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+\section{Inner Limit in Hausdorff Space by Set Closures}
+Tags: Set Theory, Topology, Measure Theory, Limits of Sequence of Sets
+
+\begin{theorem}
+Let $\struct {\XX, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
+Let $\sequence {C_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]] in $\XX$.
+Then:
+:$\displaystyle \liminf_n C_n = \bigcap_{N \mathop \in \NN_\infty^\#} \map \cl {\bigcup_{n \mathop \in N} C_n}$
+where:
+:$\cl$ denotes [[Definition:Closure (Topology)|set closure]]
+:$\NN_\infty^\#$ denotes the [[Definition:Set|set]] of [[Definition:Cofinal Subset|cofinal subsets]] of $\N$.
+\end{theorem}
+
+\begin{proof}
+$(1)$: Let:
+:$\displaystyle x \in \liminf_n \ C_n$
+Let:
+:$\Sigma \in \NN_\infty^\#$
+Let $W$ be a [[Definition:Open Neighborhood|neighborhood]] of $x$. 
+Then there exists $N_0 \in \N$ such that for all $n \ge N_0$ such that $n \in \Sigma$:
+:$W \cap C_n \ne \O$
+Thus:
+:$\displaystyle x \in \map \cl {\bigcup_{n \mathop \in \Sigma} C_n}$
+$(2)$: Let:
+:$\displaystyle x \notin \liminf_n C_n$
+Then there exists an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+Let $\map \mho x := \set {V \in \tau: x \in V}$ denote the [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] of $x$.
+Let $W \in \map \mho x$ such that:
+:$\Sigma_0 := \set {n \in \N: W \cap C_n = \O}$
+is [[Definition:Cofinal Subset|cofinal]].
+Then:
+:$\displaystyle x \notin \map \cl {\bigcup_{n \mathop \in \Sigma_0} C_n}$
+This completes the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{At Most Two Horizontal Asymptotes}
+Tags: Limits of Functions, Analytic Geometry
+
+\begin{theorem}
+The [[Definition:Graph of Mapping|graph]] of a [[Definition:Real Function|real function]] has at most two [[Definition:Horizontal Asymptote|horizontal asymptotes]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of a [[Definition:Horizontal Asymptote|horizontal asymptote]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordinal is Transitive}
+Tags: Ordinals, Transitive Classes
+
+\begin{theorem}
+Every [[Definition:Ordinal|ordinal]] (by [[Definition:Ordinal/Definition 3|Definition 3]]) is a [[Definition:Transitive Set|transitive set]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be an [[Definition:Ordinal|ordinal]] by [[Definition:Ordinal/Definition 3|Definition 3]].
+By definition:
+:$\forall a \in S: a = S_a \subseteq S$
+where $S_a$ denotes the [[Definition:Initial Segment|initial segment]] of $S$ determined by $a$.
+That is, $S$ is a [[Definition:Transitive Set|transitive set]].
+{{qed}}
+[[Category:Ordinals]]
+[[Category:Transitive Classes]]
+7pw4myves5i5yz0bituyrotteg1v09k
+\end{proof}<|endoftext|>
+\section{Foundational Relation has no Relational Loops}
+Tags: Set Theory
+
+\begin{theorem}
+Let $\prec$ be a [[Definition:Foundational Relation|foundational relation]] on $A$ and let $x_1, x_2, \ldots, x_n \in A$.
+Then:
+: $\neg \left({x_1 \prec x_2 \land x_3 \prec x_4 \cdots \land x_n \prec x_1}\right)$
+That is, there are no [[Definition:Relational Loop|relational loops]] within $A$.
+\end{theorem}
+
+\begin{proof}
+Since $x_1, x_2, \ldots, x_n \in A$, there exists a $y$ such that $y = \left\{{x_1, x_2, \ldots, x_n}\right\}$.
+Then $y$ is a [[Definition:Non-Empty Set|nonempty]] [[Definition:Subset|subset]] of $A$.
+So, by the definition of a [[Definition:Foundational Relation|foundational relation]]:
+: $\exists w \in y: \forall z \in y: \neg w \prec z$
+Now, suppose $x_1 \prec x_2 \land x_2 \prec x_3 \cdots \land x_n \prec x_1$.
+But since the elements of $y$ are $x_1, x_2, \ldots, x_n$, then this contradicts the previous statement, since:
+: $\forall w \in y: \exists z \in y: w \prec z$
+Thus a founded relation has no [[Definition:Relational Loop|relational loops]].
+{{explain|is a "founded relation" the same as a "foundational relation"? If so, use the same terminology throughout the entire page; if not, provide a definition for the former.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Epsilon is Foundational}
+Tags: Class Theory, Axiom of Foundation
+
+\begin{theorem}
+Let $\Epsilon$ denote the [[Definition:Epsilon Relation|epsilon relation]].
+Then $\Epsilon$ is a [[Definition:Foundational Relation|foundational relation]] on every [[Definition:Class (Class Theory)|class]] $A$.
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+By the [[Axiom:Axiom of Foundation|axiom of foundation]]:
+:$\forall S: \paren {\exists x: x \in S \implies \exists y \in S: \forall x \in S: \neg x \in y}$
+That is, by [[Nonempty Class has Members]]:
+:$\forall S: \paren {S \ne \O \implies \exists y \in S: \forall x \in S: \neg x \in y}$ 
+This holds for all [[Definition:Set|sets]] $S$ whose construction is based on the [[Definition:Zermelo-Fraenkel Axioms|Zermelo-Fraenkel axioms]].
+We can weaken the antecedent of the above statement with this statement:
+:$\forall S: \paren {\paren {S \ne \O \land S \subseteq A} \implies \exists y \in S: \forall x \in S: \neg x \in y}$
+Note that this step does not require that $A$ be a [[Definition:Set|set]]: it can be any [[Definition:Class (Class Theory)|class]], even a [[Definition:Proper Class|proper class]].
+By definition, it follows that $\Epsilon$ is a [[Definition:Foundational Relation|foundational relation]] on every class $A$.
+[[Category:Class Theory]]
+[[Category:Axiom of Foundation]]
+695zkccvjz04mph4p8ung9e7zxklqcn
+\end{proof}<|endoftext|>
+\section{Inner Limit in Normed Spaces by Open Balls}
+Tags: Limits of Sequence of Sets
+
+\begin{theorem}
+Let $\left \langle{C_n}\right \rangle_{n \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Set|sets]] in a [[Definition:Normed Vector Space|normed vector space]] $\left({\mathcal X, \left\Vert{\cdot}\right\Vert}\right)$.
+Then the [[Definition:Inner Limit|inner limit]] of $\left \langle{C_n}\right \rangle_{n \in \N}$ is:
+: $\displaystyle \liminf_n \ C_n = \left\{{x: \forall \epsilon > 0: \exists N \in \mathcal N_\infty: \forall n \in N: x \in C_n + \epsilon B}\right\}$
+where $B$ denotes the [[Definition:Open Ball|open unit ball]] of the space.
+{{explain|What are $N$ and $\mathcal N_\infty$ in this context? Also, what point is at the center of $B$? And what does $C_n + \epsilon B$ mean? For the latter one suspects $\cup$, but this needs to be checked. This page might need to be rewritten from a new perspective, as the original author was touchy about symbols used and departed ProofWiki in a rage when his notation was changed.}}
+\end{theorem}
+
+\begin{proof}
+The proof is an immediate result of [[Inner Limit in Hausdorff Space by Open Neighborhoods]] since the arbitrary [[Definition:Open Set (Topology)|open sets]] can be here replaced by [[Definition:Open Ball|open balls]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inner Limit is Closed Set}
+Tags: Limits of Sequence of Sets
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff]] [[Definition:Topological Space|topological space]].
+Let $\sequence {C_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of sets in $S$.
+Then the [[Definition:Inner Limit|inner limit]] $\liminf_n C_n$ is a [[Definition:Closed Set (Topology)|closed set]].
+\end{theorem}
+
+\begin{proof}
+According to [[Inner Limit in Hausdorff Space by Set Closures]], the [[Definition:Inner Limit|inner limit]] is given by an arbitrary [[Definition:Set Intersection|intersection]] of [[Definition:Closed Set (Topology)|closed sets]] which is [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Topology|topology]] $\tau$.
+{{qed}}
+[[Category:Limits of Sequence of Sets]]
+olfc9bqbfl4j70ofagsfnhu9jetdors
+\end{proof}<|endoftext|>
+\section{Local Basis of Topological Vector Space}
+Tags: 
+
+\begin{theorem}
+Let $\struct {\XX, \tau}$ be a [[Definition:Topological Vector Space|topological vector space]].
+Let $0_\XX$ denote the [[Definition:Zero Vector|zero vector]] of $\XX$.
+Then there exists a [[Definition:Local Basis|local basis]] $\BB$ of $0_\XX$ with the following properties:
+:$(1): \quad \forall W \in \BB: \exists V \in \BB$ such that $V + V \subseteq W$ (where the addition $V + V$ is meant in the sense of the [[Definition:Minkowski Sum|Minkowski sum]])
+:$(2): \quad$ Every $W \in \BB$ is [[Definition:Star Shaped_Set|star-shaped (balanced)]]
+:$(3): \quad$ Every $W \in \BB$ is [[Definition:Absorbent Set|absorbent]].
+:$(4): \quad \displaystyle \bigcap_\BB W = \set {0_\XX}$.
+\end{theorem}
+
+\begin{proof}
+The proof will be carried out in various steps.
+We will construct a collection of [[Definition:Star Shaped_Set|star-shaped]] neighborhoods of $0_\XX$.
+Then we will show that it is indeed a local basis with the required properties.
+Firstly we define the following [[Definition:Set|set]]:
+:$\BB_0 := \set {W \in \tau: 0 \in W, W \text{ is star-shaped} }$
+=== $\BB_0$ is a local basis for $0_\XX$ ===
+Let $U \ni 0_\XX$ be an [[Definition:Open Set (Topology)|open set]].
+Notice that:
+:$0 \cdot 0_\XX = 0_\XX$
+This way we have proved that $\BB_0$ is a local basis for $0_\XX$.
+Since $\cdot: K \times \XX \to \XX$ is a continuous mapping, there is a neighborhood of $\tuple {0, 0_\XX} \in K \times \XX$ in the form:
+:$\openint {-\epsilon} \epsilon \times W$
+such that:
+:$\map \cdot {\openint {-\epsilon} \epsilon \times W} \subseteq U$
+which means that
+:$\openint {-\epsilon} \epsilon \cdot W \subseteq U$
+Let:
+:$\displaystyle G := \bigcup_{\lambda \mathop \in \openint {-\epsilon} \epsilon, \lambda \mathop \ne 0} \lambda W \subseteq U$
+We have that:
+:$(1): \quad$ For every $\lambda \ne 0$, the set $\lambda \cdot W$ is open.
+Hence $G$ is open as union of open sets.
+:$(2): \quad 0_\XX \in G$ since $0 \in W$.
+:$(3): \quad$ If $x \in G$ then $-x \in G$.
+Therefore, $G$ is open, star-shaped, contains $0_\XX$ and $G \subseteq U$.
+{{qed|lemma}}
+=== Condition $(1)$ is satisfied by $\BB_0$ ===
+Let $V \in \BB_0$.
+We have that $+\tuple {0_\XX, 0_\XX} = 0_\XX \in W$.
+Since $+$ is a continuous mapping and $\BB_0$ is a local basis of $0_\XX$, there exists a $V \in \BB_0$ such that:
+:$+\tuple {V, V} = V + V \subseteq W$
+{{qed|lemma}}
+=== $\BB_0$ consists of Absorbent Sets ===
+We notice that:
+:$\tuple {\dfrac 1 n, x} \overset n \to \tuple {0, x}$
+and:
+:$\map \cdot {\dfrac 1 n, x} \overset n \to \map \cdot {0, x} = 0_\XX$
+Thus, there is a $n \in \N$ such that:
+:$\frac 1 n x \in V$
+or what amounts to the same thing:
+:$x \in n V$
+{{qed|lemma}}
+=== Condition $(4)$ is satisfied by $\BB_0$ ===
+It suffices that we find a $V \in \BB_0$ such that $x \notin V$.
+This is possible since $\tau$ is a Hausdorff topology.
+{{qed|lemma}}
+The proof is now complete.
+{{qed}}
+tse15mbihzw34nxocvkoaninbfxns2p
+\end{proof}<|endoftext|>
+\section{No Membership Loops}
+Tags: Class Theory, Axiom of Foundation
+
+\begin{theorem}
+For any [[Definition:Proper Class|proper classes]] or [[Definition:Set|sets]] $A_1, A_2, \ldots, A_n$:
+:$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+Either $A_1, A_2, \ldots, A_n$ are all [[Definition:Set|sets]], or there exists a [[Definition:Proper Class|proper class]] $A_m$ such that $1 \le m \le n$.
+Suppose there exists a [[Definition:Proper Class|proper class]] $A_m$.
+Then, by the definition of a [[Definition:Proper Class|proper class]], $\neg A_m \in A_{m+1}$, since it is not a member of any [[Definition:Class (Class Theory)|class]].
+The result then follows directly.
+Otherwise it follows that all $A_1, A_2, \ldots, A_n$ are [[Definition:Set|sets]].
+Then, by the fact that [[Epsilon is Foundational]] and a [[Foundational Relation has no Relational Loops]], it follows that: 
+:$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composite of Isomorphisms is Isomorphism/Algebraic Structure}
+Tags: Isomorphisms
+
+\begin{theorem}
+Let:
+: $\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$
+: $\left({S_2, *_1, *_2, \ldots, *_n}\right)$
+: $\left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$
+be [[Definition:Algebraic Structure|algebraic structures]].
+Let:
+: $\phi: \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_2, *_1, *_2, \ldots, *_n}\right)$
+: $\psi: \left({S_2, *_1, *_2, \ldots, *_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$
+be [[Definition:Isomorphism (Abstract Algebra)|isomorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+If $\phi$ and $\psi$ are both [[Definition:Isomorphism (Abstract Algebra)|isomorphisms]], then they are by definition:
+* [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]]
+* [[Definition:Bijection|bijections]].
+So:
+* From [[Composite of Homomorphisms on Algebraic Structure is Homomorphism]] we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]]
+* From [[Composite of Bijections is Bijection]] we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both [[Definition:Bijection|bijections]]
+and hence by definition also [[Definition:Isomorphism (Abstract Algebra)|isomorphisms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composite of Isomorphisms is Isomorphism/R-Algebraic Structure}
+Tags: Isomorphisms
+
+\begin{theorem}
+Let:
+: $\left({S_1, \ast_1}\right)_R$
+: $\left({S_2, \ast_2}\right)_R$
+: $\left({S_3, \ast_3}\right)_R$
+be [[Definition:R-Algebraic Structure|$R$-algebraic structures]] with the same number of operations.
+Let:
+: $\phi: \left({S_1, \ast_1}\right)_R \to \left({S_2, \ast_2}\right)_R$
+: $\psi: \left({S_2, \ast_2}\right)_R \to \left({S_3, \ast_3}\right)_R$
+be [[Definition:R-Algebraic Structure Isomorphism|isomorphisms]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also an [[Definition:R-Algebraic Structure Isomorphism|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+If $\phi$ and $\psi$ are both [[Definition:R-Algebraic Structure Isomorphism|isomorphisms]], then they are by definition:
+* [[Definition:R-Algebraic Structure Homomorphism|homomorphisms]]
+* [[Definition:Bijection|bijections]].
+So:
+* From [[Composite of Homomorphisms is Homomorphism/R-Algebraic Structure|Composite of Homomorphisms for R-Algebraic Structures is Homomorphism]] we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both [[Definition:R-Algebraic Structure Isomorphism|homomorphisms]]
+* From [[Composite of Bijections is Bijection]] we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both [[Definition:Bijection|bijections]];
+and hence by definition also [[Definition:R-Algebraic Structure Isomorphism|isomorphisms]].
+{{qed}}
+[[Category:Isomorphisms]]
+5alihydv9np3orurmuk1lfl63nhxvnb
+\end{proof}<|endoftext|>
+\section{Element of Transitive Class}
+Tags: Class Theory, Transitive Classes
+
+\begin{theorem}
+Let $B$ be a [[Definition:Transitive Class|transitive class]].
+Then:
+:$A \in B \implies A \subsetneq B$
+where $\subsetneq$ denotes a [[Definition:Proper Subset|proper subset]]).
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+By the definition of a [[Definition:Transitive Class|transitive class]]:
+:$A \in B \implies A \subseteq B$
+But $A \ne B$ because $\paren {A = B \land A \in B} \implies A \in A$, which by [[No Membership Loops]] is a contradiction.
+Therefore $A \subsetneq B$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent Line to Convex Graph}
+Tags: Differential Calculus, Analytic Geometry, Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] that is [[Definition:Continuous on Interval|continuous]] on some [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$ and [[Definition:Differentiable on Interval|differentiable]] and [[Definition:Convex Real Function|convex]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$. 
+Then all the [[Definition:Tangent to Curve|tangent lines]] to $f$ are below the [[Definition:Graph of Mapping|graph]] of $f$.
+{{explain|"below"}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Concaveup.png|500px]]
+Let $\mathcal T$ be the [[Definition:Tangent to Curve|tangent line]] to $f$ at some [[Definition:Point|point]] $\left({c, f \left({c}\right)}\right)$, $c \in \left({a \,.\,.\, b}\right)$.
+From the [[Gradient-Intercept Form]] of a line, given any point $\left({x_1, y_1}\right)$ and the [[Definition:Gradient|gradient]] $m$, the equation of such a line is:
+{{begin-eqn}}
+{{eqn | l = y - y_1
+      | r = m \left({x - x_1}\right)
+      | c = 
+}}
+{{eqn | ll=\implies
+      | l = y
+      | r = m \left({x - x_1}\right) + y_1
+      | c = 
+}}
+{{end-eqn}}
+For $\mathcal T$:
+* $y = \mathcal T \left({x}\right)$
+* $y_1 = f \left({c}\right)$
+* $m = f' \left({c}\right)$
+* $x = x$
+* $x_1 = c$
+so
+:$\mathcal T \left({x}\right) = f' \left({c}\right) \left({x - c}\right) + f \left({c}\right)$
+Consider the graph of $f$ to the right of $\left({c, f \left({c}\right)}\right)$, that is, any $x$ in $\left({c \,.\,.\, b}\right)$. 
+Let $d$ be the directed vertical distance from  $\mathcal T$ to the graph of $f$.
+That is, if $f$ is above $\mathcal T$ then $d > 0$.
+If $f$ is below $\mathcal T$, then $d < 0$. 
+(From the diagram, it is apparent that $\mathcal T$ is below $f$, but we shall prove it analytically.)
+$d$ can be evaluated by:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = f \left({x}\right) - \mathcal T \left({x}\right)
+      | c = 
+}}
+{{eqn | r = f \left({x}\right) - f' \left({c}\right)\left({x - c}\right) - f \left({c}\right)
+      | c = 
+}}
+{{eqn | r = f \left({x}\right) - f \left({c}\right) - f' \left({c}\right) \left({x - c}\right)
+      | c = 
+}}
+{{end-eqn}}
+By the [[Mean Value Theorem]], there exists some constant $k$ in $\left({c \,.\,.\, b}\right)$ such that:
+{{begin-eqn}}
+{{eqn | l = f' \left({k}\right)
+      | r = \frac {f \left({x}\right) - f \left({c}\right)} {x - c}
+      | c = 
+}}
+{{eqn | ll=\implies
+      | l = f' \left({k}\right) \left({x - c}\right)
+      | r = f \left({x}\right) - f \left({c}\right)
+      | c = 
+}}
+{{end-eqn}}
+Substitute this into the formula for $d$:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = f' \left({k}\right) \left({x - c}\right) - f' \left({c}\right) \left({x - c}\right)
+      | c = 
+}}
+{{eqn | r = \left({f' \left({k}\right) - f' \left({c}\right)}\right) \left({x - c}\right)
+      | c = 
+}}
+{{end-eqn}}
+Recall that $x$ lies in the [[Definition:Open Real Interval|interval]] $\left({c \,.\,.\, b}\right)$.
