+\begin{theorem}
+Let $f_c: \R \to \R$ be the [[Definition:Constant Mapping|constant mapping]]:
+:$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$
+Then $f_c$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous on $\R$]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Constant Function is Uniformly Continuous/Metric Space|Constant Function is Uniformly Continuous: Metric Space]]
+:[[Real Number Line is Metric Space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Complex Integrals}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $f, g: \closedint a b \to \C$ be [[Definition:Complex Riemann Integral|complex Riemann integrable]] [[Definition:Real Function|functions]] over $\closedint a b$.
+Let $\lambda, \mu \in \C$ be [[Definition:Complex Number|complex]] [[Definition:Constant|constants]].
+Then:
+:$\displaystyle \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$
+\end{theorem}
+
+\begin{proof}
+First, we prove the result for addition only without multiplying by $\lambda, \mu$:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \map f t + \map g t \rd t
+      | r = \int_a^b \map \Re {\map f t + \map g t} \rd t + i \int_a^b \map \Im {\map f t + \map g t} \rd t
+      | c = {{Defof|Complex Riemann Integral}}
+}}
+{{eqn | r = \int_a^b \map \Re {\map f t} + \map \Re {\map g t} \rd t
+}}
+{{eqn | o =
+      | ro= +
+      | r = i \int_a^b \map \Im {\map f t} + \map \Im {\map g t} \rd t
+      | c = [[Addition of Real and Imaginary Parts]]
+}}
+{{eqn | r = \int_a^b \map \Re {\map f t} \rd t + \int_a^b \map \Re {\map g t} \rd t
+}} 
+{{eqn | o =
+      | ro= +
+      | r = i \paren {\int_a^b \map \Im {\map f t} \rd t + \int_a^b \map \Im {\map g t} \rd t}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int_a^b \map f t \rd t + \int_a^b \map g t \rd t
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Next, we show the result for only one complex integral multiplied by a constant $\lambda$.
+By [[Definition:Complex Number|definition of complex number]], we can find $\lambda_x, \lambda_y \in \R$ so $\lambda = \lambda_x + i \lambda_y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \lambda \map f t \rd t
+      | r = \int_a^b \map \Re {\lambda \map f t} \rd t + i \int_a^b \map \Im {\lambda \map f t} \rd t
+      | c = {{Defof|Complex Riemann Integral}}
+}}
+{{eqn | r = \int_a^b \map \Re {\paren {\lambda_x + i \lambda_y} \map f t} \rd t
+}}
+{{eqn | o =
+      | ro= +
+      | r = i \int_a^b \map \Im {\paren {\lambda_x + i \lambda_y} \map f t} \rd t
+}}
+{{eqn | r = \int_a^b \map \Re {\lambda_x \map f t} + \map \Re {i \lambda_y \map f t} \rd t
+}}
+{{eqn | o =
+      | ro= +
+      | r = i \int_a^b \map \Im {\lambda_x \map f t} + \map \Im {i \lambda_y \map f t} \rd t
+      | c = [[Addition of Real and Imaginary Parts]]
+}}
+{{eqn | r = \int_a^b \lambda_x \map \Re {\map f t} - \lambda_y \map \Im {\map f t} \rd t
+}}
+{{eqn | o =
+      | ro= +
+      | r = i \int_a^b \lambda_x \map \Im {\map f t} + \lambda_y \map \Re {\map f t} \rd t
+      | c = [[Multiplication of Real and Imaginary Parts]]
+}}
+{{eqn | r = \lambda_x \int_a^b \map \Re {\map f t} \rd t - \lambda_y \int_a^b \map \Im {\map f t} \rd t
+}}
+{{eqn | o =
+      | ro= +
+      | r = i \lambda_x \int_a^b \map \Im {\map f t} \rd t + i \lambda_y \int_a^b \map \Re {\map f t} \rd t
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \lambda_x \paren {\int_a^b \map \Re {\map f t} \rd t + i \int_a^b \map \Im {\map f t} \rd t}
+}}
+{{eqn | o =
+      | ro= +
+      | r = i \lambda_y \paren {i \int_a^b \map \Im {\map f t} \rd t + \int_a^b \map \Re {\map f t} \rd t}
+}}
+{{eqn | r = \lambda_x \int_a^b \map f t \rd t + i \lambda_y \int_a^b \map f t \rd t
+      | c = {{Defof|Complex Riemann Integral}}
+}}
+{{eqn | r = \lambda \int_a^b \map f t \rd t
+}}
+{{end-eqn}}
+{{qed|lemma}}
+It follows from the results above that:
+{{begin-eqn}}
+{{eqn | l = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t
+      | r = \int_a^b \lambda \map f t \rd t + \int_a^b \mu \map g t \rd t
+}}
+{{eqn | r = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus Larger than Real Part and Imaginary Part}
+Tags: Complex Modulus
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\operatorname{Re} \left({z}\right)$ denote the [[Definition:Real Part|real part]] of $z$, and $\operatorname{Im} \left({z}\right) $ the [[Definition:Imaginary Part|imaginary part]] of $z$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Pointwise Difference of Simple Functions is Simple Function}
+Tags: Simple Functions
+
+\begin{theorem}
+Let $\left({X, \Sigma}\right)$ be a [[Definition:Measurable Space|measurable space]].
+Let $f,g : X \to \R$ be [[Definition:Simple Function|simple functions]].
+Then $f - g: X \to \R, \left({f - g}\right) \left({x}\right) := f \left({x}\right) - g \left({x}\right)$ is also a [[Definition:Simple Function|simple function]].
+\end{theorem}
+
+\begin{proof}
+By [[Scalar Multiple of Simple Function is Simple Function]], $- g = -1 \cdot g$ is a [[Definition:Simple Function|simple function]].
+By [[Pointwise Sum of Simple Functions is Simple Function]], so is $f - g$.
+{{qed}}
+[[Category:Simple Functions]]
+mpoikn4qgke6a5sf5msokunhl9flqe3
+\end{proof}<|endoftext|>
+\section{Modulus and Argument of Complex Exponential}
+Tags: Complex Analysis, Complex Modulus
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\hointr a {a + 2 \pi}$ be a [[Definition:Half-Open Real Interval|half open interval]] of [[Definition:Length of Real Interval|length]] $2 \pi$.
+Let $r \in \hointr 0 {+\infty}$ and $\theta \in \hointr a {a + 2 \pi}$.
+Then:
+:$r = \cmod z$ and $\theta = \map \arg z$
+{{iff}}:
+:$z = r e^{i \theta}$
+where:
+:$\cmod z$ denotes the [[Definition:Modulus of Complex Number|modulus]] of $z$
+:$\map \arg z$ denotes the [[Definition:Argument of Complex Number|argument]] of $z$
+:$x \mapsto e^x$ is the [[Definition:Complex Exponential Function|complex exponential function]].
+If $z = 0$ or $r = 0$, then $\theta$ may be any number in $\hointr a {a + 2 \pi}$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary condition ===
+Let $r = \cmod z$.
+If $z = 0$, we have:
+:$z = 0e^{i \theta} = re^{i \theta}$
+Suppose $z \ne 0$ and $\theta = \map \arg z$.
+By definition of [[Definition:Argument of Complex Number|argument]], the following two equations hold:
+:$(1): \quad \dfrac {\map \Re z} r = \cos \theta$
+:$(2): \quad \dfrac {\map \Im z} r = \sin \theta$
+where:
+:$\map \Re z$ denotes the [[Definition:Real Part|real part]] of $z$
+:$\map \Im z$ denotes the [[Definition:Imaginary Part|imaginary part]] of $z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \map \Re z + i \map \Im z
+      | c = {{Defof|Complex Number}}
+}}
+{{eqn | r = r \cos \theta + i r \sin \theta
+      | c = from $(1)$ and $(2)$
+}}
+{{eqn | r = r \paren {\cos \theta + i \sin \theta}
+}}
+{{eqn | r = re^{i \theta}
+      | c = [[Euler's Formula]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient condition ===
+Let $z = re^{i \theta}$.
+From the equations above, we find:
+:$\map \Re {re^{i \theta} } = r \cos \theta$
+:$\map \Im {re^{i \theta} } = r \sin \theta$
+Then:
+{{begin-eqn}}
+{{eqn | l = \cmod {r e^{i \theta} }
+      | r = \sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2}
+      | c = {{Defof|Modulus of Complex Number}}
+}}
+{{eqn | r = r \sqrt {\cos^2 \theta + \sin^2 \theta}
+      | c = as $r \ge 0$
+}}
+{{eqn | r = r  
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+If $r \ne 0$, we find $\map \arg {r e^{i \theta} }$ by solving the two equations by definition of [[Definition:Argument of Complex Number|argument]]:
+:$(1): \quad \dfrac {r \cos \theta} r = \map \cos {\map \arg {r e^{i \theta} } }$
+:$(2): \quad \dfrac {r \sin \theta} r = \map \sin {\map \arg {r e^{i \theta} } }$
+We find:
+:$\map \arg {r e^{i \theta} } = \theta$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bases of Vector Space have Equal Cardinality}
+Tags: Bases of Vector Spaces
+
+\begin{theorem}
+Let $k$ be a [[Definition:Division Ring|division ring]].
+Let $V$ be a [[Definition:Vector Space|vector space]] over $k$.
+Let $X$ and $Y$ be [[Definition:Basis of Vector Space|bases]] of $V$.
+Then $X$ and $Y$ are [[Definition:Equinumerous|equinumerous]].
+\end{theorem}
+
+\begin{proof}
+We will first prove that there is an [[Definition:Injection|injection]] from $X$ to $Y$.
+Let $x \in X$.
+By [[Expression of Vector as Linear Combination from Basis is Unique/General Result|Expression of Vector as Linear Combination from Basis is Unique: General Result]], there is a unique [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $C_x$ of $R \times Y$ such that:
+:$\displaystyle x = \sum_{\tuple {r, v} \mathop \in C_x} r \cdot v$ and
+:$\forall \tuple {r, v} \in C_x: r \ne 0_R$
+Define $\Phi: X \to \powerset Y$ by:
+:$\map \Phi x := \Img {C_x}$
+We next show that $\sequence {\map \Phi x}_{x \mathop \in X}$ satisfies the '''marriage condition'''. 
+That is, for every [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $F$ of $X$:
+:$\displaystyle \card F \le \card {\bigcup \map \Phi F}$
+Since $X$ is a [[Definition:Basis of Vector Space|basis]], it is [[Definition:Linearly Independent Set|linearly independent]].
+By [[Subset of Linearly Independent Set is Linearly Independent]], $F$ is also [[Definition:Linearly Independent Set|linearly independent]].
+By the definition of $\Phi$:
+:$\displaystyle F \subseteq \operatorname {span} \bigcup \map \Phi F$.
+By [[Finite Union of Finite Sets is Finite]], $\displaystyle \bigcup \map \Phi F$ is [[Definition:Finite Set|finite]].
+Thus by [[Size of Linearly Independent Subset is at Most Size of Finite Generator]]:
+:$\displaystyle \size F \le \size {\bigcup \map \Phi F}$
+By [[Hall's Marriage Theorem/General Set|Hall's Marriage Theorem]], there is an [[Definition:Injection|injection]] from $X$ into $Y$.
+Precisely the same argument with $X$ and $Y$ interchanged shows that there is an injection from $Y$ into $X$ as well.
+Thus by the [[Cantor-Bernstein-Schröder Theorem]], $X$ and $Y$ are [[Definition:Equinumerous|equinumerous]].
+{{qed}}
+{{BPI|Hall's Marriage Theorem/General Set|3}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Symmetric Difference/(1) iff (2)}
+Tags: Equivalence of Definitions of Symmetric Difference
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+{{TFAENocat|def = Symmetric Difference|view = symmetric difference $S * T$ between $S$ and $T$}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = S * T
+      | r = \paren {S \setminus T} \cup \paren {T \setminus S}
+      | c = {{Defof|Symmetric Difference|index = 1}}
+}}
+{{eqn | r = \paren {\paren {S \cup T} \setminus T} \cup \paren {\paren {S \cup T} \setminus S}
+      | c = [[Set Difference with Union is Set Difference]]
+}}
+{{eqn | r = \paren {S \cup T} \setminus \paren {T \cap S}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \paren {S \cup T} \setminus \paren {S \cap T}
+      | c = [[Intersection is Commutative]]
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Symmetric Difference/(1) iff (3)}
+Tags: Equivalence of Definitions of Symmetric Difference
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+{{TFAENocat|def = Symmetric Difference|view = symmetric difference $S * T$ between $S$ and $T$}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = S * T
+      | r = \paren {S \setminus T} \cup \paren {T \setminus S}
+      | c = {{Defof|Symmetric Difference|index = 1}}
+}}
+{{eqn | r = \paren {S \cap \overline T} \cup \paren {\overline S \cap T}
+      | c = [[Set Difference as Intersection with Complement]]
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Symmetric Difference/(2) iff (4)}
+Tags: Equivalence of Definitions of Symmetric Difference
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+{{TFAENocat|def = Symmetric Difference|view = symmetric difference $S * T$ between $S$ and $T$}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = S * T
+      | r = \paren {S \cup T} \setminus \paren {S \cap T}
+      | c = {{Defof|Symmetric Difference|index = 2}}
+}}
+{{eqn | r = \paren {S \cup T} \cap \paren {\overline {S \cap T} }
+      | c = [[Set Difference as Intersection with Complement]]
+}}
+{{eqn | r = \paren {S \cup T} \cap \paren {\overline S \cup \overline T}
+      | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Subset of Countable Set is Countable}
+Tags: Countable Sets, Subsets
+
+\begin{theorem}
+A [[Definition:Subset|subset]] of a [[Definition:Countable Set|countable set]] is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Countable Set|countable set]].
+Let $T \subseteq S$.
+By definition, there exists an [[Definition:Injection|injection]] $f: S \to \N$.
+Let $i: T \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]].
+We have that [[Inclusion Mapping is Injection|$i$ is an injection]].
+Because the [[Composite of Injections is Injection|composite of injections is an injection]], it follows that $f \circ i: T \to \N$ is an [[Definition:Injection|injection]].
+Hence, $T$ is [[Definition:Countable Set|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine Exponential Formulation}
+Tags: Sine Function, Sine Exponential Formulation
+
+\begin{theorem}
+:$\sin z = \dfrac {\map \exp {i z} - \map \exp {-i z} } {2 i}$
+\end{theorem}
+
+\begin{proof}
+Recall the definition of the [[Definition:Complex Sine Function|sine function]]:
+{{begin-eqn}}
+{{eqn | l = \sin z
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = 
+}}
+{{eqn | r = z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \cdots + \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Recall the definition of the [[Definition:Exponential Function/Complex/Sum of Series|exponential as a power series]]:
+{{begin-eqn}}
+{{eqn | l = \exp z
+      | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}
+      | c = 
+}}
+{{eqn | r = 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Then, starting from the {{RHS}}:
+{{begin-eqn}}
+{{eqn | l = \frac {\exp \paren {i z} - \exp \paren {-i x} } {2 i}
+      | r = \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n - \paren {-i z}^n} {n!} }
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} - \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }
+      | c = split into [[Definition:Even Integer|even]] and [[Definition:Odd Integer|odd]] $n$
+}}
+{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = as $\paren {-i z}^{2 n} = \paren {i z}^{2 n}$
+}}
+{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = as $\paren {-1}^{2 n + 1} = -1$
+}}
+{{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = cancel $2$
+}}
+{{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n z^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = as $i^{2 n + 1} = i \paren {-1})^n $
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1!} }
+      | c = cancel $i$
+}}
+{{eqn | r = \sin z
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Recall [[Euler's Formula]]:
+:$\exp \paren {i z} = \cos z + i \sin z$
+Then, starting from the {{RHS}}:
+{{begin-eqn}}
+{{eqn | l = \frac {\exp \paren {i z} - \exp \paren {-i z} }{2 i}
+      | r = \frac {\paren {\cos z + i \sin z} - \paren {\cos \paren {-z} + i \sin \paren {-z} } } {2 i}
+}}
+{{eqn | r = \frac {\paren {\cos z + i \sin z - \cos z - i \sin \paren {-z} } } {2 i}
+      | c = [[Cosine Function is Even]]
+}}
+{{eqn | r = \frac {i \sin z - i \sin \paren {-z} } {2 i}
+      | c = 
+}}
+{{eqn | r = \frac {i \sin z - i \paren {-\sin \paren {-z} } } {2 i}
+      | c = [[Sine Function is Odd]]
+}}
+{{eqn | r = \frac {2 i \sin z} {2 i}
+      | c = 
+}}
+{{eqn | r = \sin z
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \exp \paren {i z}
+      | r = \cos z + i \sin z
+      | c = [[Euler's Formula]]
+}}
+{{eqn | n = 2
+      | l = \exp \paren {-i z}
+      | r = \cos z - i \sin z
+      | c = [[Euler's Formula/Corollary|Euler's Formula: Corollary]]
+}}
+{{eqn | ll= \leadsto
+      | l = \exp \paren {i z} - \exp \paren {-i z}
+      | r = \paren {\cos z + i \sin z} - \paren {\cos z - i \sin z}
+      | c = $(1) - (2)$
+}}
+{{eqn | r = 2 i \sin z
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\exp \paren {i z} - \exp \paren {-i z} } {2 i}
+      | r = \sin z
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine and Cosine Exponential Formulation}
+Tags: Sine Function, Cosine Function, Exponential Function
+
+\begin{theorem}
+For any [[Definition:Complex Number|complex number]] $x$:
+\end{theorem}<|endoftext|>
+\section{Sine of Sum}
+Tags: Sine Function, Sine of Sum, Compound Angle Formulas
+
+\begin{theorem}
+:$\map \sin {a + b} = \sin a \cos b + \cos a \sin b$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {a + b} + i \, \map \sin {a + b}
+      | r = e^{i \paren {a + b} }
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = e^{i a} e^{i b}
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \paren {\cos a + i \sin a} \paren {\cos b + i \sin b}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \paren {\cos a \cos b - \sin a \sin b} + i \paren {\sin a \cos b + \cos a \sin b}
+      | c = [[Complex Numbers form Field]]
+}}
+{{end-eqn}}
+By equating the [[Definition:Imaginary Part|imaginary]] parts, the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}} 
+{{eqn | l = \sin a \cos b + \cos a \sin b
+      | r = \paren {\frac {e^{i a} - e^{-i a} }{2 i} } \cos b + \cos a \paren {\frac {e^{i b} - e^{-i b} }{2 i} }
+      | c = [[Sine Exponential Formulation]]
+}}
+{{eqn | r = \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} + e^{-i b} } 2}
+          + \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} - e^{-i b} } {2 i} }
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} - e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}
+      | c = expanding
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} + e^{-i a} e^{i b} - e^{-i a} e^{-i b} } {4 i}
+}}
+{{eqn | r = \frac {e^{i a} e^{i b} - e^{-i a} e^{-i b} } {2 i}
+      | c = sinplifying
+}}
+{{eqn | r = \frac {e^{i \paren {a + b} } - e^{-i \paren {a + b} } } {2 i}
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \map \sin {a + b}
+      | c = [[Sine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin \left({a + b}\right)
+      | r = \cos \left({\frac \pi 2 - \left({a + b}\right)}\right)
+      | c = [[Cosine of Complement equals Sine]]
+}}
+{{eqn | r = \cos \left({\left({\frac \pi 2 - a}\right) - b}\right)
+}}
+{{eqn | r = \cos \left({\frac \pi 2 - a}\right) \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b
+      | c = [[Cosine of Difference]]
+}}
+{{eqn | r = \sin a \cos b + \sin \left({\frac \pi 2 - a}\right) \sin b
+      | c = [[Cosine of Complement equals Sine]]
+}}
+{{eqn | r = \sin a \cos b + \cos a \sin b
+      | c = [[Sine of Complement equals Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Recall the analytic definitions of [[Definition:Complex Sine Function|sine]] and [[Definition:Complex Cosine Function|cosine]]:
+:$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1}} {\left({2 n + 1}\right)!}$
+:$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n}} {\left({2 n}\right)!}$
+Let:
+{{begin-eqn}}
+{{eqn | l = g \left({a}\right)
+      | r = \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b
+}}
+{{eqn | l = h \left({a}\right)
+      | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b
+}}
+{{end-eqn}}
+Let us [[Definition:Differentiation With Respect To|differentiate these with respect to]] $a$, keeping $b$ constant.
+Then from [[Derivative of Sine Function]] and [[Derivative of Cosine Function]], we have:
+{{begin-eqn}}
+{{eqn | l = g' \left({a}\right)
+      | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right)
+}}
+{{eqn | l = h' \left({a}\right)
+      | r = - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right)
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = D_a \left({\left({g \left({a}\right)}\right)^2 + \left({h \left({a}\right)}\right)^2}\right)
+      | r = 2 g \left({a}\right) g' \left({a}\right) + 2 h \left({a}\right) h' \left({a}\right)
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Thus from [[Derivative of Constant]]:
+:$\forall a \in \R: g \left({a}\right)^2 + h \left({a}\right)^2 = c$
+In particular, it is true for $a = 0$, and so:
+:$g \left({0}\right)^2 + h \left({0}\right)^2 = 0$
+So:
+:$g \left({a}\right)^2 + h \left({a}\right)^2 = 0$
+But from [[Square of Real Number is Non-Negative]]:
+:$g \left({a}\right)^2 \ge 0$ and $h \left({a}\right)^2 \ge 0$
+So it follows that:
+:$g \left({a}\right) = 0$
+and:
+:$h \left({a}\right) = 0$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = 2 \sin a \cos b
+      | r = \sin \paren {a + b} + \sin \paren {a - b}
+      | c = [[Simpson's Formulas/Sine by Cosine/Proof 2|Simpson's Formula for Sine by Cosine: Proof 2]]
+}}
+{{eqn | n = 2
+      | l = 2 \cos a \sin b
+      | r = \sin \paren {a + b} - \sin \paren {a - b}
+      | c = [[Simpson's Formulas/Cosine by Sine/Proof 2|Simpson's Formula for Cosine by Sine: Proof 2]]
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \sin \paren {a + b}
+      | r = 2 \sin a \cos b + 2 \cos a \sin b
+      | c = $(1) + (2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \sin \paren {a + b}
+      | r = \sin a \cos b + \cos a \sin b
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hyperbolic Sine Function is Odd}
+Tags: Hyperbolic Sine Function, Hyperbolic Sine Function is Odd, Examples of Odd Functions
+
+\begin{theorem}
+:$\map \sinh {-x} = -\sinh x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sinh {-x}
+      | r = \frac {e^{-x} - e^{-\paren {-x} } } 2
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \frac {e^{-x} - e^x} 2
+}}
+{{eqn | r = -\frac {e^x - e^{-x} } 2
+}}
+{{eqn | r = -\sinh x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \map \sinh {-x}
+	| r = -i \, \map \sin {-i x}
+	| c = [[Hyperbolic Sine in terms of Sine]]
+}}
+{{eqn	| r = i \, \map \sin {i x}
+	| c = [[Sine Function is Odd]]
+}}
+{{eqn	| r = -\sinh x
+	| c = [[Hyperbolic Sine in terms of Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential of Sum/Complex Numbers}
+Tags: Exponential of Sum
+
+\begin{theorem}
+Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
+Let $\exp z$ be the [[Definition:Complex Exponential Function|exponential of $z$]].
+Then:
+:$\map \exp {z_1 + z_2} = \paren {\exp z_1} \paren {\exp z_2}$
+\end{theorem}
+
+\begin{proof}
+This proof is based on the [[Definition:Exponential Function/Complex/Differential Equation|definition of the complex exponential]] as the [[Definition:Unique|unique]] [[Definition:Solution of Differential Equation|solution]] of the [[Definition:Differential Equation|differential equation]]:
+:$\dfrac \d {\d z} \exp = \exp$
+which satisfies the [[Definition:Initial Condition|initial condition]] $\map \exp 0 = 1$.
+Define the [[Definition:Complex Function|complex function]]  $f: \C \to \C$ by:
+:$\map f z = \map \exp z \, \map \exp {z_1 + z_2 - z}$
+Then find its [[Definition:Derivative of Complex Function|derivative]]:
+{{begin-eqn}}
+{{eqn | l = D_z \, \map f z
+      | r = \paren {D_z \, \map \exp z} \map \exp {z_1 + z_2 - z} + \map \exp z \paren {D_z \map \exp {z_1 + z_2 - z} }
+      | c = [[Derivative of Complex Composite Function]]
+}}
+{{eqn | r = \map \exp z \, \map \exp {z_1 + z_2 - z} + \map \exp z \, \map \exp {z_1 + z_2 - z} \map {D_z} {z_1 + z_2 - z}
+      | c = as $\exp$ is its own derivative
+}}
+{{eqn | r = \map \exp z \, \map \exp {z_1 + z_2 - z} - \map \exp z \, \map \exp {z_1 + z_2 - z}
+      | c = [[Derivative of Complex Power Series]]
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+From [[Zero Derivative implies Constant Complex Function]], it follows that $f$ is [[Definition:Constant Mapping|constant]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \exp {z_1} \, \map \exp {z_2}
+      | r = \map f {z_1}
+}}
+{{eqn | r = \map f 0
+      | c = as $f$ is constant
+}}
+{{eqn | r = \map \exp 0 \, \map \exp {z_1 + z_2}
+}}
+{{eqn | r = \map \exp {z_1 + z_2}
+      | c = as $\map \exp 0 = 1$ 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hyperbolic Cosine Function is Even}
+Tags: Hyperbolic Cosine Function, Hyperbolic Cosine Function is Even, Examples of Even Functions
+
+\begin{theorem}
+:$\map \cosh {-x} = \cosh x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cosh {-x}
+      | r = \frac {e^{-x} + e^{-\paren {-x} } } 2
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \frac {e^{-x} + e^x} 2
+}}
+{{eqn | r = \frac {e^x + e^{-x} } 2
+}}
+{{eqn | r = \cosh x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \map \cosh {-x}
+	| r = \map \cos {-i x}
+	| c = [[Hyperbolic Cosine in terms of Cosine]]
+}}
+{{eqn	| r = \map \cos {i x}
+	| c = [[Cosine Function is Even]]
+}}
+{{eqn	| r = \cosh x
+	| c = [[Hyperbolic Cosine in terms of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hyperbolic Tangent Function is Odd}
+Tags: Hyperbolic Tangent Function, Examples of Odd Functions
+
+\begin{theorem}
+Let $\tanh: \C \to \C$ be the [[Definition:Hyperbolic Tangent|hyperbolic tangent function]] on the [[Definition:Complex Number|set of complex numbers]].
+Then $\tanh$ is [[Definition:Odd Function|odd]]:
+:$\map \tanh {-x} = -\tanh x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \tanh {-x}
+      | r = \frac {\map \sinh {-x} } {\map \cosh {-x} } 
+      | c = {{Defof|Hyperbolic Tangent|index = 2}}
+}}
+{{eqn | r = \frac {-\sinh x} {\map \cosh {-x} } 
+      | c = [[Hyperbolic Sine Function is Odd]]
+}}
+{{eqn | r = \frac {-\sinh x} {\cosh x} 
+      | c = [[Hyperbolic Cosine Function is Even]]
+}}
+{{eqn | r = -\tanh x
+      | c = {{Defof|Hyperbolic Tangent|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hyperbolic Sine in terms of Sine}
+Tags: Sine Function, Hyperbolic Sine Function
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Then:
+:$i \sinh z = \map \sin {i z}$
+where:
+:$\sin$ denotes the [[Definition:Complex Sine Function|complex sine]]
+:$\sinh$ denotes the [[Definition:Hyperbolic Sine|hyperbolic sine]]
+:$i$ is the [[Definition:Imaginary Unit|imaginary unit]]: $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {i z}
+      | r = \frac {e^{i \paren {i z} } - e^{i \paren {-i z} } } {2 i}
+      | c = [[Sine Exponential Formulation]]
+}}
+{{eqn | r = \paren {-i} \frac {e^{-z} - e^z} 2
+      | c = $i^2 = -1$
+}}
+{{eqn | r = i \frac {e^z - e^{-z} } 2
+      | c = $i^2 = -1$
+}}
+{{eqn | r = i \sinh z
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hyperbolic Cosine in terms of Cosine}
+Tags: Cosine Function, Hyperbolic Cosine Function
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Then:
+:$\cosh z = \map \cos {i z}$
+where:
+:$\cos$ denotes the [[Definition:Complex Cosine Function|complex cosine]]
+:$\cosh$ denotes the [[Definition:Hyperbolic Cosine|hyperbolic cosine]]
+:$i$ is the [[Definition:Imaginary Unit|imaginary unit]]: $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}} 
+{{eqn | l = \map \cos {i z}
+      | r = \frac {e^{i \paren {i z} } + e^{-i \paren {i z} } } 2
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac {e^{-z} + e^z} 2
+      | c = $i^2 = -1$
+}}
+{{eqn | r = \cosh z
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hyperbolic Tangent in terms of Tangent}
+Tags: Tangent Function, Hyperbolic Tangent Function
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Then:
+:$i \tanh z = \map \tan {i z}$
+where:
+:$\tan$ denotes the [[Definition:Complex Tangent Function|tangent function]]
+:$\tanh$ denotes the [[Definition:Hyperbolic Tangent|hyperbolic tangent]]
+:$i$ is the [[Definition:Imaginary Unit|imaginary unit]]: $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}} 
+{{eqn | l = \map \tan {i z}
+      | r = \frac {\map \sin {i z} } {\map \cos {i z} }
+      | c = {{Defof|Complex Tangent Function}}
+}}
+{{eqn | r = \frac {i \sinh z} {\map \cos {i z} }
+      | c = [[Hyperbolic Sine in terms of Sine]]
+}}
+{{eqn | r = \frac {i \sinh z} {\cosh z}
+      | c = [[Hyperbolic Cosine in terms of Cosine]]
+}}
+{{eqn | r = i \tanh z
+      | c = {{Defof|Hyperbolic Tangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order-Extension Principle}
+Tags: Set Theory, Order Theory, Named Theorems, Order-Extension Principle
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\preceq$ be an [[Definition:Ordering|ordering]] on $S$.
+Then there exists a [[Definition:Total Ordering|total ordering]] $\le$ on $S$ such that:
+:$\forall a, b \in S: \left({a \preceq b \implies a \le b}\right)$
+\end{theorem}
+
+\begin{proof}
+Let $\preceq$ be an [[Definition:Ordering|ordering]] on the [[Definition:Set|set]] $S$.
+If $\preceq$ is a [[Definition:Total Ordering|total ordering]], the result is complete.
+Suppose, then, that $\preceq$ is not a [[Definition:Total Ordering|total ordering]].
+Let $T$ be the [[Definition:Set|set]] of [[Definition:Ordering|orderings]] on $S$ that [[Definition:Extension of Relation|extend]] $\preceq$, [[Subset Relation is Ordering|ordered by inclusion]].
+Let $C$ be a [[Definition:Chain (Set Theory)|chain]] in $T$.
+By [[Union of Nest of Orderings is Ordering]], $\bigcup C$ is an [[Definition:Ordering|ordering]].
+Thus every [[Definition:Chain (Set Theory)|chain]] in $T$ has an [[Definition:Upper Bound of Set|upper bound]] in $T$.
+By [[Zorn's Lemma]], $T$ has a [[Definition:Maximal Element|maximal element]], $\le$.
+$\le$ is seen to be a [[Definition:Total Ordering|total ordering]], as follows:
+Suppose that $a, b \in S$, $a \not \le b$, and $b \not \le a$.
+Let $\le'$ be the [[Definition:Relation|relation]] defined as:
+:$\le' \mathop {:=} \le \cup \left\{{(a, b)}\right\}$
+Let $\le'^-$ be the [[Definition:Transitive Closure (Relation Theory)|transitive closure]] of
+$\le'$.
+Then by [[Ordering can be Expanded to compare Additional Pair]], $\le'^-$ is an [[Definition:Ordering|ordering]].
+But $\le'^- \mathop {\supsetneq} \le$, contradicting the [[Definition:Maximal Element|maximality]] of $\le$.
+Thus, $\le$ is a [[Definition:Total Ordering|total ordering]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preceq$.
+By [[Reflexive Reduction of Ordering is Strict Ordering]], $\prec$ is a [[Definition:Strict Ordering|strict ordering]].
+By the [[Order-Extension Principle/Strict|strict form of the Order-Extension Principle]], there exists a [[Definition:Strict Total Ordering|strict total ordering]] $<$ on $S$ such that:
+:$\forall a, b \in S: \left({a \prec b \implies a < b}\right)$
+Let $\le$ be the [[Definition:Reflexive Closure|reflexive closure]] of $<$.
+Let $a, b \in S$.
+If $a \preceq b$, then by [[Law of Excluded Middle]] either $a \prec b$ or $a = b$.
+If $a = b$, then by the definition of [[Definition:Reflexive Closure|reflexive closure]], $a \le b$.
+If $a \prec b$, then by the choice of $<$, $a < b$ so $a \le b$.
+Thus for all $a, b \in S$, $a \preceq b \implies a < b$.
+By [[Reflexive Closure of Strict Total Ordering is Total Ordering]], $\le$ is a [[Definition:Total Ordering|total ordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Nest of Orderings is Ordering}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $S$ be a set.
+Let $C$ be a nonempty [[Definition:Nest|nest]] of [[Definition:Ordering|orderings]] on $S$.
+Then $\bigcup C$ is an ordering on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\preceq$ be an arbitrary element of $C$.
+Let $\sim \, = \, \bigcup C$.
+Checking in turn each of the criteria for an [[Definition:Ordering|ordering]]:
+Let $a, b \in S$.
+Let $a \sim b$ and $b \sim a$.
+Since $a \sim b$ and $b \sim a$, there exist $\preceq_1, \preceq_2 \, \in \, C$ such that $a \preceq_1 b$ and $b \preceq_2 a$.
+Since $C$ is a chain:
+:$\preceq_1 \, \subset \, \preceq_2$
+or:
+:$\preceq_2 \, \subset \, \preceq_1$
+{{WLOG}}, suppose $\preceq_1 \, \subset \, \preceq_2$.
+Then it must hold that $a \preceq_2 b$.
+Since:
+:$a \preceq_2 b$
+:$b \preceq_2 a$
+:$\preceq_2 \, \in \, T$ is an ordering
+it follows that:
+:$a = b$
+and so $\sim$ is [[Definition:Antisymmetric Relation|antisymmetric]].
+{{qed|lemma}}
+=== Reflexivity ===
+Let $a \in S$.
+Because $\preceq$ is an ordering:
+:$a \preceq a$
+As $\preceq \, \subseteq \, \sim$:
+: $a \sim a$
+and so $\sim$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $a, b, c \in S$.
+Let $a \sim b$ and $b \sim c$.
+Thus there are elements $\preceq_1$ and $\preceq_2$ of $C$ such that:
+:$a \preceq_1 b$ and $b \preceq_2 c$
+Since $C$ is a chain, it must hold (as above) that:
+:$a \preceq_2 b$ or $b \preceq_1 c$
+{{WLOG}}, suppose $a \preceq_2 b$
+Since:
+:$a \preceq_2 b$
+:$b \preceq_2 c$
+:$\preceq_2 \, \in \, T$ is an ordering
+it follows that:
+:$a \preceq_2 c$
+Since $\preceq_2 \, \subseteq \, \sim$:
+:$a \sim c$
+So $\sim$ has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+$\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
+Hence by definition it is an [[Definition:Ordering|ordering]].
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Order Theory]]
+qytyf4e5w8hyzajk2ab51r9gy6muw1n
+\end{proof}<|endoftext|>
+\section{Ordering can be Expanded to compare Additional Pair}
+Tags: Order Theory, Ordering can be Expanded to compare Additional Pair
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a$ and $b$ be [[Definition:Non-Comparable|non-comparable]] elements of $S$.
+That is, let:
+:$a \not\preceq b$
+and:
+:$b \not\preceq a$
+Let ${\preceq'} = {\preceq} \cup \left\{ {\left({a, b}\right)} \right\}$.
+Let $\preceq'^+$ be the [[Definition:Transitive Closure (Relation Theory)|transitive closure]] of
+$\preceq'$.
+Then:
+:$\preceq'^+$ is an [[Definition:Ordering|ordering]].
+$\preceq'^+$ can be defined by letting $p \preceq'^+ q$ {{iff}}:
+:$p \preceq q$ or
+:$p \preceq a$ and $b \preceq q$.
+\end{theorem}
+
+\begin{proof}
+$\preceq'^+$ is a [[Definition:Superset|superset]] of $\preceq$.
+By [[Relation Contains Diagonal Relation iff Reflexive]]:
+:$\Delta_S \subseteq \preceq$
+where $\Delta_S$ denotes the [[Definition:Diagonal Relation|diagonal relation]].
+By [[Subset Relation is Transitive]] it follows that:
+:$\Delta_S \subseteq \preceq'^+$
+By [[Relation Contains Diagonal Relation iff Reflexive]], it follows that $\preceq'^+$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+$\preceq'^+$ is [[Definition:Transitive Relation|transitive]] by the definition of [[Definition:Transitive Closure (Relation Theory)|transitive closure]] as the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Transitive Relation|transitive]] [[Definition:Superset|superset]].
+{{qed|lemma}}
+Next we will establish that if $x \preceq'^+ y$ then either $x \preceq y$ or $x \preceq a$ and $b \preceq y$.
+Let $x, y \in S$ and let $x \preceq'^+ y$.
+By the finite chain definition of [[Definition:Transitive Closure (Relation Theory)/Finite Chain|transitive closure]], for some $n \in \N_{>0}$ there are $x_0, x_1, \dots, x_n$ such that: 
+:$x_0 = x$
+:$x_n = y$
+:$(1): \quad x_0 \preceq' x_1 \preceq' \cdots \preceq' x_n$
+Assume that this [[Definition:Finite Sequence|sequence]] between $x_0$ and $x_n$ is as short as possible.
+That is, let $n$ be the [[Definition:Smallest Element|smallest]] number such that such a [[Definition:Finite Sequence|sequence]] exists.
+In $(1)$, at most one of the relations can be $a \preceq' b$:
+If there were a segment of the form $a \preceq' b \preceq' \cdots \preceq' a \preceq' b$, then the sequence could be shortened by replacing that with $a \preceq' b$.
+Thus $(1)$ takes one of these forms:
+:$(a): \quad x_0 \preceq x_1 \preceq \cdots \preceq x_n$
+:$(b): \quad x_0 \preceq \cdots \preceq x_j \preceq a \preceq' b \preceq x_{j+3} \preceq \cdots \preceq x_n$
+If $(a)$, [[Transitive Chaining]] shows that:
+:$x_0 \preceq x_n$
+That is, $x \preceq y$.
+If $(b)$, [[Transitive Chaining]] shows that:
+:$x_0 \preceq a \preceq' b \preceq x_n$
+That is, $x \preceq a$ and $b \preceq y$.
+So whenever $x \preceq'^+ y$, either:
+:$x \preceq y$ or both
+:$x \preceq a$ and $b \preceq y$
+{{qed|lemma}}
+Suppose [[Proof by Contradiction|for the sake of contradiction]] that $x \preceq'^+ y$, $y \preceq'^+ x$, and $x \ne y$.
+By the above, we must have:
+:$(1a): \quad x \preceq y$ or both
+:$(1b): \quad x \preceq a$ and $b \preceq y$
+:$(2a): \quad y \preceq x$ or both
+:$(2b): \quad y \preceq a$ and $b \preceq x$.
+If $(1a)$ and $(2a)$ hold, then $x \preceq y$ and $y \preceq x$.
+Thus $x = y$ by antisymmetry of $\preceq$, contradicting the assumption that $x \ne y$.
+If $(1b)$ and $(2a)$ hold, then $b \preceq y$, and $y \preceq x$, and $x \preceq a$.
+By [[Transitive Chaining]], $b \preceq a$.
+This contradicts the premise that $a$ and $b$ are not [[Definition:Comparable|comparable]].
+A similar contradiction results if $(1a)$ and $(2b)$ hold.
+If $(1b)$ and $(2b)$ hold, then $b \preceq x$ and $x \preceq a$.
+Thus by transitivity, $b \preceq a$.
+This contradicts the premise that $a$ and $b$ are not comparable.
+We have shown that assuming $\preceq'^+$ is not antisymmetric leads to a contradiction, so
+:$\preceq'^+$ is antisymmetric.
+{{qed|lemma}}
+Since $\preceq'^+$ is transitive and antisymmetric, it is an [[Definition:Ordering|ordering]].
+We must now show that for $p, q \in S$, $p \preceq'^+ q$ {{iff}}:
+:$p \preceq q$ or
+:$p \preceq a$ and $b \preceq q$.
+We have already shown above that $p \preceq'^+ q$ implies one of these conditions.
+We now prove the converse.
+Suppose that $p \preceq q$.
+Then $p \preceq' q$ by [[Set is Subset of Union]].
+By the definition of [[Definition:Transitive Closure (Relation Theory)/Smallest Transitive Superset|transitive closure]]:
+:$p \preceq'^+ q$
+Suppose instead that $p \preceq a$ and $a \preceq q$.
+Then by [[Set is Subset of Union]] and the definition of $\preceq'$:
+:$p \preceq' a$
+:$a \preceq' b$
+:$b \preceq' q$
+Thus by the definition of [[Definition:Transitive Closure (Relation Theory)/Smallest Transitive Superset|transitive closure]]:
+:$p \preceq'^+ a$
+:$a \preceq'^+ b$
+:$b \preceq'^+ q$
+Since the transitive closure is transitive, [[Transitive Chaining]] shows that $p \preceq'^+ q$.
+Thus the statement that $p \preceq'^+ q$ is equivalent to the conditions described in the theorem statement.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preceq$.
+Let $\prec' = {\prec} \cup \set {\tuple {a, b} }$.
+By [[Reflexive Reduction of Ordering is Strict Ordering]], $\prec$ is a [[Definition:Strict Ordering|strict ordering]].
+Define a relation $\prec'_2$ by letting $p \prec'_2 q$ {{iff}}:
+:$p \prec q$ or
+:$p \preceq a$ and $b \preceq q$
+By [[Strict Ordering can be Expanded to Compare Additional Pair]], $\prec'_2$ is a [[Definition:Strict Ordering|strict ordering]] and is the [[Definition:Transitive Closure of Relation|transitive closure]] of $\prec'$.
+Then the [[Definition:Reflexive Closure|reflexive closure]] of $\prec'_2$, ${\prec'_2}^=$ is the [[Definition:Transitive Reflexive Closure|transitive reflexive closure]] of $\prec'$.
+From [[Equivalence of Definitions of Reflexive Transitive Closure]], ${\prec'_2}^=$ is the [[Definition:Transitive Closure of Relation|transitive closure]] of the [[Definition:Reflexive Closure|reflexive closure]] of $\prec'$.
+The [[Definition:Reflexive Closure|reflexive closure]] of $\prec'$ is $\preceq'$, so ${\prec'_2}^=$ is the [[Definition:Transitive Closure of Relation|transitive closure]] of $\preceq'$.
+But ${\prec'_2}^=$ is clearly the relation defined by the two conditions in the theorem statement.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Recursive Construction of Transitive Closure}
+Tags: Transitive Closures
+
+\begin{theorem}
+Given the [[Definition:Relation|relation]] $\mathcal R$, its [[Definition:Transitive Closure of Relation|transitive closure]] $\mathcal R^+$ can be constructed as follows:
+Let:
+:$\mathcal R_n := \begin{cases}
+\mathcal R & : n = 0 \\
+\mathcal R_{n-1} \cup \left\{{\left({x_1, x_3}\right): \exists x_2: \left({x_1, x_2}\right) \in \mathcal R_{n-1} \land \left({x_2, x_3}\right) \in \mathcal R_{n-1}}\right\} & : n > 0
+\end{cases}$
+Finally, let:
+:$\displaystyle \mathcal R^+ = \bigcup_{i \in \N} \mathcal R_i$.
+Then $R^+$ is the transitive closure of $R$.