+So $x > c$, and the quantity $x - c$ is [[Definition:Strictly Positive Real Number|(strictly) positive]].
+$k$ is also in the [[Definition:Open Real Interval|interval]] $\left({c \,.\,.\, b}\right)$ and so $k > c$.
+By construction, $f$ is [[Definition:Convex Real Function|convex]]. 
+By the definition of [[Definition:Convex Real Function|convex]]:
+:$k > c \implies f' \left({k}\right) > f' \left({c}\right)$
+which means that:
+:$\left({f' \left({k}\right) - f' \left({c}\right)}\right) > 0$
+Then $d$ is the product of two [[Definition:Strictly Positive Real Number|(strictly) positive]] quantities and is itself [[Definition:Strictly Positive Real Number|(strictly) positive]]. 
+Similarly, consider the graph of $f$ to the left of $\left({c, f \left({c}\right)}\right)$, that is, any $x$ in $\left({a \,.\,.\, c}\right)$.
+By the same process as above, we will have:
+:$d = \left({f' \left({k}\right) - f' \left({c}\right)}\right) \left({x - c}\right)$
+This time, $x < c$ and the quantity $x - c$ is [[Definition:Strictly Negative Real Number|(strictly) negative]].
+Further, $k < c$, and so by a similar argument as above:
+:$k < c \implies f' \left({k}\right) < f' \left({c}\right)$
+and the quantity $\left({f' \left({k}\right) - f' \left({c}\right)}\right)$ is also [[Definition:Strictly Negative Real Number|(strictly) negative]].
+Thus $d$ will be the product of two [[Definition:Strictly Negative Real Number|(strictly) negative]] quantities, and will again be [[Definition:Strictly Positive Real Number|(strictly) positive]]. 
+{{qed}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Characterization of Lower Semicontinuity}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f: S \to \overline \R$ be an [[Definition:Extended Real-Valued Function|extended real valued function]].
+Let $S$ be endowed with a [[Definition:Topology|topology]] $\tau$.
+The following are equivalent:
+:$(1): \quad$ $f$ is [[Definition:Lower Semicontinuous on Subset|lower semicontinuous (LSC)]] on $S$.
+:$(2): \quad$ The [[Definition:Epigraph|epigraph]] $\map {\operatorname{epi}} f$ of $f$ is a [[Definition:Closed Set (Topology)|closed set]] in $S \times \R$ with the [[Definition:Product Topology|product topology]].
+:$(3): \quad$ All [[Definition:Lower Level Set|lower level sets]] of $f$ are [[Definition:Closed Set (Topology)|closed]] in $S$.
+\end{theorem}
+
+\begin{proof}
+The proof is carried out in the following three steps:
+=== LSC implies Closed Epigraph ===
+Let $f: S \to \overline \R$ be [[Definition:Lower Semicontinuous on Subset|LSC]] on $S$.
+Let $\operatorname{epi} \left({f}\right)$ denote the [[Definition:Epigraph|epigraph]] of $f$.
+Take a [[Definition:Sequence|sequence]] $\sequence{\tuple{x_n, a_n}}_{n \mathop \in \N} \in \map {\operatorname{epi}} f$ such that:
+:$\tuple{x_n, a_n} \to \tuple{\bar x, \bar a} \in S \times \R$
+as $n \to \infty$.
+This implies that $x_n \to \bar x$ and $a_n \to \bar a$ by definition of the [[Definition:Product Topology|product topology]] on $S \times \R$ and because a [[Continuous Mapping is Sequentially Continuous]].
+Consequently:
+{{begin-eqn}}
+{{eqn | l =  \map f {\bar x}
+      | r = \liminf_{x \mathop \to \bar x} \map f x
+      | c = Definition of [[Definition:Lower Semicontinuous|Lower Semicontinuous]]
+}}
+{{eqn | o = \le
+      | r = \liminf_{n \mathop \to \infty} \map f {x_n}
+      | c = [[Relationship between Limit Inferior and Lower Limit]]
+}}
+{{eqn | o = \le
+      | r = \liminf_{n \mathop \to \infty} a_n
+      | c = because $\tuple{x_n, a_n} \in \map {\operatorname{epi} } f$
+}}
+{{eqn | r = \bar a
+      | c = because $a_n \to \bar a$
+}}
+{{end-eqn}}
+Thus:
+:$\tuple{\bar x, \bar a} \in \map {\operatorname{epi}} f$
+{{proofread}}
+{{questionable|This only proves sequential closedness.}}
+{{qed|lemma}}
+=== Closed Epigraph implies Closed Level Sets ===
+Let $\map {\operatorname{epi}} f$ be a [[Definition:Closed Set (Topology)|closed set]] in $S \times \R$.
+Let $a \in \R$.
+Then the [[Definition:Lower Level Set|$\alpha$-lower level set]]:
+:$\displaystyle \operatorname{lev} \limits_{\mathop \le a} = \map {\operatorname{epi}} f \cap S \times \set{a}$
+{{questionable|The identity claimed here is not true}}
+is [[Definition:Closed Set (Topology)|closed]] in $S$ because:
+: [[Definition:Closed Set (Topology)|closedness]] is preserved under [[Definition:Set|Intersection|intersection]]
+and:
+: $S \times \set{a}$ is a [[Definition:Closed Set (Topology)|closed set]] with respect with the [[Definition:Product Topology|product topology]] of $S \times \R$.
+{{qed|lemma}}
+=== Closed Level Sets implies LSC ===
+Let all [[Definition:Lower Level Set|lower level sets]] of $f$ be [[Definition:Closed Set (Topology)|closed]].
+Let $x \in S$.
+Let $\displaystyle a: = \liminf_{t \mathop \to x} \map f t$.
+In order to prove that $f$ is [[Definition:Lower Semicontinuous|LSC]] it suffices to show that $a = \map f x$. 
+We know already that $a \le \map f x$.
+Thus it suffices to show that $\map f x \le a$.
+Let $a = \infty$.
+As $\infty \le \infty$ the proof is complete.
+Let $a < \infty$.
+Then we can find a [[Definition:Sequence|sequence]] $\sequence{x_n}_{n \mathop \in \N}$ such that:
+: $x_n \to x$ and $\map f {x_n} \to a$
+as $n \to \infty$.
+For any $b > a$:
+:$\map f {x_n} \le b$
+or equivalently:
+:$x_n \in \displaystyle \operatorname{lev} \limits_{\mathop \le b} f$
+which [[Definition:By Hypothesis|by hypothesis]] is [[Definition:Closed Set (Topology)|closed]].
+Therefore $\displaystyle \operatorname{lev} \limits_{\mathop \le b} f$ contains the [[Definition:Limit Point of Sequence|limit point]] $x$.
+We have that for all $b > a$:
+:$\map f x \le b$
+Therefore:
+:$\map f x \le a$
+Hence the result.
+{{qed|lemma}}
+So:
+: $(1) \implies (2) \implies (3)$
+and the proof is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Restriction of Foundational Relation is Foundational}
+Tags: Foundational Relations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\RR \subseteq S \times S$ be a [[Definition:Foundational Relation|foundational relation]] on $S$.
+Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Let $\RR {\restriction_T} \subseteq T \times T$ be the [[Definition:Restriction of Relation|restriction]] of $\RR$ to $T$.
+Then $\RR {\restriction_T}$ is a [[Definition:Foundational Relation|foundational relation]] on $T$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Foundational Relation|foundational relation]], every [[Definition:Non-Empty Set|nonempty]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Minimal Element|minimal element]].
+By [[Subset Relation is Transitive]], every [[Definition:Subset|subset]] of $T$ is also a [[Definition:Subset|subset]] of $S$.
+Therefore every [[Definition:Non-Empty Set|nonempty]] [[Definition:Subset|subset]] of $T$ has a [[Definition:Minimal Element|minimal element]].
+Thus by definition, $\RR$ is a [[Definition:Foundational Relation|foundational relation]] on $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Restriction of Strict Well-Ordering is Strict Well-Ordering}
+Tags: Well-Orderings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Strict Well-Ordering|strict well-ordering]] of $A$.
+Let $B \subseteq A$.
+Then $R$ is a [[Definition:Strict Well-Ordering|strict well-ordering]] of $B$.
+\end{theorem}
+
+\begin{proof}
+By [[Restriction of Foundational Relation is Foundational]], $R$ is a [[Definition:Foundational Relation|foundational relation]] on $B$.
+By [[Restriction of Total Ordering is Total Ordering]], $R$ is a [[Definition:Total Ordering|total ordering]] on $B$.
+By the above two statements, $R$ is a [[Definition:Strict Well-Ordering|strict well-ordering]] of $B$.
+{{explain|Indicate the specific definition which identifies that a foundation relation which is a total ordering is a well-ordering.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Alternative Definition of Ordinal}
+Tags: Ordinals
+
+\begin{theorem}
+A [[Definition:Set|set]] $S$ is an [[Definition:Ordinal|ordinal]] {{iff}} $S$ is [[Definition:Transitive Set|transitive]] and is [[Definition:Strict Well-Ordering|strictly well-ordered]] by the [[Definition:Epsilon Relation|$\in$-relation]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $S$ is an [[Definition:Ordinal|ordinal]].
+Then [[Ordinal is Transitive|$S$ is transitive]].
+By definition, the [[Definition:Strict Well-Ordering|strict well-ordering]] on $S$ is given by the [[Definition:Epsilon Relation|$\in$-relation]].
+Hence, the [[Definition:Necessary Condition|necessary condition]] is satisfied.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $S$ is a [[Definition:Transitive Set|transitive set]] that is [[Definition:Strict Well-Ordering|strictly well-ordered]] by the [[Definition:Epsilon Relation|$\in$-relation]].
+Let $a \in S$. Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = S \cap a
+      | c = $S$ is [[Definition:Transitive Set|transitive]], $a \subseteq S$; then apply [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = \left\{ {x \in S: x \in a} \right\}
+}}
+{{eqn | r = S_a
+      | c = Definition of [[Definition:Initial Segment|Initial Segment]]
+}}
+{{end-eqn}}
+That is, $S$ is an [[Definition:Ordinal|ordinal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordinal Class is Ordinal}
+Tags: Ordinal Class
+
+\begin{theorem}
+The [[Definition:Ordinal Class|Ordinal Class]] $\On$ is an [[Definition:Ordinal|ordinal]].
+\end{theorem}
+
+\begin{proof}
+{{questionable|This whole proof probably needs to be rewritten from scratch.}}
+{{NotZFC}}
+The [[Definition:Epsilon Relation|epsilon relation]] is equivalent to the strict subset relation when [[Definition:Restriction of Relation|restricted]] to [[Definition:Ordinal|ordinals]] by [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]].
+It follows that:
+:$\forall x \in \On: x \subset \On$
+{{explain|We have a problem here since $\On$ is a proper class. The notation $\in $ an $\subseteq$ can naturally be extended to classes, but one must be careful with their use.}}
+The [[Definition:Initial Segment|initial segment]] of the [[Definition:Ordinal Class|class of ordinals]] is:
+:$\set {x \in \On : x \subset \On}$
+{{explain|What is meant by $\subset$ here: $\subseteq$ or $\subsetneqq$? <s>Initial segment makes no sense here because $\On$ is not a set.</s><br/>Sorry, but "makes no sense" is a bit strong: "initial segment" has been defined on {{ProofWiki}} in the context of a well-ordered set, but the term makes perfect sense if defined in the context of a class.}}
+This [[Definition:Class (Class Theory)|class]] is [[Definition:Class Equality|equal]] to $\On$.
+Therefore, by the definition of [[Definition:Ordinal|ordinal]], $\On$ is an [[Definition:Ordinal|ordinal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transitive Set is Proper Subset of Ordinal iff Element of Ordinal}
+Tags: Ordinals, Transitive Classes
+
+\begin{theorem}
+Let $A$ be an [[Definition:Ordinal|ordinal]].
+Let $B$ be a [[Definition:Transitive Set|transitive set]].
+Then:
+:$B \subsetneq A \iff B \in A$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $B \in A$.
+From [[Ordinal is Transitive]], it follows that $B \subseteq A$.
+Also, $B \ne A$ by [[Ordinal is not Element of Itself]].
+Therefore, $B \subsetneq A$, as desired.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $B \subsetneq A$.
+By the definition of [[Definition:Set Equality|set equality]], the [[Definition:Set Difference|set difference]] $A \setminus B$ is non-[[Definition:Empty Set|empty]].
+By the definition of a [[Definition:Strict Well-Ordering|strict well-ordering]], there exists a [[Definition:Minimal Element|minimal element]] $x$ of $A \setminus B$.
+As $x \in A$, it follows from [[Ordinal is Transitive]] that $x \subseteq A$.
+Next, since the [[Definition:Strict Well-Ordering|strict well-ordering]] on $A$ is given by the [[Definition:Epsilon Restriction|epsilon restriction]] $\Epsilon \! \restriction_A$, it follows by the definition of a [[Definition:Minimal Element|minimal element]] that:
+:$\forall y \in A \setminus B: y \notin x$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \O
+      | r = \paren {A \setminus B} \cap x
+}}
+{{eqn | r = \paren {A \cap x} \setminus B
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = x \setminus B
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \subseteq
+      | r = B
+      | c = by [[Set Difference with Superset is Empty Set]]
+}}
+{{end-eqn}}
+Suppose that $z \in B$.
+Since $B \subset A$, it follows that $z \in A$.
+Recall that the [[Definition:Strict Well-Ordering|strict well-ordering]] on $A$ is given by the [[Definition:Epsilon Restriction|epsilon restriction]] $\Epsilon \! \restriction_A$.
+From the [[Trichotomy Law (Ordering)]], it follows that $z \in x$ or $x = z$ or $x \in z$.
+If $x = z$ or $x \in z$, then it follows by the [[Definition:Transitive Set|transitivity]] of $B$ that $x \in B$.
+This contradicts the definition of $x$.
+Hence, $z \in x$.
+That is, $B \subseteq x$.
+We have shown that $x \subseteq B$ [[Definition:Conjunction|and]] $B \subseteq x$.
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$B = x \in A$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Arcsine Function}
+Tags: Integral Calculus
+
+\begin{theorem}
+:$\displaystyle \int \arcsin x \rd x = x \arcsin x + \sqrt {1 - x^2} + C$
+for $x \in \closedint {-1} 1$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \arcsin x \rd x
+      | r = \int 1 \cdot \arcsin x \rd x
+}}
+{{eqn | r = x \arcsin x - \int x \paren {\frac \rd {\rd x} \arcsin x} \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = x \arcsin x - \int \frac x {\sqrt {1 - x^2} } \rd x
+      | c = [[Derivative of Arcsine Function]]
+}}
+{{end-eqn}}
+[[Integration by Substitution|Substitute]]:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = 1 - x^2
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = -2 x
+      | c = differentiating {{WRT|Differentiation}} $x$
+}}
+{{eqn | ll = \leadsto
+      | l = -\frac 1 2 \frac {\d u} {\d x}
+      | r = x
+}}
+{{eqn | l = \int \arcsin x \rd x 
+      | r = x \arcsin x - \int \frac {- \frac 1 2} {\sqrt u} \frac {\d u} {\d x} \rd x
+}}
+{{eqn | l = \int \arcsin x \rd x 
+      | r = x \arcsin x + \frac 1 2 \int u^{-1/2} \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = x \arcsin x + u^{1/2} + C
+      | c = [[Integral of Power]]
+}}
+{{eqn | r = x \arcsin x + \sqrt {1 - x^2} + C
+      | c = from $u = 1 - x^2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Derivative of Convex Real Function is Non-Negative}
+Tags: Differential Calculus, Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice differentiable]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Then $f$ is [[Definition:Convex Real Function|convex]] on $\left({a \,.\,.\, b}\right)$ {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' \ge 0$ on $\left({a \,.\,.\, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Function is Convex iff Derivative is Increasing]], $f$ is [[Definition:Convex Real Function|convex]] {{iff}} $f'$ is [[Definition:Increasing Real Function|increasing]].
+From [[Derivative of Monotone Function]], $f'$ is [[Definition:Increasing Real Function|increasing]] {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' \ge 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordinal Membership is Trichotomy}
+Tags: Ordinals
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Ordinal|ordinals]].
+Then:
+:$\left({A = B}\right) \lor \left({A \in B}\right) \lor \left({B \in A}\right)$
+where $\lor$ denotes [[Definition:Disjunction|logical or]].
+\end{theorem}
+
+\begin{proof}
+By [[Relation between Two Ordinals]], it follows that:
+:$\left({A = B}\right) \lor \left({A \subset B}\right) \lor \left({B \subset A}\right)$
+By [[Transitive Set is Proper Subset of Ordinal iff Element of Ordinal]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Derivative of Concave Real Function is Non-Positive}
+Tags: Differential Calculus, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then $f$ is [[Definition:Concave Real Function|concave]] on $\openint a b$ {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' \le 0$ on $\openint a b$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Function is Concave iff Derivative is Decreasing]], $f$ is [[Definition:Concave Real Function|concave]] {{iff}} $f'$ is [[Definition:Decreasing Real Function|decreasing]].
+From [[Derivative of Monotone Function]], $f'$ is [[Definition:Decreasing Real Function|decreasing]] {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' \le 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordinal is Member of Ordinal Class}
+Tags: Ordinal Arithmetic
+
+\begin{theorem}
+Let $A$ be an [[Definition:Ordinal|ordinal]].
+Then:
+:$A \in \On \lor A = \On$
+where $\On$ denote the [[Definition:Ordinal Class|class of ordinals]] .
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]] $A$ is an [[Definition:Ordinal|ordinal]]
+From [[Ordinal Class is Ordinal]] and [[Ordinal Membership is Trichotomy]]:
+:$A \in \On \lor A = \On \lor \On \in A$
+But by the [[Burali-Forti Paradox]] $\On$ is a [[Definition:Proper Class|proper class]].
+Therefore:
+:$A \in \On \lor A = \On$
+{{qed}}
+{{explain|please say or link to why being a proper class means that $\On \notin A$}}
+\end{proof}<|endoftext|>
+\section{Ordinal is Subset of Ordinal Class}
+Tags: Ordinal Class
+
+\begin{theorem}
+Suppose $A$ is an [[Definition:Ordinal|ordinal]].
+Then:
+:$A \subseteq \On$
+where $\On$ represents the [[Definition:Ordinal Class|class of all ordinals]].
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+By [[Ordinal is Member of Ordinal Class]]:
+:$A \in \On \lor A = \On$.
+In either case:
+:$A \subseteq \On$
+since $\On$ is [[Definition:Transitive Class|transitive]].
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+{{:De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection}}
+\end{theorem}<|endoftext|>
+\section{Classical Probability is Probability Measure}
+Tags: Probability Theory
+
+\begin{theorem}
+The [[Definition:Classical Probability Model|classical probability model]] is a [[Definition:Probability Measure|probability measure]].
+\end{theorem}
+
+\begin{proof}
+We check all the [[Axiom:Kolmogorov Axioms|Kolmogorov axioms]] in turn:
+=== First Axiom ===
+From [[Empty Set is Subset of All Sets]] and from the definitions of the [[Definition:Event Space|event space]] and [[Definition:Sample Space|sample space]]:
+:$\varnothing \subseteq \Sigma \subseteq \Omega $
+From [[Cardinality of Empty Set]] and [[Cardinality of Subset of Finite Set]]:
+:$0 \le \card \Sigma \le \card \Omega$
+Dividing all terms by $\card \Omega$:
+:$0 \le \dfrac {\card \Sigma} {\card \Omega} \le 1$
+The middle term is the [[Definition:Probability|asserted definition of $\map \Pr {\, \cdot \,}$]].
+{{explain|Link to a page expanding on that statement -- perhaps better to say that it's the definition of the image of Pr?}}
+{{qed|lemma}}
+=== Second Axiom ===
+[[Definition:By Hypothesis|By hypothesis]]:
+:$\map \Pr \Omega = \dfrac {\card \Omega} {\card \Omega} = 1$
+{{qed|lemma}}
+=== Third Axiom ===
+Follows from [[Cardinality is Additive Function]] and the [[Inclusion-Exclusion Principle/Corollary|corollary to the Inclusion-Exclusion Principle]].
+{{qed}}
+{{proofread}}
+[[Category:Probability Theory]]
+o3l502qow7dfw5nt0ovn2rjryjdeq5x
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/General Case}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $\mathcal P \left({T}\right)$ be the [[Definition:Power Set|power set]] of $T$.
+Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.
+Then:
+==== [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|Difference with Intersection]] ====
+{{:De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection}}
+==== [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|Difference with Union]] ====
+{{:De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union}}
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+{{:De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection}}
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+{{:De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection}}
+\end{theorem}<|endoftext|>
+\section{Image of Set Difference under Mapping/Corollary 1}
+Tags: Image of Set Difference under Mapping
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $S_1 \subseteq S_2 \subseteq S$.
+Then:
+:$\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
+where $\complement$ (in this context) denotes [[Definition:Relative Complement|relative complement]].
+\end{theorem}
+
+\begin{proof}
+From [[Image of Set Difference under Relation/Corollary 1|Image of Set Difference under Relation: Corollary 1]] we have:
+:$\relcomp {\mathcal R \sqbrk {S_2} } {\mathcal R \sqbrk {S_1} } \subseteq \mathcal R \sqbrk {\relcomp {S_2} {S_1} }$
+where $\mathcal R \subseteq S \times T$ is a [[Definition:Relation|relation]] on $S \times T$.
+As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], it follows directly that:
+:$\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
+{{qed}}
+[[Category:Image of Set Difference under Mapping]]
+6vparbmf8toamnwm6qzwws9aegj6j36
+\end{proof}<|endoftext|>
+\section{Image of Set Difference under Mapping/Corollary 2}
+Tags: Image of Set Difference under Mapping
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
+where:
+:$\Img f$ denotes the [[Definition:Image of Mapping|image]] of $f$
+:$\complement_{\Img f}$ denotes the [[Definition:Relative Complement|complement relative to $\Img f$]].
+This can be expressed in the language and notation of [[Definition:Direct Image Mapping|direct image mappings]] as:
+:$\forall X \in \powerset S: \relcomp {\Img f} {\map {f^\to} X} \subseteq \map {f^\to} {\relcomp S X}$
+That is:
+:$\forall X \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } X \subseteq \map {\paren {f^\to \circ \complement_S} } X$
+where $\circ$ denotes [[Definition:Composition of Mappings|composition of mappings]].
+\end{theorem}
+
+\begin{proof}
+From [[Image of Set Difference under Relation/Corollary 2|Image of Set Difference under Relation: Corollary 2]] we have:
+:$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk X} \subseteq \mathcal R \sqbrk {\relcomp S X}$
+where $\mathcal R \subseteq S \times T$ is a [[Definition:Relation|relation]] on $S \times T$.
+As $f$, being a [[Definition:Mapping|mapping]], is also a [[Definition:Relation|relation]], it follows directly that:
+:$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Set Difference under Mapping/Corollary 1}
+Tags: Preimages under Mappings, Set Difference, Set Complement
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $T_1 \subseteq T_2 \subseteq T$.
+Then:
+:$\complement_{f^{-1} \left[{T_2}\right]} \left({f^{-1} \left[{T_1}\right]}\right) = f^{-1} \left[{\complement_{T_2} \left({T_1}\right)}\right]$
+where:
+: $\complement$ (in this context) denotes [[Definition:Relative Complement|relative complement]]
+: $f^{-1} \left[{T_1}\right]$ denotes [[Definition:Preimage of Subset under Mapping|preimage]].
+\end{theorem}
+
+\begin{proof}
+From [[One-to-Many Image of Set Difference#Corollary 1|One-to-Many Image of Set Difference: Corollary 1]] we have:
+:$\complement_{\mathcal R \left[{T_2}\right]} \left({\mathcal R \left[{T_1}\right]}\right) = \mathcal R \left[{\complement_{T_2} \left({T_1}\right)}\right]$
+where $\mathcal R \subseteq T \times S$ is a [[Definition:One-to-Many Relation|one-to-many relation]] on $T \times S$.
+Hence as [[Inverse of Mapping is One-to-Many Relation|$f^{-1}: T \to S$ is a one-to-many relation]]:
+:$\complement_{f^{-1} \left[{T_2}\right]} \left({f^{-1} \left[{T_1}\right]}\right) = f^{-1} \left[{\complement_{T_2} \left({T_1}\right)}\right]$
+{{qed}}
+[[Category:Preimages under Mappings]]
+[[Category:Set Difference]]
+[[Category:Set Complement]]
+kzmeqg8okrg6o116vd5hzaexl3zv0cn
+\end{proof}<|endoftext|>
+\section{Complement of Preimage equals Preimage of Complement}
+Tags: Mapping Theory, Relative Complement
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $T_1$ be a [[Definition:Subset|subset]] of $T$.
+Then:
+:$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
+where:
+:$\complement_S$ (in this context) denotes [[Definition:Relative Complement|relative complement]]
+:$f^{-1} \sqbrk {T_1}$ denotes [[Definition:Preimage of Subset under Mapping|preimage]].
+\end{theorem}
+
+\begin{proof}
+From [[One-to-Many Image of Set Difference/Corollary 2|One-to-Many Image of Set Difference: Corollary 2]] we have:
+:$\relcomp {\Img \RR} {\RR \sqbrk {S_1} } = \RR \sqbrk {\relcomp S {S_1} }$
+where:
+:$S_1 \subseteq S$
+:$\RR \subseteq T \times S$ is a [[Definition:One-to-Many Relation|one-to-many relation]] on $T \times S$.
+Hence as [[Inverse of Mapping is One-to-Many Relation|$f^{-1}: T \to S$ is a one-to-many relation]]:
+:$\relcomp {\Preimg f} {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
+But from [[Preimage of Mapping equals Domain]], we have that:
+:$\Preimg f = S$
+Hence:
+:$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{One-to-Many Image of Set Difference/Corollary 1}
+Tags: Set Difference, Relation Theory
+
+\begin{theorem}
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] which is [[Definition:One-to-Many Relation|one-to-many]].
+Let $A \subseteq B \subseteq S$.
+Then:
+:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$
+where $\complement$ (in this context) denotes [[Definition:Relative Complement|relative complement]].
+\end{theorem}
+
+\begin{proof}
+We have that $A \subseteq B$.
+Then by definition of [[Definition:Relative Complement|relative complement]]:
+:$\relcomp B A = B \setminus A$
+:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk B \setminus \RR \sqbrk A$
+Hence, when $A \subseteq B$:
+:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$
+means exactly the same thing as:
+:$\RR \sqbrk B \setminus \RR \sqbrk A = \RR \sqbrk {B \setminus A}$
+Hence the result from [[One-to-Many Image of Set Difference]].
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Relation Theory]]
+kidx68wz3t76abdkb6vnfv5eq6tkopc
+\end{proof}<|endoftext|>
+\section{One-to-Many Image of Set Difference/Corollary 2}
+Tags: Set Difference, Relation Theory
+
+\begin{theorem}
+Let $\mathcal R \subseteq S \times T$ be a [[Definition:Relation|relation]] which is [[Definition:One-to-Many Relation|one-to-many]].
+Let $A$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} = \mathcal R \sqbrk {\relcomp S A}$
+where $\complement$ (in this context) denotes [[Definition:Relative Complement|relative complement]].
+In the language of [[Definition:Direct Image Mapping|direct image mappings]] this can be presented as:
+:$\forall A \in \powerset S: \map {\paren {\complement_{\Img {\mathcal R} } \circ \mathcal R^\to} } A = \map {\paren {\mathcal R^\to \circ \complement_S} } A$
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Image of Relation|image of $\mathcal R$]]:
+:$\Img {\mathcal R} = \mathcal R \sqbrk S$
+So, when $B = S$ in [[One-to-Many Image of Set Difference/Corollary 1|One-to-Many Image of Set Difference: Corollary 1]]:
+:$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} = \relcomp {\mathcal R \sqbrk S} {\mathcal R \sqbrk A}$
+Hence:
+:$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} = \mathcal R \sqbrk {\relcomp S A}$
+means exactly the same thing as:
+:$\relcomp {\mathcal R \sqbrk S} {\mathcal R \sqbrk A} = \mathcal R \sqbrk {\relcomp S A}$
+that is:
+:$\mathcal R \sqbrk S \setminus \mathcal R \sqbrk A = \mathcal R \sqbrk {S \setminus A}$
+Hence the result from [[One-to-Many Image of Set Difference]].
+{{Qed}}
+[[Category:Set Difference]]
+[[Category:Relation Theory]]
+9adga48l5jrmaq0u5x1l3ojd7d37hzr
+\end{proof}<|endoftext|>
+\section{Integer Power Function is Bijective iff Index is Odd}
+Tags: Real Functions, Powers, Bijections
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Let $f_n: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map {f_n} x = x^n$
+Then $f_n$ is a [[Definition:Bijection|bijection]] {{iff}} $n$ is [[Definition:Odd Integer|odd]].
+\end{theorem}
+
+\begin{proof}
+=== Even Index ===
+Suppose $n$ is [[Definition:Even Integer|even]].
+Let $x \ne 0$.
+Then $1^n = \paren {-1}^n$ by [[Power of Ring Negative]], so $f_n$ is not [[Definition:Injection|injective]].
+Also, by [[Even Power is Non-Negative]], $f_n$ is not [[Definition:Surjective|surjective]].
+By definition, a [[Definition:Bijection|bijection]] is both [[Definition:Injection|injective]] and [[Definition:Surjective|surjective]].
+It follows that for [[Definition:Even Integer|even]] $n$, $f_n$ is not [[Definition:Bijection|bijective]].
+{{qed|lemma}}
+=== Odd Index ===
+Now suppose $n$ is [[Definition:Odd Integer|odd]].
+From [[Odd Power Function is Strictly Increasing]], $f_n$ is [[Definition:Injection|injective]].
+From [[Odd Power Function is Surjective]], $f^n$ is [[Definition:Surjective|surjective]].
+So when $n$ is [[Definition:Odd Integer|odd]], $f_n$ is both [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]], and so by definition [[Definition:Bijection|bijective]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Infinite Sequences is Uncountable}
+Tags: Countable Sets, Diagonal Arguments
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] which contains more than one [[Definition:Element|element]].
+Let $S^\infty$ denote the set of all [[Definition:Sequence|sequences]] of elements of $S$.
+Then $S^\infty$ is [[Definition:Uncountable Set|uncountable]].
+\end{theorem}
+
+\begin{proof}
+As $S$ has more than one element, it must have at least two.
+So, let $a, b \in S$ be those two elements.
+Let $Z$ be the set of all sequences from $\set {a, b}$.
+Suppose $S^\infty$ were [[Definition:Countable Set|countable]].
+From [[Subset of Countably Infinite Set is Countable]], $Z$ is likewise [[Definition:Countable Set|countable]].
+So by definition, it would be possible to set up a [[Definition:Bijection|bijection]] $\phi: Z \leftrightarrow \N$ between $Z$ and the set $\N$ of [[Definition:Natural Number|natural numbers]]:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \leftrightarrow
+      | r = \sequence {z_0} = \tuple {z_{00}, z_{01}, z_{02}, \ldots, z_{0n}, \ldots}
+}}
+{{eqn | l = 1
+      | o = \leftrightarrow
+      | r = \sequence {z_1} = \tuple {z_{10}, z_{11}, z_{12}, \ldots, z_{1n}, \ldots}
+}}
+{{eqn | l = 
+      | o = \vdots
+      | r = 
+}}
+{{eqn | l = n
+      | o = \leftrightarrow
+      | r = \sequence {z_n} = \tuple {z_{n0}, z_{n1}, z_{n2}, \ldots, z_{nn}, \ldots}
+}}
+{{eqn | l = 
+      | o = \vdots
+      | r =  
+}}
+{{end-eqn}}
+where each of $z_{rs}$ is:
+: an element of $\set {a, b}$
+: the $s$'th element of $\sequence {z_r}$, the particular sequence which is in correspondence with $r \in \N$.
+Now we create the sequence $Q = \sequence {q_{dd} }$ where:
+:$q_{dd} = \begin{cases} a & : z_{dd} = b \\ b & : z_{dd} = a \end{cases}$
+Thus:
+:$\forall n \in \N: q_{nn} \notin \sequence {z_n}$
+and so:
+:$\forall n \in \N: Q \ne \sequence {z_n}$
+So $Q \notin Z$.
+That is, we have constructed a sequence which has not been placed into correspondence with an element of $\N$.
+Thus by definition $Z$ is [[Definition:Uncountable Set|uncountable]].
+Hence, by the [[Rule of Transposition]], $S^\infty$ is likewise [[Definition:Uncountable Set|uncountable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Topologies is not necessarily Topology}
+Tags: Topology
+
+\begin{theorem}
+Let $\left({\tau_i}\right)_{i \in I}$ be an arbitrary [[Definition:Indexing Set|indexed set]] of [[Definition:Topology|topologies]] for a [[Definition:Set|set]] $S$.
+Then $\tau := \displaystyle \bigcup_{i \mathop \in I} {\tau_i}$ is not necessarily also a [[Definition:Topology|topology]] for $S$.
+\end{theorem}
+
+\begin{proof}
+Let $S := \left\{{0, 1, 2}\right\}$ be a [[Definition:Set|set]].
+Let:
+: $\tau_1 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{0, 1}\right\}, S}\right\}$ 
+: $\tau_2 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{2}\right\}, \left\{{0, 2}\right\}, S}\right\}$
+be topologies on $S$. 
+Then:
+: $\tau := \tau_1 \cup \tau_2 = \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{2}\right\}, \left\{{0, 1}\right\}, \left\{{0, 2}\right\}, S}\right\}$
+For $\tau$ to be a [[Definition:Topology|topology]] the [[Definition:Set Union|union]] of any number of elements of $\tau$ should also be in $\tau$.
+But:
+:$\left\{{1}\right\} \cup \left\{{2}\right\} = \left\{{1, 2}\right\} \not \in \tau$ 
+Therefore $\tau$ is not a topology on $S$.
+Hence the result. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Niven's Theorem}
+Tags: Trigonometry
+
+\begin{theorem}
+Consider the [[Definition:Angle|angles]] $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.
+The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are [[Definition:Rational Number|rational]] are:
+:$\theta = 0: \sin \theta = 0$
+:$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
+:$\theta = \dfrac \pi 2: \sin \theta = 1$
+\end{theorem}
+
+\begin{proof}
+We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are [[Definition:Rational Number|rational]] then:
+:$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$
+=== [[Niven's Theorem/Lemma|Lemma]] ===
+{{:Niven's Theorem/Lemma}}
+{{refactor|Extract this lemma into its own page.}}
+For any [[Definition:Integer|integer]] $n \ge 1$, there exists a [[Definition:Polynomial over Real Numbers|polynomial]] $F_n \left({x}\right)$ such that:
+:$F_n \left({2 \cos t}\right) = 2 \cos n t$
+In addition:
+:$\deg F_n = n$
+and $F_n$ is a [[Definition:Monic Polynomial|monic polynomial]] with [[Definition:Integer|integer]] [[Definition:Polynomial Coefficient|coefficients]].
+=== Proof of Lemma ===
+The proof proceeds by [[Proof by Mathematical Induction|induction]].
+For $n = 1$, it is seen that:
+:$F_1 \left({x}\right) = x$
+fulfils the propositions.
+For $n = 2$:
+:$F_2 \left({x}\right) = x^2 - 2$
+For $n > 2$:
+{{begin-eqn}}
+{{eqn | l = 2 \cos \left({n - 1}\right) t \cos t
+      | r = \cos n t + \cos \left({n - 2}\right) t
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = 2 \cos n t
+      | r = \left({2 \cos \left({n - 1}\right) t}\right) \left({2 \cos t}\right) - 2 \cos \left({n - 2}\right) t
+      | c = 
+}}
+{{eqn | r = 2 \cos t F_{n - 1} \left({2 \cos t}\right) - F_{n - 2} \left({2 \cos t}\right)
+      | c = 
+}}
+{{end-eqn}}
+so:
+: $F_n \left({x}\right) = x F_{n - 1} \left({x}\right) - F_{n - 2} \left({x}\right) \in \Z \left[{x}\right]$
+will fulfil:
+: $F_n \left({2 \cos t}\right) = 2 \cos n t$
+Because $\deg F_{n - 1} = n - 1$ and $\deg F_{n - 2} = n - 2$, we can conclude that:
+:$\deg F_n = \deg \left({x F_{n - 1} \left({x}\right) - F_{n - 2} \left({x}\right)}\right) = n$
+In addition, the [[Definition:Leading Coefficient of Polynomial|leading coefficient]] of $F_n$ is equal to the [[Definition:Leading Coefficient of Polynomial|leading coefficient]] of $F_{n - 1}$, which is $1$.
+Hence the lemma.
+{{qed|lemma}}
+Suppose that $\dfrac \theta \pi$ is [[Definition:Rational Number|rational]], meaning:
+:$\theta = \dfrac {2 \pi k} n$
+where $k, n \in \Z$ and $n \ge 1$.
+Suppose also that $\cos \theta \in \Q$.
+Denoting $c = 2 \cos \theta \in \Q$, we get:
+:$F_n \left({c}\right) = F_n \left({2 \cos \dfrac {2 \pi k} n}\right) = 2 \cos \left({2 \pi k}\right) = 2$
+So $c$ is a [[Definition:Rational Number|rational]] [[Definition:Root of Polynomial|root]] of $F_n \left({x}\right) - 2$, which is a [[Definition:Monic Polynomial|monic polynomial]] with [[Definition:Integer|integer]] [[Definition:Polynomial Coefficient|coefficients]].
+By [[Rational Root Theorem]], $c$ must be an [[Definition:Integer|integer]].
+But:
+:$\left\vert{c}\right\vert = \left\vert{2 \cos \theta}\right\vert \le 2$
+so:
+:$c \in \left\{ {-2, -1, 0, 1, 2}\right\}$
+Assuming that $0 \le \theta \le \dfrac \pi 2$, we get that:
+:$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$
+Thus for any $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$ such that both $\dfrac \theta \pi$ and $\cos \theta$ are [[Definition:Rational Number|rational]], then:
+:$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$
+Instead of the above, suppose that:
+:$0 \le \alpha \le \dfrac \pi 2$
+and both of $\dfrac \alpha \pi$ and $\sin \alpha$ are [[Definition:Rational Number|rational]].
+Then we can denote $\theta = \dfrac \pi 2 - \alpha$ and get that:
+:$0 \le \theta \le \dfrac \pi 2$
+:$\dfrac \theta \pi \in Q$
+:$\cos \theta \in \Q$
+So:
+:$\dfrac \pi 2 - \alpha = \theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$
+therefore:
+:$\alpha \in \left\{ {0, \dfrac \pi 6, \dfrac \pi 2}\right\}$
+{{qed}}
+{{Namedfor|Ivan Morton Niven|cat = Niven}}
+It is suspected that this result is considerably older, and may date back as far as {{AuthorRef|Charles Hermite}}.