+\end{theorem}
+
+\begin{proof}
+We must show that:
+* $\mathcal R \subseteq \mathcal R^+$
+* $\mathcal R^+$ is [[Definition:Transitive Relation|transitive]]
+* $\mathcal R^+$ is the smallest [[Definition:Relation|relation]] with both of those characteristics.
+=== Proof of Subset ===
+$\mathcal R \subseteq \mathcal R^+$: $\mathcal R^+$ [[Definition:Subset|contains]] all of the $\mathcal R_i$, so in particular $\mathcal R^+$ contains $\mathcal R$.
+=== Proof of Transitivity ===
+Every [[Definition:Element|element]] of $\mathcal R^+$ is in one of the $\mathcal R_i$.
+From the method of construction of $\mathcal R^+$, we have that $\forall i, j \in \N: \mathcal R_i \subseteq \mathcal R_{\max \left({i, j}\right)}$.
+Suppose $\left({s_1, s_2}\right) \in \mathcal R_j$ and $\left({s_2, s_3}\right) \in \mathcal R_k$.
+Then as:
+: $\mathcal R_j \subseteq \mathcal R_{\max \left({j, k}\right)}$
+and:
+: $\mathcal R_k \subseteq \mathcal R_{\max \left({j, k}\right)}$
+it follows that:
+: $\left({s_1, s_2}\right) \in \mathcal R_{\max \left({j, k}\right)}$ and $\left({s_2, s_3}\right) \in \mathcal R_{\max \left({j, k}\right)}$
+It follows from the method of construction that $\left({s_1, s_3}\right) \in \mathcal R_{\max \left({j, k}\right)}$.
+Hence as $\mathcal R_{\max \left({j, k}\right)} \subseteq \mathcal R^+$, it follows that $\mathcal R^+$ is [[Definition:Transitive Relation|transitive]].
+=== Proof of being the smallest such relation ===
+Let $\mathcal R'$ be any [[Definition:Transitive Relation|transitive relation]] containing $\mathcal R$.
+We want to show that $\mathcal R^+ \subseteq \mathcal R'$.
+It is sufficient to show that $\forall i \in \N: \mathcal R_i \subseteq \mathcal R'$.
+Since $\mathcal R \subseteq \mathcal R'$, we have that $\mathcal R_0 \subseteq \mathcal R'$.
+Suppose $\mathcal R_i \subseteq \mathcal R'$.
+From the method of construction, as $\mathcal R'$ is transitive, $\mathcal R_{i+1} \subseteq \mathcal R'$.
+Therefore, by [[Principle of Mathematical Induction|induction]], $\forall i \in \N: \mathcal R_i \subseteq \mathcal R'$.
+So $\mathcal R^+ \subseteq \mathcal R'$, and hence the result.
+{{qed}}
+[[Category:Transitive Closures]]
+3idvvcqj8zmsoq42p8z4rh1hhhpr1m4
+\end{proof}<|endoftext|>
+\section{Contour Integral is Well-Defined}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C$ be a [[Definition:Contour (Complex Plane)|contour]] defined by a [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$ of [[Definition:Directed Smooth Curve|directed smooth curves]] in the [[Definition:Complex Plane|complex plane]]$\C$.
+Let $C_i$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Let $f: \operatorname{Im} \left({C}\right) \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex function]], where $\operatorname{Im} \left({C}\right)$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C$.
+Suppose that $\sigma_i: \left[{c_i \,.\,.\, d_i}\right] \to \C$ is a [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|reparameterization]] of $C_i$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Then:
+:$\displaystyle \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} f \left({\sigma_i \left({t}\right) }\right) \sigma_i' \left({t}\right) \rd t$
+and all [[Definition:Complex Riemann Integral|complex Riemann integrals]] in the equation are defined.
+\end{theorem}
+
+\begin{proof}
+Define $g_i: \left[{a_i\,.\,.\,b_i}\right] \to \C$ by $g_i \left({t}\right) = f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right)$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+By [[Definition:Smooth Path (Complex Analysis)|definition of smooth path]], it follows that $\gamma_i$ and $\gamma_i'$ are [[Definition:Continuous Complex Function|continuous]] for all $i \in \left\{ {1, \ldots, n}\right\}$.
+From [[Continuity of Composite Mapping/Corollary]] and [[Sum Rule for Continuous Functions]], it follows that $g_i$ is continuous.
+From [[Continuous Complex Function is Complex Riemann Integrable]], we find that $\displaystyle \int_{a_i}^{b_i} g_i \left({t}\right) \rd t$ is defined.
+Similarly, it can be shown that $\displaystyle \int_{c_i}^{d_i} f \left({\sigma_i \left({t}\right) }\right) \sigma_i' \left({t}\right) \rd t$ is defined for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Hence, all [[Definition:Complex Riemann Integral|complex Riemann integrals]] in the theorem are defined.
+The equality now follows from [[Contour Integral is Independent of Parameterization]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boolean Prime Ideal Theorem}
+Tags: Boolean Prime Ideal Theorem, Axiom of Choice, Boolean Algebras
+
+\begin{theorem}
+Let $\left({S, \le}\right)$ be a [[Definition:Boolean Algebra|Boolean algebra]].
+Let $I$ be an [[Definition:Ideal (Order Theory)|ideal]] in $S$.
+Let $F$ be a [[Definition:Filter|filter]] on $S$.
+Let $I \cap F = \varnothing$.
+Then there exists a [[Definition:Prime Ideal (Order Theory)|prime ideal]] $P$ in $S$ such that:
+: $I \subseteq P$
+and:
+: $P \cap F = \varnothing$
+\end{theorem}<|endoftext|>
+\section{Central Limit Theorem}
+Tags: Probability Theory
+
+\begin{theorem}
+Let $X_1, X_2, \ldots$ be a sequence of [[Definition:Independent Random Variables|independent]] identically distributed random variables with:
+:Mean $E \left[{X_i}\right] = \mu \in \left({-\infty \,.\,.\, \infty}\right)$
+:Variance $V \left({X_i}\right) = \sigma^2 > 0$
+Let:
+: $\displaystyle S_n = \sum_{i \mathop = 1}^n X_i$
+Then:
+:$\displaystyle \frac {S_n - n \mu} {\sqrt {n \sigma^2} } \xrightarrow {D} N \left({0, 1}\right)$ as $n \to \infty$
+that is, converges in distribution to a standard normal.
+\end{theorem}
+
+\begin{proof}
+Let $Y_i = \dfrac {X_i - \mu} {\sigma}$.
+We have that:
+: $E \left[{Y_i}\right] = 0$
+and:
+: $E \left[{Y_i^2}\right] = 1$
+Then by [[Taylor's Theorem]] the [[Definition:Characteristic Function of Random Variable|characteristic function]] can be written:
+:$\phi_{Y_i} = 1 - \dfrac {t^2} 2 + o \left({t^2}\right)$
+Now let:
+{{begin-eqn}}
+{{eqn | l = U_n 
+      | r = \frac {S_n - n \mu} {\sqrt {n \sigma^2} }
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \frac {X_i - \mu} {\sqrt {n \sigma^2} } 
+}}
+{{eqn | r = \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n \left({\frac {X_i - \mu} {\sigma} }\right) 
+}}
+{{eqn | r = \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n Y_i 
+}}
+{{end-eqn}}
+Then its characteristic function is given by:
+{{begin-eqn}}
+{{eqn | l = \phi_{U_n} \left({t}\right) 
+      | r = E \left[{e^{i t U_n} }\right]
+}}
+{{eqn | r = E \left[{\exp \left({\frac {i t} {\sqrt n} \sum_n^{i \mathop = 1} Y_i}\right)}\right] 
+}}
+{{eqn | r = \prod_n^{i \mathop = 1} E \left[{\exp \left({\frac{i t} {\sqrt n} Y_i}\right)}\right]
+      | c = since $Y_i$ are independent identically distributed
+}}
+{{eqn | r = \prod_n^{i \mathop = 1} \phi_{Y_i} \left({\frac t {\sqrt n} }\right)
+}}
+{{eqn | r = \left({\phi_{Y_i} \left({\frac t {\sqrt n} }\right)}\right)^n
+      | c = since $Y_i$ are independent identically distributed
+}}
+{{eqn | r = \left({1 - \frac {t^2} {2 n} + o \left({t^2}\right)}\right)^n
+}}
+{{end-eqn}}
+{{explain|What is the significance of the limits going from $n$ to $i = 1$ in the summations and products above, or is this a mistake?}}
+Recall that the characteristic equation of a standard normal is given by:
+:$e^{-\frac 1 2 t^2}$.
+Indeed the characteristic equations of the series converges to the standard normal characteristic equation:
+:$\left({1 - \dfrac {t^2} {2 n} + o \left({t^2}\right)}\right)^n \to e^{-\dfrac 1 2 t^2}$ as $n \to \infty$
+Then [[Lévy's Continuity Theorem]] applies.
+In particular, the convergence in distribution of the $U_n$ to some random variable with standard normal distribution is equivalent to continuity of the limiting characteristic equation at $t = 0$.
+But, $e^{-\frac 1 2 t^2}$ is clearly continuous at $0$.
+So we have that $\dfrac{S_n - n \mu} {\sqrt{n \sigma^2} }$ converges in distribution to a standard normal random variable.
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Tangent Exponential Formulation}
+Tags: Tangent Function, Tangent Exponential Formulation
+
+\begin{theorem}
+Let $z$ be a [[Definition:Complex Number|complex number]].
+Let $\tan z$ denote the [[Definition:Complex Tangent Function|tangent function]] and $i$ denote the [[Definition:Imaginary Unit|imaginary unit]]: $i^2 = -1$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Arcsine Logarithmic Formulation}
+Tags: Arcsine Function
+
+\begin{theorem}
+For any [[Definition:Real Number|real number]] $x: -1 \le x \le 1$:
+:$\arcsin x = \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} }$
+where $\arcsin x$ is the [[Definition:Arcsine|arcsine]] and $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+Assume $y \in \R$ where $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \arcsin x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \sin y
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \frac 1 {2 i} \paren {e^{i y} - e^{-i y} }
+      | c = [[Sine Exponential Formulation]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 2 i x
+      | r = e^{i y} - e^{-i y}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 2 i x e^{i y}
+      | r = e^{2 i y} - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y} -2 i x e^{i y}
+      | r = -1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y} - 2 i x e^{i y} - x^2
+      | r = 1 - x^2
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {e^{i y} - i x}^2
+      | r = 1 - x^2
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{i y} - i x
+      | r = \sqrt {1 - x^2}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i y
+      | r = \map \ln {i x + \sqrt {1 - x^2} }
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y
+      | r = \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arccosine Logarithmic Formulation}
+Tags: Arccosine Function
+
+\begin{theorem}
+For any [[Definition:Real Number|real number]] $x: -1 \le x \le 1$:
+:$\arccos x = \dfrac 1 i \map \ln {x + \sqrt {x^2 - 1} }$
+where $\arccos x$ is the [[Definition:Arccosine|arccosine]] and $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+Assume $y \in \R$ such that $0 \le y \le \pi$.
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \arccos x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \cos y
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \frac 1 2 \paren {e^{-i y} + e^{i y} }
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 2 x
+      | r = e^{-i y} + e^{i y}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 2 x e^{i y}
+      | r = 1 + e^{2 i y}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y} - 2 x e^{i y}
+      | r = -1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y} - 2 x e^{i y} + x^2
+      | r = x^2 - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {e^{i y} - x}^2
+      | r = x^2 - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{i y} - x
+      | r = \sqrt {x^2 - 1}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i y
+      | r = \map \ln {x + \sqrt {x^2 - 1} }
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y
+      | r = \dfrac 1 i \map \ln {x + \sqrt {x^2 - 1} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arctangent Logarithmic Formulation}
+Tags: Arctangent Function
+
+\begin{theorem}
+For any [[Definition:Real Number|real number]] $x$:
+:$\arctan x = \dfrac 1 2 i \, \map \ln {\dfrac {1 - i x} {1 + i x} }$
+where $\arctan x$ is the [[Definition:Arctangent|arctangent]] and $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+Assume $y \in \R$, $ -\dfrac \pi 2 \le y \le \dfrac \pi 2 $.
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \arctan x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \tan y
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = i \frac {1 - e^{2 i y} } {1 + e^{2 i y} }
+      | c = [[Tangent Exponential Formulation]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i x
+      | r = \frac {e^{2 i y} - 1} {e^{2 i y} + 1}
+      | c = $ i^2 = -1 $
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i x \paren {e^{2 i y} + 1}
+      | r = e^{2 i y} - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i x e^{2 i y} + i x
+      | r = e^{2 i y} - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y} - i x e^{2 i y}
+      | r = 1 + i x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y}
+      | r = \frac {1 + i x} {1 - i x}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{-2 i y}
+      | r = \frac {1 - i x} {1 + i x}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -2 i y
+      | r = \map \ln {\frac {1 - i x} {1 + i x} }
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y
+      | r = \frac 1 2 i \map \ln {\frac {1 - i x} {1 + i x} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Contour is Well-Defined}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C_1, \ldots, C_n$ be [[Definition:Directed Smooth Curve|directed smooth curves]].
+Let $C_i$ be [[Definition:Directed Smooth Curve/Parameterization|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$.
+Let $C$ be the [[Definition:Contour (Complex Plane)|contour]] defined by the [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$.
+Suppose that $\sigma_i: \closedint {c_i} {d_i} \to \C$ is a [[Definition:Directed Smooth Curve/Parameterization|reparameterization]] of $C_i$ for all $i \in \set {1, \ldots, n}$
+Then:
+:$\displaystyle \sum_{i \mathop = 1}^n \int_{a_i }^{b_i} \size {\map {\gamma_i'} t} \rd t = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \size {\map {\sigma_i'} t} \rd t$
+and all [[Definition:Definite Integral|real integrals]] in the equation are defined.
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Directed Smooth Curve|definition of directed smooth curve]], it follows that  $\sigma_i = \gamma_i \circ \phi_i$ for all $i \in \set {1, \ldots, n}$.
+Here, $\phi_i: \closedint {c_i} {d_i} \to \closedint {a_i} {b_i}$ is a [[Definition:Bijection|bijective]] [[Definition:Differentiable on Interval|differentiable]] [[Definition:Strictly Increasing Real Function|strictly increasing function]]. 
+For all $i \in set {1, \ldots, n}$, $\gamma_i$ and $\sigma_i$ are [[Definition:Continuous Complex Function|continuous]].
+From [[Complex Modulus Function is Continuous]] and [[Continuity of Composite Mapping/Corollary|Continuity of Composite Mapping: Corollary]], it follows that $\size {\gamma_i'}$ and $\size {\sigma_i'}$ are [[Definition:Continuous Real Function on Subset|continuous]].
+From [[Continuous Real Function is Darboux Integrable]], we find that $\displaystyle \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t$ and $\displaystyle \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \size {\map {\sigma_i'} t} \rd t$ are defined.
+Hence, all [[Definition:Definite Integral|real integrals]] in the theorem are defined.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n \int_{a_i }^{b_i} \size {\map {\gamma_i'} t} \rd t
+      | r = \sum_{i \mathop = 1}^n \int_{\map {\phi_i^{-1} } {a_i} }^{\map {\phi_i^{-1} } {b_i} } \size {\map {\gamma_i'} {\map \phi u} } \map {\phi'} u \rd u
+      | c = [[Integration by Substitution|substitution]] with $t = \map \phi u$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{\map {\phi_i^{-1} } {a_i} }^{\map {\phi_i^{-1} } {b_i} } \size {\map {\gamma_i'} {\map \phi u} \map {\phi'} u}\rd u
+      | c = as $\map {\phi'} u > 0$ by [[Derivative of Monotone Function]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{\map {\phi_i^{-1} } {a_i} }^{\map {\phi_i^{-1} } {b_i} } \size {\map {\sigma'} u} \rd u
+      | c = [[Derivative of Complex Composite Function]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \size {\map {\sigma'} u} \rd u
+      | c = [[Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Contour Integration]]
+ijjqjztb5xqpzf8c0vs04d5aqltuys9
+\end{proof}<|endoftext|>
+\section{Two Ring is Boolean Ring}
+Tags: Boolean Rings
+
+\begin{theorem}
+Let $2$ be the [[Definition:Two Ring|two ring]].
+Then $2$ is a [[Definition:Boolean Ring|Boolean ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Integers Modulo m is Ring]], $2$ is a [[Definition:Ring with Unity|ring with unity]].
+Furthermore, the identities:
+:$0 \cdot 0 = 0$
+:$1 \cdot 1 = 1$
+show that $2$ is also an [[Definition:Idempotent Ring|idempotent ring]].
+Hence the result, by definition of [[Definition:Boolean Ring|Boolean ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Idempotent Ring is Commutative}
+Tags: Idempotent Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be an [[Definition:Idempotent Ring|idempotent ring]].
+Denote with $0_R$ the [[Definition:Ring Zero|zero]] of $R$.
+Then $\struct {R, +, \circ}$ is a [[Definition:Commutative Ring|commutative ring]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x + y
+      | r = \paren {x + y}^2
+      | c = {{Defof|Idempotent Ring}}
+}}
+{{eqn | r = x^2 + x \circ y + y \circ x + y^2
+      | c = [[Binomial Theorem/Ring Theory|Binomial Theorem: Ring Theory]]
+}}
+{{eqn | r = x + x \circ y + y \circ x + y
+      | c = {{Defof|Idempotent Ring}}
+}}
+{{end-eqn}}
+Subtracting $x + y$ from both sides yields:
+{{begin-eqn}}
+{{eqn | l = x \circ y + y \circ x
+      | r = 0_R
+}}
+{{eqn | r = y \circ x + y \circ x
+      | c = [[Idempotent Ring has Characteristic Two]]
+}}
+{{end-eqn}}
+Finally, subtracting $y \circ x$ from both sides, we obtain:
+:$x \circ y = y \circ x$
+and conclude $R$ is a [[Definition:Commutative Ring|commutative ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Idempotent Ring has Characteristic Two}
+Tags: Idempotent Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be an [[Definition:Idempotent Ring|idempotent]] [[Definition:Non-Null Ring|non-null ring]].
+Denote with $0_R$ the [[Definition:Ring Zero|zero]] of $R$.
+Then $\struct {R, +, \circ}$ has [[Definition:Characteristic of Ring|characteristic]] $2$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x + x
+      | r = \paren {x + x}^2
+      | c = {{Defof|Idempotent Ring}}
+}}
+{{eqn | r = \paren {x + x} \paren {x + x}
+}}
+{{eqn | r = x \paren {x + x} + x \paren {x + x}
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = x^2 + x^2 + x^2 + x^2
+      | c = {{Ring-axiom|D}} again
+}}
+{{eqn | r = x + x + x + x
+      | c = {{Defof|Idempotent Ring}}
+}}
+{{eqn | ll= \leadsto
+      | l = 0_R
+      | r = x + x
+      | c = $\struct {\R, +}$ is a [[Definition:Group|group]], so [[Cancellation Laws]] apply
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two-Valued Functions form Boolean Ring}
+Tags: Boolean Rings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]], and let $2$ be the [[Definition:Two Ring|two ring]].
+Let $2^S$ be the [[Definition:Set of All Mappings|set of all]] [[Definition:Two-Valued Function|$2$-valued functions]] on $S$.
+Denote with $+$ and $\cdot$ the [[Definition:Pointwise Operation|pointwise operations induced]] on $2^S$ by $+_2$ and $\times_2$, respectively.
+Then $\struct {2^S, +, \cdot}$ is a [[Definition:Boolean Ring|Boolean ring]].
+\end{theorem}
+
+\begin{proof}
+By [[Structure Induced by Ring Operations is Ring]], $\struct {2^S, +, \cdot}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+By [[Unity of Induced Structure]], $\struct {2^S, +, \cdot}$ also has a [[Definition:Unity of Ring|unity]].
+By [[Induced Structure is Idempotent]], $\cdot$ is an [[Definition:Idempotent Operation|idempotent operation]].
+Hence $\struct {2^S, +, \cdot}$ is a [[Definition:Boolean Ring|Boolean ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Integration by Substitution}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right]$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\phi: \left[{a \,.\,.\, b}\right] \to \R$ be a [[Definition:Real Function|real function]] which has a [[Definition:Derivative|derivative]] on $\left[{a \,.\,.\, b}\right]$.
+Let $f: A \to \C$ be a [[Definition:Continuous Complex Function|continuous complex function]], where $A$ is a [[Definition:Subset|subset]] of the [[Definition:Image of Mapping|image]] of $\phi$.
+If $\phi \left({a}\right) \le \phi \left({b}\right)$, then:
+:$\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \, \mathrm d t = \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u$
+If $\phi \left({a}\right) > \phi \left({b}\right)$, then:
+:$\displaystyle \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} f \left({t}\right) \, \mathrm  dt = -\int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) ,\ \mathrm d u$
+\end{theorem}
+
+\begin{proof}
+Let $\operatorname{Re}$ and $\operatorname{Im}$ denote [[Definition:Real Part|real parts]] and [[Definition:Imaginary Part|imaginary parts]] respectively.
+Let $\phi \left({a}\right) \le \phi \left({b}\right)$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \, \mathrm d t
+      | r = \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t
+      | c = Definition of [[Definition:Complex Riemann Integral|Complex Riemann Integral]]
+}}
+{{eqn | r = \int_a^b \operatorname{Re} \left({ f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u + i \int_a^b \operatorname{Im} \left({ f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int_a^b \operatorname{Re} \left({ f \left({ \phi \left({u}\right) }\right) \phi' \left({u}\right) }\right) \, \mathrm d u + i \int_a^b \operatorname{Im} \left({ f \left({ \phi \left({u}\right) }\right) \phi' \left({u}\right) }\right) \, \mathrm d u
+      | c = [[Multiplication of Real and Imaginary Parts]]
+}}
+{{eqn | r = \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u
+}}
+{{end-eqn}}
+Let $\phi \left({a}\right) > \phi \left({b}\right)$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} f \left({t}\right) \, \mathrm d t
+      | r = \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} \operatorname{Re} \left({ f \left({t}\right) }\right) \, \mathrm d t + i \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} \operatorname{Im} \left({ f \left({t}\right) }\right) \, \mathrm d t
+      | c = Definition of [[Definition:Complex Riemann Integral|Complex Riemann Integral]]
+}}
+{{eqn | r = \int_b^a \operatorname{Re} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u + i \int_b^a \operatorname{Im} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\int_a^b \operatorname{Re} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u - i \int_a^b \operatorname{Im} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u
+      | c = Definition of [[Definition:Definite Integral|Definite Integral]]
+}}
+{{eqn | r = -\int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Injection of Interval is Strictly Monotone}
+Tags: Continuous Functions
+
+\begin{theorem}
+Let $I$ be a [[Definition:Real Interval|real interval]].
+Let $f: I \to \R$ be an [[Definition:Injection|injective]] [[Definition:Continuous on Interval|continuous real function]].
+Then $f$ is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+\end{theorem}
+
+\begin{proof}
+Suppose that $f$ is not [[Definition:Strictly Monotone Real Function|strictly monotone]].
+That is, there exist $x, y, z \in I$ with $x < y < z$ such that either $f \left({x}\right) \le f \left({y}\right)$ and $f \left({y}\right) \ge f \left({z}\right)$, or $f \left({x}\right) \ge f \left({y}\right)$ and $f \left({y}\right) \le f \left({z}\right)$.
+Suppose $f \left({x}\right) \le f \left({y}\right)$ and $f \left({y}\right) \ge f \left({z}\right)$.
+If $f \left({x}\right) = f \left({y}\right)$, or $f \left({y}\right) = f \left({z}\right)$, or $f \left({x}\right) = f \left({z}\right)$, $f$ is not [[Definition:Injection|injective]], which is a contradiction.
+Thus, $f \left({x}\right) < f \left({y}\right)$ and $f \left({y}\right) > f \left({z}\right)$.
+Suppose $f \left({x}\right) < f \left({z}\right)$.
+That is:
+: $f \left({x}\right) < f \left({z}\right) < f \left({y}\right)$
+As $f$ is [[Definition:Continuous on Interval|continuous]] on $I$, the [[Intermediate Value Theorem]] can be applied.
+Hence there exists $c \in \left({x \,.\,.\, y}\right)$ such that $f \left({c}\right) = f \left({z}\right)$.
+As $z \notin \left({x \,.\,.\, y}\right)$, we have $c \ne z$.
+So $f$ is not [[Definition:Injection|injective]], which is a contradiction.
+Suppose instead $f \left({x}\right) > f \left({z}\right)$.
+That is:
+: $f \left({z}\right) < f \left({x}\right) < f \left({y}\right)$
+Again, as $f$ is [[Definition:Continuous on Interval|continuous]] on $I$, the [[Intermediate Value Theorem]] can be applied.
+Then, there exists $c \in \left({y \,.\,.\, z}\right)$ such that $f \left({c}\right) = f \left({x}\right)$.
+So $f$ is not [[Definition:Injection|injective]], which is a contradiction.
+If we suppose $f \left({x}\right) \ge f \left({y}\right)$ and $f \left({y}\right) \le f \left({z}\right)$, we reach a similar contradiction.
+By [[Proof by Contradiction]], $f$ is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+{{qed}}
+[[Category:Continuous Functions]]
+rq33x2iks7zqezdi3sp7eax9o3txvzb
+\end{proof}<|endoftext|>
+\section{Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice}
+Tags: Order Theory, Lattice Theory
+
+\begin{theorem}
+Let $L$ be a [[Definition:Distributive Lattice|distributive lattice]].
+{{explain|Full notational specification needed}}
+Let $F$ be a [[Definition:Filter|filter]] in $L$.
+Let $M$ be an [[Definition:Ideal (Order Theory)|ideal]] in $L$ which is disjoint from $F$ such that no ideal in $L$ larger than $M$ is disjoint from $F$.
+Then $M$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+\end{theorem}
+
+\begin{proof}
+Suppose, for the sake of contradiction, that $M$ is not a prime ideal.
+Then by [[Prime Ideal in Lattice]], there are elements $a$ and $b$ of $L$ such that
+:$a \wedge b \in M$
+:$a \notin M$
+:$b \notin M$
+=== Lemma ===
+There do not exist $m$ and $n$ in $M$ such that $m \vee a \in F$ and $n \vee b \in F$.
+=== Proof ===
+Suppose such exist.
+Since:
+:$m \vee \left({n \vee b}\right) \ge n \vee b$
+:$n \vee b \in F$
+:$F$ is a filter
+it follows that:
+:$m \vee \left({n \vee b}\right) \in F$
+Applying associativity yields:
+:$\left({m \vee n}\right) \vee b \in F$
+By the same argument:
+:$\left({m \vee n}\right) \vee a \in F$
+By the definition of a filter:
+:$\left({\left({m \vee n}\right) \vee b}\right) \wedge \left({\left({n \vee m}\right) \vee a}\right) \in F$
+Distributing:
+:$\left({m \vee n}\right) \wedge \left({b \vee a}\right) \in F$
+But by assumption:
+:$b \vee a \in M$
+and by the definition of an ideal:
+:$m \vee n \in M$
+so again by the definition of an ideal:
+:$\left({m \vee n}\right) \wedge \left({b \vee a}\right) \in M$
+contradicting the supposition that $M$ is disjoint from $F$.
+{{qed|lemma}}
+{{WLOG}}, we can thus suppose that:
+:$m \vee a \notin F$ for all $m \in M$
+Let $N = \left\{ {x \in L: x \le m \vee a \text{ for some } m \in M}\right\}$.
+=== Lemma ===
+$N$ is an ideal in $L$.
+=== Proof ===
+Let:
+:$x \in N$
+:$y \in L$
+:$y \le x$
+Then by the definition of $N$ there exists an $m \in M$ such that:
+:$x \le m \vee a$
+Since $y \le x$, it follows that:
+:$y \le m \vee a$
+so $y \in N$.
+Let:
+:$x \in N$
+:$y \in N$
+Then there exist $m_x$ and $m_y$ in $M$ such that:
+:$x \le m_x \vee a$
+:$y \le m_y \vee a$
+Then:
+: $x \vee y \le \left({m_x \vee a}\right) \vee \left({m_y \vee a}\right) = \left({m_x \vee m_y}\right) \vee a$
+But $m_x \vee m_y \in M$, so:
+:$x \vee y \in N$
+{{qed|lemma}}
+=== Lemma ===
+$M \subsetneq N$
+=== Proof ===
+Let $m \in M$.
+Then:
+:$m \le \left({m \vee a}\right)$
+so $m \in N$.
+Thus $M \subseteq N$.
+We have:
+:$a \le \left({m \vee a}\right)$
+so:
+:$a \in N$
+but:
+:$a \notin M$
+Thus $M \subsetneq N$.
+{{qed|lemma}}
+=== Lemma ===
+$N \cap F = \varnothing$
+=== Proof ===
+Suppose $x \in N \cap F$.
+Then:
+:$x \in N$
+so for some $m \in M$:
+:$x \le m \vee a$
+Furthermore, $x \in F$.
+So by the definition of a filter:
+:$m \vee a \in F$
+But this contradicts our assumption that $m \vee a \notin F$ for all $m \in M$.
+{{qed|lemma}}
+By assuming that $M$ is not a prime ideal, we have constructed an ideal $N$ properly containing $M$ that is disjoint from $F$, contradicting the maximality of $M$.
+Thus $M$ is a prime ideal.
+{{qed}}
+[[Category:Order Theory]]
+[[Category:Lattice Theory]]
+ii74u4scsol96px9414dsbisl09sn58
+\end{proof}<|endoftext|>
+\section{Prime Ideal in Lattice}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({L, \le}\right)$ be a [[Definition:Lattice|lattice]].
+Let $I$ be an [[Definition:Ideal (Order Theory)|ideal]] in $L$.
+Then $I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]] {{iff}}:
+:$\forall a, b \in L: a \wedge b \in I \implies a \in I \text{ or } b \in I$
+where $a \wedge b$ denotes $\min \left\{{a, b}\right\}$, the [[Definition:Meet (Order Theory)|meet]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $I$ be a [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+Let $a, b \in L$ such that $a, b \notin I$.
+Then $a, b \in L \setminus I$.
+By the definition of [[Definition:Prime Ideal (Order Theory)|prime ideal]] $L \setminus I$ is a [[Definition:Filter|filter]].
+By the definition of a [[Definition:Filter|filter]]:
+:$\exists c \in L \setminus I: c \le a, c \le b$
+By the definition of [[Definition:Meet (Order Theory)|meet]]:
+:$c \le a \wedge b$
+Since $c \in L \setminus I$, the definition of a filter implies that
+: $a \wedge b \in L \setminus I$
+{{refactor|The above argument should be extracted into a separate page which may or may not be instantiated as a separate definition of a filter. As it is, we haven't even got the definition of "meet" written yet - although whether there is any point in doing so as $a \wedge b :{{=}} \inf \left\{ {a, b}\right\}$ should be sufficient.}}
+Thus $a \wedge b \notin I$.
+By [[Rule of Transposition|contraposition]] it follows that $a \wedge b \in I$ implies that $a \in I$ or $b \in I$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+For all $a$ and $b$ in $L$, let $a \wedge b \in I$ imply that $a \in I$ or $b \in I$.
+Let $x, y \in L \setminus I$.
+Then $x, y \notin I$.
+By supposition, $x \wedge y \notin I$, so:
+:$x \wedge y \in L \setminus I$
+Let $p \in L \setminus I, q \in L$ such that $p \le q$.
+Let $q \in I$.
+Then by definition of [[Definition:Ideal (Order Theory)|ideal]], $p \in I$.
+This contradicts $p \in L \setminus I$.
+Thus $q \notin I$.
+Thus $q \in L \setminus I$.
+Therefore, $L \setminus I$ is a [[Definition:Filter|filter]].
+{{qed}}
+[[Category:Order Theory]]
+nbzym84t11zdrzd9hov1qc77usdkscr
+\end{proof}<|endoftext|>
+\section{Inverse of Increasing Bijection need not be Increasing}
+Tags: Increasing Mappings, Bijections
+
+\begin{theorem}
+Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $\phi: S \to T$ be a [[Definition:Bijection|bijection]] which is [[Definition:Increasing Mapping|increasing]].
+Then $\phi^{-1}: T \to S$ is not necessarily also [[Definition:Increasing Mapping|increasing]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $S := \mathcal P \left({\left\{{a, b}\right\}}\right)$.
+Let $T := \left\{{1, 2, 3, 4}\right\}$.
+From [[Subset Relation on Power Set is Partial Ordering]], we have that $\left({S, \subseteq}\right)$ is an [[Definition:Ordered Set|ordered set]].
+Clearly so is $\left({T, \le}\right)$ (although its [[Definition:Ordering|ordering]] is in fact [[Definition:Total Ordering|total]], it is still technically an [[Definition:Ordered Set|ordered set]]).
+Let $\phi: S \to T$ be defined as:
+: $\phi \left({\varnothing}\right) = 1$
+: $\phi \left({\left\{{a}\right\}}\right) = 2$
+: $\phi \left({\left\{{b}\right\}}\right) = 3$
+: $\phi \left({\left\{{a, b}\right\}}\right) = 4$
+By inspection it can be seen that $\phi$ is a [[Definition:Bijection|bijection]].
+Also by inspection it can be seen that $\phi$ is [[Definition:Increasing Mapping|increasing]].
+But note that while $2 \le 3$, it is not the case that $\left\{{a}\right\} \subseteq \left\{{b}\right\}$.
+That is, $\phi^{-1}$ is ''not'' [[Definition:Increasing Mapping|increasing]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order Isomorphic Sets are Equivalent}
+Tags: Equivalence Relations, Order Isomorphisms
+
+\begin{theorem}
+Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be [[Definition:Order Isomorphism|order isomorphic]].
+Then $S$ and $T$ are [[Definition:Set Equivalence|equivalent]].
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Order Isomorphism|order isomorphism]] is a [[Definition:Bijection|bijection]] $\phi$ such that:
+: $\phi: S \to T$ is [[Definition:Order-Preserving|order-preserving]]
+: $\phi^{-1}: T \to S$ is [[Definition:Order-Preserving|order-preserving]].
+So, by definition, there exists a [[Definition:Bijection|bijection]] between $S$ and $T$.
+The result follows by definition of [[Definition:Set Equivalence|set equivalence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Contour Integrals}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C$ be a [[Definition:Contour (Complex Plane)|contour in $\C$]].
+Let $f, g: \Img C \to \C$ be [[Definition:Continuous Complex Function|continuous complex functions]], where $\Img C$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C$.
+Let $\lambda, \mu \in \C$ be [[Definition:Complex Number|complex]] [[Definition:Constant|constants]].
+Then:
+:$\displaystyle \int_C \paren {\lambda \map f z + \mu \map g z} \rd z = \lambda \int_C \map f z \rd z + \mu \int_C \map g z \rd z$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Contour (Complex Plane)|contour]], $C$ is a [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$ of [[Definition:Directed Smooth Curve|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 = \int_C \paren {\lambda \map f z + \mu \map g z} \rd z
+      | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \paren {\lambda \map f {\map {\gamma_i} t} + \mu \map g {\map {\gamma_i} t} } \map {\gamma_i'} t \rd t
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {\lambda \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t + \mu \int_{a_i}^{b_i} \map g {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}
+      | c = [[Linear Combination of Complex Integrals]]
+}}
+{{eqn | r = \lambda \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t + \mu \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map g {\map {\gamma_i} t} \map {\gamma_i'} t \rd t
+}}
+{{eqn | r = \lambda \int_C \map f z \rd z + \mu \int_C \map g z \rd z 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Product Metric Induces Product Topology}
+Tags: Metric Spaces, P-Product Metrics
+
+\begin{theorem}
+Let $M_A = \struct{A, d_A}$ and $M_B = \struct{B, d_B}$ be [[Definition:Metric Space|metric spaces]].
+Let $\tau_A$ and $\tau_B$ be the [[Definition:Topology Induced by Metric|topologies on $A$ and $B$ induced by $d_A$ and $d_B$]], respectively.
+Let $p \ge 1$ be an [[Definition:Extended Real Number Line|extended real number]].
+Let $M = \struct{A \times B, d}$ be the [[Definition:P-Product Metric|$p$-product of $M_A$ and $M_B$]].
+We have that [[P-Product Metric is Metric|$M$ is a metric space]].
+Let $\tau$ be the [[Definition:Topology Induced by Metric|topology on $A \times B$ induced by $d$]].
+Then $\struct{A \times B, \tau}$ is the [[Definition:Product Space (Topology)|product space]] of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$.
+\end{theorem}
+
+\begin{proof}
+By [[P-Product Metrics are Lipschitz Equivalent|$p$-Product Metrics are Lipschitz Equivalent]] and [[Lipschitz Equivalent Metrics are Topologically Equivalent]], it suffices to consider the case $p = \infty$.
+Let $\struct{A \times B, \tau'}$ be the [[Definition:Product Space (Topology)|product space]] of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$.
+By the definition of $d$, it follows that an [[Definition:Open Ball|open ball]] in $M$ is the [[Definition:Cartesian Product|(Cartesian) product]] of an [[Definition:Open Ball|open ball]] in $M_A$ and an [[Definition:Open Ball|open ball]] in $M_B$.
+Since an [[Open Ball of Metric Space is Open Set|open ball is an open set]], it follows from [[Equivalence of Definitions of Analytic Basis]] that $\tau \subseteq \tau'$.
+Let $W \in \tau'$, $\tuple{x, y} \in W$.
+By the definition of $\tau'$, it follows from [[Equivalence of Definitions of Analytic Basis]] that there exists a $U$ [[Definition:Open Set (Metric Space)|open]] in $M_A$ and a $V$ [[Definition:Open Set (Metric Space)|open]] in $M_B$ such that $\tuple{x, y} \in U \times V \subseteq W$.
+By the definition of an [[Definition:Open Set (Metric Space)|open set]]:
+:There exists a [[Definition:Strictly Positive Real Number|strictly positive real number]] $\alpha \in \R_{>0}$ such that the [[Definition:Open Ball|open $\alpha$-ball]] of $x$ in $M_A$ is [[Definition:Subset|contained]] in $U$
+:There exists a [[Definition:Strictly Positive Real Number|strictly positive real number]] $\beta \in \R_{>0}$ such that the [[Definition:Open Ball|open $\beta$-ball]] of $y$ in $M_B$ is [[Definition:Subset|contained]] in $V$
+Let $\epsilon = \min \set{\alpha, \beta}$.
+Then $\epsilon \in \R_{>0}$, and the [[Definition:Open Ball|open $\epsilon$-ball]] of $\tuple{x, y}$ in $M$ is [[Definition:Subset|contained]] in $U \times V$.
+Hence, $W$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+That is, $\tau' \subseteq \tau$.
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+: $\tau = \tau'$
+as desired.
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:P-Product Metrics]]
+lqibvwe21ff2d5d9bz5iuk99rslkxwi
+\end{proof}<|endoftext|>
+\section{P-Product Metrics are Lipschitz Equivalent}
+Tags: Lipschitz Equivalence, P-Product Metrics
+
+\begin{theorem}
+Let $M_A = \left({A, d_A}\right)$ and $M_B = \left({B, d_B}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\tau_A$ and $\tau_B$ be the [[Definition:Topology Induced by Metric|topologies on $A$ and $B$ induced by $d_A$ and $d_B$]], respectively.
+For all [[Definition:Extended Real Number Line|extended real numbers]] $p \ge 1$, let $M_p = \left({A \times B, d_p}\right)$ be the [[Definition:P-Product Metric|$p$-product of $M_A$ and $M_B$]].
+Then all the [[Definition:Metric|metrics]] $\left\{{d_p: p \ge 1}\right\}$ are [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
+\end{theorem}
+
+\begin{proof}
+From [[P-Product Metric is Metric|$p$-Product Metric is Metric]], $M_p$ is a [[Definition:Metric Space|metric space]].
+For all [[Definition:Real Number|real numbers]] $p \ge 1$, it follows from the definition of $d_p$ that:
+:$\forall x, y \in A \times B: d_{\infty} \left({x, y}\right) \le d_p \left({x, y}\right) \le 2^{1/p} d_{\infty} \left({x, y}\right)$
+Hence, $d_p$ and $d_{\infty}$ are [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
+The result follows from [[Lipschitz Equivalence is Equivalence Relation]].
+{{qed}}
+[[Category:Lipschitz Equivalence]]
+[[Category:P-Product Metrics]]
+e5q0v5e89hcpltsp1iq4vys99oipf9s
+\end{proof}<|endoftext|>
+\section{Ordered Sum of Tosets is Totally Ordered Set/General Result}
+Tags: Total Orderings, Ordered Sums
+
+\begin{theorem}
+Let $S_1, S_2, \ldots, S_n$ all be [[Definition:Toset|tosets]].
+Let $T_n$ be the [[Definition:Ordered Sum|ordered sum]] of $S_1, S_2, \ldots, S_n$:
+:$\forall n \in \N_{>0}: T_n = \begin{cases}
+S_1 & : n = 1 \\
+T_{n-1} + S_n & : n > 1
+\end{cases}$
+Then $T_n$ is a [[Definition:Toset|toset]].
+\end{theorem}
+
+\begin{proof}
+From [[Ordered Sum of Tosets is Totally Ordered Set]], $S_1 + S_2$ is a [[Definition:Toset|toset]].
+Suppose $T_{n-1}$ is a [[Definition:Toset|toset]].
+Given that $S_n$ is a [[Definition:Toset|toset]], it follows from [[Ordered Sum of Tosets is Totally Ordered Set]] that $T_{n-1} + S_n$ is also a [[Definition:Toset|toset]].
+The result follows by [[Principle of Mathematical Induction|induction]].
+{{qed}}
+[[Category:Total Orderings]]
+[[Category:Ordered Sums]]
+422v3esqiz5ug0v76ygkcvqvi1lkwuq
+\end{proof}<|endoftext|>
+\section{Supremum of Suprema}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $\mathbb T \subseteq \powerset S$, where $\powerset S$ is the [[Definition:Power Set|power set]] of $S$.
+Suppose all $T \in \mathbb T$ admit a [[Definition:Supremum of Set|supremum]] $\sup T$ in $S$.
+Then:
+:$\sup \bigcup \mathbb T = \sup {\set {\sup T: T \in \mathbb T} }$
+if one of these two quantities exists (in $S$).
+\end{theorem}
+
+\begin{proof}
+Suppose that $s = \sup \bigcup \mathbb T \in S$.
+By [[Set is Subset of Union]], $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
+Hence by [[Supremum of Subset]]:
+:$\forall T \in \mathbb T: \sup T \preceq s$
+Suppose now that $a \in S$ satisfies:
+:$\forall T \in \mathbb T: \sup T \preceq a$
+Then by [[Definition:Transitive Relation|transitivity]] of $\preceq$:
+:$\forall t \in T: t \preceq a$
+Since this holds for any $T \in \mathbb T$, also:
+:$\forall t \in \bigcup \mathbb T: t \preceq a$
+Hence $s \preceq a$, by definition of [[Definition:Supremum of Set|supremum]].
+That is, $s = \sup {\set {\sup T: T \in \mathbb T} }$.
+{{qed|lemma}}
+Suppose now that $r = \sup {\set {\sup T: T \in \mathbb T} } \in S$.
+By definition of [[Definition:Supremum of Set|supremum]], for all $T \in \mathbb T$ and $t \in T$:
+:$t \preceq \sup T$
+By [[Definition:Transitive Relation|transitivity]] of $\preceq$:
+:$\forall T \in \mathbb T: \forall t \in T: t \preceq r$
+Hence for all $t \in \bigcup \mathbb T$:
+:$t \preceq r$
+Suppose that $a \in S$ satisfies:
+:$\forall t \in \bigcup \mathbb T: t \preceq a$
+In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$:
+:$\sup T \preceq a$
+and therefore by definition of [[Definition:Supremum of Set|supremum]], also:
+:$r \preceq a$
+That is, $r = \sup \bigcup \mathbb T$.