+[[Category:Trigonometry]]
+m1u0lyspoe8aleud92enyws7wdojzv7
+\end{proof}<|endoftext|>
+\section{Relative Frequency is Probability Measure}
+Tags: Probability Theory, Proofs by Induction, Statistics
+
+\begin{theorem}
+The [[Definition:Relative Frequency Model|relative frequency model]] is a [[Definition:Probability Measure|probability measure]].
+\end{theorem}
+
+\begin{proof}
+We check all the [[Axiom:Kolmogorov Axioms|Kolmogorov axioms]] in turn:
+=== First Axiom ===
+Let $n$ be the number of times a certain [[Definition:Event|event]] $\omega$ is observed to happen.
+Let $n'$ be the number of times $\omega$ is observed not to happen.
+By [[Law of Excluded Middle]] and [[Principle of Non-Contradiction]], $\omega$ either happened or did not, and not both.
+Therefore $n + n'$ is the total number of observations.
+It is supposed that at least one observation is actually made, so that $n + n' \ne 0$.
+The [[Definition:Relative Frequency Model|relative frequency model]] says that the probability of $\omega$ occurring can be defined as:
+:$\map \Pr \omega = \dfrac n {n + n'}$
+the [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] of which are [[Definition:Positive Integer|positive integers]] such that $n + n' \ge n$.
+If $\omega$ is observed never to happen, then $n = 0$ and $\map \Pr \omega = \dfrac 0 {0 + n'} = 0$. 
+If $\omega$ is observed to always happen, then $n' = 0$ and $\map \Pr \omega = \dfrac n {n + 0} = 1$.
+Otherwise, if $n, n' \ne 0$, from [[Mediant is Between]]:
+:$0 = \dfrac 0 {0 + n'} <  \dfrac n {n + n'} < \dfrac n {n + 0} = 1$
+Thus $\Pr$ is [[Definition:Bounded Mapping|bounded]]:
+:$0 \le \map \Pr {\cdot} \le 1$
+{{qed|lemma}}
+=== Second Axiom ===
+[[Definition:By Hypothesis|By hypothesis]]:
+:$\map \Pr \Omega = \dfrac {n + n'} {n + n'} = 1$
+{{qed|lemma}}
+=== Third Axiom ===
+This is a proof by [[Principle of Mathematical Induction|induction]].
+==== Basis for the Induction ====
+The case $j = 2$ is verified as follows:
+Let $A$ and $B$ be two [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Event|events]].
+Let $p$ and $q$ be the number of times $A$ and $B$ have been observed, respectively.
+Let $n$ be the total number of trials observed. 
+By the definition of [[Definition:Pairwise Disjoint|pairwise disjoint]], $A$ and $B$ never happened at the same time.
+By the same reasoning as the proof for the [[Axiom:Kolmogorov Axioms#First Axiom|first axiom]], in all $n$ observations:
+:$A$ happened $p$ times
+:$B$ happened $q$ times
+:$A \lor B$ happened $p + q$ times.
+{{MissingLinks|There's a result that can be linked to so as to replace all these words. In fact, now I think of it, there's a result which allows us to replace this entire inductive argument, we just need to find it.}}
+By hypothesis:
+{{begin-eqn}}
+{{eqn | l = \map \Pr {A \cup B}
+      | r = \dfrac {p + q} n
+}}
+{{eqn | l =
+      | r = \dfrac p n + \dfrac q n
+}}
+{{eqn | l = 
+      | r = \map \Pr A + \map \Pr B
+}}
+{{end-eqn}}
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+==== Induction Hypothesis ====
+Let $A_1, A_2, \ldots, A_j$ be $j$ [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Event|events]].
+By the definition of the [[Definition:Relative Frequency Model|relative frequency model]], $j$ is [[Definition:Finite|finite]].
+Assume:
+:$\displaystyle \map \Pr {\bigcup_{i \mathop = 1}^j A_i} = \map \Pr {A_1} + \map \Pr {A_2} + \cdots + \map \Pr {A_j}$
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+==== Induction Step ====
+This is our [[Definition:Induction Step|induction step]]:
+Let $A_1, A_2, \ldots, A_j, A_{j + 1}$ be $j+1$ [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Event|events]].
+Define $C = A_1 \lor A_2 \lor A_3 \lor \cdots \lor A_j$.
+Then $C$ and $A_{j+1}$ are also pairwise disjoint.
+By the [[Principle of Mathematical Induction#Basis For the Induction|base case]]: 
+:$\map \Pr {C \cup A_{j + 1} } = \map \Pr C + \map \Pr {A_{j + 1} }$
+By the definition of $C$, this equation is [[Definition:Logical Equivalence|logically equivalent]] to:
+:$\displaystyle \map \Pr {\bigcup_{i \mathop = 1}^{j + 1} A_i} = \sum_{i \mathop = 1}^{j + 1} \map \Pr {A_i}$
+By the definition of the [[Definition:Relative Frequency Model|relative frequency model]], $j + 1$ is [[Definition:Finite|finite]].
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+{{LEM}}
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line is Topology}
+Tags: Sorgenfrey Line
+
+\begin{theorem}
+The [[Definition:Sorgenfrey Line|Sorgenfrey Line]] is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+We have to check that $\BB = \set {\hointr a b: a, b \in \R}$ fulfills the axioms of being a [[Definition:Basis (Topology)|basis]] for a [[Definition:Topology|topology]].
+From [[Equivalent Definitions of Synthetic Basis]] we only have to check that:
+:$(1): \quad \bigcup \BB = \R$
+:$(2): \quad \forall B_1, B_2 \in \BB: \exists V \in \BB: V \subseteq B_1 \cap B_2$
+We have that:
+:$\forall n \in \Z: \hointr n {n + 1} \in \BB$
+:$\R = \displaystyle \bigcup_{n \mathop \in \Z} \hointr n {n + 1} \subseteq \bigcup \BB$
+Hence $\R = \bigcup \BB$ and condition $(1)$ is fulfilled.
+Now take $ B_1, B_2 \in \BB$ where:
+:$B_1 = \hointr {a_1} {b_1}$
+:$B_2 = \hointr {a_2} {b_2}$
+Let $B_3$ be constructed as:
+:$B_3 := \hointr {\max \set {a_1, a_2} } {\min \set {b_1, b_2} } \in \BB$
+From the method of construction, it is clear that $B_3 = B_1 \cap B_2$.
+Thus taking $V = B_3$, condition $(2)$ is fulfilled.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line is Hausdorff}
+Tags: Sorgenfrey Line, Hausdorff Spaces
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+Then $T$ is [[Definition:Hausdorff Space|Hausdorff]].
+\end{theorem}
+
+\begin{proof}
+Take $x, y \in \R$ such that $x \ne y$.
+{{WLOG}}, assume that $x < y$.
+From [[Real Numbers are Close Packed]]:
+:$\exists t \in \R: x < t < y$
+Then:
+:$\hointr x t \cap \hointr t {y + 1} = \O$
+We also have that $x \in \hointr x t$ by definition of [[Definition:Half-Open Real Interval|half-open interval]].
+Also, as $t < y < y+1$ it is clear that $y \in \hointr t {y + 1}$.
+By definition of the [[Definition:Sorgenfrey Line|Sorgenfrey line]], both are [[Definition:Open Set (Topology)|open]] in $T$.
+Thus we have found two [[Definition:Disjoint Sets|disjoint subsets]] of $\R$ which are [[Definition:Open Set (Topology)|open]] in $T$, such that one contains $x$ and the other contains $y$.
+Hence the [[Definition:Sorgenfrey Line|Sorgenfrey Line]] is [[Definition:Hausdorff Space|Hausdorff]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Sequence in Set of Integers}
+Tags: Sequences, Convergence
+
+\begin{theorem}
+Let $\left \langle {x_n}\right \rangle_{n \in \N}$ be a [[Definition:Sequence|sequence]] in the set $\Z$ of [[Definition:Integer|integers]] considered as a [[Definition:Topological Subspace|subspace]] of the [[Definition:Real Number Line|real number line]] $\R$ under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]].
+Then $\left \langle {x_n}\right \rangle_{n \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] in $\R$ to a [[Definition:Limit Point of Sequence|limit]] {{iff}}:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
+That is, {{iff}} the sequence reaches some value of $\Z$ and "stays there".
+\end{theorem}
+
+\begin{proof}
+Suppose $\left \langle {x_n}\right \rangle_{n \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] to a [[Definition:Limit Point of Sequence|limit]] $l$.
+Consider the [[Definition:Open Set (Topology)|open set]] in $\R$:
+:$U := \left({l - \dfrac 1 2 \,.\,.\, l + \dfrac 1 2}\right)$
+Then $\forall x \in \Z: x \in U \implies x = l$
+It follows by definition of [[Definition:Convergent Sequence (Topology)|convergence]] that:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
+where $x_k = l$.
+Now suppose that:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
+Then trivially $\left \langle {x_n}\right \rangle_{n \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] to the [[Definition:Limit Point of Sequence|limit]] $x_k$.
+{{qed}}
+[[Category:Sequences]]
+[[Category:Convergence]]
+3xlxzoi9mutdaomqia6slko422gsgqr
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line is Perfectly Normal}
+Tags: Sorgenfrey Line, Perfectly Normal Spaces
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+Then $T$ is [[Definition:Perfectly Normal Space|perfectly normal]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Perfectly Normal Space|perfectly normal]] space, it is necessary to prove that $T$ is a [[Definition:T1 Space|$T_1$ space]] and that any closed set is [[Definition:G-Delta Set|$G_\delta$]].
+From [[T2 Space is T1 Space|$T_2$ Space is $T_1$ Space]] and [[Sorgenfrey Line is Hausdorff]]:
+:the [[Definition:Sorgenfrey Line|Sorgenfrey line]] is a [[Definition:T1 Space|$T_1$ space]].
+From [[Complement of F-Sigma Set is G-Delta Set|Complement of $F_\sigma$ Set is $G_\delta$ Set]] it is sufficient to prove that an open set of $T$ is [[Definition:F-Sigma Set|$F_\sigma$]].
+Let $W$ be any open set in $T$.
+Let $O \subseteq W$ be the interior of $W$ with respect to the metric space:
+:$\R = \struct {\R, d}$
+where $d$ is the [[Definition:Usual Metric|usual metric]] on $\R$.
+From the [[Definition:Sorgenfrey Line|definition of $T$]], for each $x \in W \setminus O$, we can choose $h_x \in W$ such that $\hointr x {h_x} \subseteq W$.
+Suppose $\hointr x {h_x} \cap \hointr y {h_y} \ne \O$ for some [[Definition:Distinct|distinct]] points $x, y \in W \setminus O$.
+Then either $x < y < h_x$ or $y < x < h_y$.
+If $x < y < h_x$, then $y \in \openint x {h_x} \subseteq O$, contradicting $y \in W \setminus O$.
+Similarly, if $y < x < h_y$, then $x \in \openint y {h_y}) \subseteq O$, contradicting $x \in W \setminus O$.
+Thus $\family {\hointr x {h_x} }_{x \mathop \in W \setminus O}$ is a [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Indexed Family of Sets|indexed family]] of open sets of $T$.
+From [[Sorgenfrey Line is Separable]] and [[Separable Space satisfies Countable Chain Condition]]:
+:$\set {\hointr x {h_x} : x \in W \setminus O}$ is countable
+and thus:
+:$W \setminus O$ is countable.
+From [[Metric Space is Perfectly T4]]:
+:$O$ is an $F_\sigma$ set in $\R$.
+Thus from [[Sorgenfrey Line is Expansion of Real Line]]:
+:$O$ is an $F_\sigma$ set in the Sorgenfrey line.
+Since $W \setminus O$ is a [[Definition:Countable Union|countable union]] of [[Definition:Singleton|singletons]] and $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ space]]:
+:$W \setminus O$ is an $F_\sigma$ set in $T$.
+Since $W = O \cup \left({W \setminus O}\right)$ and [[F-Sigma Sets Closed under Union|$F_\sigma$ sets are closed under unions]]:
+:$W$ is an $F_\sigma$ set in $T$.
+{{qed}}
+[[Category:Sorgenfrey Line]]
+[[Category:Perfectly Normal Spaces]]
+sxfiiej1t8wvwxi5jz5xw9wqkc7doyz
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line is Expansion of Real Line}
+Tags: Sorgenfrey Line
+
+\begin{theorem}
+Let $\R = \struct {\R, d}$ be the [[Definition:Metric Space|metric space]] defined in [[Real Number Line is Metric Space]].
+Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+Then $T$ is an [[Definition:Expansion of Topology|expansion]] of $\R$ as a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+It is enough to prove that any [[Definition:Open Set (Topology)|open set]] in $\R$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Let $a, b \in \R$.
+Then:
+:$\displaystyle \openint a b = \bigcup_{\epsilon \mathop > 0} \hointr {a + \epsilon} b$
+Since $\hointr {a + \epsilon} b$ are [[Definition:Open Set (Topology)|open]] in $T$, $\openint a b$ is also [[Definition:Open Set (Topology)|open]] in $T$.
+{{qed}}
+[[Category:Sorgenfrey Line]]
+5lkzxxu4seeyk0z1uuijn6e1df02vcl
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line satisfies all Separation Axioms}
+Tags: Sorgenfrey Line
+
+\begin{theorem}
+Let $T = \left({\R, \tau}\right)$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+Then $T$ satisfies all [[Definition:Separation Axioms|separation axioms]].
+\end{theorem}
+
+\begin{proof}
+We have [[Sorgenfrey Line is Perfectly Normal]].
+The result follows from [[Sequence of Implications of Separation Axioms]].
+{{qed}}
+[[Category:Sorgenfrey Line]]
+bf26hxhpvaaf8ewwt7bpfv4w79sbih9
+\end{proof}<|endoftext|>
+\section{Argument of Product equals Sum of Arguments}
+Tags: Argument of Complex Number, Complex Multiplication
+
+\begin{theorem}
+Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
+Let $\arg$ be the [[Definition:Argument of Complex Number|argument operator]].
+Then:
+:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
+where $k$ can be $0$, $1$ or $-1$.
+\end{theorem}
+
+\begin{proof}
+Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$.
+Then the [[Definition:Polar Form of Complex Number|polar forms]] of $z_1, z_2$ are:
+:$z_1 = \cmod {z_1} \paren {\cos \theta_1 + i \sin \theta_1}$
+:$z_2 = \cmod {z_2} \paren {\cos \theta_2 + i \sin \theta_2}$
+By the definition of [[Definition:Complex Multiplication|complex multiplication]], factoring $\cmod {z_1} \cmod {z_2}$ from all terms, we have:
+:$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\paren {\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }$
+Using [[Sine of Sum]] and [[Cosine of Sum]], we have:
+:$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\map \cos {\theta_1 + \theta_2} + i \, \map \sin {\theta_1 + \theta_2} }$
+The theorem follows from the [[Definition:Argument of Complex Number|definition of $\map arg z$]], which says that $\map \arg {z_1 z_2}$ satisfies the equations:
+:$(1): \quad \dfrac {\cmod {z_1} \cmod {z_2} \map \cos {\theta_1 + \theta_2} } {\cmod {z_1} \cmod {z_2} } = \map \cos {\map \arg {z_1 z_2} }$
+:$(2): \quad \dfrac {\cmod {z_1} \cmod {z_2} \map \sin {\theta_1 + \theta_2} } {\cmod {z_1} \cmod {z_2} } = \map \sin {\map \arg {z_1 z_2} }$ 
+which in turn means that:
+:$\map \cos {\theta_1 + \theta_2} = \map \cos {\map \arg {z_1 z_2} }$
+:$\map \sin {\theta_1 + \theta_2} = \map \sin {\map \arg {z_1 z_2} }$
+There are $3$ possibilities for the size of $\theta_1 + \theta_2$:
+:$(1): \quad \theta_1 + \theta_2 > \pi$
+Then:
+:$-\pi < \theta_1 + \theta_2 - 2 \pi \le \pi$
+and we have:
+{{begin-eqn}}
+{{eqn | l = \map \cos {\theta_1 + \theta_2}
+      | r = \map \cos {\theta_1 + \theta_2 - 2 \pi}
+      | c = [[Cosine of Angle plus Full Angle]]
+}}
+{{eqn | l = \map \sin {\theta_1 + \theta_2}
+      | r = \map \sin {\theta_1 + \theta_2 - 2 \pi}
+      | c = [[Sine of Angle plus Full Angle]]
+}}
+{{end-eqn}}
+and so $\theta_1 + \theta_2 - 2 \pi$ is the [[Definition:Argument of Complex Number|argument]] of $z_1 z_2$ within its [[Definition:Principal Range|principal range]].
+:$(2): \quad \theta_1 + \theta_2 \le -\pi$
+Then:
+: $-\pi < \theta_1 + \theta_2 + 2 \pi \le \pi$
+and we have:
+{{begin-eqn}}
+{{eqn | l = \map \cos {\theta_1 + \theta_2}
+      | r = \map \cos {\theta_1 + \theta_2 + 2 \pi}
+      | c = [[Cosine of Angle plus Full Angle]]
+}}
+{{eqn | l = \map \sin {\theta_1 + \theta_2}
+      | r = \map \sin {\theta_1 + \theta_2 + 2 \pi}
+      | c = [[Sine of Angle plus Full Angle]]
+}}
+{{end-eqn}}
+and so $\theta_1 + \theta_2 + 2 \pi$ is within the [[Definition:Principal Range|principal range]] of $z_1 z_2$.
+:$(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$
+Then $\theta_1 + \theta_2$ is already within the [[Definition:Principal Range|principal range]] of $z_1 z_2$.
+Therefore:
+:$\map \arg {z_1 z_2} = \theta_1 + \theta_2 = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
+where $k$ can be $0$, $1$ or $-1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergence of Sequence in Discrete Space}
+Tags: Sequences, Discrete Topology, Convergence
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Discrete Space|discrete topological space]].
+Let $H = \left \langle{x_n}\right \rangle_{n \in \N}$ be a [[Definition:Sequence|sequence]] in $S$.
+Then $H$ [[Definition:Convergent Sequence (Topology)|converges]] in $T$ to a [[Definition:Limit Point of Sequence|limit]] [[Definition:Iff|iff]]:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
+That is, [[Definition:Iff|iff]] the sequence reaches some value of $S$ and "stays there".
+\end{theorem}
+
+\begin{proof}
+Suppose $H = \left \langle {x_n}\right \rangle_{n \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] to a [[Definition:Limit Point of Sequence|limit]] $L$.
+As $T$ is a [[Definition:Discrete Space|discrete space]], $\left\{{L}\right\}$ is an [[Definition:Open Set (Topology)|open set]] in $T$.
+Then trivially:
+:$\forall x \in H: x \in \left\{{L}\right\} \implies x = L$
+So for $H$ to be [[Definition:Convergent Sequence (Topology)|convergent]], it is necessary that:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = L$
+It follows by definition of [[Definition:Convergent Sequence (Topology)|convergence]] that:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
+where $l = L$.
+Now suppose that:
+:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
+Then trivially $\left \langle {x_n}\right \rangle_{n \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] to the [[Definition:Limit Point of Sequence|limit]] $L$.
+{{qed}}
+[[Category:Sequences]]
+[[Category:Discrete Topology]]
+[[Category:Convergence]]
+31dkxcqscqe5s3wd912otkzfhixnj47
+\end{proof}<|endoftext|>
+\section{Discrete Subspace of Fortissimo Space}
+Tags: Fortissimo Space, Discrete Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau_p}\right)$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Let $T' = \left({S \setminus \left\{{p}\right\}, \tau_p}\right)$ be the [[Definition:Topological Subspace|topological subspace]] induced on $T$ by the [[Definition:Subset|subset]] $S \setminus \left\{{p}\right\}$.
+Then $T'$ is a [[Definition:Discrete Space|discrete topological space]].
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Fortissimo Space|Fortissimo space]], any $A \subset S \setminus \left\{{p}\right\}$ is [[Definition:Open Set (Topology)|open in $S$]], because $p \notin A$.
+Thus by definition of [[Definition:Topological Subspace|topological subspace]], $A \subset S \setminus \left\{{p}\right\}$ is [[Definition:Open Set (Topology)|open in $S \setminus \left\{{p}\right\}$]].
+The result follows by the definition of [[Definition:Discrete Topology|discrete space]].
+{{qed}}
+[[Category:Fortissimo Space]]
+[[Category:Discrete Topology]]
+lvele0fmw6hgzfeibimfmewtfwl2i3m
+\end{proof}<|endoftext|>
+\section{Power Reduction Formulas}
+Tags: Trigonometric Identities, Hyperbolic Functions
+
+\begin{theorem}
+=== [[Power Reduction Formulas/Sine Squared|Square of Sine]] ===
+{{:Power Reduction Formulas/Sine Squared}}
+=== [[Power Reduction Formulas/Cosine Squared|Square of Cosine]] ===
+{{:Power Reduction Formulas/Cosine Squared}}
+=== [[Power Reduction Formulas/Tangent Squared|Square of Tangent]] ===
+{{:Power Reduction Formulas/Tangent Squared}}
+=== [[Power Reduction Formulas/Sine Cubed|Cube of Sine]] ===
+{{:Power Reduction Formulas/Sine Cubed}}
+=== [[Power Reduction Formulas/Cosine Cubed|Cube of Cosine]] ===
+{{:Power Reduction Formulas/Cosine Cubed}}
+=== [[Power Reduction Formulas/Sine to 4th|Fourth Power of Sine]] ===
+{{:Power Reduction Formulas/Sine to 4th}}
+=== [[Power Reduction Formulas/Cosine to 4th|Fourth Power of Cosine]] ===
+{{:Power Reduction Formulas/Cosine to 4th}}
+=== [[Power Reduction Formulas/Sine to 5th|Fifth Power of Sine]] ===
+{{:Power Reduction Formulas/Sine to 5th}}
+=== [[Power Reduction Formulas/Cosine to 5th|Fifth Power of Cosine]] ===
+{{:Power Reduction Formulas/Cosine to 5th}}
+\end{theorem}<|endoftext|>
+\section{Countable Space is Sigma-Compact}
+Tags: Sigma-Compact Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]], where $S$ is a [[Definition:Countable Set|countable set]].
+Then $T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Finite Space Satisfies All Compactness Properties]], for every $p \in S$, $\left\{{p}\right\}$ is [[Definition:Compact Topological Space|compact]].
+Then $\displaystyle S = \bigcup_{p \mathop \in S} \left\{{p}\right\}$ is a [[Definition:Countable Union|countable union]] of [[Definition:Compact Topological Space|compact sets]].
+Thus by definition $T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+{{qed}}
+[[Category:Sigma-Compact Spaces]]
+916oy2s4pz8ar082wzktb4juasfaeyu
+\end{proof}<|endoftext|>
+\section{Closed Sets of Fortissimo Space}
+Tags: Closed Sets, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $H \subseteq S$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}}:
+:$p \in H$
+or
+:$H$ is [[Definition:Countable Set|countable]]
+or both.
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Fortissimo Space|Fortissimo space]], $U \subseteq S$ is [[Definition:Open Set (Topology)|open]]  in $T$ {{iff}}:
+:$p \in \relcomp S U$
+or
+:$\relcomp S U$ is [[Definition:Countable Set|countable]]
+or both.
+The result follows from the definition of [[Definition:Closed Set (Topology)|closed set]].
+{{qed}}
+[[Category:Closed Sets]]
+[[Category:Fortissimo Space]]
+565de7e0p1d6wni18ezbm0minu19t5d
+\end{proof}<|endoftext|>
+\section{Union of Topologies on Singleton or Doubleton is Topology}
+Tags: Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] containing either exactly one or exactly two [[Definition:Element|elements]].
+Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexing Set|indexed set]] of [[Definition:Topology|topologies]] for a [[Definition:Set|set]] $S$.
+Then $\tau := \displaystyle \bigcup_{i \mathop \in I} {\tau_i}$ is also a [[Definition:Topology|topology]] for $S$.
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Set|set]] containing exactly one element.
+Suppose $S = \set x$ for a certain object $x$. 
+From [[Topology on Singleton is Indiscrete Topology]], the only possible topology on $S$ is the [[Definition:Indiscrete Topology|indiscrete topology]]. 
+From [[Union is Idempotent]], the [[Definition:Set Union|union]] of any number of [[Definition:Indiscrete Topology|indiscrete topologies]] on $S$ is the [[Definition:Indiscrete Topology|indiscrete topology]].
+Thus the union of any number of topologies on a set with exactly one element is a topology on that set.
+{{qed|lemma}}
+Let $S$ be a [[Definition:Set|set]] containing exactly two elements.
+Suppose $S = \set {x, y}$ for certain objects $x$ and $y$. 
+Then the [[Definition:Power Set|power set]] of $S$ is the set:
+: $\map {\mathcal P} S = \set{\varnothing, \set{x}, \set{y}, \set{x, y}}$
+Since all [[Definition:Topology|topologies]] $\tau$ on $S$ are subsets of $\tau \subseteq \mathcal P \left({S}\right)$, one of the following must hold:
+: $\tau_1 = \O$
+: $\tau_2 = \set{\O}$
+: $\tau_3 = \set{\set{x}}$
+: $\tau_4 = \set{\set{y}}$
+: $\tau_5 = \set{\O, \set{x}}$
+: $\tau_6 = \set{\O, \set{y}}$
+: $\tau_7 = \set{\set{x}, \set{y}}$
+: $\tau_8 = \set{\O, \set{x}, \set{y}}$
+: $\tau_9 = \set{\set{x}, \set{x, y}}$
+: $\tau_{10} = \set{\set{y}, \set{x, y}}$
+: $\tau_{11} = \set{\set{x}, \set{y}, \set{x, y}}$
+: $\tau_{12} = \set{\set{x, y}}$
+: $\tau_{13} = \set{\O, \set{x, y}}$
+: $\tau_{14} = \set{\O, \set{x}, \set{x, y}}$
+: $\tau_{15} = \set{\O, \set{y}, \set{x, y}}$
+: $\tau_{16} = \set{\O, \set{x}, \set{y}, S}$.
+By definition of a topology, $S$ must be an element of the topology.
+Thus $\tau_1$ up to $\tau_8$ are not topologies on $S$. 
+By [[Empty Set is Element of Topology]], for $\tau$ to be a topology for $S$, it is necessary that $\varnothing \in \tau$. 
+Therefore $\tau_9$ up to $\tau_{12}$ are also not topologies on $S$.
+By [[Indiscrete Topology is Topology]], $\tau_{13}$ is a topology on $S$.
+By [[Discrete Topology is Topology]], $\tau_{16}$ is a topology on $S$.
+By [[Particular Point Topology is Topology]], both $\tau_{14}$ and $\tau_{15}$ are topologies.
+Finally, it remains to show that any union of elements in $\{\tau_{13}, \tau_{14}, \tau_{15}, \tau_{16}\}$ gives a topology on $S$.
+By inspection, the following statements are seen to hold:
+: $\tau_{13} \cup \tau_{14} = \tau_{14}$
+: $\tau_{13} \cup \tau_{15} = \tau_{15}$
+: $\tau_{13} \cup \tau_{16} = \tau_{16}$
+: $\tau_{14} \cup \tau_{15} = \tau_{16}$
+: $\tau_{14} \cup \tau_{16} = \tau_{16}$
+: $\tau_{15} \cup \tau_{16} = \tau_{16}$
+Now let $\paren{\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty [[Definition:Indexing Set|indexed set]] of [[Definition:Topology|topologies]] for a [[Definition:Set|set]] $S$.
+By the above, if $\tau_{16}$ is one of the $\tau_i$:
+: $\displaystyle \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_{16}$
+So assume that $\tau_{16}$ is not one of the $\tau_i$.
+If both $\tau_{14}$ and $\tau_{15}$ are in $\paren{\tau_i}_{i \mathop \in I}$, then by the above:
+: $\displaystyle \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_{16}$
+Now suppose that $\tau_{15}$ is in $\paren{\tau_i}_{i \mathop \in I}$ and $\tau_{14}$ is not.
+Then by the above it follows that:
+: $\tau = \tau_{15}$
+Now suppose that $\tau_{14}$ is in $\paren{\tau_i}_{i \mathop \in I}$ and $\tau_{15}$ is not.
+Then by the above it follows that:
+: $\tau = \tau_{14}$
+Finally, assume that $\tau_{14}$ and $\tau_{15}$ are not in $\paren{\tau_i}_{i \mathop \in I}$.
+Then the only topologies in $\paren{\tau_i}_{i \mathop \in I}$ are $\tau_{13}$. 
+In the first case, we find:
+:$\tau = \tau_{13}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meager Sets in Arens-Fort Space}
+Tags: Meager Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Let $A \subseteq S$.
+Then $A$ is [[Definition:Meager Space|meager]] {{iff}} $A = \set {\tuple {0, 0} }$.
+\end{theorem}
+
+\begin{proof}
+First let $A = \set {\tuple {0, 0} }$.
+From the definition of [[Definition:Arens-Fort Space|Arens-Fort space]], $\set {\tuple {0, 0} }$ is [[Definition:Closed Set (Topology)|closed]] because $S \setminus \set {\tuple {0, 0} }$ is [[Definition:Open Set (Topology)|open]].
+From [[Closed Set Equals its Closure]]:
+:$\set {\tuple {0, 0} }^- = \set {\tuple {0, 0} }$
+where $\set {\tuple {0, 0} }^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\set {\tuple {0, 0} }$.
+From the definition of [[Definition:Arens-Fort Space|Arens-Fort space]], $\set {\tuple {0, 0} }$ is not [[Definition:Open Set (Topology)|open]].
+Therefore the smallest [[Definition:Open Set (Topology)|open set]] contained in $\set {\tuple {0, 0} }$ is $\O$.
+Hence:
+:$\set {\tuple {0, 0} }^\circ = \O$ 
+where $\set {\tuple {0, 0} }^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $\set {\tuple {0, 0} }$.
+Thus we have that:
+:$\paren {\set {\tuple {0, 0} }^-}^\circ = \O$
+and so by definition $\set {\tuple {0, 0} }$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+From [[Union of Singleton]] it trivially follows that $\set {\tuple {0, 0} }$ is the [[Definition:Set Union|union]] of a [[Definition:Countable Set|countable set]] of [[Definition:Subset|subsets]] of $S$ (that is: just the one) which are [[Definition:Nowhere Dense|nowhere dense]] in $S$.
+Hence by definition $\set {\tuple {0, 0} }$ is [[Definition:Meager Space|meager]].
+{{qed|lemma}}
+Now assume $A \ne \set {\tuple {0, 0} }$.
+Then $\exists x \in A: x \ne \tuple {0, 0}$. 
+By definition of the [[Definition:Arens-Fort Space|Arens-Fort space]], $\set x$ is [[Definition:Open Set (Topology)|open]] in $T$. 
+Thus, from [[Interior of Open Set]] $\set x^\circ = \set x$.
+Hence $\set x \subseteq \paren {\set x^-}^\circ$ and by definition is not [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+So $A$ is not the [[Definition:Set Union|union]] of a [[Definition:Countable Set|countable set]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $S$.
+So, by definition, $A$ is not [[Definition:Meager Space|meager]].
+{{qed}}
+[[Category:Meager Spaces]]
+[[Category:Arens-Fort Space]]
+07ys17kihokqi6du8kytry68idscq0f
+\end{proof}<|endoftext|>
+\section{Right and Left Regular Representations in Topological Group are Homeomorphisms}
+Tags: Homeomorphisms, Regular Representations, Topological Groups
+
+\begin{theorem}
+Let $\left({G, \cdot, \tau}\right)$ be a [[Definition:Topological Group|topological group]].
+Let $g \in G$ be an [[Definition:Element|element]] of $G$.
+Then the [[Definition:Left Regular Representation|left]] and [[Definition:Right Regular Representation|right]] [[Definition:Regular Representations|regular representations]] with respect to $g$:
+:$L_g: \left({G, \tau}\right) \to \left({G, \tau}\right)$
+and:
+:$R_g: \left({G, \tau}\right) \to \left({G, \tau}\right)$
+are [[Definition:Homeomorphism|homeomorphisms]].
+\end{theorem}
+
+\begin{proof}
+Let:
+:$m: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$
+be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \left({x, y}\right) \in G \times G: m \left({x, y}\right) = x \cdot y$
+From the definition of [[Definition:Topological Group|topological group]], $m$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
+Let $\phi_1: \left({G, \tau}\right) \to \left({G, \tau}\right) \times \left({G, \tau}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\phi_1 \left({x}\right) = \left({g, x}\right)$
+Let $\phi_2: \left({G, \tau}\right) \to \left({G, \tau}\right) \times \left({G, \tau}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\phi_2 \left({x}\right) = \left({x, g}\right)$
+Let $\operatorname{pr}_1: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the [[Definition:First Projection|first projection]] on $\left({G, \tau}\right) \times \left({G, \tau}\right)$:
+:$\forall \left({x, g}\right) \in G \times G: \operatorname{pr}_1 \left({x, g}\right) = x$
+Let $\operatorname{pr}_2: \left({G, \tau}\right) \times \left({G, \tau}\right) \to \left({G, \tau}\right)$ be the [[Definition:Second Projection|second projection]] on $\left({G, \tau}\right) \times \left({G, \tau}\right)$:
+:$\forall \left({g, x}\right) \in G \times G: \operatorname{pr}_2 \left({g, x}\right) = x$
+By inspection, it is seen that:
+:$\phi_1$ is the [[Definition:Injective Restriction|injective restriction]] to $g \times G$ of the [[Definition:Inverse of Mapping|inverse]] of $\operatorname{pr}_2$.
+:$\phi_2$ is the [[Definition:Injective Restriction|injective restriction]] to $G \times g$ of the [[Definition:Inverse of Mapping|inverse]] of $\operatorname{pr}_1$.
+Let $\phi_1 \left({x}\right) = \phi_1 \left({y}\right)$ for some $x, y \in G$.
+Then by definition of $\phi_1$:
+:$\left({g, x}\right) = \left({g, y}\right)$
+By [[Equality of Ordered Pairs]]:
+:$x = y$
+and so $\phi_1$ is [[Definition:Injection|injective]].
+Similarly iet $\phi_2 \left({x}\right) = \phi_2 \left({y}\right)$ for some $x, y \in G$.
+Then by definition of $\phi_2$:
+:$\left({x, g}\right) = \left({y, g}\right)$
+By [[Equality of Ordered Pairs]]:
+:$x = y$
+and so $\phi_2$ is [[Definition:Injection|injective]].
+Let $\left({g, y}\right) \in G \times G$.
+Then:
+:$\exists y \in G: \phi_1 \left({y}\right) = \left({g, y}\right)$
+and so $\phi_1$ is [[Definition:Surjection|surjective]].
+Similarly let $\left({y, g}\right) \in G \times G$.
+Then:
+:$\exists y \in G: \phi_2 \left({y}\right) = \left({y, g}\right)$
+and so $\phi_2$ is [[Definition:Surjection|surjective]].
+Thus as $\phi_1$ and $\phi_2$ are both [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]], they are by definition both [[Definition:Bijection|bijections]].
+From [[Projection from Product Topology is Open]], both $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are [[Definition:Open Mapping|open]].
+From [[Bijection is Open iff Inverse is Continuous]]:
+:$\phi_1$ and $\phi_2$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+From [[Composite of Continuous Mappings is Continuous]]:
+:$m \circ \phi_1: \left({G, \tau}\right) \to \left({G, \tau}\right)$
+and:
+:$m \circ \phi_2: \left({G, \tau}\right) \to \left({G, \tau}\right)$
+are [[Definition:Continuous Mapping (Topology)|continuous mappings]].
+By definition of [[Definition:Left Regular Representation|left regular representation]]:
+:$\forall x \in G: L_g \left({x}\right) = g \cdot x = m \left({g, x}\right) = m \circ \phi_1 \left({x}\right)$
+Similarly, by definition of [[Definition:Right Regular Representation|right regular representation]]:
+:$\forall x \in G: R_g \left({x}\right) = x \cdot g = m \left({x, g}\right) = m \circ \phi_2 \left({x}\right)$
+Thus $L_g$ and $R_g$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+By definition of [[Definition:Inverse Element|inverse element]] of the [[Definition:Group|group]] $G$:
+:$L_{g^{-1} } \left({x}\right) = g^{-1} \cdot x$
+:$R_{g^{-1} } \left({x}\right) = x \cdot g^{-1}$
+Thus for all $x \in G$:
+:$L_g \circ L_{g^{-1} } \left({x}\right) = g \cdot \left({g^{-1} \cdot x}\right) = x$
+:$L_{g^{-1} } \circ L_g \left({x}\right) = g^{-1} \cdot \left({g \cdot x}\right) = x$
+and similarly:
+:$R_g \circ R_{g^{-1} } \left({x}\right) = \left({x \cdot g^{-1} }\right) \cdot g = x$
+:$R_{g^{-1} } \circ R_g \left({x}\right) = \left({x \cdot g}\right) \cdot g^{-1} = x$
+So:
+:$L_g \circ L_{g^{-1} } = L_{g^{-1} } \circ L_g = I_G$
+:$R_g \circ R_{g^{-1} } = R_{g^{-1} } \circ R_g = I_G$
+where $I_G$ is the [[Definition:Identity Mapping|identity mapping]] on $G$.
+Thus by definition, $L_{g^{-1} }$ and $R_{g^{-1} }$ are the [[Definition:Inverse Mapping|inverses]] of $L_g$ and $R_g$ respectively.
+As $L_{g^{-1} }$ and $R_{g^{-1} }$ are themselves [[Definition:Regular Representations|regular representations]] with respect to elements of $G$, they are themselves [[Definition:Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijections]].
+So $L_g$ and $R_g$ are [[Definition:Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijections]] whose [[Definition:Inverse Mapping|inverses]] are likewise [[Definition:Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijections]].
+Hence, by definition, $L_g$ and $R_g$ are [[Definition:Homeomorphism|homeomorphisms]].
+{{qed}}
+[[Category:Homeomorphisms]]
+[[Category:Regular Representations]]
+[[Category:Topological Groups]]
+8uw0cp2tyyx5oa1q6pm3eroocis6btm
+\end{proof}<|endoftext|>
+\section{Center of Group is Subgroup}
+Tags: Subgroups, Direct Proofs, Centers of Groups, Center of Group is Subgroup
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+The [[Definition:Center of Group|center]] $\map Z G$ of $G$ is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+We have the result [[Center is Intersection of Centralizers]].
+That is, $\map Z G$ is the [[Definition:Set Intersection|intersection]] of all the [[Definition:Centralizer of Group Element|centralizers]] of $G$.
+All of these are [[Definition:Subgroup|subgroups]] of $G$ by [[Centralizer of Group Element is Subgroup]].
+Thus from [[Intersection of Subgroups is Subgroup]], $\map Z G$ is also a [[Definition:Subgroup|subgroup]] of $G$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$).
+Apply the [[Two-Step Subgroup Test]]:
+=== Condition $(1)$ ===
+By the definition of [[Definition:Identity Element|identity]], $e g = g e = g$ for all $g \in G$.
+So, $e \in \map Z G$, meaning $\map Z G$ is [[Definition:Non-Empty Set|non-empty]].
+{{qed|lemma}}
+=== Condition $(2)$ ===
+Suppose $a, b \in \map Z G$.