+{{qed}}
+[[Category:Order Theory]]
+2fhnbjumyyvbt8a4ykp6kutfy4l6h0w
+\end{proof}<|endoftext|>
+\section{Ordered Product of Tosets is Totally Ordered Set/General Result}
+Tags: Total Orderings, Ordered Products
+
+\begin{theorem}
+Let $S_1, S_2, \ldots, S_n$ all be [[Definition:Toset|tosets]].
+Let $T_n$ be the [[Definition:Ordered Product|ordered product]] of $S_1, S_2, \ldots, S_n$:
+:$\forall n \in \N_{>0}: T_n = \begin{cases}
+S_1 & : n = 1 \\
+T_{n-1} \cdot S_n & : n > 1
+\end{cases}$
+Then $T_n$ is a [[Definition:Toset|toset]].
+\end{theorem}
+
+\begin{proof}
+From [[Ordered Product of Tosets is Totally Ordered Set]], $S_1 \cdot S_2$ is a [[Definition:Toset|toset]].
+Suppose $T_{n-1}$ is a [[Definition:Toset|toset]].
+Given that $S_n$ is a [[Definition:Toset|toset]], it follows from [[Ordered Product of Tosets is Totally Ordered Set]] that $T_{n-1} \cdot S_n$ is also a [[Definition:Toset|toset]].
+The result follows by [[Principle of Mathematical Induction|induction]].
+{{qed}}
+[[Category:Total Orderings]]
+[[Category:Ordered Products]]
+1902bsmwyduw1aka6zdb5jqe3xfbpig
+\end{proof}<|endoftext|>
+\section{Derivative of Complex Composite Function}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: D \to \C$ be a [[Definition:Complex-Differentiable Function|complex-differentiable function]], where $D \subseteq \C$ is an [[Definition:Open Set (Complex Analysis)|open set]].
+Let $g: \operatorname{Im} \left({f}\right) \to \C$ be a [[Definition:Complex-Differentiable Function|complex-differentiable function]], where $\operatorname{Im} \left({f}\right)$ denotes the [[Definition:Image of Mapping|image]] of $f$.
+Define $h = f \circ g: D \to C$ as the [[Definition:Composite Function|composite]] of $f$ and $g$.
+Then $h$ is [[Definition:Complex-Differentiable Function|complex-differentiable]], and its [[Definition:Derivative of Complex Function|derivative]] is defined as:
+:$\forall z \in D: h' \left({z}\right) =  f' \left({g \left({z}\right)}\right) g' \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+{{MissingLinks|among general reference to results, maybe also some links to details on this "method of variations" approach}}
+Put $y = g \left({z}\right)$.
+Let $\delta z \in \C \setminus \left\{ {0} \right\}$.
+Put $\delta y = g \left({z + \delta z}\right) - y$, so:
+:$g \left({z + \delta z}\right) = y + \delta y$
+As $\delta z \to 0$, we have:
+:$(1): \quad \delta y \to 0$ by definition of [[Definition:Continuous Complex Function|continuity]], as $g$ is [[Definition:Continuous Complex Function|continuous]].
+:$(2): \quad \dfrac {\delta y} {\delta z} \to g'\left({z}\right)$ by definition of [[Definition:Complex-Differentiable Function|complex-differentiability]].
+There are two cases to consider:
+=== Case 1 ===
+Suppose $g' \left({z}\right) \ne 0$.
+When $\delta z \to 0$, we have $\delta y \ne 0$ from $(2)$, if $\left\vert{\delta z}\right\vert$ is sufficiently small.
+Then:
+{{begin-eqn}}
+{{eqn | l = \lim_{\delta z \mathop \to 0} \frac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z}
+      | r = \lim_{\delta z \mathop \to 0} \frac {f \left({g \left({z + \delta z}\right)}\right) - f \left({g \left({z}\right)}\right)} {g \left({z + \delta z}\right) - g \left({z}\right)} \frac {g \left({z + \delta z}\right) - g \left({z}\right)} {\delta z}
+      | c = 
+}}
+{{eqn | r = \lim_{\delta z \mathop \to 0} \frac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \frac {\delta y} {\delta z}
+      | c = 
+}}
+{{eqn | r = f' \left({y}\right) g^{\prime} \left({z}\right)
+      | c = As $\delta y \to 0$, then $\dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \to f^{\prime} \left({y}\right)$
+}}
+{{end-eqn}}
+hence the result.
+{{qed|lemma}}
+=== Case 2 ===
+Now suppose $g' \left({z}\right) = 0$.
+When $\delta z \to 0$, there are two possibilities:
+=== Case 2a ===
+If $\delta y = 0$, then:
+:$\dfrac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} = 0 = f' \left({y}\right) g' \left({z}\right)$
+Hence the result.
+{{qed|lemma}}
+=== Case 2b ===
+If $\delta y \ne 0$, then:
+:$\dfrac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} = \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \dfrac {\delta y} {\delta z}$
+As $\delta y \to 0$:
+: $(1): \quad \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \to f' \left({y}\right)$
+: $(2): \quad \dfrac {\delta y} {\delta z} \to 0$
+Thus:
+: $\displaystyle \lim_{\delta z \mathop \to 0} \frac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} \to 0 = f' \left({y}\right) g' \left({z}\right)$
+Again, hence the result.
+{{qed|lemma}}
+All cases have been covered, so by [[Proof by Cases]], the result is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Idempotents is Idempotent Ring}
+Tags: Ring Theory, Idempotent Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\struct {A, \oplus, \circ}$ be its [[Definition:Ring of Idempotents|ring of idempotents]].
+Then $\struct {A, \oplus, \circ}$ is an [[Definition:Idempotent Ring|idempotent ring]].
+\end{theorem}
+
+\begin{proof}
+First, it is to be established that $\struct {A, \oplus, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]] in the first place.
+This we do by verifying the [[Definition:Ring Axioms|ring axioms]].
+=== Axiom $(\text A 0)$: Closure for $\oplus$ ===
+Let $x, y \in A$.
+It is to be shown that $x \oplus y \in A$, i.e. that $x \oplus y$ is an [[Definition:Idempotent Element|idempotent element]] of $R$.
+Compute as follows:
+{{begin-eqn}}
+{{eqn | l = \paren {x \oplus y} \circ \paren {x \oplus y}
+      | r = \paren {x + y - 2 x \circ y} \circ \paren {x + y - 2 x \circ y}
+      | c = [[Definition:Ring of Idempotents|Definition of $\oplus$]]
+}}
+{{eqn | r = \paren {x + y - 2 x \circ y} \circ x + \paren {x + y - 2 x \circ y} \circ y - \paren {x + y - 2 x \circ y} \circ \paren {2 x \circ y}
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = x \circ x + y \circ x - \paren {2 x \circ y} \circ x + x \circ y + y \circ y - \paren {2 x \circ y} \circ y
+}}
+{{eqn | o = 
+      | ro= -
+      | r = x \circ \paren {2 x \circ y} - y \circ \paren {2 x \circ y} + \paren {2 x \circ y} \circ \paren {2 x \circ y}
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = x \circ x + y \circ x - 2 x \circ y \circ x + x \circ y + y \circ y - 2 x \circ y \circ y
+}}
+{{eqn | o = 
+      | ro= -
+      | r = 2 x \circ x \circ y - 2 y \circ x \circ y + 4 x \circ y \circ x \circ y
+      | c = [[Product of Integral Multiples]]
+}}
+{{eqn | r = x + x \circ y - 2 x \circ y + x \circ y + y - 2 x \circ y - 2 x \circ y - 2 x \circ y + 4 x \circ y
+      | c = $\circ$ is [[Definition:Commutative Operation|commutative]], $x, y$ are [[Definition:Idempotent Element|idempotent]]
+}}
+{{eqn | r = x + y - 2 x \circ y
+      | c = [[Integral Multiple Distributes over Ring Addition]]
+}}
+{{eqn | r = x \oplus y
+      | c = [[Definition:Ring of Idempotents|Definition of $\oplus$]]
+}}
+{{end-eqn}}
+Hence $x \oplus y \in A$, as desired.
+{{qed|lemma}}
+=== Axiom $(\text A 1)$: Associativity of $\oplus$ ===
+Let $x, y, z \in A$.
+It is to be shown that $\oplus$ is [[Definition:Associative|associative]], that is:
+:$\paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z}$
+This is shown by the following computation:
+{{begin-eqn}}
+{{eqn | l = \paren {x \oplus y} \oplus z
+      | r = \paren {x \oplus y} + z - 2 \paren {x \oplus y} \circ z
+}}
+{{eqn | r = x + y - 2 x \circ y + z - 2 \paren {x + y - 2 x \circ y} \circ z
+}}
+{{eqn | r = x + y + z - 2 x \circ y - 2 x \circ z - 2 y \circ z + 4 x \circ y \circ z
+}}
+{{eqn | r = x + \paren {y + z - 2 y \circ z} - 2 x \circ \paren {y + z - 2 y \circ z}
+}}
+{{eqn | r = x + \paren {y \oplus z} - 2 x \circ \paren {y \oplus z}
+}}
+{{eqn | r = x \oplus \paren {y \oplus z}
+}}
+{{end-eqn}}
+{{finish|justify equalities}}
+{{qed|lemma}}
+=== Axiom $(\text A 2)$: Commutativity of $\oplus$ ===
+Let $x, y \in A$.
+It is to be shown that $\oplus$ is [[Definition:Commutative|commutative]], that is:
+:$x \oplus y = y \oplus x$
+This is shown by the following computation:
+{{begin-eqn}}
+{{eqn | l = x \oplus y
+      | r = x + y - 2 x \circ y
+}}
+{{eqn | r = y + x - 2 y \circ x
+      | c = $+, \circ$ are [[Definition:Commutative Operation|commutative]]
+}}
+{{eqn | r = y \oplus x
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Axiom $(\text A 3)$: Identity for $\oplus$ ===
+Let $x \in A$, and let $0_R$ be the [[Definition:Ring Zero|zero]] of $R$.
+By [[Ring Zero is Idempotent]], $0_R \in A$.
+It suffices to show that:
+:$x \oplus 0_R = x$
+since {{Ring-axiom|A2}} was already verified above.
+Now:
+{{begin-eqn}}
+{{eqn | l = x \oplus 0_R
+      | r = x + 0_R - 2 x \circ 0_R
+}}
+{{eqn | r = x
+      | c = [[Ring Product with Zero]]
+}}
+{{end-eqn}}
+as desired.
+{{qed|lemma}}
+=== Axiom $(\text A 4)$: Inverses for $\oplus$ ===
+Let $x \in A$.
+It is to be shown that there exists $y \in A$ such that:
+:$x \oplus y = y \oplus x = 0_R$
+by $(A3)$ above.
+In fact, one has:
+{{begin-eqn}}
+{{eqn | l = x \oplus x
+      | r = x + x - 2 x \circ x
+}}
+{{eqn | r = 2 x - 2 x
+      | c = $x$ is an [[Definition:Idempotent Element|idempotent element]]
+}}
+{{eqn | r = 0_R
+}}
+{{end-eqn}}
+so that each $x \in A$ is its own [[Definition:Inverse Element|inverse]] for $\oplus$.
+{{qed|lemma}}
+=== Axiom $(\text M 0)$: Closure for $\circ$ ===
+Let $x, y \in A$.
+It is to be shown that:
+:$x \circ y \in A$
+that is, that $x \circ y$ is [[Definition:Idempotent Element|idempotent]].
+We have the following computation:
+{{begin-eqn}}
+{{eqn | l = \paren {x \circ y} \circ \paren {x \circ y}
+      | r = x \circ x \circ y \circ y
+      | c = $\circ$ is [[Definition:Associative|associative]] and [[Definition:Commutative Operation|commutative]]
+}}
+{{eqn | r = x \circ y
+      | c = $x, y$ are [[Definition:Idempotent Element|idempotent elements]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Axiom $(\text M 1)$: Associativity of $\circ$ ===
+Immediate from [[Restriction of Associative Operation is Associative]].
+{{qed|lemma}}
+=== Axiom $(\text D)$: Distributivity ===
+By [[Restriction of Commutative Operation is Commutative]], $\circ$ is [[Definition:Commutative Operation|commutative]] on $A$.
+Thus to establish [[Definition:Distributive Operation|distributivity]], it suffices to verify, for $x, y, z \in A$:
+:$x \circ \paren {y \oplus z} = \paren {x \circ y} \oplus \paren {x \circ z}$
+To this end, we compute as follows:
+{{begin-eqn}}
+{{eqn | l = x \circ \paren {y \oplus z}
+      | r = x \circ \paren {y + z - 2 y \circ z}
+}}
+{{eqn | r = x \circ y + x \circ z - 2 x \circ y \circ z
+}}
+{{eqn | r = x \circ y + x \circ z - 2 x \circ x \circ y \circ z
+      | c = $x$ is an [[Definition:Idempotent Element|idempotent element]]
+}}
+{{eqn | r = x \circ y + x \circ z - 2 x \circ y \circ x \circ z
+      | c = $\circ$ is [[Definition:Commutative Operation|commutative]]
+}}
+{{eqn | r = \paren {x \circ y} \oplus \paren {x \circ z}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Therefore, having verified all [[Definition:Ring Axioms|ring axioms]], we conclude $\struct {A, \oplus, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+By assumption all $x \in A$ are [[Definition:Idempotent Element|idempotent elements]] for $\circ$.
+Thus $\circ$ is an [[Definition:Idempotent Operation|idempotent operation]] on $A$.
+Consequently, $\struct {A, \oplus, \circ}$ is an [[Definition:Idempotent Ring|idempotent ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kelley's Theorem}
+Tags: Set Theory, Nets (Topology)
+
+\begin{theorem}
+Let $\struct {D, \preceq}$ be a [[Definition:Directed Set|directed set]],
+Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
+Let $n: D \to S$ be a [[Definition:Net (Set Theory)|net]] in $S$.
+Then $n$ has a [[Definition:Universal Net|universal]] [[Definition:Subnet|subnet]].
+\end{theorem}<|endoftext|>
+\section{Reparameterization of Contour is Contour}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $\closedint a b$ and $\closedint c d$ be [[Definition:Closed Real Interval|closed real intervals]].
+Let $\gamma: \closedint a b \to \C$ be a [[Definition:Contour (Complex Plane)|contour]] in $\C$.
+That is, there exists a [[Definition:Subdivision (Real Analysis)|subdivision]] $a_0, a_1 , \ldots, a_n$ of $\closedint a b$ such that $\gamma \restriction_{I_i}$ is a [[Definition:Smooth Path (Complex Analysis)|smooth path]] for all $i \in \set {1, \ldots, n}$, where $I_i = \closedint {a_{i - 1} } {a_i}$.
+Here, $\gamma \restriction_{I_i}$ denotes the [[Definition:Restriction of Mapping|restriction]] of $\gamma$ to $I_i$.
+Let $\phi: \closedint c d \to \closedint a b$  be a [[Definition:Differentiable on Interval|differentiable]] [[Definition:Real Function|real function]] with $\map \phi c = a$, and $\map \phi d = b$.
+Suppose that $\phi$ is either [[Definition:Bijection|bijective]] or [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Define $\sigma: \closedint c d \to \C$ as a [[Definition:Composition of Mappings|composite function]] by $\sigma = \gamma \circ \phi$.
+Then $\sigma$ is a [[Definition:Contour (Complex Plane)|contour]].
+For all $i \in \set {1, \ldots, n}$, let $J_i := \closedint {\map {\phi^{-1} } {a_{i - 1} } } {\map {\phi^{-1} } {a_i} }$.
+Then:
+:$\map {\sigma' \restriction_{J_i} } t = \map {\gamma' \restriction_{I_i} } {\map \phi t} \map {\phi'} t$
+\end{theorem}
+
+\begin{proof}
+$\phi$ is either [[Definition:Bijection|bijective]] or [[Definition:Strictly Increasing Real Function|strictly increasing]].
+From [[Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone]], it follows that in both cases $\phi$ is both [[Definition:Bijection|bijective]] and [[Definition:Strictly Monotone Real Function|strictly monotone]].
+As $\map \phi c < \map \phi d$, $\phi$ must be [[Definition:Strictly Increasing Real Function|strictly increasing]].
+As $\phi$ is [[Definition:Strictly Increasing Real Function|strictly increasing]], we have $c = \map {\phi^{-1} } {a_0} < \map {\phi^{-1} } {a_1} < \cdots < \map {\phi^{-1} } {a_n} = d$.
+Hence, $\map {\phi^{-1} } {a_0}, \map {\phi^{-1} } {a_1}, \dotsc, \map {\phi^{-1} } {a_n}$ form a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$.
+Define $J_i := \closedint {\map {\phi^{-1} } {a_{i - 1} } } {\map {\phi^{-1} } {a_i} }$ as a [[Definition:Closed Real Interval|closed real interval]].
+From [[Derivative of Complex Composite Function]], it follows that $\sigma$ [[Definition:Restriction of Mapping|restricted]] to $J_i$ is [[Definition:Complex-Differentiable Function|complex differentiable]].
+Its [[Definition:Derivative of Complex Function|derivative]] is given by:
+:$\map {\sigma' \restriction_{J_i} } t = \map {\gamma' \restriction_{I_i} } {\map \phi t} \map {\phi'} t$
+From [[Derivative of Monotone Function]], it follows that for all $t \in \closedint c d$, we have $\map {\phi'} t > 0$.
+As $\map {\gamma' \restriction_{I_i} } {\map \phi t} \ne 0$ by definition of [[Definition:Contour (Complex Plane)|contour]], we have $\map {\sigma' \restriction_{J_i} } t \ne 0$.
+Then $\sigma$ fulfils both requirement $(1)$ and $(2)$ from the definition of [[Definition:Contour (Complex Plane)|contour]], so $\sigma$ is a contour.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Contour Integral is Independent of Parameterization}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C$ be a [[Definition:Contour (Complex Plane)|contour]] defined by a [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$ of [[Definition:Directed Smooth Curve|directed smooth curves]].
+Let $C_i$ be [[Definition:Directed Smooth Curve/Parameterization|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Let $f: \operatorname{Im} \left({C}\right) \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex function]], where $\operatorname{Im} \left({C}\right)$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C$.
+Suppose that $\sigma_i: \left[{c_i \,.\,.\, d_i}\right] \to \C$ is a [[Definition:Directed Smooth Curve/Parameterization|reparameterization]] of $C_i$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Then:
+:$\displaystyle \int_C f \left({z}\right) \rd z = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} f \left({\sigma_i \left({t}\right) }\right) \sigma_i' \left({t}\right) \rd t$
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Directed Smooth Curve/Parameterization|definition of parameterization]], $\gamma_i = \sigma_i \circ \phi_i$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Here, $\phi_i: \left[{c_i \,.\,.\, d_i}\right] \to \left[{a_i \,.\,.\, b_i}\right]$ is a [[Definition:Bijection|bijective]] [[Definition:Differentiable on Interval|differentiable]] [[Definition:Strictly Increasing Real Function|strictly increasing function]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_C f \left({z}\right) \rd z 
+      | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{\phi_i^{-1} \left({a_i}\right) }^{\phi_i^{-1} \left({b_i}\right)} f \left({\gamma_i \left({\phi_i \left({u}\right) }\right) }\right) \gamma_i' \left({\phi_i \left({u}\right) }\right) \phi_i' \left({u}\right) \rd u
+      | c = [[Complex Integration by Substitution|substitution]] with $t = \phi \left({u}\right)$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{\phi_i^{-1} \left({a_i}\right) }^{\phi_i^{-1} \left({b_i}\right) } f \left({ \sigma_i \left({u}\right) }\right) \sigma_i' \left({u}\right) \rd u
+      | c = [[Derivative of Complex Composite Function]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} f \left({ \sigma_i \left({u}\right) }\right) \sigma_i' \left({u}\right) \rd u
+      | c = [[Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unity is Unity in Ring of Idempotents}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative and unitary ring]] whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $\left({A, \oplus, \circ}\right)$ be the [[Definition:Ring of Idempotents|ring of idempotents]] of $R$.
+Then $1_R$ is also a [[Definition:Unity of Ring|unity]] for $\left({A, \oplus, \circ}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Unity of Ring is Idempotent]], $1_R$ is an [[Definition:Idempotent Element|idempotent element]] of $R$.
+Hence $1_R \in A$.
+Recall that the [[Definition:Ring Product|ring product]] of $A$ is a [[Definition:Restriction of Operation|restriction]] from that of $R$.
+Hence, for each $x \in A$:
+:$x \circ 1_R = x = 1_R \circ x$
+so that $1_R$ is a [[Definition:Unity of Ring|unity]] for $A$, as desired.
+{{qed}}
+[[Category:Ring Theory]]
+nspjhifbhflpwipy9lij3cnpga97hy1
+\end{proof}<|endoftext|>
+\section{Ring of Idempotents of Commutative and Unitary Ring is Boolean Ring}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative and unitary ring]].
+Let $\struct {A, \oplus, \circ}$ be its [[Definition:Ring of Idempotents|ring of idempotents]].
+Then $\struct {A, \oplus, \circ}$ is a [[Definition:Boolean Ring|Boolean ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Idempotents is Idempotent Ring]], $\struct {A, \oplus, \circ}$ is an [[Definition:Idempotent Ring|idempotent ring]].
+By [[Unity is Unity in Ring of Idempotents]], $\struct {A, \oplus, \circ}$ is also a [[Definition:Unitary Ring|unitary ring]].
+Hence, by definition, $\struct {A, \oplus, \circ}$ is a [[Definition:Boolean Ring|Boolean ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Boolean Group}
+Tags: Definitions: Group Theory
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Then $\struct {G, \circ}$ is a '''Boolean group''' {{iff}} all its [[Definition:Element|elements]], other than the [[Definition:Identity Element|identity]], have [[Definition:Order of Group Element|order]] $2$.
+\end{theorem}<|endoftext|>
+\section{Boolean Ring has Proper Zero Divisor}
+Tags: Boolean Rings
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Boolean Ring|Boolean ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Suppose that $R$ has more than two elements.
+Then $R$ has a [[Definition:Proper Zero Divisor|proper zero divisor]].
+\end{theorem}
+
+\begin{proof}
+Since $R$ has more than two elements, there exist distinct non-[[Definition:Ring Zero|zero]] elements $x, y \in R$.
+Note that $x + y \ne 0_R$ since $x$ and $y$ are distinct (by [[Idempotent Ring has Characteristic Two]]).
+If $x \circ y = 0$, $x$ is a [[Definition:Proper Zero Divisor|proper zero divisor]].
+If $x \circ y \ne 0$, then:
+{{begin-eqn}}
+{{eqn|l = \left({x \circ y}\right) \circ \left({x + y}\right)
+     |r = x \circ y \circ x + x \circ y \circ y
+}}
+{{eqn|r = x \circ y + x \circ y
+     |c = $R$ is an [[Definition:Idempotent Ring|idempotent ring]], [[Idempotent Ring is Commutative]]
+}}
+{{eqn|r = 0_R
+     |c = [[Idempotent Ring has Characteristic Two]]
+}}
+{{end-eqn}}
+Hence $x \circ y$ is a [[Definition:Proper Zero Divisor|proper zero divisor]].
+The result follows, from [[Proof by Cases]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone}
+Tags: Continuous Functions, Real Functions, Monotone Mappings, Bijections
+
+\begin{theorem}
+Let $\closedint a b$ and $\closedint c d$ be [[Definition:Closed Real Interval|closed real intervals]].
+Let $f: \closedint c d \to \closedint a b$ be a [[Definition:Continuous Real Function|continuous real function]].
+Let $\map f c, \map f d \in \set {a, b}$.
+Then $f$ is [[Definition:Bijection|bijective]] {{iff}} $f$ is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary condition ===
+Let $f$ be a [[Definition:Bijection|bijection]].
+From [[Continuous Injection of Interval is Strictly Monotone]], it follows that $f$ is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+{{qed|lemma}}
+=== Sufficient condition ===
+Let $f$ be [[Definition:Strictly Monotone Real Function|strictly monotone]].
+From [[Strictly Monotone Real Function is Bijective]], it follows that $f$ is a [[Definition:Bijection|bijection]] on its [[Definition:Image of Mapping|image]].
+From [[Image of Interval by Continuous Function is Interval]], it follows that the [[Definition:Image of Mapping|image]] of $f$ is a [[Definition:Real Interval|real interval]].
+Suppose that $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Then by definition $\map f c < \map f d$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \map f c
+      | r = a
+}}
+{{eqn | l = \map f d
+      | r = b
+}}
+{{end-eqn}}
+It follows that for all $x \in \closedint c d$:
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | o = \ge
+      | r = \map f c
+}}
+{{eqn | r = a
+}}
+{{eqn | l = \map f x
+      | o = \le
+      | r = \map f d
+}}
+{{eqn | r = b
+}}
+{{end-eqn}}
+Thus, the image of $f$ is $\closedint a b$.
+{{qed|lemma}}
+Suppose instead that $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+Then by definition $\map f c > \map f d$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \map f c
+      | r = b
+}}
+{{eqn | l = \map f d
+      | r = a
+}}
+{{end-eqn}}
+It follows that for all $x \in \closedint c d$:
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | o = \le
+      | r = \map f c
+}}
+{{eqn | r = b
+}}
+{{eqn | l = \map f x
+      | o = \ge
+      | r = \map f d
+}}
+{{eqn | r = a
+}}
+{{end-eqn}}
+Thus the image of $f$ is again $\closedint a b$.
+{{qed}}
+[[Category:Continuous Functions]]
+[[Category:Real Functions]]
+[[Category:Monotone Mappings]]
+[[Category:Bijections]]
+9qldbiiz1uw3zwfxzb3epczq8pqb78p
+\end{proof}<|endoftext|>
+\section{Concatenation of Contours is Contour}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C$ and $D$ be [[Definition:Contour (Complex Plane)|contours]].
+That is, $C$ is a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Directed Smooth Curve|directed smooth curves]] $C_1, \ldots, C_n$.
+Let $C_i$ be [[Definition:Directed Smooth Curve/Parameterization|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_i: \left[{a_i\,.\,.\,b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Similarly, $D$ is a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Directed Smooth Curve|directed smooth curves]] $D_1, \ldots, D_m$.
+Let $D_i$ be [[Definition:Directed Smooth Curve/Parameterization|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\sigma_i: \left[{c_i\,.\,.\,d_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, m}\right\}$.
+Suppose $\gamma_n \left({b_n}\right) = \sigma_1 \left({c_1}\right)$.
+Then the finite sequence:
+:$C_1, \ldots, C_n, D_1, \ldots, D_m$
+defines a [[Definition:Contour (Complex Plane)|contour]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Contour (Complex Plane)|contour]], each $C_i$ and $D_j$ is a [[Definition:Directed Smooth Curve|directed smooth curve]] for all $i \in \left\{ {1, \ldots, n}\right\}, j \in \left\{ {1, \ldots, m}\right\}$.
+By definition of [[Definition:Contour (Complex Plane)|contour]], $\gamma_i \left({b_i}\right) = \gamma_{i+1} \left({a_{i+1} }\right)$ and $\sigma_j \left({d_j}\right) = \sigma_{j+1} \left({c_{j+1} }\right)$ for all $i \in \left\{ {1, \ldots, n-1}\right\}, j \in \left\{ {1, \ldots, m-1}\right\}$.
+By assumption, $\gamma_n \left({b_n}\right) = \sigma_1 \left({c_1}\right)$.
+Hence, $C_1, \ldots, C_n, D_1, \ldots, D_m$ defines a [[Definition:Contour (Complex Plane)|contour]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Contour Integral of Concatenation of Contours}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C$ and $D$ be [[Definition:Contour (Complex Plane)|contours]] in $\C$.
+Suppose that the [[Definition:End Point of Contour (Complex Plane)|end point]] of $C$ is equal to the [[Definition:Start Point of Contour (Complex Plane)|start point]] of $D$, so the [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] $C \cup D$ is defined.
+Let $f: \operatorname{Im} \left({C \cup D}\right) \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex function]], where $\operatorname{Im} \left({C \cup D}\right)$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C \cup D$.
+Then:
+:$\displaystyle \int_{C \cup D} f \left({z}\right) \ \mathrm dz = \int_C f \left({z}\right) \rd z + \int_D f \left({z}\right) \rd z$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Contour (Complex Plane)|contour]], $C$ is 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]] $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+Similarly, $D$ is a [[Definition:Finite Sequence|finite sequence]] $D_1, \ldots, D_m$ of [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]].
+Let $D_j$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\sigma_j: \left[{c_j \,.\,.\, d_j}\right] \to \C$ for all $j \in \left\{ {1, \ldots, m}\right\}$.
+Put $\gamma_i = \sigma_{i - n}, a_i = c_{i - n}$ and $b_i = d_{i - n}$ for all $i \in \left\{ {n + 1, \ldots, n + m}\right\}$.
+Then $C \cup D$ is a sequence of $n + m$ [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curves]] which are [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by $\gamma_1, \ldots, \gamma_{n + m}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \displaystyle \int_{C \cup D} f \left({z}\right) \rd z
+      | r = \sum_{i \mathop = 1}^{n + m} \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right) \rd t + \sum_{j \mathop = 1}^m \int_{c_j}^{d_j} f \left({\sigma_j \left({t}\right) }\right) \sigma_j' \left({t}\right) \rd t
+}}
+{{eqn | r = \int_C f \left({z}\right) \rd z + \int_D f \left({z}\right) \rd z
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Regular Expression is Accepted by Finite State Machine}
+Tags: Formal Systems
+
+\begin{theorem}
+Let $R$ be a [[Definition:Regular Expression|regular expression]].
+Then there exists a [[Definition:Finite State Machine|finite state machine]] $F$ such that its accepted language $L \left({F}\right)$ is exactly $L \left({R}\right)$, the language defined by $R$.
+\end{theorem}
+
+\begin{proof}
+This proof proceeds by structural induction.
+=== Case $1$ ===
+Let $R$ be the [[Definition:Empty Set|empty-set]] [[Definition:Regular Expression|regular expression]], $\varnothing$.
+Then:
+:$L \left({R}\right) = \varnothing$
+Consider the [[Definition:Finite State Machine|finite state machine]] $F_\varnothing$ defined as:
+:$F_\varnothing = \left({S_\varnothing, A_\varnothing, I_\varnothing, \Sigma, T_\varnothing}\right)$
+where:
+: $S_\varnothing = \left\{ { \mathsf{Rej} }\right\}$
+: $A_\varnothing = \varnothing$
+: $I_\varnothing = \mathsf{Rej}$
+: $T_\varnothing \left({ s, \sigma }\right) = \mathsf{Rej}$ for all $s \in S_\varnothing, \sigma \in \Sigma$
+This machine is always in a rejecting state and never leaves it.
+So no word is in $L \left({ F_\varnothing }\right)$.
+Therefore:
+:$L \left({ F_\varnothing }\right) = \varnothing = L \left({R}\right)$
+{{qed|lemma}}
+=== Case $2$ ===
+Let $R$ be the [[Definition:Null String|empty-word]] [[Definition:Regular Expression|regular expression]], $\epsilon$.
+Then:
+:$L \left({R}\right) = \left\{ { \left[\right] }\right\}$
+Consider the [[Definition:Finite State Machine|finite state machine]] $F_\epsilon$ defined as:
+: $F_\epsilon = \left({ S_\epsilon, A_\epsilon, I_\epsilon, \Sigma, T_\epsilon }\right)$
+where:
+: $S_\epsilon = \left\{ { \mathsf{Acc}, \mathsf{Rej} }\right\}$
+: $A_\epsilon = \left\{ { \mathsf{Acc} }\right\}$
+: $I_\epsilon = \mathsf{Acc}$
+: $T_\epsilon \left({ s, \sigma }\right) = \mathsf{Rej}$ for all $s \in S_\epsilon, \sigma \in \Sigma$
+This machine starts out in an accepting state.
+So $\left[\right]$ (the [[Definition:Null String|empty word]] is in $L \left({ F_\epsilon }\right)$.
+Furthermore, any symbol moves the machine to a rejecting state and never back.
+So no other word is in $L \left({ F_\epsilon }\right)$.
+Therefore:
+:$L \left({ F_\epsilon }\right) = \left\{ { \left[\right] }\right\} = L \left({R}\right)$
+{{qed|lemma}}
+=== Case $3$ ===
+Let $R$ be a ''literal'' $\sigma$.
+Then:
+:$L \left({R}\right) = \left\{ { \left[{\sigma}\right] }\right\}$
+Consider the [[Definition:Finite State Machine|finite state machine]] $F_\sigma$ defined as:
+: $F_\sigma = \left({ S_\sigma, A_\sigma, I_\sigma, \Sigma, T_\sigma }\right)$
+where:
+: $S_\sigma = \left\{ { \mathsf{Start}, \mathsf{Acc}, \mathsf{Rej} }\right\}$
+: $A_\sigma = \left\{ { \mathsf{Acc} }\right\}$
+: $I_\sigma = \mathsf{Start}$
+: $T_\sigma \left({ \mathsf{Start}, \sigma }\right) = \mathsf{Acc}$
+: $T_\sigma \left({ s', \sigma' }\right) = \mathsf{Rej}$ for all other $s' \in S_\sigma, \sigma' \in \Sigma$
+This machine starts out in a rejecting state.
+So $\left[{}\right]$ (the [[Definition:Null String|empty word]]) is not in $L \left({ F_\sigma }\right)$.
+After receiving the symbol $\sigma$ at the start, this machine moves to an accepting state.
+So $\left[{\sigma}\right]$ is in $L \left({ F_\sigma }\right)$.
+Any other initial symbol, and any symbol after the initial, moves the machine to a rejecting state and never back.
+So no other word is in $L \left({ F_\sigma }\right)$.
+Therefore:
+:$L \left({ F_\sigma }\right) = \left\{ { \left[{\sigma}\right] }\right\} = L \left({R}\right)$
+{{qed|lemma}}
+=== Case $4$ ===
+Let $R$ be a concatenation:
+:$R = R_1 R_2$
+By the induction hypothesis, there exist [[Definition:Finite State Machine|finite state machine]]s:
+: $F_1 = \left({ S_1, A_1, I_1, \Sigma, T_1 }\right): L \left({F_1}\right) = L \left({R_1}\right)$
+: $F_2 = \left({ S_2, A_2, I_2, \Sigma, T_2 }\right): L \left({F_2}\right) = L \left({R_2}\right)$
+Define a new [[Definition:Finite State Machine|finite state machine]] $F_c$ as:
+: $F_c = \left({ S_c, A_c, I_c, \Sigma, T_c }\right) $
+where:
+: $S_c = S_1 \times \mathcal P \left({ S_2 }\right)$
+: $A_c = \left\{ { \left({ s_1, s_2 }\right) : s_1 \in S_1 \land s_2 \cap A_2 \neq \varnothing }\right\}$
+: $I_c = \begin{cases} \left({ I_1, \varnothing }\right) & : I_1 \notin A_1 \\ \left({ I_1, \left\{ {I_2} \right\} }\right) & : I_1 \in A_1 \end{cases}$
+: $\displaystyle T_c \left({ \left({ s_1, s_2 }\right), \sigma }\right) = \begin{cases} \left({ T_1 \left({ s_1, \sigma }\right), \bigcup_{s \in s_2} \left\{ { T_2 \left({ s, \sigma }\right) }\right\} }\right) & : T_1 \left({ s_1, \sigma }\right) \notin A_1 \\ \left({ T_1 \left({ s_1, \sigma }\right), \bigcup_{s \in s_2} \left\{ { T_2 \left({ s, \sigma }\right) }\right\} \cup \left\{ {I_2} \right\} }\right) & : T_1 \left({ s_1, \sigma }\right) \in A_1 \end{cases}$
+where:
+: $\times$ denotes the [[Definition:Cartesian Product|Cartesian Product]]
+: $\mathcal P$ the [[Definition:Power Set|Power Set]].
+This machine $F_c$ effectively simulates one copy of $F_1$ and any number of copies of $F_2$.
+Every time the simulated $F_1$ encounters an accepting state, a new copy of $F_2$ is run.
+The combined $F_c$ reaches an accepting state if any one of the simulated $F_2$s do.
+Therefore, the language accepted by this state machine is the concatenation of the accepted languages of $F_1$ and $F_2$.
+{{qed|lemma}}
+=== Case $5$ ===
+Let $R$ be an alternation:
+:$R = R_1 \mid R_2$
+By the induction hypothesis, there exist [[Definition:Finite State Machine|finite state machine]]s:
+: $F_1 = \left({ S_1, A_1, I_1, \Sigma, T_1 }\right): L \left({F_1}\right) = L \left({R_1}\right)$
+: $F_2 = \left({ S_2, A_2, I_2, \Sigma, T_2 }\right): L \left({F_2}\right) = L \left({R_2}\right)$
+Define a new [[Definition:Finite State Machine|finite state machine]] $F_a$ as:
+: $F_a = \left({ S_a, A_a, I_a, \Sigma, T_a }\right)$
+where:
+: $S_a = S_1 \times S_2$
+: $A_a = \left\{ { \left({ s_1, s_2 }\right) : s_1 \in A_1 \lor s_2 \in A_2 }\right\}$
+: $I_a = \left({ I_1, I_2 }\right)$
+: $T_a \left({ \left({ s_1, s_2 }\right), \sigma }\right) = \left({ T_1 \left({ s_1, \sigma }\right), T_2 \left({ s_2, \sigma }\right) }\right)$
+where $\times$ denotes the [[Definition:Cartesian Product|Cartesian Product]].
+This machine $F_a$ effectively simulates $F_1$ and $F_2$ in parallel.
+$F_a$ reaches an accepting state if any one of the simulated machines do.
+Therefore, the language accepted by this state machine is the union of the accepted languages of $F_1$ and $F_2$.
+{{qed|lemma}}
+=== Case $6$ ===
+Let $R$ be a Kleene star:
+:$R = R_1^*$
+By the induction hypothesis, there exists a [[Definition:Finite State Machine|finite state machine]]:
+: $F_1 = \left({ S_1, A_1, I_1, \Sigma, T_1 }\right): L \left({F_1}\right) = L \left({R_1}\right)$
+Define a new [[Definition:Finite State Machine|finite state machine]] $F_k$ as:
+: $F_k = \left({ S_k, A_k, I_k, \Sigma, T_k }\right)$
+where:
+: $S_k = \mathcal P \left({ S_1 }\right)$
+: $A_k = \left\{ { S \subseteq S_k : I_1 \in S }\right\}$
+: $I_k = \left\{ {I_1} \right\} $
+: $\displaystyle T_k \left({ S, \sigma }\right) = \begin{cases} U_k \left({ S, \sigma }\right) & : U_k \left({ S, \sigma }\right) \cap A_1 = \varnothing \\ U_k \left({ S, \sigma }\right) \cup \left\{ {I_1} \right\} & : U_k \left({ S, \sigma }\right) \cap A_1 \neq \varnothing \end{cases}$
+: $U_k \left({ S, \sigma }\right) = \bigcup_{s \in S} \left\{ { T_1 \left({ s, \sigma }\right) }\right\}$
+where $\mathcal P$ denotes the [[Definition:Power Set|Power Set]].
+This machine $F_k$ effectively simulates any number of copies of $F_1$ simultaneously.
+Every time any of the simulated machines reaches an accepting state, a new copy is run.
+$F_k$ reaches an accepting state whenever $I_1$ is in its state.
+This occurs in two situations:
+: at the beginning
+and:
+: when any of the simulated machines reaches an accepting state.
+Therefore, the language accepted by $F_k$ consists of arbitrary numbers of concatenations of strings accepted by $F_1$.
+By structural induction, the result follows.
+{{qed}}
+[[Category:Formal Systems]]
+tpfx4zb7heqe54c057iv27g1p7h5kye
+\end{proof}<|endoftext|>
+\section{Directed Smooth Curve Relation is Equivalence}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\sim$ denote a [[Definition:Endorelation|relation]] on the [[Definition:Set|set]] of all [[Definition:Smooth Path (Complex Analysis)|smooth paths]]: $\left\{ {\gamma: I \to \C : \text{$I$ is a closed real interval, $\gamma$ is a smooth path} }\right\}$.
+Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ and $\sigma : \left[{c \,.\,.\, d}\right] \to \C$ be smooth paths.
+Define $\sim$ as follows: 
+:$\gamma \sim \sigma$ {{iff}} there exists a [[Definition:Bijection|bijective]] [[Definition:Differentiable on Interval|differentiable]] [[Definition:Strictly Increasing Real Function|strictly increasing]] [[Definition:Real Function|real function]] $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ such that $\sigma = \gamma \circ \phi$.
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]] on the set of all smooth paths.
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for an [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+Let $\gamma: \left[{ a \,.\,.\, b }\right] \to \C$ be a [[Definition:Smooth Path (Complex Analysis)|smooth path]].
+Define $\phi: \left[{ a \,.\,.\, b }\right] \to \left[{ a \,.\,.\, b }\right]$ as the [[Definition:Identity Mapping|identity function]], so :
+$\phi \left({t}\right) = t $ for all $t \in \left[{ a \,.\,.\, b }\right]$
+From [[Identity Mapping is Bijection]], it follows that $\phi$ is [[Definition:Bijection|bijective]].
+From [[Derivative of Identity Function]], it follows that: $\phi' \left({t}\right) = 1 > 0$
+for all $t \in \left[{ a \,.\,.\, b }\right]$.
+Then [[Derivative of Monotone Function]] shows that $\phi$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Hence, $\gamma \sim \gamma$, so $\sim$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$ and $\sigma : \left[{ c \,.\,.\, d }\right] \to \C$ be [[Definition:Smooth Path (Complex Analysis)|smooth paths]] such that $\gamma \sim \sigma$.
+That is, $\sigma = \gamma \circ \phi$ where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is [[Definition:Bijection|bijective]], [[Definition:Differentiable on Interval|differentiable]] and [[Definition:Strictly Increasing Real Function|strictly increasing]].
+From [[Bijection iff Inverse is Bijection]], it follows that $\phi$ has an [[Definition:Inverse Mapping|inverse function]] $\phi^{-1}: \left[{a \,.\,.\, b}\right] \to \left[{c \,.\,.\, d}\right]$.
+Then $\gamma = \sigma \circ \phi^{-1}$.
+From [[Derivative of Inverse Function]]:
+: $D \phi^{-1} \left({t}\right) = \dfrac 1 {\phi' \left({\phi^{-1} \left({t}\right) }\right)}$
+for all $t \in \left[{a \,.\,.\, b}\right]$.
+From [[Derivative of Monotone Function]]:
+:$\phi' \left({\phi^{-1} \left({t}\right) }\right) > 0$
+Then:
+:$D \phi^{-1} \left({t}\right) > 0$
+so $\phi^{-1}$ is strictly increasing.
+Hence, $\sigma \sim \gamma$, so $\sim$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+Let $\gamma : \left[{ a \,.\,.\, b }\right] \to \C$, $\sigma : \left[{ c \,.\,.\, d }\right] \to \C$ and $\rho : \left[{ g \,.\,.\, h }\right] \to \C$ be [[Definition:Smooth Path (Complex Analysis)|smooth paths]] such that $\gamma \sim \sigma$ and $\sigma \sim \rho$.
+That is:
+: $\sigma = \gamma \circ \phi$ and $\rho = \sigma \circ \psi$
+where $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ and $\psi: \left[{g \,.\,.\, h}\right] \to \left[{c \,.\,.\, d}\right]$ are [[Definition:Bijection|bijective]], [[Definition:Differentiable on Interval|differentiable]] and [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Then:
+:$\rho = \gamma \circ \left({\phi \circ \psi}\right)$
+From [[Composite of Bijections is Bijection]], it follows that $\phi \circ \psi$ is bijective.
+From [[Derivative of Composite Function]], it follows that:
+:$D \left({\phi \circ \psi}\right) \left({t}\right) = \phi' \left({\psi \left({t}\right) }\right) \psi' \left({t}\right)$
+for all $t \in \left[{ g \,.\,.\, h }\right]$.
+As $\phi'\left({\psi \left({t}\right) }\right) > 0$ and $\psi' \left({t}\right) > 0$, this implies that:
+:$D \left({\phi \circ \psi}\right) \left({t}\right) > 0$
+so $\phi \circ \psi$ is strictly increasing.
+Hence, $\gamma \sim \rho$, so $\sim$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+Therefore, $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Max Operation Representation on Real Numbers}
+Tags: Analysis, Max and Min Operations
+
+\begin{theorem}
+Let $x, y \in \R$.
+Then:
+:$\max \left\{{x, y}\right\} = \frac 1 2 \left({x + y + \left\vert{x - y}\right\vert}\right)$
+where $\operatorname{max}$ denotes the [[Definition:Max Operation|max operation]].
+\end{theorem}
+
+\begin{proof}
+From the [[Trichotomy Law for Real Numbers]] exactly one of the following holds:
+:* $x < y$ and so $\max \left\{{x, y}\right\} = y$
+:* $x = y$ and so $\max \left\{{x, y}\right\} = x = y$
+:* $y < x$ and so $\max \left\{{x, y}\right\} = x$
+By the definition of the [[Definition:Absolute Value|absolute value]] function for each case respectively we have:
+:* $\left\vert{x - y}\right\vert = y - x$
+:* $\left\vert{x - y}\right\vert = 0$
+:* $\left\vert{x - y}\right\vert = x - y$
+Thus the equation holds by $+$ being [[Definition:Commutative Operation|commutative]] and [[Definition:Associative|associative]] as for each case:
+:* $\frac 1 2 \left({x + y + y - x} \right) = y$
+:* $\frac 1 2 \left({x + y + 0} \right) = x = y$
+:* $\frac 1 2 \left({x + y + x - y} \right) = x$
+{{qed}}
+[[Category:Analysis]]
+[[Category:Max and Min Operations]]
+noh4h8i0dl1dakz0153irzt311b6pr9
+\end{proof}<|endoftext|>
+\section{Reparameterization of Directed Smooth Curve Preserves Image}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right]$ and $\left[{c\,.\,.\, d}\right]$ be [[Definition:Closed Real Interval|closed real intervals]].
+Let $\gamma : \left[{a \,.\,.\, b}\right] \to \C$ be a [[Definition:Smooth Path (Complex Analysis)|smooth path]].
+Let $C$ be a [[Definition:Directed Smooth Curve|directed smooth curve]] with [[Definition:Directed Smooth Curve/Parameterization|parameterization]] $\gamma$.
+Suppose that $\sigma : \left[{c \,.\,.\, d}\right] \to \C$ is a [[Definition:Directed Smooth Curve/Parameterization|reparameterization]] of $C$.
+Then $\operatorname{Im} \left({\gamma}\right) = \operatorname{Im} \left({\sigma}\right)$, where $\operatorname{Im} \left({\gamma}\right)$ denotes the [[Definition:Image of Mapping|image]] of $\gamma$.
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Directed Smooth Curve|definition of directed smooth curve]], there exists a [[Definition:Bijection|bijective]] [[Definition:Differentiable on Interval|differentiable]] [[Definition:Strictly Increasing Real Function|strictly increasing]] [[Definition:Real Function|function]] $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ such that $\sigma = \gamma \circ \phi$.
+From [[Surjection by Restriction of Codomain]], it follows that there exists a [[Definition:Complex Function|function]] $\tilde{\gamma}: \left[{a \,.\,.\, b}\right] \to \operatorname{Im} \left({\gamma}\right)$ that is the [[Definition:Surjective Restriction|surjective restriction]] of $\gamma$ to $\operatorname{Im} \left({\gamma}\right)$.
+Put $\tilde{\sigma} := \tilde{\gamma} \circ \phi : \left[{c \,.\,.\, d}\right] \to \operatorname{Im} \left({\gamma}\right)$.
+From [[Composite of Surjections is Surjection]], it follows that $\tilde{\sigma}$ is [[Definition:Surjection|surjective]].
+Then:
+:$\operatorname{Im} \left({\tilde{\sigma} }\right) = \operatorname{Im} \left({\gamma}\right)$
+Now:
+: $\sigma \left({t}\right) = \tilde{\sigma} \left({t}\right)$ for all $t \in \left[{c \,.\,.\, d}\right] = \operatorname{Dom} \left({\sigma}\right) = \operatorname{Dom} \left({\tilde{\sigma} }\right)$
+where $\operatorname{Dom} \left({\sigma}\right)$ denotes the [[Definition:Domain of Mapping|domain]] of $\sigma$.
+It follows that:
+:$\operatorname{Im} \left({\sigma}\right) = \operatorname{Im} \left({\tilde{\sigma} }\right)$
+Hence:
+:$\operatorname{Im} \left({\tilde{\gamma} }\right) = \operatorname{Im} \left({\sigma}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\R^n$ be a [[Definition:Real Cartesian Space|real cartesian space]] of [[Definition:Dimension of Vector Space|$n$ dimensions]].
+Let $\left[{a \,.\,.\, b}\right]$ and $\left[{c \,.\,.\, d}\right]$ be [[Definition:Closed Real Interval|closed real intervals]].
+Let $\gamma: \left[{a \,.\,.\, b}\right] \to \R^n$ be a [[Definition:Smooth Path (Real Cartesian Space)|smooth path]] in $\R^n$.
+Let $C$ be a [[Definition:Directed Smooth Curve|directed smooth curve]] with [[Definition:Parameterization of Directed Smooth Curve|parameterization]] $\gamma$.
+Suppose that $\sigma: \left[{c \,.\,.\, d}\right] \to \C$ is a [[Definition:Parameterization of Directed Smooth Curve|reparameterization]] of $C$.
+Then the [[Definition:Start Point of Directed Smooth Curve|start points]] and [[Definition:End Point of Directed Smooth Curve|end points]] of $\gamma$ and $\sigma$ are identical:
+:$\gamma \left({a}\right) = \sigma \left({c}\right)$
+:$\gamma \left({b}\right) = \sigma \left({d}\right)$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Parameterization of Directed Smooth Curve|reparameterization]], there exists a [[Definition:Bijection|bijective]] [[Definition:Differentiable on Interval|differentiable]] [[Definition:Strictly Increasing Real Function|strictly increasing]] [[Definition:Real Function|real function]] $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ such that $\sigma = \gamma \circ \phi$.
+As $\phi^{-1} \left({a}\right) \in \left[{c \,.\,.\, d}\right]$:
+: $c \le \phi^{-1} \left({a}\right)$
+As $\phi$ is [[Definition:Strictly Increasing Real Function|strictly increasing]]:
+: $\phi \left({c}\right) \le \phi \left({\phi^{-1} \left({a}\right) }\right) = a$
+As $\phi \left({c}\right) \in \left[{a \,.\,.\, b}\right]$:
+: $\phi \left({c}\right) = a$
+Hence:
+:$\sigma \left({c}\right) = \gamma \circ \phi \left({c}\right) = \gamma \left({a}\right)$
+As $\phi^{-1} \left({b}\right) \in \left[{c \,.\,.\, d}\right]$:
+: $d \ge \phi^{-1} \left({b}\right)$
+As $\phi$ is [[Definition:Strictly Increasing Real Function|strictly increasing]]:
+: $\phi \left({d}\right) \ge \phi \left({\phi^{-1} \left({b}\right) }\right) = b$
+As $\phi \left({d}\right) \in \left[{a \,.\,.\, b}\right]$:
+: $\phi \left({d}\right) = b$
+Hence:
+:$\sigma \left({d}\right) = \gamma \circ \phi \left({d}\right) = \gamma \left({b}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Ring Negatives}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Then:
+:$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ \paren {-y} = x \circ y$
+where $\paren {-x}$ denotes the [[Definition:Ring Negative|negative]] of $x$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {-x} \circ \paren {-y}
+      | r = -\paren {x \circ \paren {-y} }
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | r = -\paren {-\paren {x \circ y} }
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | r = x \circ y
+      | c = [[Negative of Ring Negative]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance Function for Distinct Elements in Metric Space is Strictly Positive}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $A$ be a [[Definition:Set|set]].
+Let $d: A \times A \to \R$ be a [[Definition:Real-Valued Function|real-valued function]] on $A$ with the following properties:
+{{begin-axiom}}
+{{axiom|n = M1'
+       |q = \forall x, y \in A
+       |m = d \left({x, y}\right) = 0 \iff x = y
+}}
+{{axiom|n = M2
+       |q = \forall x, y, z \in A
+       |m = d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)
+}}
+{{axiom|n = M3
+       |q = \forall x, y \in A
+       |m = d \left({x, y}\right) = d \left({y, x}\right)
+}}
+{{end-axiom}}
+which can be considered as being an alternative formulation of the [[Definition:Metric Space Axioms|metric space axioms]].
+Then:
+:$\forall x, y \in A: x \ne y \implies d \left({x, y}\right) > 0$
+which is [[Definition:Metric Space Axioms|metric space axiom $(M4)$]].
+Thus $d$ is a [[Definition:Distance Function|distance function]], so making $M := \left({A, d}\right)$ a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | lo=\forall x, y \in A:
+      | l=d \left({x, y}\right) + d \left({y, x}\right)
+      | o=\ge
+      | r=d \left({x, x}\right)
+      | c=from $(M2)$
+}}
+{{eqn | ll=\implies
+      | l=2 d \left({x, y}\right)
+      | o=\ge
+      | r=0
+      | c=from $(M1')$ and $(M3)$
+}}
+{{eqn | ll=\implies
+      | l=d \left({x, y}\right)
+      | o=\ge
+      | r=0
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Metric Spaces]]
+7kxzz1voxpwwkchjinbb4k00ozu82bi
+\end{proof}<|endoftext|>
+\section{Transpose of Upper Triangular Matrix is Lower Triangular}
+Tags: Triangular Matrices, Transposes of Matrices
+
+\begin{theorem}
+The [[Definition:Transpose of Matrix|transpose]] of an [[Definition:Upper Triangular Matrix|upper triangular matrix]] is a [[Definition:Lower Triangular Matrix|lower triangular matrix]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf U = \sqbrk a_{m n}$ be an [[Definition:Upper Triangular Matrix|upper triangular matrix]].
+By definition:
+:$\forall a_{i j} \in \mathbf U: i > j \implies a_{i j} = 0$
+Let $\mathbf U^\intercal = \sqbrk b_{n m}$ be the [[Definition:Transpose of Matrix|transpose]] of $\mathbf U$.
+That is:
+:$\mathbf U^\intercal = \sqbrk b_{n m}: \forall i \in \closedint 1 n, j \in \closedint 1 n: b_{i j} = a_{j i}$
+Thus:
+:$\forall b_{j i} \in \mathbf U^\intercal: i > j \implies b_{j i} = 0$
+By exchanging $i$ and $j$ in the notation of the above:
+:$\forall b_{i j} \in \mathbf U^\intercal: i < j \implies b_{i j} = 0$
+Thus by definition it is seen that $\mathbf U^\intercal$ is a [[Definition:Lower Triangular Matrix|lower triangular matrix]].
+{{qed}}
+[[Category:Triangular Matrices]]
+[[Category:Transposes of Matrices]]
+8brtoc9pbbp6vnhh6u25d3syuohyc91
+\end{proof}<|endoftext|>
+\section{Derivative iff Right and Left Derivative}
+Tags: Analysis
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]].
+Then $f$ is [[Definition:Differentiable Real Function|differentiable]] {{iff}} it has both [[Definition:Right-Hand Derivative|right-]] and [[Definition:Left-Hand Derivative|left-hand derivatives]] which [[Definition:Agreement of Mappings|agree]].
+\end{theorem}
+
+\begin{proof}
+[[Definition:Derivative of Real Function|Derivatives]] are defined in terms of [[Definition:Limit of Real Function|limits]].
+The result follows from [[Limit iff Limits from Left and Right]].
+{{qed}}
+[[Category:Analysis]]
+mp510lwkzpr8y4j9e78aa5dcv5wyvvk
+\end{proof}<|endoftext|>
+\section{Composition of Ring Homomorphisms is Ring Homomorphism}
+Tags: Ring Homomorphisms, Composition of Ring Homomorphisms is Ring Homomorphism
+
+\begin{theorem}
+Let:
+:$\struct {R_1, +_1, \circ_1}$
+:$\struct {R_2, +_2, \circ_2}$
+:$\struct {R_3, +_3, \circ_3}$
+be [[Definition:Ring (Abstract Algebra)|rings]].
+Let:
+:$\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$
+:$\psi: \struct {R_2, +_2, \circ_2} \to \struct {R_3, +_3, \circ_3}$
+be [[Definition:Ring Homomorphism|homomorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also a [[Definition:Ring Homomorphism|homomorphism]].
+\end{theorem}
+
+\begin{proof}
+A specific instance of [[Composite of Homomorphisms on Algebraic Structure is Homomorphism]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+So as to alleviate possible confusion over notation, let the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.
+Then what we are trying to prove is denoted:
+:$\paren {\psi \bullet \phi}: \struct {R_1, +_1, \circ_1} \to \struct {R_3, +_3, \circ_3}$ is a [[Definition:Ring Homomorphism|homomorphism]].
+To prove the above is the case, we need to demonstrate that the [[Definition:Morphism Property|morphism property]] is held by $+_1$ and $\circ_1$ under $\psi \bullet \phi$.
+We take two elements $x, y \in R_1$, and put them through the following wringer with respect to $+_1$:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\psi \bullet \phi} } {x +_1 y}
+      | r = \map \psi {\map \phi {x +_1 y} }
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map \psi {\map \phi x +_2 \map \phi y}
+      | c = [[Definition:Morphism Property|Morphism Property]] applied to $+_1$ under $\phi$
+}}
+{{eqn | r = \map \psi {\map \phi x} +_3 \map \psi {\map \phi y}
+      | c = [[Definition:Morphism Property|Morphism Property]] applied to $+_2$ under $\psi$
+}}
+{{eqn | r = \map {\paren {\psi \bullet \phi} } x +_3 \map {\paren {\psi \bullet \phi} } y
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{end-eqn}}
+The same applies to $\circ_1$:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\psi \bullet \phi} } {x \circ_1 y}
+      | r = \map \psi {\map \phi {x \circ_1 y} }
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map \psi {\map \phi x \circ_2 \map \phi y}
+      | c = [[Definition:Morphism Property|Morphism Property]] applied to $\circ_1$ under $\phi$
+}}
+{{eqn | r = \map \psi {\map \phi x} \circ_3 \map \psi {\map \phi y}
+      | c = [[Definition:Morphism Property|Morphism Property]] applied to $\circ_2$ under $\psi$
+}}
+{{eqn | r = \map {\paren {\psi \bullet \phi} } x \circ_3 \map {\paren {\psi \bullet \phi} } y
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{end-eqn}}
+Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the [[Definition:Morphism Property|morphism property]] is indeed held by both $+_1$ and $\circ_1$ under $\psi \bullet \phi$.
+Thus $\paren {\psi \bullet \phi}: \struct {R_1, +_1, \circ_1} \to \struct {R_3, +_3, \circ_3}$ is a [[Definition:Group Homomorphism|homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Ring Epimorphisms is Ring Epimorphism}
+Tags: Ring Epimorphisms
+
+\begin{theorem}
+Let:
+:$\struct {R_1, +_1, \circ_1}$
+:$\struct {R_2, +_2, \circ_2}$
+:$\struct {R_3, +_3, \circ_3}$
+be [[Definition:Ring (Abstract Algebra)|rings]].
+Let:
+:$\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$
+:$\psi: \struct {R_2, +_2, \circ_2} \to \struct {R_3, +_3, \circ_3}$
+be [[Definition:Ring Epimorphism|(ring) epimorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also a [[Definition:Ring Epimorphism|(ring) epimorphism]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Ring Epimorphism|ring epimorphism]] is a [[Definition:Ring Homomorphism|ring homomorphism]] which is also a [[Definition:Surjection|surection]].
+From [[Composition of Ring Homomorphisms is Ring Homomorphism]], $\psi \circ \phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+From [[Composite of Surjections is Surjection]], $\psi \circ \phi$ is a [[Definition:Surjection|surection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Ring Monomorphisms is Ring Monomorphism}
+Tags: Ring Monomorphisms
+
+\begin{theorem}
+Let:
+* $\left({R_1, +_1, \circ_1}\right)$
+* $\left({R_2, +_2, \circ_2}\right)$
+* $\left({R_3, +_3, \circ_3}\right)$
+be [[Definition:Ring (Abstract Algebra)|rings]].
+Let:
+* $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$
+* $\psi: \left({R_2, +_2, \circ_2}\right) \to \left({R_3, +_3, \circ_3}\right)$
+be [[Definition:Ring Monomorphism|(ring) monomorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also a [[Definition:Ring Monomorphism|(ring) monomorphism]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Ring Monomorphism|ring monomorphism]] is a [[Definition:Ring Homomorphism|ring homomorphism]] which is also an [[Definition:Injection|injection]].
+From [[Composition of Ring Homomorphisms is Ring Homomorphism]], $\psi \circ \phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+From [[Composite of Injections is Injection]], $\psi \circ \phi$ is an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Ring Isomorphisms is Ring Isomorphism}
+Tags: Ring Isomorphisms
+
+\begin{theorem}
+Let:
+* $\left({R_1, +_1, \circ_1}\right)$
+* $\left({R_2, +_2, \circ_2}\right)$
+* $\left({R_3, +_3, \circ_3}\right)$
+be [[Definition:Ring (Abstract Algebra)|rings]].
+Let:
+* $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$
+* $\psi: \left({R_2, +_2, \circ_2}\right) \to \left({R_3, +_3, \circ_3}\right)$
+be [[Definition:Ring Isomorphism|(ring) isomorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also a [[Definition:Ring Isomorphism|(ring) isomorphism]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Ring Isomorphism|ring isomorphism]] is a [[Definition:Ring Homomorphism|ring homomorphism]] which is also a [[Definition:Bijection|bijection]].
+From [[Composition of Ring Homomorphisms is Ring Homomorphism]], $\psi \circ \phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+From [[Composite of Bijections is Bijection]], $\psi \circ \phi$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Ring Endomorphisms is Ring Endomorphism}
+Tags: Ring Endomorphisms
+
+\begin{theorem}
+Let $R$ be a [[Definition:Set|set]].
+Let:
+: $\left({R, +_1, \circ_1}\right)$
+: $\left({R, +_2, \circ_2}\right)$
+: $\left({R, +_3, \circ_3}\right)$
+be [[Definition:Ring (Abstract Algebra)|rings]].
+Let:
+: $\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
+: $\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$
+be [[Definition:Ring Endomorphism|(ring) endomorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also a [[Definition:Ring Endomorphism|(ring) endomorphism]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Ring Endomorphism|ring endomorphism]] is a [[Definition:Ring Homomorphism|ring homomorphism]] $f$ from a [[Definition:Set|set]] to itself.
+That is:
+:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
+:$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$
+From [[Composition of Ring Homomorphisms is Ring Homomorphism]], $\psi \circ \phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$
+Thus:
+:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$
+and so $\psi \circ \phi$ is a [[Definition:Ring Endomorphism|ring endomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Ring Automorphisms is Ring Automorphism}
+Tags: Ring Automorphisms
+
+\begin{theorem}
+Let $R$ be a [[Definition:Set|set]].
+Let:
+* $\left({R, +_1, \circ_1}\right)$
+* $\left({R, +_2, \circ_2}\right)$
+* $\left({R, +_3, \circ_3}\right)$
+be [[Definition:Ring (Abstract Algebra)|rings]].
+Let:
+* $\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
+* $\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$
+be [[Definition:Ring Automorphism|(ring) automorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also a [[Definition:Ring Automorphism|(ring) automorphism]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Ring Automorphism|ring automorphism]] is a [[Definition:Ring Isomorphism|ring isomorphism]] $f$ from a [[Definition:Set|set]] to itself.
+That is:
+:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
+:$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$
+From [[Composition of Ring Isomorphisms is Ring Isomorphism]], $\psi \circ \phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]].
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$
+Thus:
+:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$
+and so $\psi \circ \phi$ is a [[Definition:Ring Automorphism|ring automorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set has Rank}
+Tags: Von Neumann Hierarchy
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then $S$ has a [[Definition:Rank (Set Theory)|rank]].
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+Let $G$ be the smallest transitive set containing $S$ as a subset.
+By [[Set Contained in Smallest Transitive Set]], $G$ must exist.
+By [[Transitive Set Contained in Von Neumann Hierarchy Level]], $G \subseteq V_i$ for some ordinal $i$.
+Therefore:
+:$G \in V_{i+1}$
+{{explain|[[Proper Well-Ordering determines Smallest Elements]] only applies once it's been shown that initial segments are sets}}
+We have that the ordinals are well-ordered.
+From [[Proper Well-Ordering determines Smallest Elements]], there exists a smallest ordinal $k$ such that $G \in V_{k+1}$.
+Hence, by definition, $G$ has rank $k$.
+{{qed}}
+[[Category:Von Neumann Hierarchy]]
+m3gvls68ke0ptcnnudc49tjecr6nq8v
+\end{proof}<|endoftext|>
+\section{Quotient Epimorphism is Epimorphism/Group}
+Tags: Quotient Groups, Group Epimorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $G / N$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $N$.
+Let $q_N: G \to G / N$ be the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to $G / N$:
+:$\forall x \in G: \map {q_N} x = x N$
+Then $q_N$ is a [[Definition:Group Epimorphism|group epimorphism]] whose [[Definition:Kernel of Group Homomorphism|kernel]] is $N$.
+\end{theorem}
+
+\begin{proof}
+The proof follows from [[Quotient Mapping on Structure is Canonical Epimorphism]].
+When $N \lhd G$, we have:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y \in G:
+      | l = \map {q_N} {x y}
+      | r = x y N
+      | c = {{Defof|Quotient Group Epimorphism}}
+}}
+{{eqn | r = \paren {x N} \paren {y N}
+      | c = {{Defof|Quotient Group}}
+}}
+{{eqn | r = \map {q_N} x \map {q_N} y
+      | c = {{Defof|Quotient Group Epimorphism}}
+}}
+{{end-eqn}}
+Therefore $q$ is a [[Definition:Group Homomorphism|homomorphism]].
+We have that:
+:$\forall x \in G: x N \in G / N = \map {q_N} x$
+so $q$ is [[Definition:Surjection|surjective]].
+Therefore $q$ is an [[Definition:Group Epimorphism|epimorphism]].
+Let $x \in G$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \map \ker {q_N}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map {q_N} x
+      | r = e_{G/N}
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x N
+      | r = e_{G/N} N
+      | c = {{Defof|Quotient Group Epimorphism}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x N
+      | r = N
+      | c = [[Coset by Identity]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = N
+      | c = [[Left Coset Equals Subgroup iff Element in Subgroup]]
+}}
+{{end-eqn}}
+thus proving that $\map \ker {q_N} = N$ from definition of [[Definition:Subset|subset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Increasing Mapping on Well-Ordered Class}
+Tags: Class Mappings, Increasing Mappings
+
+\begin{theorem}
+Let $ \left({S, \prec}\right)$ be a strictly well-ordered class.
+Let $\left({T, <}\right)$ be a strictly ordered class.
+Let $f$ be a [[Definition:Mapping|mapping]] from $S$ to $T$.
+For each $i \in S$ such that $i$ is not maximal in $S$, let:
+: $f \left({i}\right) < f \left({\operatorname{succ} \left({i}\right)}\right)$
+where $\operatorname{succ} \left({i}\right)$ is the [[Definition:Immediate Successor Element|immediate successor element]] of $i$.
+Let:
+:$\forall i, j \in S: i \preceq j \implies f \left({i}\right) \le f \left({j}\right)$
+Then for each $i, j \in S$ such that $i \prec j$:
+:$f \left({i}\right) < f \left({j}\right)$
+{{MissingLinks|In the above, link to definitions of increasing mapping and strictly increasing mapping, in order to correlate with page title.}}
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+By [[Non-Maximal Element of Well-Ordered Class has Immediate Successor]], $\operatorname{succ} \left({i}\right)$ is guaranteed to exist.
+Let $i \prec j$.
+Let $S_i := \left\{{q: i \prec q}\right\}$.
+{{explain|Important to specify what the domain of $q$ actually is -- presumably $S$.}}
+Then $\operatorname{succ} \left({i}\right)$ is the minimal element of $S_i$.
+By supposition, $j \in S_i$.
+Thus:
+:$j \not \prec \operatorname{succ} \left({i}\right)$
+Since [[Well-Ordering is Total Ordering]]:
+:$\operatorname{succ} \left({i}\right) \preceq j$
+Thus by supposition:
+:$f \left({\operatorname{succ} \left({i}\right)}\right) \le f \left({j}\right)$
+Since $f \left({i}\right) < \left({\operatorname{succ} \left({i}\right)}\right)$:
+:$f \left({i}\right) < f \left({j}\right)$
+by transitivity.
+{{qed}}
+[[Category:Class Mappings]]
+[[Category:Increasing Mappings]]
+p00orr6p1txznme27vyt9qae4lud9ij
+\end{proof}<|endoftext|>
+\section{Non-Maximal Element of Well-Ordered Class has Immediate Successor}
+Tags: Well-Orderings
+
+\begin{theorem}
+Let $\left({C, \le}\right)$ be a [[Definition:Well-Ordered Class|well-ordered class]].
+Let $x \in C$.
+Suppose that $x$ is maximal in $C$.
+{{mistake|maximal?}}
+Then $x$ has an [[Definition:Immediate Successor Element|immediate successor]] in $C$.
+\end{theorem}
+
+\begin{proof}
+Let $x$ be a non-maximal element of $C$.
+Let $S$ be the class of successors of $x$ in $C$.
+$S$ is non-empty because $x$ is not maximal.
+By [[Proper Well-Ordering Determines Smallest Elements]], $S$ has a minimal element, $y$.
+Then $x < y$ by the definition of $S$.
+Suppose that for some $z \in C$, $x < z < y$.
+Then by the definition of $S$, $z \in S$.
+This contradicts the minimality of $y$.
+Thus $y$ is the immediate successor element of $x$.
+{{qed}}
+[[Category:Well-Orderings]]
+ciuv64qtc1uddocrcxqcqjjcgfadvyu
+\end{proof}<|endoftext|>
+\section{Set Contained in Smallest Transitive Set}
+Tags: 
+
+\begin{theorem}
+Let $S$ be a set.
+Then there exists a transitive set $G$ such that:
+:$S \subseteq G$
+and:
+:if $Q$ is any transitive set such that $S \subseteq Q$, then $G \subseteq Q$.
+\end{theorem}
+
+\begin{proof}
+=== Construction of $G$ ===
+Let $U$ be the [[Definition:Class of all Sets|class of all sets]].
+{{explain|Perhaps [[Definition:Universal Class|Universal Class]] can be used here}}
+Define the [[Definition:Mapping|mapping]] $F: \N \to U$ recursively:
+:$\map F 0 = S$
+:$\map F {n + 1} = \bigcup \map F n$
+Applying the axiom of union inductively, $\map F n$ is a set for each $n \in \N$.
+Let $\displaystyle G = \bigcup_{i \mathop = 0}^\infty \map F i$.
+By the [[Axiom:Axiom of Unions|axiom of unions]], $G$ is a [[Definition:Set|set]].
+=== Transitivity ===
+It is to be proved that $G$ is transitive.
+That is:
+:$a \in b, b \in G \implies a \in G$
+Let $a \in b$ and $b \in G$.
+By the definition of $G$, there exists $n \in \N$, $b \in \map F n$.
+By the definition of $F$:
+:$\map F {n + 1} = \bigcup \map F n$
+Then by the definition of union:
+:$a \in \map F {n + 1}$
+Thus by the definition of $G$:
+:$a \in G$
+{{qed|lemma}}
+=== Minimality ===
+It is to be proved that if $Q$ is a transitive set and $S \subseteq Q$ then $G \subseteq Q$.
+{{proof wanted|Only an outline so far}}
+Let $Q$ be transitive and $S \subseteq Q$.
+Define $F$ as above.
+Prove by induction that $\map F n \subseteq Q$ for each $n$.
+{{qed}}
+p0rijby2lh6rzihk2xa1yutgmiriaxe
+\end{proof}<|endoftext|>
+\section{Transitive Set Contained in Von Neumann Hierarchy Level}
+Tags: Von Neumann Hierarchy
+
+\begin{theorem}
+Let $G$ be a transitive set.
+Then for some ordinal $i$, $G \subseteq V_i$.
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+Suppose for the sake of contradiction that for each ordinal $i$ the set $G\setminus V_i$ is non-empty. 
+Let $i$ be any ordinal.
+Then by the [[Axiom:Axiom of Foundation|axiom of foundation]]:
+$\exists x: x \in G\setminus V_i \text{ and } x \cap \left({G \setminus V_i}\right) = \varnothing$
+Since $G$ is transitive, $x \subseteq G$.
+Since $x \subseteq G$ and $x \cap \left({G \setminus V_i}\right) = \varnothing$:
+:$x \subseteq V_i$
+By the definition of $V$, $x \in V_{i+1}$, so:
+:$x \notin G \setminus V_{i+1}$
+Thus $G \setminus V_{i+1} \subsetneqq G \setminus V_i$ for each ordinal $i$.
+Let $H: \operatorname{On} \to P \left({G}\right)$ be defined by:
+:$H \left({i}\right) = G \setminus V_i$ for each ordinal $i$
+Since the class of ordinals is well-ordered, applying [[Strictly Increasing Mapping on Well-Ordered Class]] proves that for any ordinals $p$ and $q$,
+:$p < q$ implies $H \left({q}\right) \subsetneqq H \left({p}\right)$.
+Since the class of ordinals is totally ordered, for any two distinct ordinals $i$ and $j$:
+: $H \left({i}\right) \ne H \left({j}\right)$
+so $H$ is injective.
+But that contradicts the fact that $P \left({G}\right)$ is a set.
+{{qed}}
+[[Category:Von Neumann Hierarchy]]
+m6vhk4yxba1sj9q1ottirbma1w99yrw
+\end{proof}<|endoftext|>
+\section{External Direct Product Closure/General Result}
+Tags: External Direct Products
+
+\begin{theorem}
+Let $\displaystyle \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the [[Definition:External Direct Product/General Definition|external direct product]] of the [[Definition:Algebraic Structure|algebraic structures]] $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
+Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ all be [[Definition:Closed Algebraic Structure|closed algebraic structures]].
+Then $\struct {S, \circ}$ is also a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[External Direct Product Closure]].
+{{finish|Fill in the detail.}}
+{{qed}}
+[[Category:External Direct Products]]
+6fqfvdj53acppdgupdqfawyltkjk0md
+\end{proof}<|endoftext|>
+\section{External Direct Product of Abelian Groups is Abelian Group/General Result}
+Tags: Abelian Groups, Group Direct Products
+
+\begin{theorem}
+The [[Definition:External Direct Product|external direct product]] of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Abelian Group|abelian groups]] is itself an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ be [[Definition:Abelian Group|abelian groups]].
+Let $\displaystyle \struct {G, \circ} = \prod_{k \mathop = 1}^n G_k$ be the [[Definition:External Direct Product|external direct product]] of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$.
+From [[External Direct Product of Groups is Group/Finite Product|External Direct Product of Groups is Group: Finite Product]] it follows that $\struct {G, \circ}$ is a [[Definition:Group|group]].
+By definition, each of $\circ_1, \circ_2, \ldots, \circ_n$ are [[Definition:Commutative Operation|commutative operations]].
+From [[External Direct Product Commutativity/General Result|External Direct Product Commutativity: General Result]] it follows that $\circ$ is [[Definition:Commutative Operation|commutative]].
+Hence, by definition, $\struct {G, \circ}$ is an [[Definition:Abelian Group|abelian group]].
+{{qed}}
+[[Category:Abelian Groups]]
+[[Category:Group Direct Products]]
+rhjtkdwnymakyv8zp1ejrymhofvjc00
+\end{proof}<|endoftext|>
+\section{Homomorphism of External Direct Products/General Result}
+Tags: Homomorphisms, Cartesian Product
+
+\begin{theorem}
+Let $n \in \N_{>0}$.
+Let:
+: $\displaystyle \left({\mathcal S_n, \circledcirc_n}\right) := \prod_{k \mathop = 1}^n S_k = \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$
+: $\displaystyle \left({\mathcal T_n, \circledast_n}\right) := \prod_{k \mathop = 1}^n T_k = \left({T_1, \ast_1}\right) \times \left({T_2, \ast_2}\right) \times \cdots \times \left({T_n, \ast_n}\right)$
+be [[Definition:External Direct Product/General Definition|external direct products]] of [[Definition:Algebraic Structure|algebraic structures]].
+Let $\Phi_n: \left({\mathcal S_n, \circledcirc_n}\right) \to \left({\mathcal T_n, \circledast_n}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\Phi_n: \left({s_1, \ldots, s_n}\right) := \begin{cases}
+\phi_1 \left({s_1}\right) & : n = 1 \\
+\left({\phi_1 \left({s_1}\right), \phi_2 \left({s_2}\right)}\right) & : n = 2 \\
+\left({\Phi_n \left({s_1, \ldots, s_{n - 1} }\right), \phi_n \left({s_n}\right)}\right) & : n > 2 \\
+\end{cases}$
+That is:
+:$\Phi_n: \left({s_1, \ldots, s_n}\right) := \left({\phi_1 \left({s_1}\right), \phi_2 \left({s_2}\right), \ldots, \phi_n \left({s_n}\right)}\right)$
+Let $\phi_k: \left({S_k, \circ_k}\right) \to \left({T_k, \ast_k}\right)$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] for each $k \in \left\{{1, 2, \ldots, n}\right\}$.
+Then $\Phi_n$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({\mathcal S_n, \circledcirc_n}\right)$ to $\left({\mathcal T_n, \circledast_n}\right)$.
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\Phi_n: \left({\mathcal S_n, \circledcirc_n}\right) \to \left({\mathcal T_n, \circledast_n}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
+=== Basis for the Induction ===
+$P \left({1}\right)$ is true, as this just says:
+:$\Phi_1 = \phi_1: \left({S_1, \circ_1}\right) \to \left({T_1, \ast_1}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]
+which holds [[Definition:By Hypothesis|by hypothesis]].
+$P \left({2}\right)$ is the case:
+:$\Phi_2 = \phi_1 \times \phi_2: \left({S_1 \times S_2, \circledcirc_2}\right) \to \left({T_1 \times T_2, \circledast_2}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]
+which has been proved in [[Homomorphism of External Direct Products]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\Phi_m: \left({\mathcal S_m, \circledcirc_m}\right) \to \left({\mathcal T_m, \circledast_m}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
+Then we need to show:
+:$\Phi_{m+1}: \left({\mathcal S_{m+1}, \circledcirc_{m+1}}\right) \to \left({\mathcal T_{m+1}, \circledast_{m+1}}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Let $\left({s_1, \ldots, s_{m+1}}\right)$ and $\left({t_1, \ldots, t_{m+1}}\right)$ be arbitrary elements of $\mathcal S_{m+1}$.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \Phi_{m+1} \left({\left({s_1, \ldots, s_{m+1} }\right) \circledcirc_{m+1} \left({t_1, \ldots, t_{m+1} }\right)}\right)
+}}
+{{eqn | r = \Phi_{m+1} \left({\left({\left({s_1, \ldots, s_m}\right), s_{m+1} }\right) \circledcirc_{m+1} \left({\left({t_1, \ldots, t_m}\right), t_{m+1} }\right)}\right)
+      | c = Definition of [[Definition:Cartesian Product|Cartesian Product]]
+}}
+{{eqn | r = \left({\Phi_m \left({\left({s_1, \ldots, s_m}\right) \circledcirc_m \left({t_1, \ldots, t_m}\right)}\right), s_{m+1} \circ_{m+1} t_{m+1} }\right)
+      | c = Definition of [[Definition:External Direct Product/General Definition|External Direct Product]]
+}}
+{{eqn | r = \left({\Phi_m \left({\left({s_1, \ldots, s_m}\right) \circledcirc_m \left({t_1, \ldots, t_m}\right)}\right), \phi_{m+1} \left({s_{m+1} }\right) \ast_{m+1} \phi_{m+1} \left({t_{m+1} }\right) }\right)
+      | c = $\phi_{m+1}$ is a [[Definition:Homomorphism (Abstract Algebra)|Homomorphism]]
+}}
+{{eqn | r = \left({\Phi_m \left({s_1, \ldots, s_m}\right) \circledast_m \Phi_m \left({t_1, \ldots, t_m}\right), \phi_{m+1} \left({s_{m+1} }\right) \ast_{m+1} \phi_{m+1} \left({t_{m+1} }\right) }\right)
+      | c = [[Homomorphism of External Direct Products/General Result#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \Phi_{m+1} \left({s_1, \ldots, s_m, s_{m+1} }\right) \circledast_{m+1} \Phi_{m+1} \left({t_1, \ldots, t_m, t_{m+1} }\right)
+      | c = Definition of [[Definition:External Direct Product/General Definition|External Direct Product]]
+}}
+{{end-eqn}}
+So $P \left({m}\right) \implies P \left({m+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N_{>0}: \Phi_n: \left({\mathcal S_n, \circledcirc_n}\right) \to \left({\mathcal T_n, \circledast_n}\right)$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
+{{qed}}
+[[Category:Homomorphisms]]
+[[Category:Cartesian Product]]
+9nsxqrm8m7xr1hby13g8lfa86vf1c0r
+\end{proof}<|endoftext|>
+\section{Cesàro Mean}
+Tags: Analysis, Topology, Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Complex Number|complex numbers]].
+Suppose that $\sequence {a_n}$ [[Definition:Convergent Sequence (Analysis)|converges]] to $\ell$ in $\C$:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = \ell$
+Then also:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac {a_1 + \dotsb + a_n} n = \ell$
+\end{theorem}
+
+\begin{proof}
+For every fixed integer $n_0$, we write:
+:$\displaystyle \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} \le \frac {\cmod {a_1 - \ell} + \dotsb + \cmod {a_n - \ell} } n \le \frac {n_0 \displaystyle \sup_{k \mathop \le n_0} \cmod {a_k - \ell} } n + \sup_{n_0 \mathop < k \mathop \le n} \cmod {a_k - \ell}$
+As $n$ tends to $+\infty$, we get:
+:$\displaystyle \limsup_{n \mathop \to \infty} \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} \le \sup_{k \mathop > n_0} \cmod {a_k - \ell}$
+As $n_0$ tends to $+\infty$, we finally conclude:
+:$\displaystyle \limsup_{n \mathop \to \infty} \cmod {\frac {a_1 + \dotsb + a_n} n - \ell} = 0$
+{{qed}}
+{{explain|More detailed justification and explanation of all the steps}}
+\end{proof}<|endoftext|>
+\section{Characterization of Minimal Element}
+Tags: 
+
+\begin{theorem}
+Let $C$ be a [[Definition:Class (Class Theory)|class]].
+Let $\prec$ be a [[Definition:Relation|relation]] on $C$.
+Let $B$ be a subclass of $C$.
+Let $x \in B$.
+Let $S_x = \left\{ {y \in C: y \prec x \text{ and } y \ne x}\right\}$ be the initial segment of $x$ in $C$.
+Then $x$ is a minimal element of $B$ {{iff}} $B \cap S_x = \varnothing$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $x$ is a minimal element of $B$.
+Then for each $z \in B$ such that $z \ne x$, $z \not \prec x$.
+Thus $S_x \cap B = \varnothing$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $x$ is not a minimal element of $B$.
+Then for some $z \in B$, $z \prec x$ and $z \ne x$.
+Thus $z \in S_x$.
+Since $z \in B$, $B \cap S_x \ne \varnothing$.
+{{qed}}
+g2hpzljsznffa7wy4aysono39arex5o
+\end{proof}<|endoftext|>
+\section{Non-Empty Class has Element of Least Rank}
+Tags: Von Neumann Hierarchy
+
+\begin{theorem}
+Let $C$ be a [[Definition:Class (Class Theory)|class]].
+Let $C \ne \varnothing$.
+Then $C$ has an [[Definition:Element|element]] of least [[Definition:Rank (Set Theory)|rank]].
+That is:
+:$\exists x \in C: \forall y \in C: \operatorname{rank} \left({x}\right) \le \operatorname {rank}\left({y}\right)$
+where $\operatorname{rank}\left({x}\right)$ is the [[Definition:Rank (Set Theory)|rank]] of $x$.
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+By [[Set has Rank]], each [[Definition:Element|element]] of $C$ has a [[Definition:Rank (Set Theory)|rank]].
+Let $R$ be the [[Definition:Class (Class Theory)|class]] of [[Definition:Rank (Set Theory)|ranks]] of [[Definition:Element|element]]s of $C$.
+$R$ is [[Definition:Non-Empty Class|non-empty]] because $C$ is [[Definition:Non-Empty Class|non-empty]].
+{{explain|This should follow from arbitrary intersections of ordinals being ordinals.}}
+Since any [[Definition:Non-Empty Class|non-empty]] [[Definition:Class (Class Theory)|class]] of [[Definition:Ordinal|ordinals]] has a [[Definition:Smallest Element|least element]], $R$ has a least element, $q$.
+By the definition of $R$:
+:$\exists x \in C: \operatorname{rank}\left({x}\right) = q$
+Then $x$ is an element of $C$ of least [[Definition:Rank (Set Theory)|rank]].
+{{qed}}
+[[Category:Von Neumann Hierarchy]]
+9zkt1txf45kfd89za25q4fr3u487f2w
+\end{proof}<|endoftext|>
+\section{Reversed Directed Smooth Curve is Directed Smooth Curve}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $C$ be a [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] in $\C$.
+Let $C$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma: \left[{a \,.\,.\, b}\right] \to \C$.
+Define $\psi: \left[{a \,.\,.\, b}\right] \to \left[{a \,.\,.\, b}\right]$ by $\psi \left({t}\right) = a + b - t$.
+Define $\rho: \left[{a \,.\,.\, b}\right] \to \C$ by $\rho = \gamma \circ \psi$.
+Then $\rho$ is a [[Definition:Smooth Path (Complex Analysis)|smooth path]] which [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterizes]] a [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] $-C$.
+The [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] $-C$ is independent of the [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterization]] $\gamma$.
+\end{theorem}
+
+\begin{proof}
+First, we prove that $\rho$ is a [[Definition:Smooth Path (Complex Analysis)|smooth path]]:
+{{begin-eqn}}
+{{eqn | l = \rho' \left({t}\right)
+      | r = \gamma' \left({\psi\left({t}\right) }\right) \psi' \left({t}\right)
+      | c = [[Derivative of Complex Composite Function]]
+}}
+{{eqn | r = -\gamma' \left({\psi\left({t}\right) }\right)
+      | c = [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]: $\psi' \left({t}\right) = -1$
+}}
+{{eqn | o = \ne
+      | r = 0
+      | c = {{Defof|Smooth Path (Complex Analysis)|Smooth Path}}: $\gamma' \left({\psi\left({t}\right)}\right) \ne 0$
+}}
+{{end-eqn}}
+As $\gamma'$ is [[Definition:Continuously Differentiable|continuously differentiable]], and $\psi$ is [[Definition:Continuous Complex Function|continuous]], it follows that $\rho'$ is [[Definition:Continuous Complex Function|continuous]].
+Then $\rho$ is a [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterization]] of a [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] $-C$. 
+{{qed|lemma}}
+Next, we prove that the definition of $-C$ is independent of the [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterization]] $\gamma$.
+Suppose $\sigma$ is another [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterization]] of $C$:
+: $\sigma = \gamma \circ \phi$
+Here $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is a [[Definition:Bijection|bijective]] [[Definition:Differentiable on Interval|differentiable]] [[Definition:Strictly Increasing Real Function|strictly increasing function]].
+Define $\tilde \psi: \left[{c \,.\,.\, d}\right] \to \left[{c \,.\,.\, d}\right]$ by $\tilde \psi \left({t}\right) = c + d - t$.