+Using the [[Definition:Associative|associative]] property and the definition of [[Definition:Center of Group|center]], we have:
+:$\forall g \in G: \paren {a b} g = a \paren {b g} = a \paren {g b} = \paren {a g} b = \paren {g a} b = g \paren {a b}$
+Thus, $a b \in \map Z G$.
+{{qed|lemma}}
+=== Condition $(3)$ ===
+Suppose $c \in \map Z G$. Then:
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \in
+      | r = \map Z G
+}}
+{{eqn | ll= \leadsto
+      | l = \forall g \in G: c g
+      | r = g c
+      | c = Definition of [[Definition:Center of Group|center]]
+}}
+{{eqn | ll= \leadsto
+      | l = c^{-1} \paren {c g} c^{-1}
+      | r = c^{-1} \paren {g c} c^{-1}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {c^{-1} c} g c^{-1}
+      | r = c^{-1} g \paren {c c^{-1} }
+}}
+{{eqn | ll= \leadsto
+      | l = e g c^{-1}
+      | r = c^{-1} g e
+}}
+{{eqn | ll= \leadsto
+      | l = g c^{-1}
+      | r = c^{-1} g
+}}
+{{eqn | ll= \leadsto
+      | l = c^{-1}
+      | o = \in
+      | r = \map Z G
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Therefore:
+: $\map Z G \le G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Center of Group is Abelian Subgroup}
+Tags: Subgroups, Abelian Groups, Centers of Groups
+
+\begin{theorem}
+The [[Definition:Center of Group|center]] $Z \left({G}\right)$ of any [[Definition:Group|group]] $G$ is a [[Definition:Subgroup|subgroup]] of $G$ which is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+By [[Center of Group is Subgroup]], $Z \left({G}\right)$ is a [[Definition:Subgroup|subgroup]] of $G$.
+The definition of the [[Definition:Center of Group|center]] $Z \left({G}\right)$ grants that all [[Definition:Element|elements]] of $Z \left({G}\right)$ [[Definition:Commute|commute]] with all [[Definition:Element|elements]] of $G$.
+In particular, all [[Definition:Element|elements]] of $Z \left({G}\right)$ [[Definition:Commute|commute]] with all [[Definition:Element|elements]] of $Z \left({G}\right)$ as $Z \left({G}\right) \subseteq G$.
+Therefore $Z \left({G}\right)$ is [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dihedral Group/Group Presentation}
+Tags: Dihedral Groups, Group Presentations
+
+\begin{theorem}
+The [[Definition:Dihedral Group|dihedral group]] $D_n$ has the [[Definition:Group Presentation|group presentation]]:
+:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
+That is, the [[Definition:Dihedral Group|dihedral group]] $D_n$ is [[Definition:Generator of Group|generated]] by two elements $\alpha$ and $\beta$ such that:
+:$(1): \quad \alpha^n = e$
+:$(2): \quad \beta^2 = e$
+:$(3): \quad \beta \alpha = \alpha^{n - 1} \beta$
+\end{theorem}
+
+\begin{proof}
+By definition, the '''[[Definition:Dihedral Group|dihedral group]]''' $D_n$ of order $2 n$ is the [[Definition:Group|group]] of [[Definition:Symmetry Mapping|symmetries]] of the [[Definition:Regular Polygon|regular $n$-gon]].
+So, let $P$ denote a [[Definition:Regular Polygon|regular polygon]] with $n$ [[Definition:Side of Polygon|sides]].
+Let $\alpha$ be a [[Definition:Plane Rotation|rotation]] of $P$ by $\dfrac {2 \pi} n$.
+Let $\beta$ be a [[Definition:Plane Reflection|reflection]] $P$ whose [[Definition:Axis of Reflection|axis of reflection]] is the [[Definition:Y-Axis|$y$ axis]].
+It takes $n$ [[Definition:Plane Rotation|rotations]] by $\dfrac {2 \pi} n$ to return $P$ to its original position.
+Thus $\alpha$ is of [[Definition:Order of Group Element|order]] $n$, and so:
+:$\alpha^n = e$
+Two [[Definition:Plane Reflection|reflections]] of $P$ in the same [[Definition:Axis of Reflection|axis of reflection]] also returns $P$ to its original position.
+Thus $\beta$ is of [[Definition:Order of Group Element|order]] $2$, and so:
+:$\beta^2 = e$
+It remains to be shown that $\beta \alpha \beta = \alpha^{-1}$.
+Let the [[Definition:Vertex of Polygon|vertices]] of $P$ be labelled in order [[Definition:Anticlockwise|anticlockwise]] from the [[Definition:Y-Axis|$y$-axis]]: $A_1, A_2, \ldots, A_n$.
+Let $P$ be arranged so that $A_1$ is on the [[Definition:Y-Axis|$y$-axis]].
+From here, $\beta$ on $P$ exchanges the places of $A_2$ and $A_n$.
+From here, $\alpha$ sends $A_2$ to the position of $A_1$, $A_1$ to the original position of $A_2$, and $A_3$ to the original position of $A_n$.
+From here, $\beta$ sends exchanges the positions of $A_3$ and $A_1$.
+So $\beta \alpha \beta$ on $P$ results in $A_1$ moving to the original position of $A_n$, $A_2$ moving to the original position of $A_1$ and $A_3$ moving to the original position of $A_2$.
+That is, $\beta \alpha \beta$ is a [[Definition:Plane Rotation|rotation]] of $P$ by $-\dfrac {2 \pi} n$, or $\alpha^{-1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection with Subgroup Product of Superset}
+Tags: Subset Products, Subgroups
+
+\begin{theorem}
+Let $X, Y, Z$ be [[Definition:Subgroup|subgroups]] of a [[Definition:Group|group]] $\struct {G, \circ}$.
+Let $Y \subseteq X$.
+Then:
+:$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$
+where $Y \circ Z$ denotes [[Definition:Subset Product|subset product]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Set Equality/Definition 2|set equality]], it suffices to prove:
+:$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
+and:
+:$Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$
+=== $X \cap \paren {Y \circ Z}$ is contained in $Y \circ \paren {X \cap Z}$ ===
+Let $s \in X \cap \paren {Y \circ Z}$.
+{{begin-eqn}}
+{{eqn | l = s
+      | o = \in
+      | r = X \cap \paren {Y \circ Z}
+      | c = 
+}}
+{{eqn | lll= \leadsto
+      | l = s
+      | o = \in
+      | r = X
+      | c = Definition of [[Definition:Set Intersection|Set Intersection]]
+}}
+{{eqn | lo= \land
+      | l = s
+      | o = \in
+      | r = Y \circ Z
+      | c = Definition of [[Definition:Set Intersection|Set Intersection]]
+}}
+{{eqn | lll= \leadsto
+      | ll= \exists y \in Y, z \in Z:
+      | l = s
+      | r = y \circ z
+      | c = Definition of [[Definition:Subset Product|Subset Product]]
+}}
+{{eqn | n = 1
+      | lll= \leadsto
+      | l = z
+      | r = y^{-1} \circ s
+      | c = [[Group has Latin Square Property]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = Y
+      | o = \subseteq
+      | r = X
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = y
+      | o = \in
+      | r = X
+      | c = {{Defof|Subset}}
+}}
+{{end-eqn}}
+So we have:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = X
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = s
+      | o = \in
+      | r = X
+      | c = {{Defof|Subset}}
+}}
+{{eqn | lll=\leadsto
+      | l = y^{-1} \circ s
+      | o = \in
+      | r = X
+      | c = [[One-Step Subgroup Test]]
+}}
+{{eqn | lll=\leadsto
+      | l = z
+      | o = \in
+      | r = X
+      | c = from $(1)$
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Set Intersection|set intersection]]:
+: $z \in X \cap Z$
+So:
+:$s = y \circ z \in Y \circ \paren {X \cap Z}$
+By definition of [[Definition:Subset|subset]]:
+: $X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
+{{qed|lemma}}
+=== $Y \circ \paren {X \cap Z}$ is contained in $X \cap \paren {Y \circ Z}$ ===
+{{begin-eqn}}
+{{eqn | l = Y \circ X
+      | o = \subseteq
+      | r = X \circ X
+      | c = [[Subset of Subset Product]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = Y \circ X
+      | o = \subseteq
+      | r = X
+      | c = [[Product of Subgroup with Itself]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = Y \circ \paren {X \cap Z}
+      | o = \subseteq
+      | r = \paren {Y \circ X} \cap \paren {Y \circ Z}
+      | c = [[Product of Subset with Intersection]]
+}}
+{{eqn | o = \subseteq
+      | r = X \cap \paren {Y \circ Z}
+      | c = from $(1)$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+We have established that:
+: $x \in Y \circ \paren {X \cap Z} \iff x \in X \cap \paren {Y \circ Z}$
+From the definition of [[Definition:Set Equality/Definition 1|set equality]]:
+: $Y \circ \paren {X \cap Z} = X \cap \paren {Y \circ Z}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Semi-Inner Product}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|vector space]] over $\Bbb F \in \left\{{\R, \C}\right\}$.
+Let $\left \langle{\cdot, \cdot}\right \rangle$ be a [[Definition:Semi-Inner Product|semi-inner product]] on $V$.
+Denote, for $x \in V$, $\left\Vert{x}\right\Vert := \left\langle{x, x}\right\rangle^{1 / 2}$.
+Then, $\forall x, y \in V, a \in \Bbb F$:
+:$(1): \quad \left\Vert{x + y}\right\Vert \le \left\Vert{x}\right\Vert + \left\Vert{y}\right\Vert$
+:$(2): \quad \left\Vert{a x}\right\Vert = \left|{a}\right| \left\Vert{x}\right\Vert$
+\end{theorem}
+
+\begin{proof}
+=== Proof of $(1)$ ===
+For $x, y \in V$, compute:
+{{begin-eqn}}
+{{eqn | l = \left\Vert{x + y}\right\Vert^2
+      | r = \left\langle{x + y, x + y}\right\rangle
+      | c = Definition of $\left\Vert{\cdot}\right\Vert$
+}}
+{{eqn | r = \left\langle{x, x}\right\rangle + \left\langle{x, y}\right\rangle + \left\langle{y, x}\right\rangle + \left\langle{y, y}\right\rangle
+      | c = [[Definition:Semi-Inner Product|Linearity]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn | o = \le
+      | r = \left\Vert{x}\right\Vert^2 + \sqrt{\left\langle{x, x}\right\rangle \left\langle{y, y}\right\rangle} + \sqrt{\left\langle{y, y}\right\rangle \left\langle{x, x}\right\rangle} + \left\Vert{y}\right\Vert^2
+      | c = [[Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces|Cauchy-Bunyakovsky-Schwarz Inequality]]
+}}
+{{eqn | r = \left\Vert{x}\right\Vert^2 + 2 \left\Vert{x}\right\Vert \left\Vert{y}\right\Vert + \left\Vert{y}\right\Vert^2
+      | c = Definition of $\left\Vert{\cdot}\right\Vert$
+}}
+{{eqn | r = \left({\left\Vert{x}\right\Vert + \left\Vert{y}\right\Vert}\right)^2
+}}
+{{end-eqn}}
+Taking square roots on either side gives the result.
+{{qed|lemma}}
+=== Proof of $(2)$ ===
+For $x \in V$, $a \in \Bbb F$, compute:
+{{begin-eqn}}
+{{eqn | l = \left\Vert{a x}\right\Vert^2
+      | r = \left\langle{a x, a x}\right\rangle
+      | c = Definition of $\left\Vert{\cdot}\right\Vert$
+}}
+{{eqn | r = a \left\langle{x, a x}\right\rangle
+      | c = [[Definition:Semi-Inner Product|Linearity]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn | r = a \overline{\left\langle{a x, x}\right\rangle}
+      | c = [[Definition:Semi-Inner Product|Conjugate symmetry]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn | r = a \overline a \overline{\left\langle{x, x}\right\rangle}
+      | c = [[Definition:Semi-Inner Product|Linearity]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn | r = \left\vert{a}\right\vert^2 \left\Vert{x}\right\Vert^2
+      | c = [[Modulus in Terms of Conjugate]]
+}}
+{{end-eqn}}
+Taking square roots on either side gives the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inner Product Norm is Norm}
+Tags: Examples of Norms, Hilbert Spaces
+
+\begin{theorem}
+Let $V$ be an [[Definition:Inner Product Space|inner product space]] over a [[Definition:Field (Abstract Algebra)|subfield]] $\Bbb F$ of $\C$.
+Let $\norm {\, \cdot \,}$ denote the [[Definition:Inner Product Norm|inner product norm]] on $V$.
+Then $\norm {\, \cdot \,}$ is a [[Definition:Norm on Vector Space|norm]] on $V$.
+\end{theorem}
+
+\begin{proof}
+Let us verify the [[Definition:Norm on Vector Space|norm]] axioms in turn.
+=== Axiom $(\text N 1)$ ===
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x = 0_V
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | o = 
+      | r = \innerprod x x = 0
+      | c = Property $(5)$ of [[Definition:Inner Product|Inner Product]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | o = 
+      | r = \innerprod x x^{1 / 2} = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | o = 
+      | r = \norm x = 0
+      | c = {{Defof|Inner Product Norm}}
+}}
+{{end-eqn}}
+{{Qed|lemma}}
+=== Axiom $(\text N 2)$ ===
+Part $(2)$ of [[Properties of Semi-Inner Product]].
+{{Qed|lemma}}
+=== Axiom $(\text N 3)$ ===
+Part $(3)$ of [[Properties of Semi-Inner Product]].
+{{Qed|lemma}}
+Hence all the properties of a [[Definition:Norm on Vector Space|norm]] have been shown to hold.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Metric Induced by Norm is Metric}
+Tags: Norm Theory, Metric Spaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\norm{\,\cdot\,}$ denote its [[Definition:Norm on Vector Space|norm]].
+Let $d$ be the [[Definition:Metric Induced by Norm|metric induced by $\norm{\,\cdot\,}$]].
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $\text M 1$ and $\text M 4$ ===
+Let $x, y \in V$. 
+Then:
+:$\map d {x, y} = \norm {x - y} \ge 0$
+and furthermore:
+{{begin-eqn}}
+{{eqn | l = \map d {x, y}
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm {x - y}
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x - y
+      | r = \mathbf 0_V
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm Axioms: Axiom $\text N 1$]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = y
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Proof of $\text M 2$ ===
+Let $x, y, z \in V$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {x, z}
+      | r = \norm {x - z}
+}}
+{{eqn | r = \norm {x - y + y - z}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \norm {x - y} + \norm {y - z}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm Axioms: Axiom $\text N 3$]]
+}}
+{{eqn | r = \map d {x, y} + \map d {y, z}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Proof of $\text M 3$ ===
+Let $x, y \in V$.
+Then: 
+{{begin-eqn}}
+{{eqn | l = \map d {x, y}
+      | r = \norm {x - y}
+}}
+{{eqn | r = \norm {-1 \paren {y - x} }
+}}
+{{eqn | r = \norm {-1} \times \norm {y - x}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm Axioms: Axiom $\text N 2$]]
+}}
+{{eqn | r = \norm {y - x}
+}}
+{{eqn | r = \map d {y, x}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+As $d$ satisfies the [[Definition:Metric Space Axioms|metric space axioms]], it is a [[Definition:Metric|metric]].
+{{qed}}
+[[Category:Norm Theory]]
+[[Category:Metric Spaces]]
+1ojmpq5wvlf7ez22c3n312r3utnvffc
+\end{proof}<|endoftext|>
+\section{Product of Sums of Four Squares/Corollary}
+Tags: Product of Sums of Four Squares, Proofs by Induction
+
+\begin{theorem}
+Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, c_1, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ be [[Definition:Integer|integers]].
+Then:
+: $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$
+That is, the [[Definition:Integer Multiplication|product]] of any number of [[Definition:Integer Addition|sums]] of four [[Definition:Square Number|squares]] is also a sum of four squares.
+\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]]:
+: $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$
+$P \left({1}\right)$ is true, as this just says:
+: $\exists w, x, y, z \in \Z: a^2 + b^2 + c^2 + d^2 = w^2 + x^2 + y^2 + z^2$
+which is trivially true.
+=== Basis for the Induction ===
+$P \left({2}\right)$ is the case:
+: $\exists w, x, y, z \in \Z: \left({a_1^2 + b_1^2 + c_1^2 + d_1^2}\right) \left({a_2^2 + b_2^2 + c_2^2 + d_2^2}\right) = w^2 + x^2 + y^2 + z^2$
+which follows from [[Product of Sums of Four Squares]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+: $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^k \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$
+Then we need to show that it directly implies:
+: $\displaystyle \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^{k+1} \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \prod_{j \mathop = 1}^{k+1} \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right)
+      | c = 
+}}
+{{eqn | r = \left({\prod_{j \mathop = 1}^k \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right)}\right) \left({a_{k+1}^2 + b_{k+1}^2 + c_{k+1}^2 + d_{k+1}^2}\right)
+      | c = 
+}}
+{{eqn | r = \left({r^2 + s^2 + t^2 + u^2}\right) \left({a_{k+1}^2 + b_{k+1}^2 + c_{k+1}^2 + d_{k+1}^2}\right)
+      | c = from the [[Product of Sums of Four Squares/Corollary#Induction Hypothesis|induction hypothesis]]: for some $r, s, t, u, \in \Z$
+}}
+{{eqn | r = w^2 + x^2 + y^2 + z^2
+      | c = from the [[Product of Sums of Four Squares/Corollary#Basis for the Induction|basis for the induction]]: for some $w, x, y, z, \in \Z$
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+: $\displaystyle \forall n \in \Z_{>0}: \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \left({a_j^2 + b_j^2 + c_j^2 + d_j^2}\right) = w^2 + x^2 + y^2 + z^2$
+{{qed}}
+[[Category:Product of Sums of Four Squares]]
+[[Category:Proofs by Induction]]
+mqkxsxvflrnnxqfi8350kswpk0gnpj9
+\end{proof}<|endoftext|>
+\section{Subgroup Subset of Subgroup Product}
+Tags: Subset Products, Subgroups
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
+Then:
+:$H \subseteq H \circ K \supseteq K$
+where $H \circ K$ denotes the [[Definition:Subset Product|subset product]] of $H$ and $K$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Subset Product|subset product]]:
+:$H \circ K = \left\{{h \circ k: h \in H, k \in K}\right\}$
+So:
+{{begin-eqn}}
+{{eqn | l=x
+      | o=\in
+      | r=H
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=x
+      | r=x \circ e
+      | c=as $e$ is the [[Definition:Identity Element|identity]] of $G$
+}}
+{{eqn | o=\in
+      | r=H \circ K
+      | c=as $e \in K$ from [[Identity of Subgroup]]
+}}
+{{eqn | ll=\implies
+      | l=H
+      | o=\subseteq
+      | r=H \circ K
+      | c=Definition of [[Definition:Subset|subset]]
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l=y
+      | o=\in
+      | r=K
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=y
+      | r=e \circ y
+      | c=as $e$ is the [[Definition:Identity Element|identity]] of $G$
+}}
+{{eqn | o=\in
+      | r=H \circ K
+      | c=as $e \in H$ from [[Identity of Subgroup]]
+}}
+{{eqn | ll=\implies
+      | l=K
+      | o=\subseteq
+      | r=H \circ K
+      | c=Definition of [[Definition:Subset|subset]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Subset Products]]
+[[Category:Subgroups]]
+dd8alail40q7bzc449ly8bqi4ahdwac
+\end{proof}<|endoftext|>
+\section{Product Space is Path-connected iff Factor Spaces are Path-connected}
+Tags: Path-Connected Spaces, Product Spaces
+
+\begin{theorem}
+Let $\SS = \family {\struct{S_i, \tau_i}}_{i \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for $i$ in some [[Definition:Indexing Set|indexing set]] $I$ such that $\forall i \in I: S_i \ne \varnothing$.
+Let $\displaystyle T = \struct{S, \tau} = \prod_{i \mathop \in I} \struct{S_i, \tau_i}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\SS$.
+Then $T$ is a [[Definition:Path-Connected Space|path-connected space]] {{iff}} each of $\struct{S_i, \tau_i}$ is a [[Definition:Path-Connected Space|path-connected space]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $S_i$ is [[Definition:Path-Connected Space|path-connected]] for each $i\in I$. 
+Let $x \ne y \in S$ be arbitrary.
+Since each $S_i$ is [[Definition:Path-Connected Space|path-connected]], we have a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]]:
+:$f_i: \closedint {0} {1} \to S_i$
+such that $\map {f_i} 0 = x_i$ and $\map {f_i} 1  = y_i$ for all $i \in I$.
+We then define:
+: $f : \closedint {0} {1} \to S$
+by setting:
+:$\forall t \in \closedint {0} {1} : \map f t_i = \map {f_i} t$
+We have that $f_i = \pr_i \circ f$ are all [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+So, by [[Continuous Mapping to Product Space]], it follows that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Also:
+: $\map f 0_i = \map {f_i} 0 = x_i$
+and:
+: $\map f 1_i = \map {f_i} 1 = y_i$
+Therefore $\map f 0 = x$ and $\map f 1 = y$.
+So there exists a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]]:
+: $f: \closedint {0} {1} \to S$ 
+such that $\map f 0 = x$ and $\map f 1 = y$.
+Since $x$ and $y$ were arbitrary, we have shown that $S$ is [[Definition:Path-Connected Space|path-connected]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Conversely, suppose $T$ is a [[Definition:Path-Connected Space|path-connected space]].
+Let $j$ be an arbitrary $j \in I$.
+Let $x_j$ and $y_j$ be points in $S_j$ such that $x_j \ne y_j$.
+We have that $\forall i \in I: S_i \ne \O$.
+Using the [[Axiom:Axiom of Choice|axiom of choice]], a [[Definition:Choice Function|choice function]] can be set up to allow the choice of some $q_i \in S_i$ for each $i \ne j$.
+We then define:
+: $\forall s \in T: \map {\pr_i} s = q_i$
+From [[Constant Mapping is Continuous]], it follows that each of $\pr_i: T \to S_i$ where $i \ne j$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]:
+Thus there exist points in $x, y \in S$ such that: 
+: $\map {\pr_i} x = q_i$
+: $\map {\pr_i} y = q_i$
+for all $i \ne j$, and:
+: $\map {\pr_j} x = x_j$
+: $\map {\pr_j} y = y_j$
+Since $T$ is [[Definition:Path-Connected Space|path-connected]], there exists a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]]:
+: $f: \closedint {0} {1} \to S$ 
+such that $\map f 0 = x$ and $\map f 1 = y$.
+By [[Continuous Mapping to Product Space]]:
+: $\pr_j \circ f: \closedint {0} {1} \to S_j$
+is also [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Furthermore:
+: $\map {\pr_j \circ f} 0 = \map {\pr_j} x = x_j$
+and:
+: $\map {\pr_j \circ f} 1 = \map {\pr_j} y = y_j$
+Thus since $x_j$ and $y_j$ were arbitrary, we have shown that $S_j$ is [[Definition:Path-Connected Space|path-connected]].
+ 
+Since $j$ was arbitrary, we have shown that all $S_i$ are [[Definition:Path-Connected Space|path-connected]].
+{{qed}}
+{{AoC}}
+\end{proof}<|endoftext|>
+\section{Normal Subgroup of Subset Product of Subgroups}
+Tags: Normal Subgroups, Subset Products
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let:
+: $H$ be a [[Definition:Subgroup|subgroup]] of $G$
+: $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then:
+: $N \lhd N H$
+where:
+: $\lhd$ denotes [[Definition:Normal Subgroup|normal subgroup]]
+: $N H$ denotes [[Definition:Subset Product|subset product]].
+\end{theorem}
+
+\begin{proof}
+From [[Subset Product with Normal Subgroup is Subgroup]]:
+:$N H = H N$ is a [[Definition:Subgroup|subgroup]] of $G$.
+By definition of [[Definition:Subset Product|subset product]] all [[Definition:Element|elements]] of $H N$ can be written in the form:
+:$h n \in H N$
+where $h \in H, n \in N$.
+Let $h n \in H N$.
+Let $n_1 \in N$.
+From [[Inverse of Group Product]]:
+:$\paren {h n} n_1 \paren {h n}^{-1} = h n n_1 n^{-1} h^{-1}$
+We have that:
+:$n n_1 n \in N$
+:$h, h^{-1} \in G$.
+Then, since $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$:
+:$\paren {h n} n_1 \paren {n^{-1} h^{-1} } \in N$
+Thus:
+: $N \lhd N H$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Products of Normal Subgroup with Normal Subgroup of Subgroup}
+Tags: Normal Subgroups, Subset Products
+
+\begin{theorem}
+Let $G$ be a group.
+Let:
+:$(1): \quad H$ be a [[Definition:Subgroup|subgroup]] of $G$
+:$(2): \quad K$ be a [[Definition:Normal Subgroup|normal subgroup]] of $H$
+:$(3): \quad N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$
+Then:
+:$N K \lhd N H$
+where:
+: $N K$ and $N H$ denote [[Definition:Subset Product|subset product]]
+: $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup|normal subgroup]].
+\end{theorem}
+
+\begin{proof}
+Consider arbitrary $x_n \in N, x_h \in H$.
+Thus:
+:$x_n x_h \in N H$
+We aim to show that:
+:$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$
+thus demonstrating $N K \lhd N H$ by the [[Normal Subgroup Test]].
+We have:
+{{begin-eqn}}
+{{eqn | l = x_n x_h N K \paren {x_n x_h}^{-1}
+      | r = x_n x_h N K {x_h}^{-1} {x_n}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = x_n N x_h K {x_h}^{-1} {x_n}^{-1}
+      | c = as $N \lhd G$ and $H \le G$
+}}
+{{eqn | r = x_n N K {x_n}^{-1}
+      | c = as $K \lhd H$
+}}
+{{eqn | r = N K {x_n}^{-1}
+}}
+{{eqn | r = K N {x_n}^{-1}
+      | c = as $N \lhd G$ and $K \le G$
+}}
+{{eqn | r = K N
+}}
+{{eqn | r = N K
+      | c = as $N \lhd G$ and $K \le G$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group}
+Tags: Quaternion Group, Groups of Order 8, Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] with the following properties:
+:$(1): \quad G$ is [[Definition:Abelian Group|non-abelian]].
+:$(2): \quad G$ is of [[Definition:Order of Structure|order $8$]].
+:$(3): \quad G$ has precisely one element of [[Definition:Order of Group Element|order $2$]].
+Then $G$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Quaternion Group|quaternion group]] $Q$.
+\end{theorem}
+
+\begin{proof}
+From [[Order of Element Divides Order of Finite Group]] all the [[Definition:Element|elements]] in $G$ have [[Definition:Order of Group Element|order]] $1, 2, 4$ or $8$.
+From [[Cyclic Group is Abelian]], $\paren 1$ and $\paren 2$, no [[Definition:Element|elements]] in $G$ have order $8$, i.e. they all have order $1, 2$ or $4$.
+Let the [[Definition:Identity Element|identity element]] be $1$ and the one with order $2$ be $-1$.
+Also denote $1 \circ a$ as $+a$, $-1 \circ a$ as $-a$, $\set {\pm a} = \set {a, -a}$ for simplicity.
+=== [[Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1|Lemma 1]] ===
+{{:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1}}
+By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]] and [[Cosets are Equivalent]], let:
+:$G / \set {\pm 1} = \set {\set {\pm 1}, \set {\pm a}, \set {\pm b}, \set {\pm c} }$
+=== [[Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2|Lemma 2]] ===
+{{:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2}}
+Now draw up the incomplete [[Definition:Cayley Table|Cayley table]] for $G / \set {\pm 1}$:
+:$\begin{array}{c|cccc}
+& \pm 1 & \pm a & \pm b & \pm c \\
+\hline
+\pm 1 & \pm 1 & \pm a & \pm b & \pm c \\
+\pm a & \pm a & \pm 1 & & \\
+\pm b & \pm b & & \pm 1 & \\
+\pm c & \pm c & & & \pm 1 \\
+\end{array}$
+By [[Group has Latin Square Property]] the [[Definition:Cayley Table|Cayley table]] can be completed:
+:$\begin{array}{c|cccc}
+& \pm 1 & \pm a & \pm b & \pm c \\
+\hline
+\pm 1 & \pm 1 & \pm a & \pm b & \pm c \\
+\pm a & \pm a & \pm 1 & \pm c & \pm b \\
+\pm b & \pm b & \pm c & \pm 1 & \pm a \\
+\pm c & \pm c & \pm b & \pm a & \pm 1 \\
+\end{array}$
+Now from the above draw up the incomplete [[Definition:Cayley Table|Cayley table]] for $G$:
+:$\begin{array}{c|cccccccc}
+& 1 & -1 & a & -a & b & -b & c & -c \\
+\hline
+1 & 1 & -1 & a & -a & b & -b & c & -c \\
+-1 & -1 & 1 & -a & a & -b & b & -c & c \\
+a & a & -a & -1 & 1 & & & & \\
+-a & -a & a & 1 & -1 & & & & \\
+b & b & -b & & & -1 & 1 & & \\
+-b & -b & b & & & 1 & -1 & & \\
+c & c & -c & & & & & -1 & 1 \\
+-c & -c & c & & & & & 1 & -1 \\
+\end{array}$
+By [[Group has Latin Square Property]] and [[Definition:Group Axioms|$\left({G1}\right):$ Associativity]], the [[Definition:Cayley Table|Cayley table]] can be completed in two ways:
+:$\begin{array}{c|cccccccc}
+& 1 & -1 & a & -a & b & -b & c & -c \\
+\hline
+1 & 1 & -1 & a & -a & b & -b & c & -c \\
+-1 & -1 & 1 & -a & a & -b & b & -c & c \\
+a & a & -a & -1 & 1 & c & -c & -b & b \\
+-a & -a & a & 1 & -1 & -c & c & b & -b \\
+b & b & -b & -c & c & -1 & 1 & a & -a \\
+-b & -b & b & c & -c & 1 & -1 & -a & a \\
+c & c & -c & b & -b & -a & a & -1 & 1 \\
+-c & -c & c & -b & b & a & -a & 1 & -1 \\
+\end{array}$
+or:
+:$\begin{array}{c|cccccccc}
+& 1 & -1 & a' & -a' & b' & -b' & c' & -c' \\
+\hline
+1 & 1 & -1 & a' & -a' & b' & -b' & c' & -c' \\
+-1 & -1 & 1 & -a' & a' & -b' & b' & -c' & c' \\
+a' & a' & -a' & -1 & 1 & -c' & c' & b' & -b' \\
+-a' & -a' & a' & 1 & -1 & c' & -c' & -b' & b' \\
+b' & b' & -b' & c' & -c' & -1 & 1 & -a' & a' \\
+-b' & -b' & b' & -c' & c' & 1 & -1 & a' & -a' \\
+c' & c' & -c' & -b' & b' & a' & -a' & -1 & 1 \\
+-c' & -c' & c' & b' & -b' & -a' & a' & 1 & -1 \\
+\end{array}$
+Referring to the [[Quaternion Group/Complex Matrices/Cayley Table|Cayley table]] of the [[Quaternion Group/Complex Matrices|Quaternion group]]:
+:$\begin{array}{r|rrrrrrrr}
+      &  \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\
+\hline
+ \mathbf 1 &  \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\
+ \mathbf i &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf 1 &  \mathbf k & -\mathbf j & -\mathbf k &  \mathbf j \\
+-\mathbf 1 & -\mathbf 1 & -\mathbf i &  \mathbf 1 &  \mathbf i & -\mathbf j & -\mathbf k &  \mathbf j &  \mathbf k \\
+-\mathbf i & -\mathbf i &  \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf k &  \mathbf j &  \mathbf k & -\mathbf j \\
+ \mathbf j &  \mathbf j & -\mathbf k & -\mathbf j &  \mathbf k & -\mathbf 1 &  \mathbf i &  \mathbf 1 & -\mathbf i \\
+ \mathbf k &  \mathbf k &  \mathbf j & -\mathbf k & -\mathbf j & -\mathbf i & -\mathbf 1 &  \mathbf i &  \mathbf 1 \\
+-\mathbf j & -\mathbf j &  \mathbf k &  \mathbf j & -\mathbf k &  \mathbf 1 & -\mathbf i & -\mathbf 1 &  \mathbf i \\
+-\mathbf k & -\mathbf k & -\mathbf j &  \mathbf k &  \mathbf j &  \mathbf i &  \mathbf 1 & -\mathbf i & -\mathbf 1
+\end{array}$
+The results follow by identifying:
+:$\tuple {1, a, b, c} = \tuple {1, c', b', a'} = \tuple {\mathbf 1, \mathbf i, \mathbf j, \mathbf k}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[Groups of Order 8]], the only [[Definition:Group|group]] satisfying all three properties is $Q = \Dic 2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quaternion Group not Dihedral Group}
+Tags: Quaternion Group, Dihedral Groups
+
+\begin{theorem}
+Let $Q$ be the [[Definition:Quaternion Group|quaternion group]].
+Then $Q$ is not [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Dihedral Group|dihedral group]] $D_4$.
+\end{theorem}
+
+\begin{proof}
+From [[Group Presentation of Dihedral Group]], $D_4$ is [[Definition:Generator of Group|generated]] by two [[Definition:Element|elements]] of [[Definition:Order of Group Element|orders]] $4$ and $2$ respectively.
+Let these [[Definition:Generator of Group|generators]] be $\alpha$ and $\beta$ where:
+:$\alpha^4 = e$
+:$\beta^2 = e$
+Hence $\alpha^2$ and $\beta$ are different [[Definition:Element|elements]] of $D_4$ which both have [[Definition:Order of Group Element|order]] $2$.
+But the [[Quaternion Group/Complex Matrices|quaternion group]]:
+:$Q = \set {\mathbf 1, -\mathbf 1, \mathbf i, -\mathbf i, \mathbf j, -\mathbf j, \mathbf k, -\mathbf k}$
+has only one [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $2$, which is $-\mathbf 1$.
+The rest have [[Definition:Order of Group Element|order]] one or four.
+Thus $Q$ and $D_4$ cannot be [[Definition:Group Isomorphism|isomorphic]] because there are not the same number of [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $2$ in both [[Definition:Group|groups]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Product of Normal Subgroups with Trivial Intersection}
+Tags: Normal Subgroups, Group Direct Products
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H, K$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Let $H \cap K = e$.
+Then $H K$ is [[Definition:Group Isomorphism|isomorphic]] to $H \times K$ where:
+: $H K$ denotes the [[Definition:Subset Product|subset product]] of $H$ and $K$
+: $H \times K$ denotes the [[Definition:Group Direct Product|direct product]] of $H$ and $K$.
+\end{theorem}
+
+\begin{proof}
+Let $G' = H K$.
+From [[Subset Product of Normal Subgroups is Normal]], $G'$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+That is $G'$ is itself a [[Definition:Group|group]].
+So by the [[Internal Direct Product Theorem]], $G'$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $H$ and $K$.
+The result follows by definition of the [[Definition:Internal Group Direct Product|internal group direct product]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection with Normal Subgroup is Normal/Examples/Subset Product of Normal Subgroup with Intersection}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $\struct G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
+Let:
+: $N_1 \lhd H_1$
+: $N_2 \lhd H_2$
+where $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup|normal subgroup]].
+Then:
+:$N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$
+\end{theorem}
+
+\begin{proof}
+Consider arbitrary $x_n \in N_1, x_h \in H_1 \cap H_2$.
+Thus:
+:$x_n x_h \in N_1 \paren {H_1 \cap H_2}$
+Note that as $x_h \in H_1 \cap H_2$ it follows that $x_h \in H_1$.
+We aim to show that:
+:$x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1} \subseteq N_1 \paren {H_1 \cap H_2}$
+thus demonstrating $N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$ by the [[Normal Subgroup Test]].
+We have:
+{{begin-eqn}}
+{{eqn | l = x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1}
+      | r = x_n x_h N_1 \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = x_n x_h N_1 {x_h}^{-1} x_h \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}
+      | c = 
+}}
+{{eqn | r = x_n N_1 x_h \paren {H_1 \cap N_2} {x_h}^{-1} {x_n}^{-1}
+      | c = as $N_1 \lhd H_1$
+}}
+{{eqn | r = x_n N_1 \paren {H_1 \cap N_2} {x_n}^{-1}
+      | c = $H_1 \cap N_2$ is [[Definition:Normal Subgroup|normal]] in $H_1$ from [[Intersection with Normal Subgroup is Normal]]
+}}
+{{eqn | r = N_1 \paren {H_1 \cap N_2} {x_n}^{-1}
+}}
+{{eqn | r = \paren {H_1 \cap N_2} N_1 {x_n}^{-1}
+      | c = [[Subset Product of Subgroups]]
+}}
+{{eqn | r = \paren {H_1 \cap N_2} N_1
+}}
+{{eqn | r = N_1 \paren {H_1 \cap N_2}
+      | c = [[Subset Product of Subgroups]]
+}}
+{{end-eqn}}
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Direct Products}
+Tags: Quotient Groups, Group Direct Products
+
+\begin{theorem}
+Let $G$ and $G'$ be [[Definition:Group|groups]].
+Let:
+:$H \lhd G$
+:$H' \lhd G'$
+where $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup|normal subgroup]].
+Then:
+:$\paren {G \times G'} / \paren {H \times H'}$ is [[Definition:Group Isomorphism|isomorphic]] to $\paren {G / H} \times \paren {G' / H'}$
+where:
+:$G \times G'$ denotes the [[Definition:Group Direct Product|group direct product]] of $H$ and $H'$
+:$G / H$ denotes the [[Definition:Quotient Group|quotient group]] of $G$ by $H$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi_1: G \to G / H$ and $\phi_2: G' \to G' / H'$ be the [[Definition:Quotient Group Epimorphism|quotient epimorphisms]] with $H$ and $H'$ as their [[Definition:Kernel of Group Homomorphism|kernels]], respectively.
+Now define a [[Definition:Group Homomorphism|homomorphism]] $\phi: G \times G' \to \paren {G / H} \times \paren {G' / H'}$ by:
+:$\phi = \phi_1 \times \phi_2$
+so:
+:$\map \phi {\tuple {x, x'} } = \tuple {\map {\phi_1} x, \map {\phi_1} {x'} }$
+The [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$ is clearly $H \times H'$, and $\phi$ is [[Definition:Surjection|surjective]].
+So by the [[First Isomorphism Theorem for Groups]]:
+:$\paren {G / H} \times \paren {G' / H'} \cong \paren {G \times G'} / \paren {H \times H'}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completion Theorem (Inner Product Space)}
+Tags: Hilbert Spaces, Named Theorems
+
+\begin{theorem}
+Let $V$ be an [[Definition:Inner Product Space|inner product space]] over a [[Definition:Field (Abstract Algebra)|subfield]] $\Bbb F$ of $\C$.
+Let $\left\langle{\cdot, \cdot}\right\rangle_V$ be the [[Definition:Inner Product|inner product]] on $V$.
+Let $d: V \times V \to \R_{\ge 0}$ be the [[Definition:Metric Induced by Norm|metric induced]] by the [[Definition:Inner Product Norm|inner product norm]].
+Let $H$ be the [[Definition:Completion (Metric Space)|completion]] of $V$ with respect to $d$.
+Then $\left\langle{\cdot, \cdot}\right\rangle_V$ can be extended to an [[Definition:Inner Product|inner product]] on $H$.