+Define $\tilde \rho: \left[{c \,.\,.\, d}\right] \to \C$ by $\tilde \rho = \sigma \circ \tilde{\psi}$.
+We now prove that $\rho$ and $\tilde \rho$ both are [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterizations]] of the same [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] $-C$.
+Both $\psi$ and $\tilde \psi$ are [[Definition:Bijection|bijective]] with $\psi^{-1} = \psi$ and $\tilde \psi^{-1} = \tilde \psi$.
+Define $\tilde \phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ by $\tilde \phi = \psi^{-1} \circ \phi \circ \tilde \psi$.
+From [[Composite of Bijections is Bijection]], it follows that $\tilde \phi$ is [[Definition:Bijection|bijective]].
+From [[Derivative of Composite Function]], it follows that $\tilde \phi$ is [[Definition:Differentiable on Interval|differentiable]] with
+{{begin-eqn}}
+{{eqn | l = \tilde \phi' \left({t}\right)
+      | r = \dfrac 1 {\psi' \left({\psi^{-1} \left({\phi \left({\tilde \psi \left({t}\right) }\right) }\right) }\right) } \phi' \left({\tilde \psi \left({t}\right) }\right) \tilde \psi' \left({t}\right)
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | r = \phi' \left({\tilde \psi \left({t}\right)}\right)
+      | c = [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]
+}}
+{{eqn | o = >
+      | r = 0
+      | c = [[Derivative of Monotone Function]]
+}}
+{{end-eqn}}
+From [[Derivative of Monotone Function]], it follows that $\tilde \phi$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+As:
+:$\rho \circ \tilde \phi = \gamma \circ \psi \circ \psi^{-1} \circ \phi \circ \tilde \psi = \gamma \circ \phi \circ \tilde \psi = \sigma \circ \tilde \phi = \tilde \rho$
+it follows that $\rho$ and $\tilde{\rho}$ are [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterizations]] of $-C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reversed Contour is Contour}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $\R^n$ be a [[Definition:Real Cartesian Space|real cartesian space]] of [[Definition:Dimension of Vector Space|$n$ dimensions]].
+Let $C$ be a [[Definition:Contour|contour in $\R^n$]] that is defined as a [[Definition:Concatenation of Contours|concatenation]] of a [[Definition:Finite Sequence|finite sequence]] $C_1, \ldots, C_n$ of [[Definition:Directed Smooth Curve|directed smooth curves]] in $\R^n$.
+Then the finite sequence of [[Definition:Reversed Directed Smooth Curve|reversed directed smooth curves]]:
+:$-C_n, -C_{n - 1}, \ldots, -C_1$
+defines a [[Definition:Contour|contour]] that is independent of the [[Definition:Parameterization of Directed Smooth Curve|parameterization]] of $C_1, \ldots, C_n$.
+\end{theorem}
+
+\begin{proof}
+Let $C_i$ be [[Definition:Parameterization of Directed Smooth Curve|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$.
+From [[Reversed Directed Smooth Curve is Directed Smooth Curve]], it follows that $-C_i$ is independent of the paraterization $\gamma_i$ of $C_i$.
+We now prove that the [[Definition:End Point of Contour (Complex Plane)|end point]] of $-C_i$ is equal to the [[Definition:Start Point of Contour (Complex Plane)|start point]] of $-C_{i-1}$ for all $i \in \left\{ {2, \ldots, n}\right\}$.
+By [[Definition:Reversed Directed Smooth Curve|definition of reversed directed smooth curve]], $-C_i$ is [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by $\rho_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$.
+Here, $\rho_i \left({t}\right) = \gamma_i \left({a_i + b_i - t}\right)$.
+From [[Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints]], it follows that the [[Definition:Endpoints of Contour|endpoints]] $\rho_i \left({a_i}\right)$ and $\rho_i \left({b_i}\right)$ are independent of the [[Definition:Parameterization of Directed Smooth Curve|parameterization]] $\rho_i$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \rho_i \left({b_i}\right)
+      | r = \phi_i \left({a_i + b_i - b_i}\right)
+}}
+{{eqn | r = \phi_i \left({a_i}\right)
+}}
+{{eqn | r = \phi_{i - 1} \left({b_{i - 1} }\right)
+      | c = {{Defof|Contour (Complex Plane)}}
+}}
+{{eqn | r = \rho_{i - 1} \left({a_{i - 1} }\right)
+}}
+{{end-eqn}}
+By definition, it follows that $-C_n, \ldots, -C_1$ can be [[Definition:Concatenation of Contours|concatenated]] to form a [[Definition:Contour|contour]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Contour Integral along Reversed Contour}
+Tags: Contour Integration
+
+\begin{theorem}
+Let $C$ be a [[Definition:Contour (Complex Plane)|contour]].
+Let $f: \Img C \to \C$ be a [[Definition:Continuous Complex Function|continuous]] [[Definition:Complex Function|complex functions]], where $\Img C$ denotes the [[Definition:Image of Contour (Complex Plane)|image]] of $C$.
+Then the [[Definition:Complex Contour Integral|contour integral]] of $f$ along the [[Definition:Reversed Contour|reversed contour]] $-C$ is:
+:$\displaystyle \int_{-C} \map f z \rd z = -\int_C \map f z \rd z$
+\end{theorem}
+
+\begin{proof}
+First, suppose that $C$ is a [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]] in $\C$.
+Let $C$ be [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by the [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\gamma: \closedint a b \to \C$.
+By definition of [[Definition:Reversed Directed Smooth Curve|reversed directed smooth curve]], $-C$ is [[Definition:Parameterization of Directed Smooth Curve (Complex Plane)|parameterized]] by a [[Definition:Smooth Path (Complex Analysis)|smooth path]] $\rho: \closedint a b \to \C$ with $\rho = \gamma \circ \psi$.
+Here, $\psi: \closedint a b \to \closedint a b$ is defined by $\map \psi t = a + b - t$.
+From [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]:
+:$\map {\psi'} t = -1$ for all $t \in \closedint a b$
+Then:
+{{begin-eqn}}
+{{eqn | l = \displaystyle \int_C \map f z \rd z 
+      | r = \int_a^b \map f {\map \rho t} \map {\rho'} t \rd t
+      | c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn | r = \int_a^b \map f {\map \gamma {\map \psi t} } \map {\gamma'} {\map \psi t} \map {\psi'} t \rd t
+      | c = [[Derivative of Complex Composite Function]]
+}}
+{{eqn | r = -\int_a^b \map f {\map \gamma u} \map {\gamma'} u \rd u
+      | c = [[Complex Integration by Substitution|substitution]] with $u = \map \psi t$, where $\map \psi a > \map \psi b$
+}}
+{{eqn | r = -\int_C \map f z \rd z
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Next, suppose that $C$ is a [[Definition:Contour (Complex Plane)|contour]].
+Then $C$ 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]].
+By definition of [[Definition:Reversed Contour|reversed contour]], $-C$ is a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of the [[Definition:Finite Sequence|finite sequence]] $-C_n, \ldots, -C_1$ of [[Definition:Directed Smooth Curve|directed smooth curves]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_C \map f z \rd z 
+      | r = \sum_{i \mathop = 1}^n \int_{-C_i} \map f z \rd z 
+      | c = [[Contour Integral of Concatenation of Contours]]
+}}
+{{eqn | r = -\sum_{i \mathop = 1}^n \int_{C_i} \map f z \rd z 
+      | c = from the result above, as each $C_i$ is a [[Definition:Directed Smooth Curve (Complex Plane)|directed smooth curve]]
+}}
+{{eqn | r = -\int_C \map f z \rd z
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Well-Founded Relation is Strongly Well-Founded/Lemma}
+Tags: Class Theory
+
+\begin{theorem}
+Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Class (Class Theory)|class]].
+Let $\prec$ be a [[Definition:Foundational Relation|foundational relation]] on $A$.
+Then $A$ has a [[Definition:Minimal Element under Relation|$\prec$-minimal element]].
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+The general strategy of the proof is as follows:
+We will recursively define a certain [[Definition:Subset|subset]], $a$, of $A$.
+We will use the fact that $\prec$ is a [[Definition:Foundational Relation|foundational relation]] to choose a [[Definition:Minimal Element under Relation|minimal element]], $m$, of $a$.
+Then we will prove that $m$ is in fact a [[Definition:Minimal Element under Relation|minimal element]] of $A$.
+For each $x \in A$, let $\map {\prec^{-1} } x$ denote the [[Definition:Preimage of Element under Relation|preimage]] under $\prec$ of $x$ in $A$.
+For each class $C$, let $\map R C$ denote the set of elements of $C$ of minimal [[Definition:Rank (Set Theory)|rank]], and let $\map R \O = \O$.
+That is:
+For a given class $C$, let $\alpha_C$ be the [[Definition:Smallest Element|smallest]] [[Definition:Ordinal|ordinal]] such that:
+:$C \cap \map V {\alpha_C} \ne \O$
+where $V$ is the [[Definition:Von Neumann Hierarchy|von Neumann hierarchy]].
+Then let $\map R C$ denote the [[Definition:Set|set]] $C \cap \map V {\alpha_C}$.
+Let $F$ be a function defined [[Principle of Recursive Definition|recursively]]:
+:$\map F 0 = \map R A$
+:$\displaystyle \map F {n + 1} = \bigcup_{y \mathop \in \map F n} \map R {\map {\prec^{-1} } y}$
+{{refactor|Extract this lemma into its own page.}}
+=== Lemma ===
+$\map F n$ is a [[Definition:Set|set]] for each $n \in \omega$.
+=== Proof ===
+Proceed by induction:
+$\map R A$ is a set, so $\map F 0$ is a set.
+Suppose that $\map R {\map F n}$ is a set.
+We know that for each $y \in \map F n$, $\map R {\map {\prec^{-1} } y}$ is a set, so by the [[Axiom:Axiom of Unions|Axiom of Unions]], $\map F {n + 1}$ is a set.
+{{qed|lemma}}
+Let $\displaystyle a = \bigcup_{n \mathop \in \omega} \map F n$.
+By the [[Axiom:Axiom of Unions|Axiom of Unions]], $a$ is a set.
+Since $\map F n \subseteq A$ for each $n \in \omega$, $a \subseteq A$.
+By [[Non-Empty Class has Element of Least Rank]], $\map F 0 \ne \O$, so $a \ne \O$.
+Since $\prec$ is [[Definition:Foundational Relation|foundational]], $a$ has a [[Definition:Minimal Element under Relation|$\prec$-minimal element]] $m$.
+{{AimForCont}} that $m$ is not a $\prec$-minimal element of $A$.
+Then, by [[Characterization of Minimal Element]], 
+:$\map {\prec^{-1} } m \ne \O$
+By [[Non-Empty Class has Element of Least Rank]], $\map {\prec^{-1} } m$ has an element $w$ of least [[Definition:Rank (Set Theory)|rank]].
+{{begin-eqn}}
+{{eqn | ll= \exists n \in \N:
+      | l = m
+      | o = \in
+      | r = \map F n
+      | c = Definition of $a$
+}}
+{{eqn | l = w
+      | o = \in
+      | r = \map F {n + 1}
+      | c = Definition of $F$
+}}
+{{eqn | l = w
+      | o = \in
+      | r = a
+      | c = Definition of $a$
+}}
+{{end-eqn}}
+Therefore, $a \cap \map {\prec^{-1} } m \ne \O$, [[Proof by Contradiction|contradicting]] the fact that $m$ is [[Definition:Minimal Element under Relation|$\prec$-minimal]] in $a$.
+Thus we conclude that $m$ is a $\prec$-minimal element of $A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stone Space is Topological Space}
+Tags: Stone Spaces
+
+\begin{theorem}
+Let $\struct {B, \preceq, \wedge, \vee}$ be a non-empty [[Definition:Boolean Algebra|Boolean algebra]].
+Let $\struct {U, \tau}$ be the [[Definition:Stone Space|Stone space]] of $B$.
+Then $\struct {U, \tau}$ is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+{{explain|generally, why the following purports to be a proof. Suggest the specific basis axioms are invoked.}}
+$U$ is non-empty by the [[Ultrafilter Lemma]].
+The topology of the Stone space is defined as the topology generated by the [[Definition:Basis (Topology)|basis]] $Q$ consisting of all sets of the form
+:$\set {x \in U: b \in x}$
+for some $b \in B$.
+We must show that $Q$ is in fact a [[Definition:Synthetic Basis|basis]].
+$Q$ is trivially a subset of the power set of $U$.
+Suppose that $x \in U$.
+Then by the definition of ultrafilter, $x \ne \O$.
+Thus $x$ has some element $b$. Then $x \in \set {x \in U: b \in x} \in Q$.
+So $Q$ covers $U$.
+Let $P, Q \in Q$. Then for some $b, c \in B$,
+:$P = \set {x \in U: b \in x}$ and
+:$Q = \set {y \in U: c \in y}$.
+Let $z \in P \cap Q$.
+Then by the definitions of $P$ and $Q$,
+:$b \in z$ and $c \in z$. So $z$ is an ultrafilter on $B$ containing $b$ and $c$.
+Since $z$ is a filter containing $b$ and $c$, it also contains $b \wedge c$.
+Let $R = \set {w \in U: b \wedge c \in w}$.
+$R$ is clearly an element of $Q$ containing $z$.
+Suppose that $v \in R$.
+Then $b \wedge c \in v$.
+Since $v$ is a filter,
+$b \in v$ and $c \in v$.
+Thus $v \in P \cap Q$.
+Since this holds for each $v \in R$, $R \subseteq P \cap Q$
+{{qed}}
+[[Category:Stone Spaces]]
+h9iz9f3u9277lm6yi1nmkt1wtydxlaq
+\end{proof}<|endoftext|>
+\section{Ring Without Unity may have Quotient Ring with Unity}
+Tags: Rings with Unity, Quotient Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $I$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\struct {R / I, +, \circ}$ be the associated [[Definition:Quotient Ring|quotient ring]].
+Then $\struct {R / I, +, \circ}$ may have a [[Definition:Unity of Ring|unity]] even if $\struct {R, +, \circ}$ has not.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Ring Direct Product|external direct product of rings]] $\Z \oplus 2 \Z$.
+From [[Integer Multiples form Commutative Ring]], $2 \Z$ does not admit a [[Definition:Unity of Ring|unity]].
+By [[Unity of External Direct Sum of Rings]], neither does $\Z \oplus 2 \Z$.
+Now consider the [[Definition:Ideal of Ring|ideal]] $\set 0 \times 2 \Z$ of $\Z \oplus 2 \Z$.
+We have for all $a \in \Z$ and $b, c \in 2 \Z$ that:
+:$\tuple {0, b} - \tuple {0, c} = \tuple {0, b - c}$
+:$\tuple {a, b} \cdot \tuple {0, c} = \tuple {0, b \cdot c}$
+so by the [[Test for Ideal]], indeed $\set 0 \times 2 \Z$ is an [[Definition:Ideal of Ring|ideal]] in $\Z \oplus 2 \Z$.
+By [[Quotient Ring of External Direct Sum of Rings]], we have:
+:$\paren {\Z \oplus 2 \Z} / \paren {\set 0 \times 2 \Z} \cong \paren {Z / \set 0} \oplus \paren {2 \Z / 2 \Z}$
+By [[Quotient Ring by Null Ideal]] and [[Quotient Ring Defined by Ring Itself is Null Ring]], this last ring is [[Definition:Ring Isomorphism|isomorphic]] to $\Z \times \set 0 \cong \Z$.
+Since $\Z$ has a [[Definition:Unity of Ring|unity]], this construction provides an example of the required kind.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Alternative Differentiability Condition}
+Tags: Analysis, Complex Analysis
+
+\begin{theorem}
+Let $\mathbb K$ be either $\R$ or $\C$.
+Let $f: D \to \mathbb K$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]], where $D \subseteq \mathbb K$ is an [[Definition:Open Set (Metric Space)|open set]].
+Let $z \in \mathbb K$.
+Then $f$ is [[Definition:Differentiable Mapping|differentiable]] at $z$ {{iff}} there exist $\alpha \in \mathbb K$ and $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:
+:$\map f {z + h} = \map f z + h \paren {\alpha + \map \epsilon h}$
+where $\map {B_r} 0$ denotes an [[Definition:Open Ball|open ball]] of $0$, and $\epsilon: \map {B_r} 0 \setminus \set 0 \to \mathbb K$ is a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]] with $\displaystyle \lim_{h \mathop \to 0} \map \epsilon h = 0$.
+If the conditions are true, then $\alpha = \map {f'} z$.
+\end{theorem}
+
+\begin{proof}
+This proof assumes that $\mathbb K = \R$.
+=== Necessary Condition ===
+Assume that $f$ is [[Definition:Differentiable Real Function|differentiable]] in $z$.
+By [[Definition:Open Set (Real Analysis)|definition of open set]], there exists $r \in \R_{>0}$ such that  $\left({z - r \,.\,.\, z + r}\right) \subseteq D$.
+We have the [[Definition:Open Ball|open ball]] $B_r \left({0}\right) = \left({-r \,.\,.\, r}\right)$.
+Define $\epsilon: \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\} \to \R$ by:
+:$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$
+If $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$, then $z+h \in \left({z - r \,.\,.\, z + r}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined.
+From [[Continuity of Composite Mapping/Corollary]], it follows that $\epsilon$ is [[Definition:Continuous Real Function|continuous]] in $\left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$.
+As $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $z$, it follows that:
+:$\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac {f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$
+If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$:
+:$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$
+{{qed|lemma}}
+=== Sufficient condition ===
+Rewrite the equation of the assumption to get:
+:$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$
+From [[Sum Rule for Limits of Functions]], it follows that:
+:$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({\alpha + \epsilon \left({h}\right)}\right) = \alpha$
+By [[Definition:Differentiable Real Function|definition of differentiability]], $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This proof assumes that $\mathbb K = \C$.
+=== Necessary Condition ===
+Assume that $f$ is [[Definition:Complex-Differentiable Function|differentiable]] in $z$.
+By [[Definition:Open Set (Complex Analysis)|definition of open set]], there exists $r \in \R_{>0}$ such that the [[Definition:Open Ball|open ball]] $B_r \left({z}\right) \subseteq D$.
+Define $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ by:
+:$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$
+If $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$, then $z+h \in B_r \left({z}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined.
+As $f$ is [[Definition:Complex-Differentiable Function|differentiable]] in $z$, it follows that:
+:$\displaystyle \lim_{h \to 0}  \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac{ f \left({z + h}\right) - f \left({z}\right) } h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$
+If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$f\left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$
+{{qed|lemma}}
+=== Sufficient condition ===
+Rewrite the equation of the assumption as:
+:$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$
+From [[Sum Rule for Limits of Functions]]:
+:$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({ \alpha + \epsilon \left({h}\right) }\right) = \alpha$
+By [[Definition:Complex-Differentiable Function|definition of differentiability]], $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Von Neumann Hierarchy is Cumulative}
+Tags: 
+
+\begin{theorem}
+For any two ordinals $x$ and $y$,
+If $x < y$ then $V(x) \subsetneqq V(y)$.
+\end{theorem}
+
+\begin{proof}
+By [[Von Neumann Hierarchy Comparison]], $V(x) \in V(y)$. (1)
+By (1) and the [[Axiom:Axiom of Foundation]], $V(x) \ne V(y)$.
+Furthermore, by (1) and  [[Von Neumann Hierarchy is Supertransitive]],
+{{ProofWanted}}
+aqsb7mcpwjsa3idzrn0l74m6ckk2kad
+\end{proof}<|endoftext|>
+\section{Limits of Real and Imaginary Parts}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]], where $D \subseteq \C$.
+Let $z_o \in D$ be a [[Definition:Complex Number|complex number]].
+Suppose $f$ is [[Definition:Continuous Complex Function|continuous]] at $z_0$.
+Then:
+:$(1): \quad \displaystyle \lim_{z \to z_o} \operatorname{Re} \left({f \left({z}\right) }\right) = \operatorname{Re} \left({ \lim_{z \to z_o} f \left({z}\right) }\right)$
+:$(2): \quad \displaystyle \lim_{z \to z_o} \operatorname{Im} \left({f \left({z}\right) }\right) = \operatorname{Im} \left({ \lim_{z \to z_o} f \left({z}\right) }\right)$
+Here, $\operatorname{Re} \left({f \left({z}\right) }\right) $ denotes the [[Definition:Real Part|real part]] of $f \left({z}\right)$, and $\operatorname{Im} \left({f \left({z}\right) }\right) $ denotes the [[Definition:Imaginary Part|imaginary part]] of $f \left({z}\right)$.
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Continuous Complex Function#Epsilon-Delta Definition|definition of continuity]]:
+:$\forall \epsilon > 0: \exists \delta > 0: \left\vert{z - z_0}\right\vert < \delta \implies \left\vert{f \left({z}\right) - f \left({z_0}\right)}\right\vert < \epsilon$
+Given $\epsilon > 0$, we find $\delta > 0$ so for all $z \in \C$ with $\left\vert{z - z_0}\right\vert < \delta$:
+{{begin-eqn}}
+{{eqn |l= \epsilon
+      |o= >
+      |r= \left\vert{f \left({z}\right) - f \left({z_0}\right) }\right\vert
+}}
+{{eqn |o= \ge
+      |r= \left\vert{ \operatorname{Re} \left({f \left({z}\right) - f \left({z_0}\right) }\right) }\right\vert
+      |c= by [[Modulus Larger than Real Part]]
+}}
+{{eqn |r= \left\vert{ \operatorname{Re} \left({f \left({z}\right) }\right) - \operatorname{Re} \left({f \left({z_0}\right) }\right) }\right\vert
+      |c= by [[Addition of Real and Imaginary Parts]]
+}}
+{{end-eqn}}
+It follows that:
+:$\forall \epsilon > 0: \exists \delta > 0: \left\vert{z - z_0}\right\vert < \delta \implies \left\vert{ \operatorname{Re} \left({ f \left({z}\right) }\right) - \operatorname{Re} \left({f \left({z_0}\right)} \right)} \right\vert < \epsilon$
+Then equation $(1)$ is proven by:
+{{begin-eqn}}
+{{eqn |l= \lim_{z \to z_o} \operatorname{Re} \left({ f \left({z}\right) }\right)
+      |r= \operatorname{Re} \left({ f \left({z_0}\right) }\right)
+      |c= by [[Definition:Limit of Complex Function|definition of limit]]
+}}
+{{eqn |r= \operatorname{Re} \left({ \lim_{z \to z_o} f \left({z}\right) }\right)
+      |c= by [[Definition:Continuous Complex Function|definition of continuity]]
+}}
+{{end-eqn}}
+The proof for equation $(2)$ with [[Definition:Imaginary Part|imaginary parts]] follows when $\operatorname{Re}$ is replaced by $\operatorname{Im}$ in the equations above.
+{{qed}}
+[[Category:Complex Analysis]]
+pp0l9bv7yjwersw3h80slmzhqm50w6x
+\end{proof}<|endoftext|>
+\section{Odd Number Theorem/Corollary}
+Tags: Odd Number Theorem
+
+\begin{theorem}
+A [[Definition:Recurrence Relation|recurrence relation]] for the [[Definition:Square Number|square numbers]] is:
+:$S_n = S_{n - 1} + 2 n - 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = S_n
+      | r = \sum_{j \mathop = 1}^n \paren {2 j - 1}
+      | c = [[Odd Number Theorem]]
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n - 1} \paren {2 j - 1} + \paren {2 n - 1}
+      | c = {{Defof|Summation}}
+}}
+{{eqn | r = S_{n - 1} + \paren {2 n - 1}
+      | c = [[Odd Number Theorem]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Odd Number Theorem]]
+e5imnpnznujgeroh2tc2wl2ur5udrfg
+\end{proof}<|endoftext|>
+\section{Effect of Sequence of Elementary Row Operations on Determinant}
+Tags: Determinants, Elementary Row Operations, Proofs by Induction
+
+\begin{theorem}
+Let $\hat o_1, \ldots, \hat o_m$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]]. 
+Here, $\hat o_i$ denotes an [[Definition:Elementary Row Operation|elementary row operation]] on a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]] over a [[Definition:Commutative Ring|commutative]] [[Definition:Ring with Unity|ring with unity]] $\struct {R, +, \circ}$.
+Here, $i \in \set {1, \ldots, m}$.
+Then there exists $c \in R$ such that for all [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $n$]] $\mathbf A$ over $R$:
+:$\map \det {\mathbf A} = c \map \det {\mathbf A'}$
+where $\mathbf A'$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]] that results from applying the [[Definition:Elementary Row Operation|elementary row operations]] $\hat o_1, \ldots, \hat o_m$ on $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]] on $m$, the number of [[Definition:Elementary Row Operation|elementary row operations]] in the [[Definition:Finite Sequence|sequence]] $\hat o_1, \ldots, \hat o_m$.
+=== Basis for the Induction ===
+Suppose $m = 1$, so there is only one [[Definition:Elementary Row Operation|elementary row operation]] $\hat o$ in the [[Definition:Finite Sequence|sequence]].
+Let $r_i$ denote the $i$'th [[Definition:Row of Matrix|row]] of $\mathbf A$.
+Suppose that $\hat o$ is of the type $r_i \to a r_i$, where $a \in R$ and $i \in \set {1, \ldots, n}$.
+From [[Effect of Elementary Row Operations on Determinant]], it follows that:
+:$\map \det {\mathbf A} = a \map \det {\mathbf A'}$
+Suppose that $\hat o$ is of the type $r_i \to r_i + ar_j$, where $a \in R$ and $i, j \in \set {1, \ldots, n}, i \ne j$.
+From [[Effect of Elementary Row Operations on Determinant]], it follows that 
+:$\map \det {\mathbf A} = \map \det {\mathbf A'} = 1_R \map \det {\mathbf A'}$
+where $1_R$ denotes the [[Definition:Identity Element|identity element]] of $\struct {R, \circ}$.
+Suppose that $\hat o$ is of the type $r_i \leftrightarrow r_j$.
+From [[Effect of Elementary Row Operations on Determinant]], it follows that 
+:$\map \det {\mathbf A} = -\map \det {\mathbf A'} = -1_R \map \det {\mathbf A'}$
+where the last equality follows from [[Product with Ring Negative/Corollary|Product with Ring Negative: Corollary]].
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+For $m \in \N$, let $\hat o_1, \ldots, \hat o_m$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]]. 
+This is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+There exists $c \in R$ such that for all [[Definition:Square Matrix|matrices]] of [[Definition:Order of Square Matrix|order $n$]] $\mathbf A$:
+:$\map \det {\mathbf A} = c \map \det {\mathbf A'}$
+where $\mathbf A'$ is the [[Definition:Square Matrix|matrix]] of [[Definition:Order of Square Matrix|order $n$]] that results from using the [[Definition:Elementary Row Operation|elementary row operations]] $\hat o_1, \ldots, \hat o_m$ on $\mathbf A$.
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+Let $\hat o_1, \ldots, \hat o_m, \hat o_{m + 1}$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]].
+Let $r_i$ denote the $i$'th [[Definition:Row of Matrix|row]] of $\mathbf A'$.
+Let $\mathbf A''$ denote the [[Definition:Square Matrix|matrix]] of [[Definition:Order of Square Matrix|order $n$]] that results from using the [[Definition:Elementary Row Operation|elementary row operation]] $\hat o_{m + 1}$ on $A'$.
+Then $\mathbf A''$ is equal to the [[Definition:Square Matrix|matrix]] that results from using the [[Definition:Elementary Row Operation|elementary row operations]] $\hat o_1, \ldots, \hat o_m, \hat o_{m + 1}$ on $A$.
+Suppose that $\hat o_{m + 1}$ is of the type $r_i \to ar_i$, where $a \in R$ and $i \in \set {1, \ldots, n}$. 
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf A}
+      | r = c \map \det {\mathbf A'}
+      | c = by the [[Effect of Sequence of Elementary Row Operations on Determinant#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | r = \paren {c a} \map \det {\mathbf A''}
+      | c = [[Effect of Elementary Row Operations on Determinant]]
+}}
+{{end-eqn}}
+Suppose that $\hat o_{m + 1}$ is of the type $r_i \to r_i + a r_j$, where $a \in R$ and $i, j \in \set {1, \ldots, n}, i \ne j$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf A}
+      | r = c \map \det {\mathbf A'}
+      | c = by the [[Effect of Sequence of Elementary Row Operations on Determinant#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | r = c 1_R \map \det {\mathbf A''}
+      | c = [[Effect of Elementary Row Operations on Determinant]]
+}}
+{{eqn | r = c \map \det {\mathbf A''}
+      | c = {{Defof|Identity Element}}, since $R$ is [[Definition:Commutative Ring|commutative]]
+}}
+{{end-eqn}}
+Suppose that $\hat o_{m + 1}$ is of the type $r_i \leftrightarrow r_j$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf A}
+      | r = c \map \det {\mathbf A'}
+      | c = by the [[Effect of Sequence of Elementary Row Operations on Determinant#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | r = c \paren {-1_R} \map \det {\mathbf A''}
+      | c = [[Effect of Elementary Row Operations on Determinant]]
+}}
+{{eqn | r = \paren {-c} \map \det {\mathbf A''}
+      | c = [[Product with Ring Negative/Corollary|Product with Ring Negative: Corollary]]
+}}
+{{end-eqn}}
+Then the [[Definition:Induction Step|induction step]] is proved for all three types of [[Definition:Elementary Row Operation|elementary row operations]].
+{{qed}}
+[[Category:Determinants]]
+[[Category:Elementary Row Operations]]
+[[Category:Proofs by Induction]]
+lxp6dfv6nef3s6nkzxs4nrdn7e7j5vo
+\end{proof}<|endoftext|>
+\section{Alternative Differentiability Condition/Proof 1}
+Tags: Analysis
+
+\begin{theorem}
+Let $\mathbb K$ be either $\R$ or $\C$.
+Let $f: D \to \mathbb K$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]], where $D \subseteq \mathbb K$ is an [[Definition:Open Set (Metric Space)|open set]].
+Let $z \in \mathbb K$.
+Then:
+: $f$ is [[Definition:Differentiable Mapping|differentiable]] at $z$
+{{iff}}:
+: there exist $\alpha \in \mathbb K$ and $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+::$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$
+where:
+: $B_r \left({0}\right)$ denotes an [[Definition:Open Ball|open ball]] of $0$
+: $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \mathbb K$ is a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]] with $\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = 0$.
+If the conditions are true, then $\alpha = f' \left({z}\right)$.
+\end{theorem}
+
+\begin{proof}
+This proof assumes that $\mathbb K = \R$.
+=== Necessary Condition ===
+Assume that $f$ is [[Definition:Differentiable Real Function|differentiable]] in $z$.
+By [[Definition:Open Set (Real Analysis)|definition of open set]], there exists $r \in \R_{>0}$ such that  $\left({z - r \,.\,.\, z + r}\right) \subseteq D$.
+We have the [[Definition:Open Ball|open ball]] $B_r \left({0}\right) = \left({-r \,.\,.\, r}\right)$.
+Define $\epsilon: \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\} \to \R$ by:
+:$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$
+If $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$, then $z+h \in \left({z - r \,.\,.\, z + r}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined.
+From [[Continuity of Composite Mapping/Corollary]], it follows that $\epsilon$ is [[Definition:Continuous Real Function|continuous]] in $\left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$.
+As $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $z$, it follows that:
+:$\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac {f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$
+If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$:
+:$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$
+{{qed|lemma}}
+=== Sufficient condition ===
+Rewrite the equation of the assumption to get:
+:$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$
+From [[Sum Rule for Limits of Functions]], it follows that:
+:$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({\alpha + \epsilon \left({h}\right)}\right) = \alpha$
+By [[Definition:Differentiable Real Function|definition of differentiability]], $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Alternative Differentiability Condition/Proof 2}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\mathbb K$ be either $\R$ or $\C$.
+Let $f: D \to \mathbb K$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]], where $D \subseteq \mathbb K$ is an [[Definition:Open Set (Metric Space)|open set]].
+Let $z \in \mathbb K$.
+Then $f$ is [[Definition:Differentiable Mapping|differentiable]] at $z$ {{iff}} there exist $\alpha \in \mathbb K$ and $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$
+where $B_r \left({0}\right)$ denotes an [[Definition:Open Ball|open ball]] of $0$, and $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \mathbb K$ is a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]] with $\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = 0$.
+If the conditions are true, then $\alpha = f' \left({z}\right)$.
+\end{theorem}
+
+\begin{proof}
+This proof assumes that $\mathbb K = \C$.
+=== Necessary Condition ===
+Assume that $f$ is [[Definition:Complex-Differentiable Function|differentiable]] in $z$.
+By [[Definition:Open Set (Complex Analysis)|definition of open set]], there exists $r \in \R_{>0}$ such that the [[Definition:Open Ball|open ball]] $B_r \left({z}\right) \subseteq D$.
+Define $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ by:
+:$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$
+If $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$, then $z+h \in B_r \left({z}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined.
+As $f$ is [[Definition:Complex-Differentiable Function|differentiable]] in $z$, it follows that:
+:$\displaystyle \lim_{h \to 0}  \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac{ f \left({z + h}\right) - f \left({z}\right) } h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$
+If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$f\left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$
+{{qed|lemma}}
+=== Sufficient condition ===
+Rewrite the equation of the assumption as:
+:$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$
+From [[Sum Rule for Limits of Functions]]:
+:$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({ \alpha + \epsilon \left({h}\right) }\right) = \alpha$
+By [[Definition:Complex-Differentiable Function|definition of differentiability]], $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Every Filter has Limit Point implies Every Ultrafilter Converges}
+Tags: Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let each [[Definition:Filter on Set|filter]] on $S$ have a [[Definition:Limit Point of Filter|limit point]] in $S$.
+Then each [[Definition:Ultrafilter on Set|ultrafilter]] on $S$ [[Definition:Convergent Filter|converges]] to a point in $S$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \left({S, \tau}\right)$ be such that each [[Definition:Filter on Set|filter]] on $S$ has a [[Definition:Limit Point of Filter|limit point]] in $S$.
+Let $\mathcal F$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $S$.
+By hypothesis, $\mathcal F$ has a [[Definition:Limit Point of Filter|limit point]] $x \in S$.
+By [[Limit Point iff Superfilter Converges]], there exists a [[Definition:Filter|filter]] $\mathcal F'$ on $S$ which [[Definition:Convergent Filter|converges]] to $x$ satisfying $\mathcal F \subseteq \mathcal F'$.
+Because $\mathcal F$ is an [[Definition:Ultrafilter on Set|ultrafilter]], $\mathcal F = \mathcal F'$. {{explain}}
+Thus $\mathcal F$ [[Definition:Convergent Filter|converges]] to $x$.
+\end{proof}<|endoftext|>
+\section{Every Ultrafilter Converges implies Every Filter has Limit Point}
+Tags: Filter Theory
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let every [[Definition:Ultrafilter on Set|ultrafilter]] on $S$ be [[Definition:Convergent Filter|convergent]].
+Then every [[Definition:Filter on Set|filter]] on $S$ has a [[Definition:Limit Point of Filter|limit point]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal F$ be a  [[Definition:Filter on Set|filter]] on $S$.
+By the [[Ultrafilter Lemma]], there exists an [[Definition:Ultrafilter on Set|ultrafilter]] $\mathcal F'$ such that $\mathcal F \subseteq \mathcal F'$.
+[[Definition:By Hypothesis|By hypothesis]], $\mathcal F'$ [[Definition:Convergent Filter|converges]] to some $x \in S$.
+This, by [[Limit Point iff Superfilter Converges]], implies that $x$ is a [[Definition:Limit Point of Filter|limit point]] of $\mathcal F$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left Distributive and Commutative implies Distributive}
+Tags: Commutativity, Distributive Operations
+
+\begin{theorem}
+Let $\struct {S, \circ, *}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let the [[Definition:Binary Operation|operation]] $\circ$ be [[Definition:Left Distributive Operation|left distributive]] over the [[Definition:Binary Operation|operation]] $*$.
+Let $\circ$ be [[Definition:Commutative Operation|commutative]].
+Then $\circ$ is [[Definition:Distributive Operation|distributive]] over $*$.
+\end{theorem}
+
+\begin{proof}
+Let $a, b, c \in S$.
+Then
+{{begin-eqn}}
+{{eqn | l = \paren {a * b} \circ c
+      | r = c \circ \paren {a * b}
+      | c = $\circ$ is [[Definition:Commutative Operation|commutative]]
+}}
+{{eqn | r = \paren {c \circ a} * \paren {c \circ b}
+      | c = $\circ$ is [[Definition:Left Distributive Operation|left distributive]] over $*$
+}}
+{{eqn | r = \paren {a \circ c} * \paren {b \circ c}
+      | c = $\circ$ is [[Definition:Commutative Operation|commutative]]
+}}
+{{end-eqn}}
+So $\circ$ is [[Definition:Right Distributive Operation|right distributive]] over $*$.
+Since $\circ$ is both [[Definition:Left Distributive Operation|left distributive]] and [[Definition:Right Distributive Operation|right distributive]] over $*$, it is [[Definition:Distributive Operation|distributive]] over $*$.
+{{qed}}
+[[Category:Commutativity]]
+[[Category:Distributive Operations]]
+fjs1jy5xpe42hhx0jeubt3pigqlnxfq
+\end{proof}<|endoftext|>
+\section{Filter on Product Space Converges iff Projections Converge}
+Tags: Topology, Filter Theory, Projections
+
+\begin{theorem}
+Let $\family{X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $\displaystyle X := \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Product Space of Topological Spaces|product space]].
+Let $\pr_i: X \to X_i$ denote the [[Definition:Projection (Mapping Theory)|projection]] from $X$ onto $X_i$.
+Let $\FF \subset \powerset X$ be a [[Definition:Filter on Set|filter]] on $X$.
+Then $\FF$ [[Definition:Convergent Filter|converges]] {{iff}} for all $i \in I$ the [[Definition:Image Filter|image filter]] $\map {\pr_i} \FF$ converges.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\FF$ converge.
+Then there is a point $x \in X$ such that $\FF$ converges to $x$.
+By [[Filter on Product Space Converges to Point iff Projections Converge to Projections of Point]]:
+:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$
+Thus
+:$\forall i \in I: \map {\pr_i} \FF$ converges.
+=== Necessary Condition ===
+Suppose that for each $i \in I$, $\map {\pr_i} \FF$ converges.
+For each $i \in I$, let $S_i$ be the set of points to which $\map {\pr_i} \FF$ converges.
+By our supposition, $S_i$ is non-empty for each $i \in I$.
+By the [[Axiom:Axiom of Choice|axiom of choice]], there is a point $x \in X$ such that:
+:$\forall i \in I: \map {\pr_i} x \in S_i$
+By [[Filter on Product Space Converges to Point iff Projections Converge to Projections of Point]], $\FF$ converges to $x$.
+{{qed}}
+{{AoC}}
+\end{proof}<|endoftext|>
+\section{Tychonoff's Theorem for Hausdorff Spaces}
+Tags: Topology
+
+\begin{theorem}
+Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Hausdorff Space|Hausdorff spaces]], where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $\displaystyle X = \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Product Space of Topological Spaces|product space]].
+Then $X$ is [[Definition:Compact Space|compact]] {{iff}} each $X_i$ is [[Definition:Compact Space|compact]].
+== Proof ==
+First assume that $X$ is [[Definition:Compact Space|compact]]. 
+From [[Projection from Product Topology is Continuous]], the [[Definition:Projection (Mapping Theory)|projections]] $\pr_i : X \to X_i$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+From [[Continuous Image of Compact Space is Compact]], it follows that the $X_i$ are [[Definition:Compact Space|compact]].
+Assume now that each $X_i$ is [[Definition:Compact Space|compact]].
+By [[Equivalent Definitions of Compactness]] it is enough to show that every [[Definition:Ultrafilter on Set|ultrafilter]] on $X$ [[Definition:Convergent Filter|converges]].
+Thus let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $X$.
+From [[Image of Ultrafilter is Ultrafilter]], for each $i \in I$, the [[Definition:Image Filter|image filter]] $\map {\pr_i} \FF$ is an [[Definition:Ultrafilter on Set|ultrafilter]] on $X_i$.
+Each $X_i$ is [[Definition:Compact Space|compact]] by assumption.
+So by [[Equivalent Definitions of Compactness]], each $\map {\pr_i} \FF$ converges.
+From [[Filter on Product of Hausdorff Spaces Converges iff Projections Converge]], $\FF$ converges.
+{{explain|Find the appropriate link for "converges" and explain why this proves the result.}}
+{{qed}}
+\end{theorem}<|endoftext|>
+\section{Filter on Product of Hausdorff Spaces Converges iff Projections Converge}
+Tags: Topology, Filter Theory, Projections
+
+\begin{theorem}
+Let $\family{X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Hausdorff Space|Hausdorff spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $\displaystyle X := \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Product Space of Topological Spaces|product space]].
+Let $\pr_i: X \to X_i$ denote the [[Definition:Projection (Mapping Theory)|projection]] from $X$ onto $X_i$.
+Let $\FF \subset \powerset X$ be a [[Definition:Filter on Set|filter]] on $X$.
+Then $\FF$ [[Definition:Convergent Filter|converges]] {{iff}} for each $i \in I$ the [[Definition:Image Filter|image filter]] $\map {\pr_i} \FF$ converges.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\FF$ converge.
+Then there is a point $x \in X$ such that $\FF$ converges to $x$.
+By [[Filter on Product Space Converges to Point iff Projections Converge to Projections of Point]]:
+:$\forall i \in I: \map {\pr_i} \FF$ converges to $x_i$.
+Thus, for each $i \in I$, $\map {\pr_i} \FF$ converges.
+=== Necessary Condition ===
+For each $i \in I$, let $\map {\pr_i} \FF$ converge.
+Since $X_i$ is a Hausdorff space for each $i \in I$, this implies that $\map {\pr_i} \FF$ converges to exactly one point for each $i \in I$.
+For each $i$, let $x_i$ be the point in $X_i$ to which $\map {\pr_i} \FF$ converges.
+Then by the definition of the product space, $x$ is an element of $X$.
+By [[Filter on Product Space Converges to Point iff Projections Converge to Projections of Point]], $\FF$ converges to $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tychonoff's Theorem/General Case}
+Tags: Topology
+
+\begin{theorem}
+Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]], where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $\displaystyle X = \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Product Space of Topological Spaces|product space]].
+Then $X$ is [[Definition:Compact Space|compact]] {{iff}} each $X_i$ is [[Definition:Compact Space|compact]].
+\end{theorem}
+
+\begin{proof}
+First assume that $X$ is [[Definition:Compact Space|compact]]. 
+From [[Projection from Product Topology is Continuous]], the [[Definition:Projection (Mapping Theory)|projections]] $\pr_i : X \to X_i$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+It follows from [[Continuous Image of Compact Space is Compact]] that the $X_i$ are [[Definition:Compact Space|compact]].
+Assume now that each $X_i$ is [[Definition:Compact Topological Space|compact]].
+By [[Equivalent Definitions of Compactness]] it is enough to show that every [[Definition:Ultrafilter on Set|ultrafilter]] on $X$ [[Definition:Convergent Filter|converges]].
+Thus let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $X$.
+From [[Image of Ultrafilter is Ultrafilter]], for each $i \in I$, the [[Definition:Image Filter|image filter]] $\map {\pr_i} \FF$ is an [[Definition:Ultrafilter on Set|ultrafilter]] on $X_i$.
+Each $X_i$ is [[Definition:Compact Space|compact]] by assumption.
+So by [[Equivalent Definitions of Compactness]], each $\map {\pr_i} \FF$ converges.
+By [[Filter on Product Space Converges iff Projections Converge]], $\FF$ converges.
+{{explain|Find the appropriate link for "converges" and explain why this proves the result.}}
+{{qed}}
+{{AoC|Filter on Product Space Converges iff Projections Converge}}
+\end{proof}<|endoftext|>
+\section{Positive Real has Real Square Root}
+Tags: Real Numbers
+
+\begin{theorem}
+Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Then:
+:$\exists y \in \R: x = y^2$
+\end{theorem}
+
+\begin{proof}
+Let $f: \R \to \R$ be defined as:
+:$\forall x \in \R: f \left({x}\right) = x^2$
+We have that $f$ is the [[Definition:Pointwise Multiplication of Real-Valued Functions|pointwise product]] of the [[Definition:Identity Mapping|identity mapping]] with itself.
+By [[Product Rule for Continuous Functions]] and [[Identity Mapping is Continuous]], $f$ is [[Definition:Continuous Real Function|continuous]].
+By [[Power Function is Unbounded Above]]:
+:$\exists q \in \R: f \left({q}\right) > x$
+Then:
+:$0^2 = 0 \le x$
+By the [[Intermediate Value Theorem]]:
+:$\exists y \in \R: 0 < y < q: y^2 = x$
+{{qed}}
+[[Category:Real Numbers]]
+ltzo78jwbsk41ccg9u01ilvme4jfeo1
+\end{proof}<|endoftext|>
+\section{Join is Idempotent}
+Tags: Join and Meet, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
+Then $\vee$ is [[Definition:Idempotent Operation|idempotent]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \vee a
+      | r = \sup \set {a, a}
+      | c = Definition of [[Definition:Join (Order Theory)|Join]]
+}}
+{{eqn | r = \sup \set a
+      | c = [[Axiom:Axiom of Extension|Axiom of Extension]]
+}}
+{{eqn | r = a
+      | c = [[Supremum of Singleton]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Positive Strictly Increasing Mappings is Strictly Increasing}
+Tags: 
+
+\begin{theorem}
+Let $A$ be an ordered set.
+Let $B$ be an ordered field.
+Let $f, g: A \to B$ be strictly increasing mappings with positive values.
+Let $h \colon A \to B$ be defined by $h \left({x}\right) = f \left({x}\right) g \left({x}\right)$.
+Then $h$ is strictly increasing.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in A$ such that $x < y$.
+If $h \left({x}\right) = 0$.
+By the definition of strictly increasing:
+:$f \left({y}\right) > f \left({x}\right) \geq 0$
+and:
+:$g \left({y}\right) > g \left({x}\right) \geq 0$
+So:
+:$h \left({y}\right) > 0 = h \left({x}\right)$
+If $h \left({x}\right) \neq 0$, then $h \left({x}\right) > 0$
+so:
+:$f \left({x}\right) > 0$
+and:
+:$g \left({x}\right) > 0$
+Also:
+:$f \left({x}\right) < f \left({y}\right)$
+and:
+:$g \left({x}\right) < g \left({y}\right)$
+by the definition of strictly increasing.
+Because $g \left({x}\right) > 0$:
+:$f \left({x}\right) g \left({x}\right) < f \left({y}\right) g \left({x}\right)$
+Because $f \left({y}\right) > 0$:
+:$f \left({y}\right) g \left({x}\right) < f \left({y}\right) g \left({y}\right)$
+By transitivity:
+:$h \left({x}\right) = f \left({x}\right) g \left({x}\right) < f \left({y}\right) g \left({y}\right) = h \left({y}\right)$
+{{qed}}
+{{expand|$B$ ordered ring?}}
+{{expand|Generalize by induction to finite products}}
+n6zhw1jc57j21aodnah26b098ud6u0w
+\end{proof}<|endoftext|>
+\section{Supremum of Singleton}
+Tags: Suprema, Singletons
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Then for all $a \in S$:
+:$\sup \set a = a$
+where $\sup$ denotes [[Definition:Supremum of Set|supremum]].
+\end{theorem}
+
+\begin{proof}
+Since $a \preceq a$, $a$ is an [[Definition:Upper Bound of Set|upper bound]] of $\set a$.
+Let $b$ be another [[Definition:Upper Bound of Set|upper bound]] of $\set a$.
+Then necessarily $a \preceq b$.
+It follows that indeed:
+:$\sup \set a = a$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Singleton}
+Tags: Infima, Singletons
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Then for all $a \in S$:
+:$\inf \set a = a$
+where $\inf$ denotes [[Definition:Infimum of Set|infimum]].
+\end{theorem}
+
+\begin{proof}
+Since $a \preceq a$, $a$ is a [[Definition:Lower Bound of Set|lower bound]] for $\set a$.
+Let $b$ be another lower bound for $\set a$.
+Then necessarily $b \preceq a$.
+It follows that indeed:
+:$\inf \set a = a$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet is Idempotent}
+Tags: Join and Meet, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Then $\wedge$ is [[Definition:Idempotent Operation|idempotent]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \wedge a
+      | r = \inf \set {a, a}
+      | c = {{Defof|Meet (Order Theory)}}
+}}
+{{eqn | r = \inf \set a
+      | c = [[Axiom:Axiom of Extension|Axiom of Extension]]
+}}
+{{eqn | r = a
+      | c = [[Infimum of Singleton]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD from Prime Decomposition}
+Tags: Greatest Common Divisor, Prime Decompositions, GCD from Prime Decomposition
+
+\begin{theorem}
+Let $a, b \in \Z$.
+From [[Expression for Integers as Powers of Same Primes]], let:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
+}}
+{{eqn | l = b
+      | r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
+}}
+{{eqn | ll= \forall i \in \set {1, 2, \dotsc, r}:
+      | l = p_i
+      | o = \divides
+      | r = a
+      | c = 
+}}
+{{eqn | lo= \lor
+      | l = p_i
+      | o = \divides
+      | r = b
+}}
+{{end-eqn}}
+That is, the [[Definition:Prime Number|primes]] given in these [[Definition:Prime Decomposition|prime decompositions]] may be [[Definition:Divisor of Integer|divisors]] of ''either'' of the numbers $a$ or $b$.
+Then:
+:$\gcd \set {a, b} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$
+where $\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+Note that if one of the [[Definition:Prime Number|primes]] $p_i$ does not appear in the [[Definition:Prime Decomposition|prime decompositions]] of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be [[Definition:Zero (Number)|zero]].
+Let $d \divides a$.
+Then:
+:$d$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le h_i \le k_i$
+:$d \divides a \iff \forall i: 1 \le i \le r, 0 \le h_i \le l_i$
+So:
+:$d \divides a \land d \divides b \iff \forall i: 1 \le i \le r, 0 \le h_i \le \min \set {k_i, l_i}$
+For $d$ to be at its greatest, we want the largest possible exponent for each of these [[Definition:Prime Number|primes]].
+So for each $i \in \closedint 1 r$, $h_i$ needs to ''equal'' $\min \set {k_i, l_i}$.
+Hence the result:
+:$\gcd \set {a, b} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{LCM from Prime Decomposition}
+Tags: Lowest Common Multiple, Prime Numbers, LCM from Prime Decomposition
+
+\begin{theorem}
+Let $a, b \in \Z$.
+From [[Expression for Integers as Powers of Same Primes]], let:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}
+}}
+{{eqn | l = b
+      | r = p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r}
+}}
+{{eqn | ll= \forall i \in \set {1, 2, \dotsc, r}:
+      | l = p_i
+      | o = \divides
+      | r = a
+      | c = 
+}}
+{{eqn | lo= \lor
+      | l = p_i
+      | o = \divides
+      | r = b
+}}
+{{end-eqn}}
+That is, the [[Definition:Prime Number|primes]] given in these [[Definition:Prime Decomposition|prime decompositions]] may be [[Definition:Divisor of Integer|divisors]] of ''either'' of the numbers $a$ or $b$.
+Then:
+:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
+where $\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lcm \set {a, b}
+      | r = \frac {a b} {\gcd \set {a, b} }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \frac {\paren {p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r} } \paren {p_1^{l_1} p_2^{l_2} \ldots p_r^{l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }
+      | c = [[GCD from Prime Decomposition]]
+}}
+{{eqn | r = \frac {\paren {p_1^{k_1 + l_1} p_2^{k_2 + l_2} \cdots p_r^{k_r + l_r} } } {p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \cdots p_r^{\min \set {k_r, l_r} } }
+      | c = 
+}}
+{{eqn | r = p_1^{k_1 + l_1 - \min \set {k_1, l_1} } p_2^{k_2 + l_2 - \min \set {k_2, l_2} } \cdots p_r^{k_r + l_r - \min \set {k_r, l_r} }
+      | c = 
+}}
+{{eqn | r = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \cdots p_r^{\max \set {k_r, l_r} }
+      | c = [[Sum Less Minimum is Maximum]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Note that if one of the [[Definition:Prime Number|primes]] $p_i$ does not appear in the [[Definition:Prime Decomposition|prime decompositions]] of either one of $a$ or $b$, then its corresponding index $k_i$ or $l_i$ will be [[Definition:Zero (Number)|zero]].
+Let $a \divides m$.
+Then:
+:$m$ is of the form $p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}, \forall i: 1 \le i \le r, 0 \le k_i \le h_i$
+:$a \divides l \iff \forall i: 1 \le i \le r, 0 \le l_i \le h_i$
+So:
+:$a \divides m \land b \divides m \iff \forall i: 1 \le i \le r, 0 \le \max \set {k_i, l_i} \le h_i$
+For $m$ to be at its smallest, we want the smallest possible exponent for each of these [[Definition:Prime Number|primes]].
+So for each $i \in \closedint 1 r$, $h_i$ needs to ''equal'' $\max \set {k_i, l_i}$.
+Hence the result:
+:$\lcm \set {a, b} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Product of Positive Element and Element Greater than One}
+Tags: Ordered Rings
+
+\begin{theorem}
+Let $\left({R, +, \circ, \le}\right)$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]] $1_R$ and [[Definition:Ring Zero|zero]] $0_R$.
+Let $x,y \in R$.
+Suppose that $x > 0_R$ and $y > 1_R$.
+Then $x \circ y > x$ and $y \circ x > x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | o = >
+      | r = 1_R
+}}
+{{eqn | l = x \circ y
+      | o = > 
+      | r = x \circ 1_R
+      | c = $x > 0_R$, [[Properties of Ordered Ring|Properties of Ordered Ring: $(6)$]]
+}}
+{{eqn | l = x \circ y
+      | o = >
+      | r = x
+      | c = {{Defof|Unity of Ring}}
+}}
+{{end-eqn}}
+A similar argument shows that $y \circ x > x$.
+{{qed}}
+[[Category:Ordered Rings]]
+3avgf1pkfjbsnrx11lyswl004xmyvda
+\end{proof}<|endoftext|>
+\section{Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element}
+Tags: Ordered Rings
+
+\begin{theorem}
+Let $\left({R, +, \circ, \le}\right)$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]].
+Let $x \in R$ with $x > 1$ and $x > 0$.
+Let $n \in \N_{>0}$.
+Then:
+: $\circ^n x \ge x$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]]:
+If $n = 1$, then $\circ^n x = x$.
+So:
+: $\circ^n x \ge x$
+Suppose that $\circ^n x \ge x$.
+Then since $x > 1$:
+: $\circ^n x > 1$
+By [[Product of Positive Element and Element Greater than One]], $x \circ \left({\circ^n x}\right) > x$, so:
+:$\circ^{n + 1} x \ge x$
+{{qed}}
+[[Category:Ordered Rings]]
+jp4f4cxys96x080ks18yg4zni6n7txu
+\end{proof}<|endoftext|>
+\section{Directed Set has Strict Successors iff Unbounded Above}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \le}\right)$ be a [[Definition:Directed Set|directed set]].
+Then every element of $S$ has a [[Definition:Strict Successor|strict successor]] in $S$ [[Definition:Iff|iff]] $S$ has no [[Definition:Upper Bound of Set|upper bound]] in $S$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that each element of $S$ has a [[Definition:Strict Successor|strict successor]] in $S$.
+If $x$ is any element of $S$, then $x$ has a [[Definition:Strict Successor|strict successor]].
+Therefore, $x$ is not an [[Definition:Upper Bound of Set|upper bound]] of $S$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $S$ has no [[Definition:Upper Bound of Set|upper bound]] in $S$.
+Let $x \in S$.
+Since $S$ has no [[Definition:Upper Bound of Set|upper bound]], $x$ is not an [[Definition:Upper Bound of Set|upper bound]] of $S$.
+Thus for some $p \in S$, $p \not\le x$.
+By the definition of a [[Definition:Directed Set|directed set]], there is a $y \in S$ such that $p \le y$ and $x \le y$.
+Since $p \not\le x$, it follows that $x \ne y$.
+Hence, since we already know that $x \le y$, in fact $x < y$.
+Thus $y$ is a [[Definition:Strict Successor|strict successor]] of $x$.
+{{qed}}
+[[Category:Order Theory]]
+64s3c8t5iwrvm98ky606f08v4dr0q00
+\end{proof}<|endoftext|>
+\section{Complex Numbers are Uncountable}
+Tags: Complex Numbers, Set Theory
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Uncountable Set|uncountably infinite]].
+\end{theorem}
+
+\begin{proof}
+For all $r \in \R$, we have $r = r + 0 i \in C$.
+Thus the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ can be considered a [[Definition:Subset|subset]] of $\C$.
+As the [[Real Numbers are Uncountable]], it follows from [[Sufficient Conditions for Uncountability]], proposition $(1)$, that $\C$ is [[Definition:Uncountable Set|uncountably infinite]].
+{{qed}}
+[[Category:Complex Numbers]]
+[[Category:Set Theory]]
+136ianqjlmsm2mh37zvydw8vz6bb46p
+\end{proof}<|endoftext|>
+\section{Infimum of Infima}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $\mathbb T$ be a collection of [[Definition:Subset|subsets]] of $S$.
+Suppose all $T \in \mathbb T$ admit an [[Definition:Infimum of Set|infimum]] $\inf T$ in $S$.
+Then:
+:$\inf \bigcup \mathbb T = \inf \left\{{\inf T: T \in \mathbb T}\right\}$
+as soon as one of these two quantities exists.
+\end{theorem}
+
+\begin{proof}
+Suppose that $s = \inf \bigcup \mathbb T \in S$.
+By [[Set is Subset of Union]], $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.
+Hence by [[Infimum of Subset]]:
+:$\forall T \in \mathbb T: s \preceq \inf T$
+Suppose now that $a \in S$ satisfies:
+:$\forall T \in \mathbb T: a \preceq \inf T$
+Then by [[Definition:Transitive Relation|transitivity]] of $\preceq$:
+:$\forall t \in T: a \preceq t$
+Since this holds for any $T \in \mathbb T$, also:
+:$\forall t \in \bigcup \mathbb T: a \preceq t$
+Hence $a \preceq s$, by definition of [[Definition:Infimum of Set|infimum]].
+That is, $s = \inf \left\{{\inf T: T \in \mathbb T}\right\}$.
+{{qed|lemma}}
+Suppose now that $r = \inf \left\{{\inf T: T \in \mathbb T}\right\} \in S$.
+By definition of [[Definition:Infimum of Set|infimum]], for all $T \in \mathbb T$ and $t \in T$:
+:$\inf T \preceq t$
+By [[Definition:Transitive Relation|transitivity]] of $\preceq$:
+:$\forall T \in \mathbb T: \forall t \in T: r \preceq t$
+Hence for all $t \in \bigcup \mathbb T$:
+:$r \preceq t$
+Suppose that $a \in S$ satisfies:
+:$\forall t \in \bigcup \mathbb T: a \preceq t$
+In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$:
+:$a \preceq \inf T$
+and therefore by definition of [[Definition:Infimum of Set|infimum]], also:
+:$a \preceq r$
+That is, $r = \inf \bigcup \mathbb T$.
+{{qed}}
+[[Category:Order Theory]]
+fq5xoc97v99rszpr7xohflxeoe0z8e5
+\end{proof}<|endoftext|>
+\section{Join Semilattice is Ordered Structure}
+Tags: Lattice Theory, Join and Meet, Join Semilattice is Ordered Structure
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Then $\left({S, \vee, \preceq}\right)$ is an [[Definition:Ordered Structure|ordered structure]].
+That is, $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\vee$.
+\end{theorem}
+
+\begin{proof}
+For $\struct {S, \vee, \preceq}$ to be an [[Definition:Ordered Structure|ordered structure]] is equivalent to, for all $a, b, c \in S$:
+:$a \preceq b \implies a \vee c \preceq b \vee c$
+:$a \preceq b \implies c \vee a \preceq c \vee b$
+Since [[Join is Commutative]], it suffices to prove the first of these implications.
+By definition of [[Definition:Join (Order Theory)|join]]:
+:$a \vee c = \sup \set {a, c}$
+where $\sup$ denotes [[Definition:Supremum of Set|supremum]].
+By [[Join Succeeds Operands]]:
+:$b \preceq b \vee c$
+:$c \preceq b \vee c$
+Now also $a \preceq b$, and by [[Definition:Transitive Relation|transitivity]] of $\preceq$ we find that:
+:$a \preceq b \vee c$
+Thus $b \vee c$ is an [[Definition:Upper Bound of Set|upper bound]] for $\set {a, c}$.
+Hence:
+:$a \vee c \preceq b \vee c$
+by definition of [[Definition:Supremum of Set|supremum]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a, b, c \in S$.
+Let $a \preceq b$.
+By the definition of [[Definition:Join Semilattice|join semilattice]]:
+: $a \vee b = b$
+Thus:
+: $\left({a \vee b}\right) \vee c = b \vee c$
+Since $\vee$ is [[Definition:Associative|associative]], [[Definition:Commutative Operation|commutative]], and [[Definition:Idempotent Operation|idempotent]]:
+:$\left({a \vee c}\right) \vee \left({b \vee c}\right) = b \vee c$
+Therefore, $a \vee c \preceq b \vee c$.
+From [[Join is Commutative]], we conclude that:
+: $c \vee a \preceq c \vee b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet Semilattice is Ordered Structure}
+Tags: Lattice Theory, Join and Meet
+
+\begin{theorem}
+Let $\left({S, \wedge, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Then $\left({S, \wedge, \preceq}\right)$ is an [[Definition:Ordered Structure|ordered structure]].
+\end{theorem}
+
+\begin{proof}
+For $\left({S, \wedge, \preceq}\right)$ to be an [[Definition:Ordered Structure|ordered structure]] is equivalent to, for all $a,b,c \in S$:
+:$a \preceq b \implies a \wedge c \preceq b \wedge c$
+:$a \preceq b \implies c \wedge a \preceq c \wedge b$
+Since [[Meet is Commutative]], it suffices to prove the first of these implications.
+By definition of [[Definition:Meet (Order Theory)|meet]]:
+:$b \wedge c = \inf \left\{{b, c}\right\}$
+where $\inf$ denotes [[Definition:Infimum of Set|infimum]].
+By [[Meet Precedes Operands]]:
+:$a \wedge c \preceq a$
+:$a \wedge c \preceq c$
+Now also $a \preceq b$, and by [[Definition:Transitive Relation|transitivity]] of $\preceq$ we find that:
+:$a \wedge c \preceq b$
+Thus $a \wedge c$ is a [[Definition:Lower Bound of Set|lower bound]] for $\left\{{b, c}\right\}$.
+Hence:
+:$a \wedge c \preceq b \wedge c$
+by definition of [[Definition:Infimum of Set|infimum]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Join Semilattice is Semilattice}
+Tags: Semilattices, Join and Meet
+
+\begin{theorem}
+Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
+Then $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Join Semilattice|join semilattice]], $\vee$ is [[Definition:Closed Operation|closed]].
+The other three defining properties for a [[Definition:Semilattice|semilattice]] follow respectively from:
+:[[Join is Commutative]]
+:[[Join is Associative]]
+:[[Join is Idempotent]]
+Hence $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet Semilattice is Semilattice}
+Tags: Semilattices, Join and Meet
+
+\begin{theorem}
+Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Then $\struct {S, \wedge}$ is a [[Definition:Semilattice|semilattice]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Meet Semilattice|meet semilattice]], $\wedge$ is [[Definition:Closed Operation|closed]].
+The other three defining properties for a [[Definition:Semilattice|semilattice]] follow respectively from:
+:[[Meet is Commutative]]
+:[[Meet is Associative]]
+:[[Meet is Idempotent]]
+Hence $\struct {S, \wedge}$ is a [[Definition:Semilattice|semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Semilattice Induces Ordering}
+Tags: Semilattices
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semilattice|semilattice]].
+Let $\preceq$ be the [[Definition:Relation|relation]] on $S$ defined by, for all $a, b \in S$:
+:$a \preceq b$ {{iff}} $a \circ b = b$
+Then $\preceq$ is an [[Definition:Ordering|ordering]].
+\end{theorem}
+
+\begin{proof}
+Let us verify that $\preceq$ satisfies the three conditions for an [[Definition:Ordering|ordering]].
+=== Reflexivity ===
+Since $\circ$ is [[Definition:Idempotent Operation|idempotent]], for all $a \in S$:
+:$a \circ a = a$
+hence $a \preceq a$.
+Thus $\preceq$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Antisymmetry ===
+Suppose that $a \preceq b$ and $b \preceq a$.
+Then from the first relation:
+:$a \circ b = b$
+and from the second:
+:$b \circ a = a$
+Since $\circ$ is [[Definition:Commutative Operation|commutative]], it follows that $a = b$.
+Hence $\preceq$ is [[Definition:Antisymmetric Relation|antisymmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Suppose that $a \preceq b$ and $b \preceq c$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \circ c
+      | r = a \circ \paren {b \circ c}
+      | c = Since $b \preceq c$
+}}
+{{eqn | r = \paren {a \circ b} \circ c
+      | c = $\circ$ is [[Definition:Associative Operation|associative]]
+}}
+{{eqn | r = b \circ c
+      | c = Since $a \preceq b$
+}}
+{{eqn | r = c
+      | c = Since $b \preceq c$
+}}
+{{end-eqn}}
+Hence, $a \preceq c$.
+Thus $\preceq$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+Having verified all three conditions, it follows that $\preceq$ is an [[Definition:Ordering|ordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elementary Row Operations Commute with Matrix Multiplication}
+Tags: Elementary Row Operations, Conventional Matrix Multiplication, Commutativity
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Operation|commutative]] [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\mathbf A = \sqbrk a_{m n}$ be an [[Definition:Matrix|$m \times n$ matrix]] over $R$.
+Let $\mathbf B = \sqbrk b_{n p}$ be an [[Definition:Matrix|$n \times p$ matrix]] over $R$.
+Let $\hat o_1, \ldots, \hat o_{\hat n}$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] that can be performed on a matrix over $R$ with $m$ [[Definition:Row of Matrix|rows]].
+Let $\mathbf A'$ denote the $m \times n$-matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n}$ on $\mathbf A$.
+Let $\mathbf C = \mathbf A \mathbf B$ be the [[Definition:Matrix Product (Conventional)|matrix product]] of $\mathbf A$ and $\mathbf B$.
+Let $\mathbf C'$ denote the $m \times p$-matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n}$ on $\mathbf C$.
+Then:
+:$\mathbf C' = \mathbf A' \mathbf B$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]] over $\hat n \in \N$, the number of [[Definition:Elementary Row Operation|elementary row operations]].
+=== Basis for the Induction ===
+Suppose that $\hat n = 0$.
+Then, the proof is trivial:
+:$\mathbf C' = \mathbf C = \mathbf A \mathbf B = \mathbf A' \mathbf B$
+=== Induction Hypothesis ===
+Let $\hat n \in \N$ be fixed.
+Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the [[Definition:Matrix|matrix]] that results from using $\hat o_1, \ldots, \hat o_{\hat n}$ on $\mathbf A$.
+Let $\mathbf C' = \sqbrk {c'}_{m p}$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{\hat n}$ on $\mathbf C = \mathbf A \mathbf B$.
+The induction hypothesis is that:
+:$ \mathbf C' = \mathbf A' \mathbf B$
+By definition of [[Definition:Matrix Product (Conventional)|matrix product]], this is equivalent to:
+:$\displaystyle \forall i \in \set {1, \ldots, m}, k \in \set {1, \ldots, p}:  c'_{i k} = \sum_{j \mathop = 1}^n a'_{i j} b_{j k}$
+=== Induction Step ===
+Let $\hat o_1, \ldots, \hat o_{\hat n}, \hat o_{\hat n + 1}$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] that can be performed on a [[Definition:Matrix|matrix]] over $R$ with $m$ [[Definition:Row of Matrix|rows]].
+Let $\mathbf A'' = \sqbrk {a''}_{m n}$ denote the [[Definition:Matrix|matrix]] that results from using $\hat o_1, \ldots, \hat o_{\hat n + 1}$ on $\mathbf A$.
+Let $\mathbf C'' = \sqbrk {c''}_{m p}$ denote the [[Definition:Matrix|matrix]] that results from using $\hat o_1, \ldots, \hat o_{\hat n + 1}$ on $\mathbf C$.
+We must show that $\mathbf C'' = \mathbf A'' \mathbf B$, which by [[Definition:Matrix Product (Conventional)|definition of matrix product]] is equivalent to:
+:$\displaystyle \forall i \in \set {1, \ldots, m}, k \in \set {1, \ldots, p}: c''_{i k} = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}$
+Let $r_{i_0}$ denote the $i_0$'th [[Definition:Row of Matrix|row of a matrix]].
+It is clear that $\hat o_{\hat n + 1}$ converts $\mathbf A'$ to $\mathbf A''$, and $\hat o_{\hat n + 1}$ converts $\mathbf C'$ to $\mathbf C''$.
+First, suppose that $\hat o_{\hat n + 1}$ is of the type $r_{i_0} \to \alpha r_{i_0}$, where $\alpha \in R$ and $i_0 \in \left\{ {1, \ldots, m}\right\}$.
+If $i \ne i_0$, then:
+{{begin-eqn}}
+{{eqn | ll= (\text i)
+      | l = c''_{i k}
+      | r = c'_{i k}
+      | c = as $\hat o_{\hat n + 1}$ has no effect on $c'_{i k}$
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n} a'_{i j} b_{j k}
+      | c = by the induction hypothesis
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}
+      | c = as $\hat o_{\hat n + 1}$ has no effect on $a'_{i j}$
+}}
+{{end-eqn}}
+If $i = i_0$, then:
+{{begin-eqn}}
+{{eqn | l = c''_{i k}
+      | r = \alpha c'_{i_0 k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{eqn | r = \alpha \sum_{j \mathop = 1}^{n} a'_{i_0 j} b_{j k}
+      | c = by the induction hypothesis
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n} \left({\alpha a'_{i_0 j} }\right) b_{j k}
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Suppose that $\hat o_{\hat n + 1}$ is of the type $r_{i_0} \to r_{i_0} + \alpha r_{i_1}$, where $\alpha \in R$ and $i_0, i_1 \in \left\{ {1, \ldots, m}\right\}, i_0 \ne i_1$.
+If $i \ne i_0$, then $\displaystyle c''_{i k} = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}$ as in the equation $(\text i )$.
+If $i = i_0$, then:
+{{begin-eqn}}
+{{eqn | l = c''_{i k}
+      | r=  c''_{i_0 k} + \alpha c''_{i_1 k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n a'_{i_0 j} b_{j k} + \alpha \sum_{j \mathop = 1}^n a'_{i_1 j} b_{j k}
+      | c = by the induction hypothesis
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n \left({a'_{i_0 j} + \alpha a'_{i_1 j} }\right) b_{j k}
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Suppose that $\hat o_{\hat n + 1}$ is of the type $r_{i_0} \leftrightarrow r_{i_1}$, where $i_0, i_1 \in \set {1, \ldots, m}, i_0 \ne i_1$.
+If $i \ne i_0$ and $i \ne i_1$, then $\displaystyle c''_{i k} = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}$ as in the equation $(\text i)$.
+If $i = i_0$, then:
+{{begin-eqn}}
+{{eqn | l = c''_{i k}
+      | r = c'_{i_1 k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n a'_{i_1 j} b_{j k}
+      | c = by the induction hypothesis
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n a''_{i j} b_{j k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{end-eqn}}
+If $i = i_1$, then:
+{{begin-eqn}}
+{{eqn | l = c''_{i k}
+      | r = c'_{i_0 k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n} a'_{i_0 j} b_{j k}
+      | c = by the induction hypothesis
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{n} a''_{i j} b_{j k}
+      | c = by $\hat o_{\hat n + 1}$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Then the induction step is proved for all three types of [[Definition:Elementary Row Operation|elementary row operations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 1}
+Tags: Hypothetical Syllogism
+
+\begin{theorem}
+:$p \implies q, q \implies r \vdash p \implies r$
+\end{theorem}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 2}
+Tags: Hypothetical Syllogism
+
+\begin{theorem}
+: $p \implies q, q \implies r, p \vdash r$
+\end{theorem}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 2/Proof 1}
+Tags: Hypothetical Syllogism
+
+\begin{theorem}
+: $p \implies q, q \implies r, p \vdash r$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies q, q \implies r, p \vdash r}}
+{{Premise|1|p \implies q}}
+{{Premise|2|q \implies r}}
+{{Premise|3|p}}
+{{SequentIntro|4|1, 2|p \implies r|1, 2|[[Hypothetical Syllogism/Formulation 1|Hypothetical Syllogism: Formulation 1]]}}
+{{ModusPonens|5|1, 2, 3|r|4|3}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 2/Proof 3}
+Tags: Truth Table Proofs, Hypothetical Syllogism
+
+\begin{theorem}
+: $p \implies q, q \implies r, p \vdash r$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the propositions in turn.
+As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} (that is, the rightmost $\land$) is $T$, that under the instance of $r$ on the {{RHS}} is also $T$:
+$\begin{array}{|ccccccccc||c|} \hline
+((p & \implies & q) & \land & (q & \implies & r)) & \land & p & r \\
+\hline
+F & T & F & T & F & T & F & F & F & F \\
+F & T & F & T & F & T & T & F & F & T \\
+F & T & T & F & T & F & F & F & F & F \\
+F & T & T & T & T & T & T & F & F & T \\
+T & F & F & F & F & T & F & F & T & F \\
+T & F & F & F & F & T & T & F & T & T \\
+T & T & T & F & T & F & F & F & T & F \\
+T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Relations Compatible with Operation is Compatible}
+Tags: Compatible Relations
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+Let $\mathcal F$ be a family of [[Definition:Endorelation|relations]] on $S$.
+Let each element of $\mathcal F$ be [[Definition:Relation Compatible with Operation|compatible with $\circ$]].
+Let $\mathcal Q = \bigcup \mathcal F$.
+Then $Q$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z \in S$.
+Let $x \mathop {\mathcal Q} y$.
+Then for some $\mathcal R \in \mathcal F$:
+:$x \mathop{\mathcal R} y$.
+Since $\mathcal R$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]]:
+:$\left({x \circ z}\right) \mathop {\mathcal R} \left({y \circ z}\right)$
+Since $\mathcal R \in \mathcal F$:
+:$\mathcal R \subseteq \bigcup \mathcal F = \mathcal Q$
+Thus 
+:$\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$
+We have shown that:
+:$\forall x, y, z \in S: x \mathop {\mathcal Q} y \implies \left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$
+A similar argument shows that:
+:$\forall x, y, z \in S: x \mathop {\mathcal Q} y \implies \left({z \circ x}\right) \mathop {\mathcal Q} \left({z \circ y}\right)$
+So $Q$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]].
+{{qed}}
+[[Category:Compatible Relations]]
+9hi6f9d88lb8x6c8dddtz5pm5zxj44r
+\end{proof}<|endoftext|>
+\section{Complement of Relation Compatible with Group is Compatible}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
+Let $\mathcal R$ be a [[Definition:Endorelation|relation]] on $G$.
+Let $\mathcal R$ be [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $\mathcal Q = \complement_{G \times G} \mathcal R$, so that:
+:$\forall a, b \in G: a \mathop {\mathcal Q} b \iff \neg \left({a \mathop {\mathcal R} b}\right)$
+Then $\mathcal Q$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z \in G$.
+Suppose that $\neg \left({\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)}\right)$.
+Then by the definition of $\mathcal Q$:
+:$\left({x \circ z}\right) \mathop {\mathcal R} \left({y \circ z}\right)$
+Since $\mathcal R$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
+:$\left({x \circ z}\right) \circ z^{-1} \mathop{\mathcal R} \left({y \circ z}\right) \circ z^{-1}$.
+By [[Definition:Group Axioms|Group Axiom $(G1)$: Associativity]] and the [[Definition:Group Axioms|Group Axiom $(G3)$: Inverses]]:
+:$x \mathop {\mathcal R} y$
+so by the definition of $\mathcal Q$:
+$\neg \left({x \mathop {\mathcal Q} y}\right)$
+By the [[Rule of Transposition]]:
+:$\forall x, y, z \in G: x \mathop {\mathcal Q} y \implies \left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$
+A similar argument shows that:
+:$\forall x, y, z \in G: x \mathop {\mathcal Q} y \implies \left({z \circ x}\right) \mathop {\mathcal Q} \left({z \circ y}\right)$
+Thus, by definition, $\mathcal Q$ is a [[Definition:Relation Compatible with Operation|relation compatible with $\circ$]].
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+h36d75mihbt885114501519gb9llmu8
+\end{proof}<|endoftext|>
+\section{Relation Compatible with Group Operation is Strongly Compatible}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+Let $\left({S,\circ}\right)$ be a [[Definition:Group|group]].
+Let $\mathcal R$ be an [[Definition:Endorelation|endorelation]] on $S$ [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $x, y, z \in S$.
+Then $\mathcal R$ is [[Definition:Relation Strongly Compatible with Operation|strongly compatible]] with $\circ$.
+That is, the following [[Definition:Biconditional|equivalences]] hold:
+:$x \mathrel{\mathcal R} y \iff x \circ z \mathrel{\mathcal R} y \circ z$
+:$x \mathrel{\mathcal R} y \iff z \circ x \mathrel{\mathcal R} z \circ y$
+\end{theorem}
+
+\begin{proof}
+Since $\mathcal R$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
+:$\forall a,b,c \in S: a \mathrel{\mathcal R} b \implies a \circ c \mathrel{\mathcal R} b \circ c$
+In particular, letting $a = x$, $b = y$, and $c = z$, we see that:
+:$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$
+On the other hand, letting $a = x \circ z$, $b = y \circ z$, and $c = z^{-1}$, we see that:
+:$x \circ z \mathrel{\mathcal R} y \circ z \implies \left({x \circ z}\right) \circ z^{-1} \mathrel{\mathcal R} \left({y \circ z}\right) \circ z^{-1}$
+By the [[Definition:Associative|associativity]] of $\circ$ and the definition of [[Definition:Inverse Element|inverse]], this reduces to:
+:$x \circ z \mathrel{\mathcal R} y \circ z \implies x \mathrel{\mathcal R} y$
+We have thus shown that:
+:$x \mathrel{\mathcal R} y \iff x \circ z \mathrel{\mathcal R} y \circ z$
+A precisely similar argument shows that:
+:$x \mathrel{\mathcal R} y \iff z \circ x \mathrel{\mathcal R} z \circ y$
+so $\mathcal R$ is [[Definition:Relation Strongly Compatible with Operation|strongly compatible]] with $\circ$.
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+0n37g0lfrnlig9tbsl2hbz1cgpxioqg
+\end{proof}<|endoftext|>
+\section{Intersection of Relations Compatible with Operation is Compatible}
+Tags: Compatible Relations
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+Let $\mathscr F$ be a [[Definition:Indexed Family|indexed family]] of [[Definition:Endorelation|relations]] on $S$.
+Suppose that each element of $\mathscr F$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $\QQ = \bigcap \mathscr F$ be the [[Definition:Intersection of Family|intersection]] of $\mathscr F$.
+Then $\QQ$ is a [[Definition:Relation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z \in S$.
+Suppose that $x \mathrel \QQ y$.
+Then for each $\RR \in \mathscr F$:
+:$x \mathrel \RR y$
+Then since $\RR$ is a [[Definition:Relation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
+:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
+Since this holds for each $\RR \in \mathscr F$:
+:$\paren {x \circ z} \mathrel \QQ \paren {y \circ z}$
+We have shown that:
+:$\forall x, y, z \in S: x \mathrel \QQ y \implies \paren {x \circ z} \mathrel \QQ \paren {y \circ z}$
+A similar argument shows that:
+:$\forall x, y, z \in S: x \mathrel \QQ y \implies \paren {z \circ x} \mathrel \QQ \paren {z \circ y}$
+so $Q$ is a [[Definition:Relation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+{{qed}}
+[[Category:Compatible Relations]]
+6ahue29b035401wxcwed9oazy028n3m
+\end{proof}<|endoftext|>
+\section{Inverse of Relation Compatible with Operation is Compatible}
+Tags: Compatible Relations
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+Let $\mathcal R$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $\mathcal Q$ be the [[Definition:Inverse Relation|inverse relation]] of $\mathcal R$.
+Then $\mathcal Q$ is compatible with $\circ$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z \in S$.
+Suppose that $x \mathop {\mathcal Q} y$.
+Then by the definition of $\mathcal Q$:
+:$y \mathop{\mathcal R} x$.
+Since $\mathcal R$ is compatible with $\circ$:
+:$\left({y \circ z}\right) \mathop {\mathcal R} \left({x \circ z}\right)$
+and
+:$\left({z \circ y}\right) \mathop {\mathcal R} \left({z \circ x}\right)$
+Thus by the definition of $\mathcal Q$:
+:$\left({x \circ z}\right) \mathop {\mathcal Q} \left({y \circ z}\right)$
+and
+:$\left({z \circ x}\right) \mathop {\mathcal R} \left({z \circ y}\right)$
+Thus $\mathcal Q$ is compatible with $\circ$.
+{{qed}}
+[[Category:Compatible Relations]]
+sbswvghz3ic34e7sfu878bm2e2p0g6p
+\end{proof}<|endoftext|>
+\section{Properties of Ordered Group/OG1}
+Tags: Ordered Groups
+
+\begin{theorem}
+{{begin-axiom}}
+{{axiom|n = \operatorname{OG}1.1
+       |m = x \preceq y \iff x \circ z \preceq y \circ z
+}}
+{{axiom|n = \operatorname{OG}1.2
+       |m = x \preceq y \iff z \circ x \preceq z \circ y
+}}
+{{axiom|n = \operatorname{OG}1.1'
+       |m = x \prec y \iff x \circ z \prec y \circ z
+}}
+{{axiom|n = \operatorname{OG}1.2'
+       |m = x \prec y \iff z \circ x \prec z \circ y
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By the definition of an [[Definition:Ordered Group|ordered group]], $\preceq$ is a relation [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Thus by [[Properties of Relation Compatible with Group Operation/CRG1]], we obtain the first two results:
+:$(\operatorname{OG}1.1):\quad x \preceq y \iff x \circ z \preceq y \circ z$
+:$(\operatorname{OG}1.2):\quad x \preceq y \iff z \circ x \preceq z \circ y$
+By [[Reflexive Reduction of Relation Compatible with Group Operation is Compatible]], $\prec$ is compatible with $\circ$.
+Thus by again [[Properties of Relation Compatible with Group Operation/CRG1]], we obtain the remaining results:
+:$(\operatorname{OG}1.1'):\quad x \prec y \iff x \circ z \prec y \circ z$
+:$(\operatorname{OG}1.2'):\quad x \prec y \iff z \circ x \prec z \circ y$
+and so the theorem is established.
+{{qed}}
+[[Category:Ordered Groups]]
+d7o5mhc9vz9d79jgkpk5orj2dbvhol0
+\end{proof}<|endoftext|>
+\section{Relations Compatible with Operation Form Complete Distributive Lattice}
+Tags: Compatible Relations, Complete Lattices, Distributive Lattices
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $C$ be the [[Definition:Set|set]] of [[Definition:Endorelation|relations]] on $S$ compatible with $\circ$.
+Then $\left({C, \cap,\cup,\subseteq}\right)$ is a [[Definition:Complete Distributive Lattice|complete distributive lattice]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Compatible Relations]]
+[[Category:Complete Lattices]]
+[[Category:Distributive Lattices]]
+hfd4uj6pfmoke74i3c7mciw8hk2knq9
+\end{proof}<|endoftext|>
+\section{Relations Compatible with Group Form Complete Boolean Algebra}
+Tags: Compatible Relations, Complete Lattices, Boolean Lattices
+
+\begin{theorem}
+Let $\left({S,\circ}\right)$ be a [[Definition:Group|group]].
+Let $C$ be the set of [[Definition:Endorelation|relations]] on $S$ which are [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Then $\left({C, \cap, \cup, \subseteq}\right)$ is a [[Definition:Complete Boolean Lattice|complete Boolean lattice]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Compatible Relations]]
+[[Category:Complete Lattices]]
+[[Category:Boolean Lattices]]
+asrktxbh6a5mdi4607z0lbuoo8d4e8u
+\end{proof}<|endoftext|>
+\section{Duality Principle (Order Theory)}
+Tags: Order Theory, Named Theorems
+
+\begin{theorem}
+Let $\Sigma$ be a [[Definition:Statement|statement]] about [[Definition:Ordered Set|ordered sets]] (whether in [[Definition:Natural Language|natural]] or a [[Definition:Formal Language|formal language]]).
+Let $\Sigma^*$ be the [[Definition:Dual Statement (Order Theory)|dual statement]] of $\Sigma$.
+\end{theorem}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 3}
+Tags: Hypothetical Syllogism
+
+\begin{theorem}
+:$\vdash \paren {\paren {p \implies q} \land \paren {q \implies r} } \implies \paren {p \implies r}$
+\end{theorem}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 4}
+Tags: Hypothetical Syllogism
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
+\end{theorem}<|endoftext|>
+\section{Hypothetical Syllogism/Formulation 5}
+Tags: Hypothetical Syllogism
+
+\begin{theorem}
+:$\vdash \paren {q \implies r} \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
+\end{theorem}<|endoftext|>
+\section{Duality Principle (Order Theory)/Global Duality}
+Tags: Order Theory, Named Theorems
+
+\begin{theorem}
+The following are [[Definition:Logical Equivalence|equivalent]]:
+:$(1): \quad \Sigma$ is true for all [[Definition:Ordered Set|ordered sets]]
+:$(2): \quad \Sigma^*$ is true for all [[Definition:Ordered Set|ordered sets]]
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $\struct {S, \succeq}$ be its [[Definition:Dual Ordered Set|dual]].
+By assumption, $\Sigma$ is true for $\struct {S, \succeq}$.
+By [[Duality Principle (Order Theory)/Local Duality|Local Duality]], this implies $\Sigma^*$ is true for $\struct {S, \preceq}$.
+Since $\struct {S, \preceq}$ was arbitrary, the result follows.
+{{qed}}
+=== $(2)$ implies $(1)$ ===
+From [[Dual of Dual Statement (Order Theory)]]:
+:$\paren {\Sigma^*}^* = \Sigma$
+The result thus follows from applying the other implication to $\Sigma^*$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Duality Principle (Order Theory)/Local Duality}
+Tags: Order Theory, Named Theorems
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]], and let $\struct {S, \succeq}$ be its [[Definition:Dual Ordered Set|dual]].
+Then the following are [[Definition:Logical Equivalence|equivalent]]:
+:$(1): \quad \Sigma$ is true for $\struct {S, \preceq}$
+:$(2): \quad \Sigma^*$ is true for $\struct {S, \succeq}$
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+By assumption, $\Sigma$ is true for $\struct {S, \preceq}$.
+By [[Dual of Dual Ordering]], the [[Definition:Dual Statement (Order Theory)|dual statement]] $\Sigma^*$ applied to $\struct {S, \succeq}$ is the same as $\Sigma$ applied to $\struct {S, \preceq}$.
+Hence $\Sigma^*$ is true for $\struct {S, \succeq}$.
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+From [[Dual of Dual Statement (Order Theory)]], $\paren {\Sigma^*}^* = \Sigma$.
+From [[Dual of Dual Ordering]], $\struct {S, \preceq}$ is the [[Definition:Dual Ordered Set|dual]] of $\struct {S, \succeq}$.
+The result thus follows from applying the other implication to $\Sigma^*$ and $\struct {S, \succeq}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modus Tollendo Tollens/Sequent Form/Proof 2}
+Tags: Modus Tollendo Tollens
+
+\begin{theorem}
+: $p \implies q, \neg q \vdash \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies q, \neg q \vdash \neg p}}
+{{Premise|1|p \implies q}}
+{{Premise|2|\neg q}}
+{{Assumption|3|p|Assume $p$ ...}}
+{{ModusPonens|4|1, 3|q|1|3|... and derive $q$ ...}}
+{{NonContradiction|5|1, 2, 3|4|2| ... demonstrating a contradiction}}
+{{Contradiction|6|1, 2|\neg p|3|5}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Complement Relation Compatible with Group Operation}
+Tags: Compatible Relations
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]].
+Let $\Delta_G$ be the [[Definition:Diagonal Relation|diagonal relation]] on $G$.
+Then $\Delta_G^c = \complement_{G \times G} \Delta_G$ is a [[Definition:Relation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+{{explain|Find another way to write the above, it's ugly}}
+In other words, $\ne$ is a relation compatible with $\circ$.
+\end{theorem}
+
+\begin{proof}
+By [[Diagonal Relation is Universally Compatible]], $\Delta_G$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+By [[Complement of Relation Compatible with Group is Compatible ]], $\Delta_G^c$ is also compatible with $\circ$.
+{{qed}}
+[[Category:Compatible Relations]]
+n07d4bm10j5n1oe24z8vocmgjasgc45
+\end{proof}<|endoftext|>
+\section{Reflexive Closure of Relation Compatible with Operation is Compatible}
+Tags: Compatible Relations, Reflexive Closures
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Let $\mathcal R$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $\mathcal R^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\prec$.
+That is, $\mathcal R^=$ is defined as the union of $\mathcal R$ with the [[Definition:Diagonal Relation|diagonal relation]] for $S$.
+Then $\mathcal R^=$ is compatible with $\circ$.
+\end{theorem}
+
+\begin{proof}
+By [[Diagonal Relation is Universally Compatible]], the [[Definition:Diagonal Relation|diagonal relation]] is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Then by [[Union of Relations Compatible with Operation is Compatible]], $\mathcal R^=$ is compatible with $\circ$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reflexive Reduction of Relation Compatible with Group Operation is Compatible}
+Tags: Compatible Relations, Reflexive Reductions
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
+Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $\RR^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\RR$.
+Then $\RR^\ne$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Reflexive Reduction|reflexive reduction]], for all $a, b \in S$:
+:$a \mathrel {\RR^\ne} b$ {{iff}} $a \mathrel \RR b$ but $a \ne b$.
+By definition of the [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$:
+:$a \ne b$ {{iff}} $\tuple {a, b} \notin \Delta_S$
+Thus, considered as [[Definition:Subset|subsets]] of $S \times S$, we have:
+:$\RR^\ne = \RR \setminus \Delta_S$
+By [[Diagonal Relation is Universally Compatible]], $\Delta_S$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Thus by [[Set Difference of Relations Compatible with Group Operation is Compatible]], $\RR^\ne$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Reflexive Reductions]]
+5bruh6rzeh2dor1v9o7f8hwoz6mv5cv
+\end{proof}<|endoftext|>
+\section{Double Negation Elimination implies Law of Excluded Middle}
+Tags: Double Negation
+
+\begin{theorem}
+Let the [[Double Negation Elimination|Law of Double Negation Elimination]] be supposed to hold:
+:$\neg \neg p \vdash p$
+Then the [[Law of Excluded Middle]] likewise holds:
+:$\vdash p \lor \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \lor \neg p}}
+{{Assumption           |1|\neg \paren {p \lor \neg p}|Assume the contrary}}
+{{Assumption           |2|p|Assume one disjunct}}
+{{Addition             |3|2|p \lor \neg p|2|1}}
+{{NonContradiction     |4|1, 2|3|1}}
+{{Contradiction        |5|1|\neg p|2|4|demonstrating a contradiction}}
+{{Addition             |6|1|p \lor \neg p|5|2}}
+{{NonContradiction     |7|1|6|1|also demonstrating a contradiction}}
+{{Contradiction        |8||\neg \neg \paren {p \lor \neg p}|1|7}}
+{{DoubleNegElimination |9||p \lor \neg p|8}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+{{BeginTableau|\vdash p \lor \neg p}}
+{{TheoremIntro|1|\neg\neg (p \lor \neg p)|[[Negation of Excluded Middle is False/Form 2]]}}
+{{DoubleNegElimination|2||p \lor \neg p|1}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Double Negation/Double Negation Introduction/Sequent Form/Formulation 1}
+Tags: Double Negation
+
+\begin{theorem}
+: $p \vdash \neg \neg p$
+\end{theorem}<|endoftext|>
+\section{Double Negation/Double Negation Introduction/Sequent Form/Formulation 2}
+Tags: Double Negation
+
+\begin{theorem}
+: $\vdash p \implies \neg \neg p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash p \implies \neg \neg p}}
+{{Assumption|1|p}}
+{{DoubleNegIntro|2|1|\neg \neg p|1}}
+{{Implication|3||p \implies \neg \neg p|1|2}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Double Negation/Double Negation Elimination/Sequent Form/Formulation 1}
+Tags: Double Negation
+
+\begin{theorem}
+: $\neg \neg p \vdash p$
+\end{theorem}<|endoftext|>
+\section{Dual of Dual Ordering}
+Tags: Order Theory, Dual Ordered Sets
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $\left({S, \succeq}\right)$ be its [[Definition:Dual Ordered Set|dual]].
+Then the [[Definition:Dual Ordered Set|dual]] of $\left({S, \succeq}\right)$ is again $\left({S, \preceq}\right)$.
+\end{theorem}
+
+\begin{proof}
+Denote with $\preceq'$ the [[Definition:Dual Ordering|dual]] of $\succeq$.
+By definition of [[Definition:Dual Ordering|dual ordering]], we thus have for all $a, b \in S$:
+:$a \preceq b$ [[Definition:Iff|iff]] $b \succeq a$
+:$b \succeq a$ [[Definition:Iff|iff]] $a \preceq' b$
+Hence $a \preceq b$ [[Definition:Iff|iff]] $a \preceq' b$.
+The result follows from [[Equality of Relations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dual of Dual Statement (Order Theory)}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\Sigma$ be a [[Definition:Statement|statement]] about [[Definition:Ordered Set|ordered sets]].
+Let $\Sigma^*$ be its [[Definition:Dual Statement (Order Theory)|dual statement]].
+Then $\Sigma$ is also the [[Definition:Dual Statement (Order Theory)|dual statement]] of $\Sigma^*$.
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Dual Statement (Order Theory)|dual statement]] $\Sigma^*$ is formed by replacing the [[Definition:Ordering|ordering]] $\preceq$ with its [[Definition:Dual Ordering|dual]] $\succeq$.
+By [[Dual of Dual Ordering]], applying this operation twice results in the original sentence $\Sigma$ again.
+{{qed}}
+[[Category:Order Theory]]
+sgenggbsvmbmw7wypp9g6hrj4lesfpe
+\end{proof}<|endoftext|>
+\section{Double Negation/Double Negation Elimination/Sequent Form/Formulation 2}
+Tags: Double Negation
+
+\begin{theorem}
+: $\vdash \neg \neg p \implies p$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \neg \neg p \implies p}}
+{{Assumption|1|\neg \neg p}}
+{{DoubleNegElimination|2|1|p|1}}
+{{Implication|3||\neg \neg p \implies p|1|2}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Double Negation/Double Negation Elimination}
+Tags: Double Negation
+
+\begin{theorem}
+The rule of '''double negation elimination''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]].
+=== [[Double Negation/Double Negation Elimination/Proof Rule|Proof Rule]] ===
+{{:Double Negation/Double Negation Elimination/Proof Rule}}
+=== [[Double Negation/Double Negation Elimination/Sequent Form|Sequent Form]] ===
+{{:Double Negation/Double Negation Elimination/Sequent Form}}
+\end{theorem}<|endoftext|>
+\section{Double Negation/Formulation 1}
+Tags: Double Negation
+
+\begin{theorem}
+:$p \dashv \vdash \neg \neg p$
+\end{theorem}<|endoftext|>
+\section{Double Negation/Formulation 1/Proof 2}
+Tags: Truth Table Proofs, Double Negation
+
+\begin{theorem}
+:$p \dashv \vdash \neg \neg p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, appropriate [[Definition:Truth Value|truth values]] match for both [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|c||ccc|} \hline
+p & \neg & \neg & p \\
+\hline
+F & F & T & F \\
+T & T & F & T \\
+\hline
+\end{array}$
+Hence $p \dashv \vdash \neg \neg p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Negation/Formulation 2}
+Tags: Double Negation
+
+\begin{theorem}
+:$\vdash p \iff \neg \neg p$
+\end{theorem}<|endoftext|>
+\section{Rule of Transposition/Formulation 1}
+Tags: Rule of Transposition
+
+\begin{theorem}
+A [[Definition:Statement|statement]] and its [[Definition:Contrapositive Statement|contrapositive]] have the same [[Definition:Truth Value|truth value]]:
+:$p \implies q \dashv \vdash \neg q \implies \neg p$
+Its abbreviation in a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] is $\textrm {TP}$.
+\end{theorem}<|endoftext|>
+\section{Rule of Transposition/Formulation 1/Proof 1}
+Tags: Rule of Transposition
+
+\begin{theorem}
+:$p \implies q \dashv \vdash \neg q \implies \neg p$
+\end{theorem}
+
+\begin{proof}
+=== [[Rule of Transposition/Formulation 1/Forward Implication/Proof|Proof of Forward Implication]] ===
+{{:Rule of Transposition/Formulation 1/Forward Implication/Proof}}
+=== [[Rule of Transposition/Formulation 1/Reverse Implication/Proof|Proof of Reverse Implication]] ===
+{{:Rule of Transposition/Formulation 1/Reverse Implication/Proof}}
+\end{proof}<|endoftext|>
+\section{Rule of Transposition/Formulation 1/Proof 2}
+Tags: Truth Table Proofs, Rule of Transposition
+
+\begin{theorem}
+:$p \implies q \dashv \vdash \neg q \implies \neg p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||ccccc|} \hline
+p & \implies & q & \neg & q & \implies & \neg & p \\
+\hline
+F & T & F & T & F & T & T & F \\
+F & T & T & F & T & T & T & F \\
+T & F & F & T & F & F & F & T \\
+T & T & T & F & T & T & F & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Positive Infinity is Maximal}
+Tags: Order Theory, Extended Real Numbers
+
+\begin{theorem}
+Let $\struct {\overline \R, \le}$ be the [[Definition:Extended Real Number Line|extended real numbers]] with the [[Definition:Ordering on Extended Real Numbers|usual ordering]].
+Then $+\infty$ is a [[Definition:Maximal Element|maximal element]] of $\overline \R$.
+\end{theorem}
+
+\begin{proof}
+By the definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on the [[Definition:Extended Real Number Line|extended real numbers]]:
+:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
+Suppose $x \in \overline \R$ and $+\infty \le x$.
+That is:
+: $\tuple {+\infty, x} \in {\le}$
+By the definition of [[Definition:Set Union|union]], $\tuple {+\infty, x}$ must lie in one of the three sets whose union forms $\le$.
+Since ${\le_\R} \subseteq \R \times \R$ and $+\infty \notin \R$:
+:$\tuple {+\infty, x} \notin {\le_\R}$
+Since $+\infty \ne -\infty$ by the definition of the [[Definition:Extended Real Number Line|extended real numbers]]:
+:$\tuple {+\infty, x} \notin \set {\tuple {-\infty, x}: x \in \overline \R}$
+Therefore:
+:$\tuple {+\infty, x} \in \set {\tuple {x, +\infty}: x \in \overline \R}$
+and we conclude that $x = +\infty$.
+That is, $+\infty$ is a [[Definition:Maximal Element|maximal element]] of $\overline \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative Infinity is Minimal}
+Tags: Order Theory, Extended Real Numbers
+
+\begin{theorem}
+Let $\left({\overline \R, \le}\right)$ be the [[Definition:Extended Real Number Line|extended real numbers]] with the [[Definition:Ordering on Extended Real Numbers|usual ordering]].
+Then $-\infty$ is a [[Definition:Minimal Element|minimal element]] of $\overline \R$.
+\end{theorem}
+
+\begin{proof}
+By the definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on the [[Definition:Extended Real Number Line|extended real numbers]]:
+:${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$
+Suppose $x \in \overline \R$ and $x \le -\infty$.
+That is:
+: $\tuple {x, -\infty} \in {\le}$
+By the definition of [[Definition:Set Union|union]], $\tuple {x, -\infty}$ must lie in one of the three sets whose union forms $\le$.
+Since ${\le_\R} \subseteq \R \times \R$ and $-\infty \notin \R$:
+:$\tuple {x, -\infty} \notin {\le_\R}$
+Since $+\infty \ne -\infty$ by the definition of the [[Definition:Extended Real Number Line|extended real numbers]]:
+:$\tuple {x, -\infty} \notin \set {\tuple {x, +\infty}: x \in \overline \R}$
+Therefore:
+:$\tuple {x, -\infty} \in \set {\tuple {-\infty, x}: x \in \overline \R}$
+and we conclude that $x = -\infty$.
+That is, $-\infty$ is a [[Definition:Minimal Element|minimal element]] of $\overline \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Ordered Group/OG2}
+Tags: Ordered Groups
+
+\begin{theorem}
+:$(\operatorname{OG}2.1):\quad x \preceq y \iff e \preceq y \circ x^{-1}$
+:$(\operatorname{OG}2.2):\quad x \preceq y \iff e \preceq x^{-1} \circ y$
+:$(\operatorname{OG}2.3):\quad x \preceq y \iff x \circ y^{-1} \preceq e$
+:$(\operatorname{OG}2.4):\quad x \preceq y \iff y^{-1} \circ x \preceq e$
+:$(\operatorname{OG}2.1'):\quad x \prec y \iff e \prec y \circ x^{-1}$
+:$(\operatorname{OG}2.2'):\quad x \prec y \iff e \prec x^{-1} \circ y$
+:$(\operatorname{OG}2.3'):\quad x \prec y \iff x \circ y^{-1} \prec e$
+:$(\operatorname{OG}2.4'):\quad x \prec y \iff y^{-1} \circ x \prec e$
+\end{theorem}<|endoftext|>
+\section{Properties of Ordered Group/OG4}
+Tags: Ordered Groups
+
+\begin{theorem}
+{{begin-axiom}}
+{{axiom|n = \operatorname{OG}4.1
+       |m = x \preceq e \iff e \preceq x^{-1}
+}}
+{{axiom|n = \operatorname{OG}4.2
+       |m = e \preceq x \iff x^{-1} \preceq e
+}}
+{{axiom|n = \operatorname{OG}4.1'
+       |m = x \prec e \iff e \prec x^{-1}
+}}
+{{axiom|n = \operatorname{OG}4.2'
+       |m = e \prec x \iff x^{-1} \prec e
+}}
+{{end-axiom}}
+\end{theorem}<|endoftext|>
+\section{Inversion Mapping Reverses Ordering in Ordered Group}
+Tags: Ordered Groups
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | n = \text{OG} 3
+      | l = x \preceq y
+      | o = \iff
+      | r = y^{-1} \preceq x^{-1}
+}}
+{{eqn | n = \text{OG}3'
+      | l = x \prec y
+      | o = \iff
+      | r = y^{-1} \prec x^{-1}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By the definition of an [[Definition:Ordered Group|ordered group]], $\preceq$ is a relation [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Thus by [[Properties of Relation Compatible with Group Operation/CRG3]], we obtain the first result:
+:$x \preceq y \iff y^{-1} \preceq x^{-1}$
+By [[Reflexive Reduction of Relation Compatible with Group Operation is Compatible]], $\prec$ is also compatible with $\circ$.
+Thus by again [[Properties of Relation Compatible with Group Operation/CRG3]], we obtain the second result:
+:$x \prec y \iff y^{-1} \prec x^{-1}$
+{{qed}}
+[[Category:Ordered Groups]]
+al7v6y3lv3z4b45o3yb0xk4d6bdtkjx
+\end{proof}<|endoftext|>
+\section{Properties of Ordered Group/OG5}
+Tags: Ordered Groups
+
+\begin{theorem}
+:$x \preceq e \implies x^n \preceq e$
+:$e \preceq x \implies e \preceq x^n$
+:$x \prec e \implies x^n \prec e$
+:$e \prec x \implies e \prec x^n$
+\end{theorem}<|endoftext|>
+\section{Operating on Transitive Relationships Compatible with Operation}
+Tags: Compatible Relations, Transitive Relations
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Let $\mathcal R$ be a [[Definition:Transitive Relation|transitive relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $\mathcal R^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\mathcal R$.
+Let $x, y, z, w \in S$.
+Then the following implications hold:
+$(1)\quad$ If $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R} w$, then $x \circ z \mathrel{\mathcal R} y \circ w$.
+$(2)\quad$ If $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R^=} w$, then $x \circ z \mathrel{\mathcal R} y \circ w$.
+$(3)\quad$ If $x \mathrel{\mathcal R^=} y$ and $z \mathrel{\mathcal R} w$, then $x \circ z \mathrel{\mathcal R} y \circ w$.
+$(4)\quad$ If $x \mathrel{\mathcal R^=} y$ and $z \mathrel{\mathcal R^=} w$, then $x \circ z \mathrel{\mathcal R^=} y \circ w$.
+\end{theorem}
+
+\begin{proof}
+We will prove $(1)$ and $(2)$. $(3)$ can be proven using precisely the same argument as $(2)$ and $(4)$ follows from $(1)$.
+Suppose that $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R} w$.
+By the definition of [[Definition:Relation Compatible with Operation|compatibility]]:
+:$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$
+:$z \mathrel{\mathcal R} w \implies y \circ z \mathrel{\mathcal R} y \circ w$
+By [[Definition:Transitive Relation|transitivity]]:
+:$(1)\quad \left({x \mathrel{\mathcal R} y}\right) \land \left({z \mathrel{\mathcal R} w}\right) \implies x \circ z \mathrel{\mathcal R} y \circ w$
+{{qed|lemma}}
+Suppose that $x \mathrel{\mathcal R} y$ and $z \mathrel{\mathcal R^=} w$.
+By [[Reflexive Closure of Relation Compatible with Operation is Compatible]], $\mathcal R^=$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+By the definition of [[Definition:Relation Compatible with Operation|compatibility]]:
+:$x \mathrel{\mathcal R} y \implies x \circ z \mathrel{\mathcal R} y \circ z$
+:$z \mathrel{\mathcal R^=} w \implies y \circ z \mathrel{\mathcal R^=} y \circ w$
+By [[Extended Transitivity]]:
+:$(2)\quad \left({x \mathrel{\mathcal R} y}\right) \land \left({z \mathrel{\mathcal R} w}\right) \implies x \circ z \mathrel{\mathcal R} y \circ w$
+{{qed|lemma}}
+$(3)$ can be proven using precisely the same argument as $(2)$.
+By [[Reflexive Closure of Relation Compatible with Operation is Compatible]], $\mathcal R^=$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+By [[Reflexive Closure of Transitive Relation is Transitive]], $\mathcal R^=$ is [[Definition:Transitive Relation|transitive]].
+Thus $(4)$ follows from $(1)$.
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Transitive Relations]]
+rpasew4agbzabt7qlr2cck5nchsk06h
+\end{proof}<|endoftext|>
+\section{Operating on Ordered Group Inequalities}
+Tags: Ordered Groups
+
+\begin{theorem}
+If $x \prec y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
+If $x \prec y$ and $z \preceq w$, then $x \circ z \prec y \circ w$.
+If $x \preceq y$ and $z \prec w$, then $x \circ z \prec y \circ w$.
+If $x \preceq y$ and $z \preceq w$, then $x \circ z \preceq y \circ w$.
+\end{theorem}
+
+\begin{proof}
+Because $\left({G, \circ, \preceq}\right)$ is a group and $\preceq$ is compatible with $\circ$, $\prec$ is compatible with $\circ$ by [[Reflexive Reduction of Relation Compatible with Group Operation is Compatible]].
+By the definition of an [[Definition:Ordering|ordering]], $\preceq$ is [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
+Therefore by [[Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering]], $\prec$ is transitive.
+Thus the theorem holds by [[Operating on Transitive Relationships Compatible with Operation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Transposition/Formulation 2}
+Tags: Rule of Transposition
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \iff \paren {\neg q \implies \neg p}$
+\end{theorem}<|endoftext|>
+\section{Rule of Transposition/Formulation 2/Proof 1}
+Tags: Rule of Transposition
+
+\begin{theorem}
+: $\vdash \left({p \implies q}\right) \iff \left({\neg q \implies \neg p}\right)$
+\end{theorem}
+
+\begin{proof}
+=== [[Rule of Transposition/Formulation 2/Forward Implication/Proof|Proof of Forward Implication]] ===
+{{:Rule of Transposition/Formulation 2/Forward Implication/Proof}}
+=== [[Rule of Transposition/Formulation 2/Reverse Implication/Proof|Proof of Reverse Implication]] ===
+{{:Rule of Transposition/Formulation 2/Reverse Implication/Proof}}
+\end{proof}<|endoftext|>
+\section{Rule of Transposition/Formulation 2/Forward Implication/Proof}
+Tags: Rule of Transposition
+
+\begin{theorem}
+:$\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau |\vdash \paren {p \implies q} \implies \paren {\neg q \implies \neg p} }}
+{{Assumption   |1|p \implies q}}
+{{Assumption   |2|\neg q}}
+{{ModusTollens |3|1, 2|\neg p|1|2}}
+{{Implication  |4|1|\neg q \implies \neg p|2|3}}
+{{Implication  |5||\paren {p \implies q} \implies \paren {\neg q \implies \neg p}|1|4}}
+{{EndTableau}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Transposition/Formulation 2/Reverse Implication/Proof}
+Tags: Rule of Transposition
+
+\begin{theorem}
+: $\vdash \left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)}}
+{{Assumption|1|\neg q \implies \neg p}}
+{{Assumption|2|p}}
+{{DoubleNegIntro|3|2|\neg \neg p|2}}
+{{ModusTollens|4|1, 2|\neg \neg q|1|3}}
+{{DoubleNegElimination|5|1, 2|q|4}}
+{{Implication|6|1|p \implies q|2|5}}
+{{Implication|7||\left({\neg q \implies \neg p}\right) \implies \left({p \implies q}\right)|1|6}}
+{{EndTableau}}
+{{Qed}}
+{{LEM|Double Negation Elimination|4}}
+\end{proof}<|endoftext|>
+\section{Complex Modulus is Norm}
+Tags: Examples of Norms, Complex Modulus
+
+\begin{theorem}
+The [[Definition:Complex Modulus|complex modulus]] is a [[Definition:Norm on Division Ring|norm]] on the [[Definition:Complex Number|set of complex numbers]] $\C$.
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Norm on Division Ring|norm axioms]].
+=== Positive Definiteness ===
+This is demonstrated in:
+: [[Complex Modulus equals Zero iff Zero]]
+: [[Complex Modulus is Non-Negative]]
+{{qed|lemma}}
+=== Multiplicativity ===
+Follows from [[Modulus of Product]].
+{{qed|lemma}}
+=== Triangle Inequality ===
+Follows from [[Triangle Inequality for Complex Numbers]].
+{{qed}}
+[[Category:Examples of Norms]]
+[[Category:Complex Modulus]]
+5a2ojdgq19ezkpwv2hj8vxlwkl81k1r
+\end{proof}<|endoftext|>
+\section{Set Difference of Relations Compatible with Group Operation is Compatible}
+Tags: Compatible Relations
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\RR, \QQ$ be [[Definition:Endorelation|relations]] on $G$ which are [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Then the [[Definition:Set Difference|difference]] $\RR \setminus \QQ$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+\end{theorem}
+
+\begin{proof}
+By [[Complement of Relation Compatible with Group is Compatible]], $\relcomp {G \times G} \QQ$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Thus by [[Intersection of Relations Compatible with Operation is Compatible]], $\RR \cap \relcomp {G \times G} \QQ$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+But:
+:$\RR \cap \relcomp {G \times G} \QQ = \RR \setminus \QQ$
+so the theorem holds.
+{{qed}}
+[[Category:Compatible Relations]]
+jyg5vn2asa1vqciut2pcdlutk31nlto
+\end{proof}<|endoftext|>
+\section{Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements}
+Tags: Compatible Relations, Semigroups
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Semigroup|semigroup]].
+Let $\mathcal R$ be a [[Definition:Transitive Relation|transitive relation]] on $S$ which is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
+Let $x, y \in S$ such that $x \mathrel{\mathcal R} y$.
+Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Then:
+: $x^n \mathrel{\mathcal R} y^n$
+where $x^n$ is the [[Definition:Power of Element of Semigroup|$n$th power]] of $x$.
+\end{theorem}
+
+\begin{proof}
+We proceed by [[Principle of Mathematical Induction|mathematical induction]].
+By definition of [[Definition:Power of Element of Semigroup|power]]:
+: $x^1 = x$
+: $y^1 = y$
+Hence, by assumption:
+: $x^1 \mathrel{\mathcal R} y^1$
+Suppose now that for $n \ge 1$:
+:$x^n \mathrel{\mathcal R} y^n$
+Recall the assumption that $x \mathrel{\mathcal R} y$.
+Applying [[Operating on Transitive Relationships Compatible with Operation]] to these relations yields:
+:$x^n \circ x \mathrel{\mathcal R} y^n \circ y$
+By definition of [[Definition:Power of Element of Semigroup|power]]:
+: $x^{n+1} = x^n \circ x$
+and:
+: $y^{n+1} = y^n \circ y$
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Semigroups]]
+oe270hq5prl06s9p8hnlb1f8xq4878z
+\end{proof}<|endoftext|>
+\section{Reflexive Closure of Transitive Antisymmetric Relation is Ordering}
+Tags: Order Theory, Reflexive Closures, Transitive Relations, Symmetric Relations
+
+\begin{theorem}
+Let $S$ be a set.
+Let $\prec$ be a [[Definition:Transitive Relation|transitive]], [[Definition:Antisymmetric Relation|antisymmetric]] [[Definition:Endorelation|relation]] on $S$.
+Let $\preceq$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\prec$.
+Then $\preceq$ is an [[Definition:Ordering|ordering]] on $S$.
+\end{theorem}
+
+\begin{proof}
+=== Reflexive ===
+Follows from [[Reflexive Closure is Reflexive]].
+{{qed|lemma}}
+=== Transitive ===
+Follows from [[Reflexive Closure of Transitive Relation is Transitive]].
+{{qed|lemma}}
+=== Antisymmetric ===
+Let $a, b \in S$.
+Suppose that $a \ne b$.
+Then by the definition of the diagonal relation,
+:$\left({(a,b)}\right) \notin \Delta_S$ and 
+:$\left({(b,a)}\right) \notin \Delta_S$.
+Since $\prec$ is antisymmetric, $\left({(a,b)}\right)$ and $\left({(b,a)}\right)$ are not both in $\prec$.
+Thus $a \not\preceq b$ or $b \not\preceq a$.
+{{qed}}
+[[Category:Order Theory]]
+[[Category:Reflexive Closures]]
+[[Category:Transitive Relations]]
+[[Category:Symmetric Relations]]
+k54ogs7jzmwk2efbsg8nhelsh36zc79
+\end{proof}<|endoftext|>
+\section{Rule of Transposition/Formulation 1/Forward Implication}
+Tags: Rule of Transposition
+
+\begin{theorem}
+:$p \implies q \vdash \neg q \implies \neg p$
+\end{theorem}<|endoftext|>
+\section{Rule of Transposition/Formulation 1/Reverse Implication}
+Tags: Rule of Transposition
+
+\begin{theorem}
+:$\neg q \implies \neg p \vdash p \implies q$
+\end{theorem}<|endoftext|>
+\section{Reflexive Closure of Transitive Relation is Transitive}
+Tags: Reflexive Closures
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\mathcal R$ be a [[Definition:Transitive Relation|transitive relation]].
+Let $\mathcal R^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\mathcal R$.
+Then $\mathcal R^=$ is also [[Definition:Transitive Relation|transitive]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b, c \in S$.
+Suppose that $a \mathrel{\mathcal R^=} b$ and $b \mathrel{\mathcal R^=} c$.
+If $a = b$, then since $b \mathrel{\mathcal R^=} c$, also $a \mathrel{\mathcal R^=} c$.
+If $b = c$, then since $a \mathrel{\mathcal R^=} b$, also $a \mathrel{\mathcal R^=} c$.
+The only case that remains is that $a \ne b$ and $b \ne c$.
+Then by the [[Definition:Reflexive Closure|definition of $\mathcal R^=$]], $a \mathrel{\mathcal R} b$ and $b \mathrel{\mathcal R} c$.
+Since $\mathcal R$ is [[Definition:Transitive Relation|transitive]], it follows that:
+:$a \mathrel{\mathcal R} c$
+and hence also $a \mathrel{\mathcal R^=} c$.
+Thus $\mathcal R^=$ is [[Definition:Transitive Relation|transitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reflexive Closure of Antisymmetric Relation is Antisymmetric}
+Tags: Reflexive Closures
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\mathcal R$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]] on $S$.
+Let $\mathcal R^=$ be the [[Definition:Reflexive Closure|reflexive closure]] of $\mathcal R$.
+Then $\mathcal R^=$ is also [[Definition:Antisymmetric Relation|antisymmetric]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$.
+Suppose that $a \mathrel{\mathcal R^=} b$ and $b \mathrel{\mathcal R^=} a$.
+By [[Definition:Reflexive Closure|definition of $\mathcal R^=$]], this means:
+:$a \mathrel{\mathcal R} b$ or $a = b$
+:$b \mathrel{\mathcal R} a$ or $b = a$
+If $a = b$ or $b = a$ we are done, by definition of an [[Definition:Antisymmetric Relation|antisymmetric relation]].
+So suppose $a \mathrel{\mathcal R} b$ and $b \mathrel{\mathcal R} a$.
+Since $\mathcal R$ is antisymmetric, it follows that $a = b$ in this case as well.
+Hence $\mathcal R^=$ is also [[Definition:Antisymmetric Relation|antisymmetric]].
+{{qed}}
+[[Category:Reflexive Closures]]
+0a3vbanfzlhyi6dvj8t7lzegrqzusl5
+\end{proof}<|endoftext|>
+\section{Positive Infinity is Greatest Element}
+Tags: Order Theory, Extended Real Numbers
+
+\begin{theorem}
+Let $\left({\overline \R, \le}\right)$ be the [[Definition:Extended Real Number Line|extended real numbers]] with their [[Definition:Ordering on Extended Real Numbers|usual ordering]].
+Then $+\infty$ is the [[Definition:Greatest Element|greatest element]] of $\overline \R$.
+\end{theorem}
+
+\begin{proof}
+We have, by definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on $\overline \R$:
+:$\forall x \in \overline \R: x \le +\infty$
+That is, $+\infty$ is the [[Definition:Greatest Element|greatest element]] of $\overline \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative Infinity is Smallest Element}
+Tags: Order Theory, Extended Real Numbers
+
+\begin{theorem}
+Let $\left({\overline \R, \le}\right)$ be the [[Definition:Extended Real Number Line|extended real numbers]] with their [[Definition:Ordering on Extended Real Numbers|usual ordering]].
+Then $-\infty$ is the [[Definition:Smallest Element|smallest element]] of $\overline \R$.
+\end{theorem}
+
+\begin{proof}
+We have, by definition of the [[Definition:Ordering on Extended Real Numbers|usual ordering]] on $\overline \R$:
+:$\forall x \in \overline \R: -\infty \le x$
+That is, $-\infty$ is the [[Definition:Smallest Element|smallest element]] of $\overline \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Relation Compatible with Group Operation/CRG1}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+:$x \mathrel {\mathcal R} y \iff x \circ z \mathrel {\mathcal R} y \circ z$
+:$x \mathrel {\mathcal R} y \iff z \circ x \mathrel {\mathcal R} z \circ y$
+\end{theorem}
+
+\begin{proof}
+By [[Relation Compatible with Group Operation is Strongly Compatible]], $\mathcal R$ is [[Definition:Relation Strongly Compatible with Operation|strongly compatible]] with $\circ$.
+Thus by the definition of strong compatibility, the theorem holds.
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+phn74ihn9gsu7nic8mwcdnr01wk6iyr
+\end{proof}<|endoftext|>
+\section{Properties of Relation Compatible with Group Operation/CRG2}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+:$(1): \quad x \mathrel {\mathcal R} y \iff e \mathrel {\mathcal R} y \circ x^{-1}$
+:$(2): \quad x \mathrel {\mathcal R} y \iff e \mathrel {\mathcal R} x^{-1} \circ y$
+:$(3): \quad x \mathrel {\mathcal R} y \iff x \circ y^{-1} \mathrel {\mathcal R} e$
+:$(4): \quad x \mathrel {\mathcal R} y \iff y^{-1} \circ x \mathrel {\mathcal R} e$
+\end{theorem}
+
+\begin{proof}
+We can apply [[Properties of Relation Compatible with Group Operation/CRG1|CRG1]] to $x$, $y$, and $x^{-1}$ to obtain the following equivalences:
+:$x \mathrel {\mathcal R} y \iff x \circ x^{-1} \mathrel {\mathcal R} y \circ x^{-1}$
+:$x \mathrel {\mathcal R} y \iff x^{-1} \circ x \mathrel {\mathcal R} x^{-1} \circ y$
+Applying [[Properties of Relation Compatible with Group Operation/CRG1|CRG1]] to $x$, $y$, and $y^{-1}$, on the other hand, yields:
+:$x \mathrel {\mathcal R} y \iff x \circ y^{-1} \mathrel {\mathcal R} y \circ y^{-1}$
+:$x \mathrel {\mathcal R} y \iff y^{-1} \circ x \mathrel {\mathcal R} y^{-1} \circ y$
+By the definition of [[Definition:Inverse Element|inverses]]:
+:$x \circ x^{-1} = x^{-1} \circ x = y \circ y^{-1} = y^{-1} \circ y = e$
+Making these substitutions proves the theorem.
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+6qzoin644p2bfszzgxuo3hyjxsolcde
+\end{proof}<|endoftext|>
+\section{Properties of Relation Compatible with Group Operation/CRG4}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+:$x \mathrel{\mathcal R} e \iff e \mathrel{\mathcal R} x^{-1}$
+:$e \mathrel{\mathcal R} x \iff x^{-1} \mathrel{\mathcal R} e$
+\end{theorem}
+
+\begin{proof}
+Applying [[User:Dfeuer/CRG2]]$(1)$ to $x$ and $e$ gives:
+:$x \mathrel{\mathcal R} e \iff e \mathrel{\mathcal R} e \circ x^{-1}$
+Applying [[User:Dfeuer/CRG2]]$(3)$ to $e$ and $x$ gives:
+:$e \mathrel{\mathcal R} x \iff e \circ x^{-1} \mathrel{\mathcal R} e$
+Since $e \circ x^{-1} = x^{-1} \circ e = x^{-1}$ for all $x \in G$, the theorem holds.
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+2s2daaweuwdvg32qu0gzsa8fkb12fkj
+\end{proof}<|endoftext|>
+\section{Dual Pairs (Order Theory)}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a, b \in S$, and let $T \subseteq S$.
+Then the following phrases about, and concepts pertaining to $\left({S, \preceq}\right)$ are [[Definition:Dual Statement (Order Theory)|dual]] to one another:
+:{| style="text-align:center"
+| $b \preceq a$
+| width = "20px" |
+| $a \preceq b$
+|-
+| $a$ [[Definition:Succeed|succeeds]] $b$
+|
+| $a$ [[Definition:Precede|precedes]] $b$
+|-
+| $a$ [[Definition:Strictly Succeed|strictly succeeds]] $b$
+|
+| $a$ [[Definition:Strictly Precede|strictly precedes]] $b$
+|-
+| $a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$
+|
+| $a$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$
+|-
+| $a$ is a [[Definition:Supremum of Set|supremum]] for $T$
+|
+| $a$ is an [[Definition:Infimum of Set|infimum]] for $T$
+|-
+| $a$ is a [[Definition:Maximal Element|maximal element]] of $T$
+|
+| $a$ is a [[Definition:Minimal Element|minimal element]] of $T$
+|-
+| $a$ is the [[Definition:Greatest Element|greatest element]]
+|
+| $a$ is the [[Definition:Smallest Element|smallest element]]
+|-
+| the [[Definition:Weak Lower Closure of Element|weak lower closure]] $a^\preceq$ of $a$
+|
+| the [[Definition:Weak Upper Closure of Element|weak upper closure]] $a^\succeq$ of $a$
+|-
+| the [[Definition:Strict Lower Closure of Element|strict lower closure]] $a^\prec$ of $a$
+|
+| the [[Definition:Strict Upper Closure of Element|strict upper closure]] $a^\succ$ of $a$
+|-
+| the [[Definition:Strict Lower Closure of Subset|strict lower closure]] $T^\prec$ of $T$
+|
+| the [[Definition:Strict Upper Closure of Subset|strict upper closure]] $T^\succ$ of $T$
+|-
+| the [[Definition:Join (Order Theory)|join]] $a \vee b$ of $a$ and $b$
+|
+| the [[Definition:Meet (Order Theory)|meet]] $a \wedge b$ of $a$ and $b$
+|-
+| $T$ is a [[Definition:Lower Set|lower set]] in $S$
+|
+| $T$ is an [[Definition:Upper Set|upper set]] in $S$
+|}
+\end{theorem}
+
+\begin{proof}
+Let $\succeq$ be the [[Definition:Dual Ordering|dual ordering]] of $\preceq$.
+By definition of [[Definition:Dual Statement (Order Theory)|dual statement]]:
+:$b \preceq a$
+is dual to:
+:$b \succeq a$
+and by definition of [[Definition:Dual Ordering|dual ordering]], this is [[Definition:Logical Equivalence|equivalent]] to:
+:$a \preceq b$
+By virtue of [[Dual of Dual Statement (Order Theory)]], the converse follows.
+The other claims are proved on the following pages, in order:
+:[[Succeed is Dual to Precede]]
+:[[Strictly Succeed is Dual to Strictly Precede]]
+:[[Upper Bound is Dual to Lower Bound]]
+:[[Supremum is Dual to Infimum]]
+:[[Maximal Element is Dual to Minimal Element]]
+:[[Greatest Element is Dual to Smallest Element]]
+:[[Weak Lower Closure is Dual to Weak Upper Closure]]
+:[[Strict Lower Closure is Dual to Strict Upper Closure]]
+:[[Join is Dual to Meet]]
+:[[Lower Set is Dual to Upper Set]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conditional is not Left Self-Distributive/Formulation 1}
+Tags: Truth Table Proofs, Conditional is not Left Self-Distributive
+
+\begin{theorem}
+While this holds:
+:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
+its converse does not:
+:$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition:
+:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
+As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
+$\begin{array}{|ccccc||ccccccc|} \hline
+(p & \implies & q) & \implies & r & (p & \implies & r) & \implies & (q & \implies & r) \\
+\hline
+F & T & F & F & F & F & T & F & T & F & T & F \\
+F & T & F & T & T & F & T & T & T & F & T & T \\
+F & T & T & F & F & F & T & F & F & T & F & F \\
+F & T & T & T & T & F & T & T & T & T & T & T \\
+T & F & F & T & F & T & F & F & T & F & F & F \\
+T & F & F & T & T & T & T & T & T & F & T & T \\
+T & T & T & F & F & T & F & F & T & T & F & F \\
+T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed|lemma}}
+The two [[Definition:Propositional Formula|formulas]] are not [[Definition:Logical Equivalence|equivalent]], as the relevant columns do not match exactly.
+For example, when $p =  q = r = F$ we have that:
+:$\paren {p \implies q} \implies r = F$
+but:
+:$\paren {p \implies r} \implies \paren {q \implies r} = T$
+Hence the result:
+:$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
+but:
+:$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$
+{{qed}}
+[[Category:Truth Table Proofs]]
+[[Category:Conditional is not Left Self-Distributive]]
+jghkmpapahabeztjswp0cgg1xrdszb5
+\end{proof}<|endoftext|>
+\section{Conditional is not Left Self-Distributive/Formulation 2}
+Tags: Conditional is not Left Self-Distributive
+
+\begin{theorem}
+While this holds:
+:$\vdash \paren {\paren {p \implies q} \implies r} \implies \paren {\paren {p \implies r} \implies \paren {q \implies r} }$
+its converse does not:
+:$\not \vdash \paren {\paren {p \implies r} \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies r}$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Can be seen to be [[Definition:Logical Equivalence|logically equivalent]] to [[Conditional is not Left Self-Distributive/Formulation 1]] by application of the [[Rule of Implication]] and [[Modus Ponendo Ponens]].}}
+\end{proof}<|endoftext|>
+\section{Cauchy-Riemann Equations}
+Tags: Complex Analysis, Cauchy-Riemann Equations
+
+\begin{theorem}
+Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
+Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
+Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be two [[Definition:Real-Valued Function|real-valued functions]] defined as:
+:$\map u {x, y} = \map \Re {\map f z}$
+:$\map v {x, y} = \map \Im {\map f z}$
+where:
+:$\map \Re {\map f z}$ denotes the [[Definition:Real Part|real part]] of $\map f z$
+:$\map \Im {\map f z}$ denotes the [[Definition:Imaginary Part|imaginary part]] of $\map f z$.
+Then $f$ is [[Definition:Complex-Differentiable Function|complex-differentiable]] in $D$ {{iff}}:
+:$u$ and $v$ are [[Definition:Differentiable Real-Valued Function on Open Set|differentiable]] in their entire [[Definition:Domain of Mapping|domain]]
+and:
+:The following two equations, known as the '''Cauchy-Riemann equations''', hold for the continuous [[Definition:Partial Derivative|partial derivatives]] of $u$ and $v$:
+:$(1): \quad \dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$
+:$(2): \quad \dfrac {\partial u} {\partial y} = -\dfrac {\partial v} {\partial x}$
+If the conditions are true, then for all $z \in D$:
+:$\map {f'} z = \map {\dfrac {\partial f} {\partial x} } z = -i \map {\dfrac {\partial f} {\partial y} } z$
+\end{theorem}
+
+\begin{proof}
+=== [[Cauchy-Riemann Equations/Necessary Condition|Necessary Condition]] ===
+{{:Cauchy-Riemann Equations/Necessary Condition}}
+=== [[Cauchy-Riemann Equations/Sufficient Condition|Sufficient Condition]] ===
+{{:Cauchy-Riemann Equations/Sufficient Condition}}
+=== [[Cauchy-Riemann Equations/Expression of Derivative|Expression of Derivative]] ===
+{{:Cauchy-Riemann Equations/Expression of Derivative}}
+{{Namedfor|Augustin Louis Cauchy|name2 = Georg Friedrich Bernhard Riemann|cat = Cauchy|cat2 = Riemann}}
+\end{proof}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Formulation 2}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+:$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
+\end{theorem}
+
+\begin{proof}
+=== [[Self-Distributive Law for Conditional/Forward Implication/Formulation 2/Proof 1|Proof of Forward Implication]] ===
+{{:Self-Distributive Law for Conditional/Forward Implication/Formulation 2/Proof 1}}
+=== [[Self-Distributive Law for Conditional/Reverse Implication/Formulation 2/Proof|Proof of Reverse Implication]] ===
+{{:Self-Distributive Law for Conditional/Reverse Implication/Formulation 2/Proof}}
+{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} } }}
+{{TheoremIntro|1|\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }|[[Self-Distributive Law for Conditional/Forward Implication/Formulation 2|Self-Distributive Law for Conditional: Forward Implication: Formulation 2]]}}
+{{TheoremIntro|2|\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }|[[Self-Distributive Law for Conditional/Reverse Implication/Formulation 2|Self-Distributive Law for Conditional: Reverse Implication: Formulation 2]]}}
+{{BiconditionalIntro|3||\paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }|1|2}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy-Riemann Equations/Necessary Condition}
+Tags: Cauchy-Riemann Equations
+
+\begin{theorem}
+Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
+Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
+Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two [[Definition:Real-Valued Function|real-valued functions]] defined as:
+:$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$
+:$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$
+where:
+: $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the [[Definition:Real Part|real part]] of $f \left({z}\right)$
+: $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the [[Definition:Imaginary Part|imaginary part]] of $f \left({z}\right)$.
+Then $f$ is [[Definition:Complex-Differentiable Function|complex-differentiable]] in $D$ {{iff}}:
+: $u$ and $v$ are [[Definition:Differentiable Real-Valued Function on Open Set|differentiable]] in their entire [[Definition:Domain of Mapping|domain]]
+and:
+: The following two equations, known as the '''Cauchy-Riemann equations''', hold for the [[Definition:Partial Derivative|partial derivatives]] of $u$ and $v$:
+:$(1): \quad \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$
+:$(2): \quad \dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}$
+If the conditions are true, then for all $z \in D$:
+:$f' \left({z}\right) = \dfrac{\partial f}{\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$
+\end{theorem}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Formulation 1}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+:$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
+\end{theorem}<|endoftext|>
+\section{Cauchy-Riemann Equations/Sufficient Condition}
+Tags: Cauchy-Riemann Equations
+
+\begin{theorem}
+Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
+Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
+Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two [[Definition:Real-Valued Function|real-valued functions]] defined as:
+:$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$
+:$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$
+where:
+: $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the [[Definition:Real Part|real part]] of $f \left({z}\right)$
+: $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the [[Definition:Imaginary Part|imaginary part]] of $f \left({z}\right)$.
+Let:
+:$u$ and $v$ be [[Definition:Differentiable Real-Valued Function on Open Set|differentiable]] in their entire [[Definition:Domain of Mapping|domain]]
+and:
+:The following two equations, known as the '''Cauchy-Riemann equations''', hold for the [[Definition:Continuous Real Function|continuous]] [[Definition:Partial Derivative|partial derivatives]] of $u$ and $v$:
+::$(1): \quad \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$
+::$(2): \quad \dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}$
+Then:
+:$f$ is [[Definition:Complex-Differentiable Function|complex-differentiable]] in $D$
+and:
+:for all $z \in D$:
+::$f' \left({z}\right) = \dfrac {\partial f} {\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+Suppose that the Cauchy-Riemann equations hold for $u$ and $v$ in their entire [[Definition:Domain of Mapping|domain]].
+Let $h, k \in \R \setminus \left\{ {0}\right\}$, and put $t = h + i k \in \C$.
+Let $\left({x, y}\right) \in \R^2$ be a point in the domain of $u$ and $v$.
+Put:
+:$a = \dfrac{\partial u}{\partial x} \left({x, y}\right) = \dfrac{\partial v}{\partial y} \left({x, y}\right)$
+and:
+:$b = - \dfrac{\partial u}{\partial y} \left({x, y}\right) = \dfrac{\partial v}{\partial x} \left({x, y}\right)$
+From the [[Alternative Differentiability Condition]], it follows that:
+:$u \left({x + h, y}\right) = u \left({x, y}\right) + h \left({a + \epsilon_{u x} \left({h}\right) }\right)$
+:$u \left({x, y + k}\right) = u \left({x, y}\right) + k \left({-b + \epsilon_{u y} \left({k}\right) }\right)$
+:$v \left({x + h, y}\right) = v \left({x, y}\right) + h \left({b + \epsilon_{v x} \left({k}\right) }\right)$
+:$v \left({x, y + k}\right) = v \left({x, y}\right) + k \left({a + \epsilon_{v y} \left({h}\right) }\right)$
+where $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are [[Definition:Continuous Real Function|continuous real functions]] that [[Definition:Convergent Function|converge to $0$]] as $h$ and $k$ tend to $0$.
+With $z = x + i y$, it follows that:
+{{begin-eqn}}
+{{eqn | l = f \left({z + t}\right)
+      | r = u \left({x + h, y + k}\right) + i v \left({x + h, y + k}\right)
+}}
+{{eqn | r = u \left({x, y}\right) + iv \left({x, y}\right) + h \left({a + i b + \epsilon_{u x} \left({h}\right) + i \epsilon_{v x} \left({h}\right)}\right) + k \left({i a - b + \epsilon_{u y} \left({k}\right) + i \epsilon_{v y} \left({k}\right)}\right)
+}}
+{{eqn | r = f \left({z}\right) + t \left({a + i b + \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)}\right)
+}}
+{{end-eqn}}
+With the assumption that the [[Definition:Partial Derivative|partial derivatives]] are [[Definition:Continuous Real Function|continuous]], we have $a + i b$,, which is a function of h, converges to itself by taking h -> 0.
+{{explain|What does that actually mean: "constant with limit sign"? Can the above statement be explained and proved with a link?}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \epsilon \left({t}\right)
+      | r = \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{\epsilon \left({t}\right) }\right\vert
+      | o = \le
+      | r = \left\vert{\dfrac h t \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{v y} \left({k}\right) }\right\vert
+      | c = [[Triangle Inequality for Complex Numbers]]
+}}
+{{eqn | r = \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert
+      | c = [[Complex Modulus of Product of Complex Numbers]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \lim_{t \mathop \to 0} \left\vert{\epsilon \left({t}\right) }\right\vert
+      | o = \le
+      | r = \lim_{t \mathop \to 0} \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert
+      | c = [[Combined Sum Rule for Limits of Functions]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert
+}}
+{{eqn | o = \le
+      | r = \lim_{t \mathop \to 0} \left\vert{ \epsilon_{ux} \left({h}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vx} \left({h}\right) }\right\vert
+      | c = as $\left\vert{\dfrac h t}\right\vert \le 1$ and $\left\vert{\dfrac k t}\right\vert \le 1$ by [[Modulus Larger than Real Part and Imaginary Part]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \lim_{t \mathop \to 0} \left\vert{ \epsilon_{uy} \left({k}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vy} \left({k}\right) }\right\vert
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+{{explain|It is not immediately apparent why $\left\vert{\dfrac h t}\right\vert \le 1$ and $\left\vert{\dfrac k t}\right\vert \le 1$ by [[Modulus Larger than Real Part and Imaginary Part]].}}
+This shows that:
+: $\displaystyle \lim_{t \mathop \to 0} \epsilon \left({t}\right) = 0$
+From [[Continuity of Composite Mapping]], it follows that $\epsilon \left({t}\right)$ is [[Definition:Continuous Complex Function|continuous]].
+Then the [[Alternative Differentiability Condition]] shows that:
+:$f' \left({z}\right) = a + i b$
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Formulation 1/Proof 2}
+Tags: Truth Table Proofs, Self-Distributive Law for Conditional
+
+\begin{theorem}
+: $p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition:
+$p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccccc|} \hline
+p & \implies & (q & \implies & r) & (p & \implies & q) & \implies & (p & \implies & r) \\
+\hline
+F & T & F & T & F & F & T & F & T & F & T & F \\
+F & T & F & T & T & F & T & F & T & F & T & T \\
+F & T & T & F & F & F & T & T & T & F & T & F \\
+F & T & T & T & T & F & T & T & T & F & T & T \\
+T & T & F & T & F & T & F & F & T & T & F & F \\
+T & T & F & T & T & T & F & F & T & T & T & T \\
+T & F & T & F & F & T & T & T & F & T & F & F \\
+T & T & T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy-Riemann Equations/Expression of Derivative}
+Tags: Cauchy-Riemann Equations
+
+\begin{theorem}
+Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
+Let $f: D \to \C$ be a [[Definition:Complex Function|complex function]] on $D$.
+Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two [[Definition:Real-Valued Function|real-valued functions]] defined as:
+:$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$
+:$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$
+where:
+: $\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the [[Definition:Real Part|real part]] of $f \left({z}\right)$
+: $\operatorname{Im} \left({f \left({z}\right)}\right) $ denotes the [[Definition:Imaginary Part|imaginary part]] of $f \left({z}\right)$.
+Then $f$ is [[Definition:Complex-Differentiable Function|complex-differentiable]] in $D$ {{iff}}:
+: $u$ and $v$ are [[Definition:Differentiable Real-Valued Function on Open Set|differentiable]] in their entire [[Definition:Domain of Mapping|domain]]
+and:
+: The following two equations, known as the '''Cauchy-Riemann equations''', hold for the [[Definition:Partial Derivative|partial derivatives]] of $u$ and $v$:
+:$(1): \quad \dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$
+:$(2): \quad \dfrac {\partial u} {\partial y} = - \dfrac {\partial v} {\partial x}$
+If the conditions are true, then for all $z \in D$:
+:$f' \left({z}\right) = \dfrac{\partial f}{\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+Let $z = x + i y$.
+Then:
+ 
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x} \left({z}\right)
+      | r = \dfrac {\partial u} {\partial x} \left({x, y}\right) + i \dfrac {\partial v} {\partial x} \left({x, y}\right)
+}}
+{{eqn | r = \operatorname{Re} \left({f' \left({z}\right) }\right) + i \operatorname{Im} \left({f' \left({z}\right) }\right)
+      | c = from the last part of the [[Cauchy-Riemann Equations/Sufficient Condition|proof for sufficient condition]]
+}}
+{{eqn | r = f' \left({z}\right)
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = -i \dfrac {\partial f} {\partial y} \left({z}\right)
+      | r = -i \left({\dfrac {\partial u} {\partial y} \left({x, y}\right) + i \dfrac {\partial v} {\partial y} \left({x, y}\right) }\right)
+}}
+{{eqn | r = -i \left({ -\operatorname{Im} \left({f' \left({z}\right) }\right) + i \operatorname{Re} \left({f' \left({z}\right) }\right) }\right)
+      | c = from the last part of the [[Cauchy-Riemann Equations/Sufficient Condition|proof for sufficient condition]]
+}}
+{{eqn | r = f' \left({z}\right)
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Forward Implication/Formulation 1/Proof}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+: $p \implies \left({q \implies r}\right) \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \implies \left({q \implies r}\right) \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)}}
+{{Premise|1|p \implies \left({q \implies r}\right)}}
+{{Assumption|2|p \implies q}}
+{{Assumption|3|p}}
+{{ModusPonens|4|1, 3|q \implies r|1|3}}
+{{ModusPonens|5|2, 3|q|2|3}}
+{{ModusPonens|6|1, 2, 3|r|4|5}}
+{{Implication|7|1, 2|p \implies r|3|6}}
+{{Implication|8|1|\left({p \implies q}\right) \implies \left({p \implies r}\right)|2|7}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Succeed is Dual to Precede}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a, b \in S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$a$ [[Definition:Succeed|succeeds]] $b$
+:$a$ [[Definition:Precede|precedes]] $b$
+\end{theorem}
+
+\begin{proof}
+By definition, $a$ [[Definition:Succeed|succeeds]] $b$ {{iff}}:
+:$b \preceq a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$a \preceq b$
+by [[Dual Pairs (Order Theory)]].
+By definition, this means $a$ [[Definition:Precede|precedes]] $b$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Succeed is Dual to Strictly Precede}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $a, b \in S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$a$ [[Definition:Strictly Succeed|strictly succeeds]] $b$
+:$a$ [[Definition:Strictly Precede|strictly precedes]] $b$
+\end{theorem}
+
+\begin{proof}
+By definition, $a$ [[Definition:Strictly Succeed|strictly succeeds]] $b$ {{iff}}:
+:$b \preceq a$ and $b \ne a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$a \succeq b$ and $b \ne a$
+by [[Dual Pairs (Order Theory)]].
+By definition, this means $a$ [[Definition:Strictly Precede|strictly precedes]] $b$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Bound is Dual to Lower Bound}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a \in S$ and $T \subseteq S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$
+:$a$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$
+\end{theorem}
+
+\begin{proof}
+By definition, $a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$ [[Definition:Iff|iff]]:
+:$\forall t \in T: t \preceq a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$\forall t \in T: a \preceq t$
+by [[Dual Pairs (Order Theory)]].
+By definition, this means $a$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum is Dual to Infimum}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a \in S$ and $T \subseteq S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$a$ is a [[Definition:Supremum of Set|supremum]] for $T$
+:$a$ is an [[Definition:Infimum of Set|infimum]] for $T$
+\end{theorem}
+
+\begin{proof}
+By definition, $a$ is a [[Definition:Supremum of Set|supremum]] for $T$ [[Definition:Iff|iff]]:
+:$a$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$
+:$a \preceq b$ for all [[Definition:Upper Bound of Set|upper bounds]] $b$ of $T$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$a$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$
+:$b \preceq a$ for all [[Definition:Lower Bound of Set|lower bounds]] $b$ of $T$
+by [[Dual Pairs (Order Theory)]].
+By definition, this means $a$ is an [[Definition:Infimum of Set|infimum]] for $T$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximal Element is Dual to Minimal Element}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$, and $a \in T$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$a$ is a [[Definition:Maximal Element|maximal element]] of $T$
+:$a$ is a [[Definition:Minimal Element|minimal element]] of $T$
+\end{theorem}
+
+\begin{proof}
+By definition, $a$ is a [[Definition:Maximal Element|maximal element]] of $T$ {{iff}}:
+:$\forall t \in T: a \preceq t$ implies $a = t$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$\forall t \in T: t \preceq a$ implies $a = t$
+by [[Dual Pairs (Order Theory)]].
+By definition, this means $a$ is a [[Definition:Minimal Element|minimal element]] of $T$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Element is Dual to Smallest Element}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $a \in S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$a$ is the [[Definition:Greatest Element|greatest element]] of $S$
+:$a$ is the [[Definition:Smallest Element|smallest element]] of $S$
+\end{theorem}
+
+\begin{proof}
+By definition, $a$ is the [[Definition:Greatest Element|greatest element]] of $S$ {{iff}}:
+:$\forall b \in S: b \preceq a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$\forall b \in S: a \preceq b$
+by [[Dual Pairs (Order Theory)]].
+By definition, this means $a$ is the [[Definition:Smallest Element|smallest element]] of $S$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Closure is Dual to Upper Closure}
+Tags: Upper Closures, Lower Closures
+
+\begin{theorem}
+Let $\left({S, \preccurlyeq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a, b \in S$.
+Let $T \subseteq S$
+The following are pairs of [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$b \in a^\preccurlyeq$, the [[Definition:Lower Closure of Element|lower closure]] of $a$
+:$b \in a^\succcurlyeq$, the [[Definition:Upper Closure of Element|upper closure]] of $a$
+:$b \in T^\preccurlyeq$, the [[Definition:Lower Closure of Subset|lower closure]] of $T$
+:$b \in T^\succcurlyeq$, the [[Definition:Upper Closure of Subset|upper closure]] of $T$
+\end{theorem}
+
+\begin{proof}
+=== Elements ===
+By definition of [[Definition:Lower Closure of Element|lower closure]], $b \in a^\preccurlyeq$ {{iff}}:
+:$b \preccurlyeq a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$a \preccurlyeq b$
+by [[Dual Pairs (Order Theory)]].
+By definition of [[Definition:Upper Closure of Element|upper closure]], this means $b \in a^\succcurlyeq$.
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed|lemma}}
+=== Sets ===
+By the definition of [[Definition:Lower Closure of Subset|lower closure]], $b \in T^\preccurlyeq$ {{iff}}:
+:$\exists a \in T: b \preccurlyeq a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$\exists a \in T: a \preccurlyeq b$
+by [[Dual Pairs (Order Theory)]].
+By the definition of [[Definition:Upper Closure of Subset|upper closure]], this means $b \in T^\succcurlyeq$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Lower Closure is Dual to Strict Upper Closure}
+Tags: Lower Closures, Upper Closures
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a, b \in S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$b \in a^\prec$, the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $a$
+:$b \in a^\succ$, the [[Definition:Strict Upper Closure of Element|strict upper closure]] of $a$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Strict Lower Closure of Element|strict lower closure]]:
+:$b \in a^\prec$
+{{iff}}
+:$b$ [[Definition:Strictly Precede|strictly precedes]] $a$
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$b$ [[Definition:Strictly Succeed|strictly succeeds]] $a$
+by [[Dual Pairs (Order Theory)]].
+By definition of [[Definition:Strict Upper Closure of Element|strict upper closure]], this means:
+: $b \in a^\succ$
+The converse follows from [[Dual of Dual Statement (Order Theory)]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Join is Dual to Meet}
+Tags: Join and Meet
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $a, b, c \in S$.
+The following are [[Definition:Dual Statement (Order Theory)|dual statements]]:
+:$c = a \vee b$, the [[Definition:Join (Order Theory)|join]] of $a$ and $b$
+:$c = a \wedge b$ the [[Definition:Meet (Order Theory)|meet]] of $a$ and $b$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Join (Order Theory)|join]], $c = a \vee b$ {{iff}}:
+:$c = \sup \left\{{a, b}\right\}$
+where $\sup$ denotes [[Definition:Supremum of Set|supremum]].
+The [[Definition:Dual Statement (Order Theory)|dual]] of this statement is:
+:$c = \inf \left\{{a, b}\right\}$
+where $\inf$ denotes [[Definition:Infimum of Set|infimum]], by [[Dual Pairs (Order Theory)]].
+By definition of [[Definition:Meet (Order Theory)|meet]], this means $c = a \wedge b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Forward Implication/Formulation 1}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+: $p \implies \left({q \implies r}\right) \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Reverse Implication/Formulation 1}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+:$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$
+\end{theorem}<|endoftext|>
+\section{Rule of Exportation/Formulation 1/Proof 2}
+Tags: Rule of Exportation, Truth Table Proofs
+
+\begin{theorem}
+:$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]] to the proposition.
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccccc||ccccc|} \hline
+(p & \land & q) & \implies & r & p & \implies & (q & \implies & r) \\
+\hline
+F & F & F & T & F & F & T & F & T & F \\
+F & F & F & T & T & F & T & F & T & T \\
+F & F & T & T & F & F & T & T & F & F \\
+F & F & T & T & T & F & T & T & T & T \\
+T & F & F & T & F & T & T & F & T & F \\
+T & F & F & T & T & T & T & F & T & T \\
+T & T & T & F & F & T & F & T & F & F \\
+T & T & T & T & T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Exportation/Formulation 1}
+Tags: Rule of Exportation
+
+\begin{theorem}
+: $\left ({p \land q}\right) \implies r \dashv \vdash p \implies \left ({q \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Square Matrix with Duplicate Columns has Zero Determinant}
+Tags: Matrix Algebra, Determinants, Square Matrix with Duplicate Columns has Zero Determinant
+
+\begin{theorem}
+If two [[Definition:Column of Matrix|columns]] of a [[Definition:Square Matrix|square matrix]] over a [[Definition:Commutative Ring|commutative ring]] $\struct {R, +, \circ}$ are identical, then its [[Definition:Determinant of Matrix|determinant]] is zero.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix]] over $R$ with two identical [[Definition:Column of Matrix|columns]].
+Let $\mathbf A^\intercal$ denote the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Then $\mathbf A^\intercal$ has two identical [[Definition:Row of Matrix|rows]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf A}
+      | r = \map \det {\mathbf A^\intercal}
+      | c = [[Determinant of Transpose]]
+}}
+{{eqn | r = 0
+      | c = [[Square Matrix with Duplicate Rows has Zero Determinant]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Relation Compatible with Group Operation/CRG3}
+Tags: Compatible Relations, Group Theory
+
+\begin{theorem}
+:$x \mathrel{\mathcal R} y \iff y^{-1} \mathrel{\mathcal R} x^{-1}$
+\end{theorem}
+
+\begin{proof}
+Let $e$ be the [[Definition:Identity Element|group identity]] of $G$.
+By [[User:Dfeuer/CRG2]] $(1)$:
+:$(1): \quad x \mathrel{\mathcal R} y \iff e \mathrel{\mathcal R} y \circ x^{-1}$
+By [[User:Dfeuer/CRG2]] $(2)$, also:
+:$(2): \quad y^{-1} \mathrel{\mathcal R} x^{-1} \iff e \mathrel{\mathcal R} \left({y^{-1}}\right)^{-1} \circ x^{-1}$
+By [[Inverse of Group Inverse]] $\left({y^{-1}}\right)^{-1} = y$.
+Thus, we can rewrite $(2)$ as:
+:$(3): \quad y^{-1} \mathrel{\mathcal R} x^{-1} \iff e \mathrel {\mathcal R} y \circ x^{-1}$
+Now note that the {{RHS}} of $(3)$ is the same as the {{RHS}} in $(1)$.
+We conclude that:
+:$x \mathrel{\mathcal R} y \iff y^{-1} \mathrel{\mathcal R} x$
+{{qed}}
+[[Category:Compatible Relations]]
+[[Category:Group Theory]]
+bxgnpz4l16tutays15dlp8v9poi7o7m
+\end{proof}<|endoftext|>
+\section{Ordering Induced by Join Semilattice}
+Tags: Order Theory, Semilattices
+
+\begin{theorem}
+Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
+By [[Join Semilattice is Semilattice]], $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]].
+By [[Semilattice Induces Ordering]], $\struct {S, \vee}$ induces an [[Definition:Ordering|ordering]] $\preceq'$ on $S$, by:
+:$a \preceq' b$ {{iff}} $a \vee b = b$
+for all $a, b \in S$.
+The [[Definition:Ordering|ordering]] $\preceq'$ coincides with the original ordering $\preceq$.
+\end{theorem}
+
+\begin{proof}
+It is to be shown that, for all $a, b \in S$:
+:$a \preceq b$ {{iff}} $b = \sup \set {a, b}$
+by definition of [[Definition:Join (Order Theory)|join]].
+Here $\sup$ denotes [[Definition:Supremum of Set|supremum]].
+Since any [[Definition:Upper Bound of Set|upper bound]] $c$ of $\set {a, b}$ must satisfy:
+:$b \preceq c$
+it suffices to verify that:
+:$a \preceq b$ {{iff}} $b$ is an [[Definition:Upper Bound of Set|upper bound]] for $\set {a, b}$
+Since $\preceq$ is [[Definition:Reflexive Relation|reflexive]], we know that:
+:$b \preceq b$
+and therefore said [[Definition:Logical Equivalence|equivalence]] is established.
+We conclude that $\preceq'$ and $\preceq$ coincide.
+{{qed}}
+{{expand|We need to start using a construct for duality. (Global) Dual is [[Ordering Induced by Meet Semilattice]]}}
+[[Category:Order Theory]]
+[[Category:Semilattices]]
+fgs0nl1upacwtukww0y2jip2lnphxmi
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Complex Derivatives}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $D$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Subset|subset]] of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$.
+Let $f, g: D \to \C$ be [[Definition:Complex-Differentiable Function|complex-differentiable functions]] on $D$
+Let $z \in D$.
+Let $w, c \in \C$ be arbitrary [[Definition:Complex Number|complex numbers]].
+Then the following results hold:
+\end{theorem}<|endoftext|>
+\section{Combination Theorem for Complex Derivatives/Sum Rule}
+Tags: Complex Analysis
+
+\begin{theorem}
+:$\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $B_r \left({0}\right)$.
+Let $z \in D$.
+By the [[Alternative Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$f\left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$
+:$g\left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$
+where $\epsilon_f, \epsilon_g: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are [[Definition:Continuous Complex Function|continuous functions]] that [[Definition:Convergent Function|converge]] to $0$ as $h$ tends to $0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left({f + g}\right) \left({z + h}\right)
+      | r = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) + g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)
+}}
+{{eqn | r = \left({f + g}\right) \left({z}\right) + h \left({f' \left({z}\right) + g' \left({z}\right) + \left({ \epsilon_f + \epsilon_g }\right) \left({h}\right) }\right)
+}}
+{{end-eqn}}
+From [[Sum Rule for Continuous Functions]], it follows that $\epsilon_f + \epsilon_g$ is a [[Definition:Continuous Complex Function|continuous function]].
+From [[Sum Rule for Limits of Functions]], it follows that $\displaystyle \lim_{h \to 0} \left({ \epsilon_f + \epsilon_g }\right) \left({h}\right) = 0$.
+Then the [[Alternative Differentiability Condition]] shows that:
+:$\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Complex Derivatives/Multiple Rule}
+Tags: Complex Analysis
+
+\begin{theorem}
+:$\left({w f}\right)' \left({z}\right) = w f' \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Open Ball|open ball]] of $0$ with [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $B_r \left({0}\right)$.
+Let $z \in D$.
+By the [[Alternative Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$f \left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon \left({h}\right) }\right)$
+where $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ is a [[Definition:Continuous Complex Function|continuous functions]] with $\displaystyle \lim_{h \mathop \to 0} \epsilon \left({h}\right) = 0$.
+Then:
+:$w f \left({z + h}\right) = w f \left({z}\right) + h \left({w f' \left({z}\right) + w \epsilon \left({h}\right) }\right)$
+From [[Multiple Rule for Continuous Functions]], it follows that $w \epsilon$ is a [[Definition:Continuous Complex Function|continuous function]].
+From [[Multiple Rule for Limits of Functions]], it follows that $\displaystyle \lim_{h \mathop \to 0} w \epsilon \left({h}\right) = \left({ \lim_{h \mathop \to 0} w }\right) \left({ \lim_{h \mathop \to 0} \epsilon \left({h}\right) }\right) = 0$.
+Then the [[Alternative Differentiability Condition]] shows that:
+:$\left({w f}\right)' \left({z}\right) = w f' \left({z}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Join Absorbs Meet}
+Tags: Join and Meet
+
+\begin{theorem}
+Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $\vee$ denote [[Definition:Join (Order Theory)|join]].
+Then $\vee$ [[Definition:Absorption Law|absorbs]] $\wedge$.
+That is, for all $a, b \in S$:
+:$a \vee \paren {a \wedge b} = a$
+\end{theorem}
+
+\begin{proof}
+By [[Ordering in terms of Join]], we have that:
+:$a \vee \left({a \wedge b}\right) = a$ {{iff}} $a \wedge b \preceq a$
+The result thus follows from [[Meet Precedes Operands]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet Absorbs Join}
+Tags: Join and Meet
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $\wedge$ denote [[Definition:Meet (Order Theory)|meet]].
+Then $\wedge$ [[Definition:Absorption Law|absorbs]] $\vee$.
+That is, for all $a, b \in S$:
+:$a \wedge \left({a \vee b}\right) = a$
+\end{theorem}
+
+\begin{proof}
+By [[Dual Pairs (Order Theory)]], we observe that the theorem statement is [[Definition:Dual Statement (Order Theory)|dual]] to that of [[Join Absorbs Meet]].
+The result follows by the [[Duality Principle (Order Theory)/Global Duality|Global Duality Principle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Exportation/Formulation 2}
+Tags: Rule of Exportation
+
+\begin{theorem}
+:$\vdash \paren {\paren {p \land q} \implies r} \iff \paren {p \implies \paren {q \implies r} }$
+\end{theorem}
+
+\begin{proof}
+=== [[Rule of Exportation/Forward Implication/Formulation 2/Proof 1|Proof of Forward Implication]] ===
+{{:Rule of Exportation/Forward Implication/Formulation 2/Proof 1}}
+=== [[Rule of Exportation/Reverse Implication/Formulation 2/Proof 1|Proof of Reverse Implication]] ===
+{{:Rule of Exportation/Reverse Implication/Formulation 2/Proof 1}}
+\end{proof}<|endoftext|>
+\section{Rule of Exportation/Reverse Implication/Formulation 2/Proof 1}
+Tags: Rule of Exportation
+
+\begin{theorem}
+:$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r} }}
+{{Assumption|1|p \implies \paren {q \implies r} }}
+{{SequentIntro|2|1|\paren {p \land q} \implies r|1|[[Rule of Exportation/Reverse Implication/Formulation 1|Rule of Exportation: Reverse Implication: Formulation 1]]}}
+{{Implication|3||\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}|1|2}}
+{{EndTableau|qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Exportation/Forward Implication/Formulation 2}
+Tags: Rule of Exportation
+
+\begin{theorem}
+: $\vdash \left({\left ({p \land q}\right) \implies r}\right) \implies \left({p \implies \left ({q \implies r}\right)}\right)$
+\end{theorem}<|endoftext|>
+\section{Rule of Exportation/Reverse Implication/Formulation 2}
+Tags: Rule of Exportation
+
+\begin{theorem}
+:$\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$
+\end{theorem}<|endoftext|>
+\section{Rule of Exportation/Forward Implication/Formulation 1}
+Tags: Rule of Exportation
+
+\begin{theorem}
+: $\left ({p \land q}\right) \implies r \vdash p \implies \left ({q \implies r}\right)$
+\end{theorem}<|endoftext|>
+\section{Rule of Exportation/Reverse Implication/Formulation 1}
+Tags: Rule of Exportation
+
+\begin{theorem}
+:$p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r$
+\end{theorem}<|endoftext|>
+\section{Rule of Commutation/Conjunction/Formulation 1/Proof 2}
+Tags: Rule of Commutation, Truth Table Proofs
+
+\begin{theorem}
+:$p \land q \dashv \vdash q \land p$
+\end{theorem}
+
+\begin{proof}
+We apply the [[Method of Truth Tables]].
+As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
+$\begin{array}{|ccc||ccc|} \hline
+p & \land & q & q & \land & p \\
+\hline
+F & F & F & F & F & F \\
+F & F & T & T & F & F \\
+T & F & F & F & F & T \\
+T & T & T & T & T & T \\
+\hline
+\end{array}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Commutation/Conjunction/Formulation 1}
+Tags: Rule of Commutation
+
+\begin{theorem}
+:$p \land q \dashv \vdash q \land p$
+\end{theorem}<|endoftext|>
+\section{Rule of Commutation/Disjunction/Formulation 1}
+Tags: Rule of Commutation
+
+\begin{theorem}
+:$p \lor q \dashv \vdash q \lor p$
+\end{theorem}<|endoftext|>
+\section{Rule of Commutation/Conjunction/Formulation 2}
+Tags: Rule of Commutation
+
+\begin{theorem}
+:$\vdash \paren {p \land q} \iff \paren {q \land p}$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\vdash \paren {p \land q} \iff \paren {q \land p} }}
+{{Assumption |1|p \land q}}
+{{Commutation|2|1|q \land p|1|Conjunction}}
+{{Implication|3||\paren {p \land q} \implies \paren {q \land p}|1|2}}
+{{Assumption |4|q \land p}}
+{{Commutation|5|4|p \land q|4|Conjunction}}
+{{Implication|6||\paren {q \land p} \implies \paren {p \land q}|4|5}}
+{{BiconditionalIntro|7||\left({p \land q}\right) \iff \paren {q \land p}|3|6}}
+{{EndTableau}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rule of Commutation/Disjunction/Formulation 2}
+Tags: Rule of Commutation
+
+\begin{theorem}
+:$\vdash \paren {p \lor q} \iff \paren {q \lor p}$
+\end{theorem}<|endoftext|>
+\section{Combination Theorem for Complex Derivatives/Product Rule}
+Tags: Complex Analysis
+
+\begin{theorem}
+:$\left({f g}\right)' \left({z}\right) = f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Open Ball|open ball]] of $0$ by [[Definition:Radius of Open Ball|radius]] $r \in \R_{>0}$ as $B_r \left({0}\right)$.
+Let $z \in D$.
+By the [[Alternative Differentiability Condition]], it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:
+:$f \left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$
+:$g \left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$
+where $\epsilon_f, \epsilon_g: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are [[Definition:Continuous Complex Function|continuous functions]] that [[Definition:Convergent Function|converge]] to $0$ as $h$ tends to $0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left({fg}\right) \left({z + h}\right) 
+      | r = f \left({z}\right) g \left({z}\right) + h \left({f \left({z}\right) g' \left({z}\right) + f \left({z}\right) \epsilon_g \left({h}\right) }\right) + h \left({g \left({z}\right) f' \left({z}\right) + g \left({z}\right) \epsilon_f \left({h}\right) }\right) + h^2 \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)
+}}
+{{eqn | r = \left({fg}\right) \left({z}\right) + h \left({f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right)
+}}
+{{end-eqn}}
+Define $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C $ by $\epsilon \left({h}\right) = h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$.
+From [[Product Rule for Continuous Functions]] and [[Combined Sum Rule for Continuous Functions]], it follows that $\epsilon$ is continuous.
+From [[Product Rule for Limits of Functions]] and [[Combined Sum Rule for Limits of Functions]], it follows that:
+:$\displaystyle \lim_{h \to 0} \epsilon \left({h}\right) = \left({\lim_{h \to 0} h}\right) \left({\lim_{h \to 0} \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) }\right) \left({\lim_{h \to 0} \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right)= 0$
+Then the [[Alternative Differentiability Condition]] shows that:
+:$\left({f g}\right)' \left({z}\right) = f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dedekind Completion is Unique up to Isomorphism}
+Tags: Order Theory
+
+\begin{theorem}
+Let $S$ be an [[Definition:Ordered Set|ordered set]].
+Let $\struct {X, f}$ and $\struct {Y, g}$ be [[Definition:Dedekind Completion|Dedekind completions]] of $S$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Order Isomorphism|order isomorphism]] $\phi: X \to Y$ such that $\phi \circ f = g$.
+\end{theorem}
+
+\begin{proof}
+By assumption, $\struct {X, f}$ is a [[Definition:Dedekind Completion|Dedekind completion]] of $S$.
+Also, $g: S \to Y$ is an [[Definition:Order Embedding|order embedding]] and $Y$ is [[Definition:Dedekind Complete|Dedekind complete]].
+Hence by definition of [[Definition:Dedekind Completion|Dedekind completion]], there exists a [[Definition:Unique|unique]] $\phi: X \to Y$ such that:
+:$\phi \circ f = g$
+It only remains to show that $\psi$ is an [[Definition:Order Isomorphism|order isomorphism]].
+By reversing the roles of $X$ and $Y$ in the above, we obtain a [[Definition:Unique|unique]] $\psi: Y \to X$ subject to:
+:$\psi \circ g = f$
+Therefore:
+:$\psi \circ \phi \circ f = f$
+:$\phi \circ \psi \circ g = g$
+By the uniqueness clause of the definition of [[Definition:Dedekind Completion|Dedekind completion]], it follows that:
+:$\psi \circ \phi = I_X$
+:$\phi \circ \psi = I_Y$
+where $I_X$ and $I_Y$ are the [[Definition:Identity Mapping|identity mappings]] on $X$ and $Y$ respectively.
+Hence $\psi$ is an [[Definition:Order Isomorphism|order isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Complex Derivatives/Combined Sum Rule}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $\dfrac \d {\d z} \left({c f + w g}\right)$ denote the [[Definition:Derivative of Complex Function|derivative]] of $c f + w g$.
+Then:
+:$\dfrac \d {\d z} \left({c f + w g}\right) \left({z}\right) = c \dfrac \d {\d z} f \left({z}\right) + w \dfrac \d {\d z} g \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = c \dfrac \d {\d z} f \left({z}\right) + w \dfrac \d {\d z} g \left({z}\right)
+      | r = \dfrac \d {\d z} \left({c f}\right) \left({z}\right) + \dfrac \d {\d z} \left({w g}\right) \left({z}\right)
+      | c = [[Combination Theorem for Complex-Differentiable Functions/Multiple Rule|Combination Theorem for Complex-Differentiable Functions: Multiple Rule]]
+}}
+{{eqn | r = \dfrac \d {\d z} \left({c f + w g}\right) \left({z}\right)
+      | c = [[Combination Theorem for Complex-Differentiable Functions/Sum Rule|Combination Theorem for Complex-Differentiable Functions:Sum Rule]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Dedekind Completion}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Then there exists a [[Definition:Dedekind Completion|Dedekind completion]] of $S$.
+That is, there exists a [[Definition:Dedekind Complete|Dedekind complete]] [[Definition:Ordered Set|ordered set]] $\tilde S$ and an [[Definition:Order Embedding|order embedding]] $\phi: S \to \tilde S$ such that:
+:For all [[Definition:Dedekind Complete|Dedekind complete]] [[Definition:Ordered Set|ordered sets]] $X$, and for all [[Definition:Order Embedding|order embeddings]] $f: S \to X$, there exists an [[Definition:Order Embedding|order embedding]] $\tilde f: \tilde S \to X$ such that:
+::$\tilde f \circ \phi = f$
+\end{theorem}
+
+\begin{proof}
+For all [[Definition:Subset|subsets]] $I \subseteq S$, define:
+:$\operatorname U \left({I}\right) = \left\{{x \in S: x}\right.$ is an [[Definition:Upper Bound of Set|upper bound]] for $\left.{I}\right\}$
+:$\operatorname L \left({I}\right) = \left\{{x \in S: x}\right.$ is a [[Definition:Lower Bound of Set|lower bound]] for $\left.{I}\right\}$
+Note that, for all $I, J \subseteq S$:
+:$I \subseteq \operatorname {LU} \left({I}\right)$
+:$I \subseteq J \implies \operatorname {LU} \left({I}\right) \subseteq \operatorname {LU} \left({J}\right)$
+:$\operatorname {ULU} \left({I}\right) = \operatorname U \left({I}\right)$
+All of the above follow directly from the relevant definitions.
+Let:
+:$\tilde S = \left\{{I \subseteq S: I = \operatorname {LU} \left({I}\right)}\right.$, $I$ is [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Above Set|bounded above]]$\left.{}\right\}$
+We have that [[Subset Relation is Ordering|$\left({\tilde S, \subseteq}\right)$ is an ordered set]].
+Let $A \subseteq \tilde S$ be [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Above Set|bounded above]].
+By [[Union is Smallest Superset/General Result|Union is Smallest Superset: General Result]], $\displaystyle \bigcup A$ is [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Above Set|bounded above]].
+It follows that:
+:$\displaystyle \operatorname{LU} \left({\bigcup A}\right) \in \tilde S$
+From [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]], it follows that $\displaystyle \operatorname{LU} \left({\bigcup A}\right)$ is an [[Definition:Upper Bound of Set|upper bound]] for $A$.
+If $I \in \tilde S$ is an [[Definition:Upper Bound of Set|upper bound]] for $A$, then, by [[Union is Smallest Superset/General Result|Union is Smallest Superset: General Result]]:
+:$\displaystyle \bigcup A \subseteq I$
+It follows that:
+:$\displaystyle \operatorname{LU} \left({\bigcup A}\right) \subseteq \operatorname{LU} \left({I}\right) = I$
+Hence:
+:$\displaystyle \operatorname{LU} \left({\bigcup A}\right) = \sup A$
+By definition, $\left({\tilde S, \subseteq}\right)$ is [[Definition:Dedekind Complete|Dedekind complete]].
+Let $\phi: S \to \tilde S$ be the [[Definition:Order Embedding|order embedding]] defined as:
+:$\forall x \in S: \phi \left({x}\right) = x^\preceq$
+where $x^\preceq$ denotes the [[Definition:Weak Lower Closure of Element|weak lower closure]] of $x$ (in $S$).
+Let $\left({X, \preccurlyeq}\right)$ be a [[Definition:Dedekind Complete|Dedekind complete]] [[Definition:Ordered Set|ordered set]], and let $f: S \to X$ be an [[Definition:Order Embedding|order embedding]].
+Let $\tilde f: \tilde S \to X$ be the [[Definition:Increasing Mapping|increasing mapping]] defined as:
+:$\forall I \in \tilde S: \tilde f \left({I}\right) = \sup f \left({I}\right)$
+where $f \left({I}\right)$ denotes the [[Definition:Image of Subset under Mapping|image]] of $I$ under $f$.
+Then $\tilde f \circ \phi = f$.
+Let $I, J \in \tilde S$ such that $\tilde f \left({I}\right) \preccurlyeq \tilde f \left({J}\right)$.
+Then $\forall x \in I: \forall u \in \operatorname U \left({J}\right): f \left({x}\right) \preccurlyeq \tilde f \left({J}\right) \preccurlyeq f \left({u}\right)$.
+Since $f$ is an [[Definition:Order Embedding|order embedding]], it follows that $I \subseteq \operatorname {LU} \left({J}\right) = J$.
+Hence, $\tilde f$ is an [[Definition:Order Embedding|order embedding]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering}
+Tags: Reflexive Reductions
+
+\begin{theorem}
+Let $S$ be a [[Definition:set|set]].
+Let $\mathcal R$ be a [[Definition:Transitive Relation|transitive]], [[Definition:Antisymmetric Relation|antisymmetric]] [[Definition:Endorelation|relation]] on $S$.
+Let $\mathcal R^\ne$ denote the [[Definition:Reflexive Reduction|reflexive reduction]] of $\mathcal R$.
+Then $\mathcal R^\ne$ is a [[Definition:Strict Ordering|strict ordering]].
+\end{theorem}
+
+\begin{proof}
+To show that $\mathcal R^\ne$ is a [[Definition:Strict Ordering|strict ordering]], it is sufficient to show that $\mathcal R^\ne$ is [[Definition:Antireflexive Relation|antireflexive]] and [[Definition:Transitive Relation|transitive]].
+=== Antireflexive ===
+Follows from [[Reflexive Reduction is Antireflexive]].
+{{qed|lemma}}
+=== Transitive ===
+Let $a, b, c \in S$.
+Let $a \mathrel{\mathcal R^\ne} b$ and $b \mathrel{\mathcal R^\ne} c$.
+By the definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
+:$a \ne b$ and $a \mathrel{\mathcal R} b$
+:$b \ne c$ and $b \mathrel{\mathcal R} c$
+Since $\mathrel{\mathcal R}$ is [[Definition:Transitive Relation|transitive]]:
+:$a \mathrel{\mathcal R} c$
+Suppose, with a view to deriving a [[Proof by Contradiction|contradiction]], that $a = c$.
+Since $a \mathrel{\mathcal R} b$ it follows that $c \mathrel{\mathcal R} b$.
+Since $c \mathrel{\mathcal R} b$, $b \mathrel{\mathcal R} c$, and $\mathrel{\mathcal R}$ is [[Definition:Antisymmetric Relation|antisymmetric]], $b = c$.
+But this [[Proof by Contradiction|contradicts]] $b \ne c$.
+The conclusion is that $a \ne c$.
+Recall that $a \mathrel{\mathcal R} c$.
+By the definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
+:$a \mathrel{\mathcal R^\ne} c$
+{{qed}}
+[[Category:Reflexive Reductions]]
+e85znurcjyy8ehftu2tbp6tfjbm71we
+\end{proof}<|endoftext|>
+\section{Reflexive Reduction of Antisymmetric Relation is Asymmetric}
+Tags: Reflexive Reductions
+
+\begin{theorem}
+Let $S$ be a set.
+Let $\mathcal R$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]] on $S$.
+Let $\mathcal R^\ne$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\mathcal R$.
+Then $\mathcal R^\ne$ is [[Definition:Asymmetric Relation|asymmetric]].
+\end{theorem}
+
+\begin{proof}
+Aiming for a [[Proof by Contradiction|contradiction]], suppose that $\mathcal R^\ne$ is not [[Definition:Asymmetric Relation|asymmetric]].
+That is:
+:$\exists a, b \in S: a \mathrel{\mathcal R^\ne} b$ and $b \mathrel{\mathcal R^\ne} a$
+Then by the definition of [[Definition:Reflexive Reduction|reflexive reduction]]:
+: $a \mathrel{\mathcal R} b$, $b \mathrel{\mathcal R} a$
+and $a \ne b$.
+But this [[Proof by Contradiction|contradicts]] the [[Definition:Antisymmetric Relation|antisymmetry]] of $\mathcal R$.
+Thus, by definition, $\mathcal R^\ne$ is an [[Definition:Asymmetric Relation|asymmetric relation]].
+{{qed}}
+[[Category:Reflexive Reductions]]
+b35ioqr8w71pfuxq9jktyw96i2tj0lb
+\end{proof}<|endoftext|>
+\section{Properties of Ordered Group}
+Tags: Ordered Groups
+
+\begin{theorem}
+Let $\left({G, \circ, \preceq}\right)$ be an [[Definition:Ordered Group|ordered group]] with [[Definition:Identity Element|identity]] $e$.
+Let $x,y,z,w \in G$.
+Then the following hold:
+\end{theorem}
\ No newline at end of file