+By definition, $H$ will be a [[Definition:Hilbert Space|Hilbert space]].
+Therefore, the theorem can alternatively be stated as:
+:Any [[Definition:Inner Product Space|inner product space]] may be [[Definition:Completion (Metric Space)|completed]] to a [[Definition:Hilbert Space|Hilbert space]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|the definition [[Definition:Completion (Inner Product Space)]] may be justified}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping is Group Endomorphism}
+Tags: Group Endomorphisms
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a [[Definition:Group Endomorphism|group endomorphism]].
+\end{theorem}
+
+\begin{proof}
+The [[Identity Mapping is Group Automorphism]].
+By definition, a [[Definition:Group Endomorphism|group endomorphism]] is a [[Definition:Group Homomorphism|group homomorphism]] from $G$ to itself.
+A [[Definition:Group Automorphism|group automorphism]] is a [[Definition:Group Isomorphism|group isomorphism]] from $G$ to itself.
+As a [[Definition:Group Isomorphism|group isomorphism]] is a [[Definition:Group Homomorphism|group homomorphism]] which is also a [[Definition:Bijection|bijection]], the result follows by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pythagoras's Theorem (Hilbert Space)}
+Tags: Hilbert Spaces, Pythagoras's Theorem
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] with [[Definition:Inner Product Norm|inner product norm]] $\norm {\, \cdot \,}$.
+Let $f_1, \ldots, f_n \in H$ be pairwise [[Definition:Orthogonal (Hilbert Space)|orthogonal]].
+Then:
+:$\displaystyle \norm {\sum_{i \mathop = 1}^n f_i}^2 = \sum_{i \mathop = 1}^n \norm {f_i}^2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \norm {\sum_{i \mathop = 1}^n f_i}^2
+      | r = \innerprod {\sum_{i \mathop = 1}^n f_i} {\sum_{j \mathop = 1}^n f_j}
+      | c = {{Defof|Inner Product Norm}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \innerprod {f_i} {f_j}
+      | c = Linearity of [[Definition:Inner Product|Inner Product]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \begin {cases} \innerprod {f_i} {f_i} & i = j \\ 0 & i \ne j \end {cases}
+      | c = The $f_i$ are pairwise [[Definition:Orthogonal (Hilbert Space)|orthogonal]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \norm {f_i}^2
+      | c = {{Defof|Inner Product Norm}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared plus a squared/Arctangent Form}
+Tags: Expressions whose Primitives are Inverse Trigonometric Functions, Primitives involving x squared plus a squared, Arctangent Function, Primitives involving Reciprocals, Primitive of Reciprocal of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$
+\end{theorem}<|endoftext|>
+\section{Endomorphism from Integers to Multiples}
+Tags: Integers, Group Endomorphisms
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be a [[Definition:Mapping|mapping]].
+Then $\phi$ is a [[Definition:Group Endomorphism|group endomorphism]] {{iff}}:
+:$\exists k \in \Z: \forall n \in \Z: \map \phi n = k n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an [[Definition:Group Endomorphism|endomorphism]].
+Let $k = \map \phi 1$.
+We have that $n = 1 + \cdots \paren n \cdots + 1$ for any [[Definition:Strictly Positive Integer|positive integer]] $n$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map \phi n
+      | r = \map \phi 1 + \cdots \paren n \cdots + \map \phi 1
+      | c = 
+}}
+{{eqn | r = k + \cdots \paren n \cdots + k
+      | c = 
+}}
+{{eqn | r = k n
+      | c = 
+}}
+{{end-eqn}}
+Also:
+{{begin-eqn}}
+{{eqn | l = \map \phi 1
+      | r = \map \phi {1 + 0}
+      | c = 
+}}
+{{eqn | r = \map \phi 1 + \map \phi 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi 0
+      | r = 0
+      | c = 
+}}
+{{eqn | r = k \cdot 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi {-1}
+      | r = -k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi {-n}
+      | r = -k n
+      | c = for all [[Definition:Strictly Positive Integer|positive integers]] $n$
+}}
+{{end-eqn}}
+Thus:
+:$\forall n \in \Z: \map \phi n = k n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $k \in \Z$ such that:
+: $\forall n \in \Z: \map \phi n = k n$
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall n, m \in \Z:
+      | l = \map \phi {n + m}
+      | r = k \paren {n + m}
+      | c = 
+}}
+{{eqn | r = k n + k m
+      | c = [[Integer Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \map \phi n + \map \phi m
+      | c = 
+}}
+{{end-eqn}}
+Thus $\phi: \struct {\Z, +} \to \struct {\Z, +}$ is a [[Definition:Group Homomorphism|group homomorphism]] from $\Z$ to $\Z$.
+Hence by definition $\phi$ is a [[Definition:Group Endomorphism|group endomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Cyclic Group is Isomorphic to Integers under Modulo Addition}
+Tags: Cyclic Groups, Modulo Arithmetic
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity element]] is $e$.
+Then $\struct {G, \circ}$ is [[Definition:Cyclic Group|cyclic]] of [[Definition:Order of Structure|order $n$]] {{iff}} $\struct {G, \circ}$ is [[Definition:Group Isomorphism|isomorphic]] with the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $n$]] $\struct {\Z_n, +_n}$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\struct {G, \circ}$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $n$]].
+From [[List of Elements in Finite Cyclic Group]]:
+:$G = \set {a^0, a^1, a^2, \ldots, a^n}$
+where $a^0 = e, a^1 = a$.
+From the definition of [[Definition:Integers Modulo m|integers modulo $n$]], $\Z_n$ can be expressed as:
+:$\Z_n = \set {\eqclass 0 n, \eqclass 1 n, \ldots, \eqclass {n - 1} n}$
+where $\eqclass x n$ is the [[Definition:Residue Class|residue class of $x$ modulo $n$]].
+Let $\phi: G \to \Z_n$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall k \in \set {0, 1, \ldots, n - 1}: \map \phi {a^k} = \eqclass k n$
+By its definition it is clear that $\phi$ is a [[Definition:Bijection|bijection]].
+Also:
+{{begin-eqn}}
+{{eqn | l = \map \phi {a^r \circ a^s}
+      | r = \map \phi {a^{r + s} }
+      | c = [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]
+}}
+{{eqn | r = \eqclass {r + s} n
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \eqclass r n +_n \eqclass s n
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \map \phi {a^r} +_n \map \phi {a^s}
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+Thus the [[Definition:Morphism Property|morphism property]] of $\phi$ is demonstrated, and $\phi$ is thus a [[Definition:Group Homomorphism|group homomorphism]].
+By definition, a [[Definition:Group Isomorphism|group isomorphism]] is a [[Definition:Group Homomorphism|group homomorphism]] which is also a [[Definition:Bijection|bijection]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose $G$ is a [[Definition:Group|group]] such that $\phi: \Z_n \to G$ is a [[Definition:Group Isomorphism|group isomorphism]].
+Let $a = \map \phi {\eqclass 1 n}$.
+Let $g \in G$.
+Then $g = \map \phi {\eqclass k n}$ for some $\eqclass k n \in \Z_n$.
+Therefore:
+{{begin-eqn}}
+{{eqn | l = g
+      | r = \map \phi {\eqclass k n}
+      | c = 
+}}
+{{eqn | r = \map \phi {\eqclass 1 n +_n \cdots (k) \cdots +_n \eqclass 1 n}
+      | c = 
+}}
+{{eqn | r = a^k
+      | c = 
+}}
+{{end-eqn}}
+So every element of $G$ is a power of $a$.
+So by definition $G$ is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parallelogram Law (Hilbert Space)}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] with associated norm $\norm {\, \cdot \,}$.
+Let $f, g \in H$ be arbitrary.
+Then:
+:$\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \norm {f + g}^2 + \norm {f - g}^2
+      | r = \innerprod {f + g} {f + g} + \innerprod {f - g} {f - g}
+      | c = {{Defof|Inner Product Norm}}
+}}
+{{eqn | r = \innerprod f f + \innerprod f g + \innerprod g f + \innerprod g g + \innerprod f f - \innerprod f g - \innerprod g f + \innerprod g g
+      | c = Linearity of [[Definition:Inner Product|Inner Product]]
+}}
+{{eqn | r = 2 \innerprod f f + 2 \innerprod g g
+}}
+{{eqn | r = 2 \paren {\norm f^2 + \norm g^2}
+      | c = {{Defof|Inner Product Norm}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unique Point of Minimal Distance}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $h \in H$.
+Let $K \subseteq H$ be a [[Definition:Closed Set (Topology)|closed]], [[Definition:Convex Set (Vector Space)|convex]], [[Definition:Non-Empty Set|non-empty]] subset of $H$.
+Then there is a unique point $k_0 \in K$ such that:
+:$\norm {h - k_0} = \map d {h, K}$
+where $d$ denotes [[Definition:Distance between Element and Subset of Metric Space|distance to a set]].
+{{refactor|Own page, and prove it}}
+Furthermore, if $K$ is a [[Definition:Closed Linear Subspace|linear subspace]], this point is characterised by:
+:$\norm {h - k_0} = \map d {h, K} \iff \paren {h - k_0} \perp K$
+where $\perp$ signifies [[Definition:Orthogonal (Hilbert Space)|orthogonality]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf 0_H$ be the [[Definition:Zero Vector|zero]] of $H$.
+Since for every $k \in K$, we have:
+:$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$
+it follows that:
+:$\map d {h, K} = \map d {\mathbf 0_H, K - h}$
+{{WLOG}}, we may therefore assume that $h = \mathbf 0_H$.
+The problem has therefore reduced to finding $k_0 \in K$ such that:
+:$\norm {k_0} = \map d {\mathbf 0_H, K} = \inf \set {\norm k : k \in K}$
+Let $d = \map d {\mathbf 0_H, K}$.
+By definition of [[Definition:Infimum of Subset of Real Numbers|infimum]], there exists a [[Definition:Sequence|sequence]] $\sequence {k_n}_{n \mathop \in \N}$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {k_n} = d$
+By the [[Parallelogram Law (Hilbert Space)|Parallelogram Law]], we have that for all $m, n \in \N$:
+:$(1): \quad \norm {\dfrac {k_n - k_m} 2 } = \dfrac 1 2 \paren {\norm {k_n}^2 + \norm {k_m}^2} - \norm {\dfrac {k_n + k_m} 2 }^2$
+Since $K$ is [[Definition:Convex Set (Vector Space)|convex]], $\dfrac {k_n + k_m} 2 \in K$.
+Hence:
+:$\norm {\dfrac {k_n + k_m} 2 }^2 \ge d^2$
+Now given $\epsilon > 0$, choose $N$ such that for all $n \ge N$:
+:$\norm {k_n}^2 < d^2 + \epsilon$
+From $(1)$, it follows that:
+:$\norm {\dfrac {k_n - k_m} 2} < d^2 + \epsilon - d^2 = \epsilon$
+and hence that $\sequence {k_n}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]].
+Since $H$ is a [[Definition:Hilbert Space|Hilbert space]] and $K$ is [[Definition:Closed Set (Metric Space)|closed]], it follows that there is a $k_0 \in K$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} k_n = k_0$
+From [[Norm is Continuous]], we infer that $\norm {k_0} = d$.
+This demonstrates existence of $k_0$.
+For uniqueness, suppose that $h_0 \in K$ has $\norm {h_0} = d$.
+Since $K$ is [[Definition:Convex Set (Vector Space)|convex]], it follows that $\dfrac {h_0 + k_0} 2 \in K$.
+This implies that $\norm {\dfrac {h_0 + k_0} 2} \ge d$.
+Now from the [[Triangle Inequality]]:
+:$\norm {\dfrac {h_0 + k_0} 2} \le \dfrac {\norm {h_0} + \norm {k_0} } 2 = d$
+meaning that $\norm {\dfrac {h_0 + k_0} 2} = d$.
+Thus, the [[Parallelogram Law (Hilbert Space)|Parallelogram Law]] implies that:
+:$d^2 = \norm {\dfrac {h_0 + k_0} 2}^2 = d^2 - \norm {\dfrac {h_0 - k_0} 2}^2$
+from which we conclude that $h_0 = k_0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form}
+Tags: Arcsine Function, Primitive of Reciprocal of Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = \arcsin \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac 1 {\sqrt {a^2 - x^2} } \rd x
+      | r = \int \frac {\rd x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } } 
+      | c = factor $a^2$ out of the [[Definition:Radicand|radicand]]
+}}
+{{eqn | r = \int \frac {\rd x} {\sqrt{a^2} \sqrt {1 - \paren {\frac x a}^2} }
+      | c = 
+}}
+{{eqn | r = \frac 1 a \int \frac {\rd x} {\sqrt {1 - \paren {\frac x a}^2} }
+}}
+{{end-eqn}}
+[[Integration by Substitution|Substitute]]:
+:$\sin \theta = \dfrac x a \iff x = a \sin \theta$
+for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
+From [[Real Sine Function is Bounded]] and [[Shape of Sine Function]], this substitution is valid for all $x / a \in \openint {-1} 1$.
+[[Definition:By Hypothesis|By hypothesis]]:
+{{begin-eqn}}
+{{eqn | l = a^2
+      | o = > 
+      | r = x^2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | o = >
+      | r = \frac {x^2} {a^2}
+      | c = dividing both terms by $a^2$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | o = >
+      | r = \paren {\frac x a}^2
+      | c = [[Powers of Group Elements]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | o = >
+      | r = \size {\frac x a}
+      | c = taking the square root of both terms
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -1
+      | o = <
+      | r = \paren {\frac x a} < 1
+      | c = [[Negative of Absolute Value]]
+}}
+{{end-eqn}}
+so this substitution will not change the domain of the integrand.
+Then:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \sin \theta
+      | c = from above
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | r = a \cos \theta \frac {\rd \theta} {\rd x}
+      | c = differentiating {{WRT|Differentiation}} $x$, [[Derivative of Sine Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | l = \frac 1 a \int \frac 1 {\sqrt {1 - \paren {\frac x a}^2 } } \rd x
+      | r = \frac 1 a \int \frac {a \cos \theta} {\sqrt {1 - \sin^2 \theta} } \frac {\rd \theta} {\rd x} \rd x
+      | c = from above
+}}
+{{eqn | r = \frac a a \int \frac {\cos \theta} {\sqrt {1 - \sin^2 \theta} } \rd \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac {\cos \theta} {\sqrt {\cos^2 \theta} } \rd \theta
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = \int \frac {\cos \theta} {\size {\cos \theta} } \rd \theta
+      | c = 
+}}
+{{end-eqn}}
+We have defined $\theta$ to be in the [[Definition:Open Real Interval|open interval]] $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
+From [[Sine and Cosine are Periodic on Reals]], $\cos \theta > 0$ for the entire interval. Therefore the [[Definition:Absolute Value|absolute value]] is unnecessary, and the integral simplifies to:
+{{begin-eqn}}
+{{eqn | l = \int \rd \theta
+      | r = \theta + C
+}}
+{{end-eqn}}
+As $\theta$ was stipulated to be in the [[Definition:Open Real Interval|open interval]] $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
+:$\sin \theta = \dfrac x a \iff \theta = \arcsin \dfrac x a$
+The answer in terms of $x$, then, is:
+:$\displaystyle \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = \arcsin \frac x a + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inner Automorphism is Automorphism}
+Tags: Inner Automorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $x \in G$.
+Let $\kappa_x$ be the [[Definition:Inner Automorphism|inner automorphism]] of $x$ in $G$.
+Then $\kappa_x$ is an [[Definition:Group Automorphism|automorphism]] of $G$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\kappa_x: G \to G$ is a [[Definition:Mapping|mapping]] defined as:
+:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
+We need to show that $\kappa_x$ is an [[Definition:Group Automorphism|automorphism]].
+First we show $\kappa_x$ is a [[Definition:Group Homomorphism|homomorphism]].
+{{begin-eqn}}
+{{eqn | ll= \forall g, h \in G:
+      | l = \map {\kappa_x} g \map {\kappa_x} h
+      | r = \paren {x g x^{-1} } \paren {x h x^{-1} }
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | r = x g \paren {x^{-1} x} h x^{-1}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = x \paren {g e h} x^{-1}
+      | c = {{GroupAxiom|3}}
+}}
+{{eqn | r = x \paren {g h} x^{-1}
+      | c = {{GroupAxiom|2}}
+}}
+{{eqn | r = \map {\kappa_x} {g h}
+      | c = Definition of $\kappa_x$
+}}
+{{end-eqn}}
+Thus the [[Definition:Morphism Property|morphism property]] is demonstrated.
+Next we show that $\kappa_x$ is [[Definition:Injection|injective]].
+{{begin-eqn}}
+{{eqn | l = \map {\kappa_x} g
+      | r = \map {\kappa_x} h
+}}
+{{eqn | ll= \leadsto
+      | l = x g x^{-1}
+      | r = x h x^{-1}
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | ll= \leadsto
+      | l = g
+      | r = h
+      | c = [[Cancellation Laws]]
+}}
+{{end-eqn}}
+So $\kappa_x$ is [[Definition:Injection|injective]].
+Finally we show that $\kappa_x$ is [[Definition:Surjection|surjective]].
+Note that $\forall h \in G: x^{-1} h x \in G$ from fact that $G$ is a [[Definition:Group|group]] and therefore [[Definition:Closure (Abstract Algebra)|closed]]. So:
+{{begin-eqn}}
+{{eqn | ll= \forall h \in G:
+      | l = \map {\kappa_x} {x^{-1} h x}
+      | r = x \paren {x^{-1} h x} x^{-1}
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | r = \paren {x x^{-1} } h \paren {x x^{-1} }
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = e h e
+      | c = {{GroupAxiom|3}}
+}}
+{{eqn | r = h
+      | c = {{GroupAxiom|2}}
+}}
+{{end-eqn}}
+Thus every element of $G$ is the image of some element of $G$ under $\kappa_x$ (that is, of $x^{-1} h x$), and [[Definition:Surjection|surjectivity]] is proved.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Order of Automorphism Group}
+Tags: Automorphism Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Structure|order]] is greater than $2$.
+Let $\Aut G$ be the [[Definition:Automorphism Group of Group|automorphism group]] of $G$.
+Then the [[Definition:Order of Structure|order]] of $\Aut G$ is greater than $1$.
+\end{theorem}
+
+\begin{proof}
+There are $3$ cases, and for each case we find an [[Definition:Group Automorphism|automorphism]] that is not the [[Definition:Identity Mapping|identity automorphism]].
+=== Case $1$: $G$ is non-[[Definition:Abelian Group|Abelian]] ===
+There exists some elements $g, h \in G$ such that $g h \ne h g$.
+Then the [[Definition:Inner Automorphism|inner automorphism]]:
+:$x \mapsto g x g^{-1}$
+gives an [[Definition:Group Automorphism|automorphism]] that is not the [[Definition:Identity Mapping|identity automorphism]], since:
+:$g h g^{-1} \ne h$
+Hence $\order {\Aut G} > 1$.
+{{qed|lemma}}
+=== Case $2$: $G$ has an element with [[Definition:Order of Group Element|order]] greater than $2$ ===
+Suppose $g \in G$ has order $k$, where $k > 2$.
+Then $g \ne g^{-1}$.
+Thus the [[Definition:Inverse Element|inverse]] [[Definition:Group Automorphism|automorphism]]:
+:$x \mapsto x^{-1}$
+is not the [[Definition:Identity Mapping|identity automorphism]].
+Hence $\order {\Aut G} > 1$.
+{{qed|lemma}}
+=== Case $3$: $G$ is [[Definition:Abelian Group|Abelian]] and has no element with [[Definition:Order of Group Element|order]] greater than $2$ ===
+By [[Fundamental Theorem of Finite Abelian Groups]], write:
+:$G = C_{a_1} \times \dots \times C_{a_k}$
+where $C_{a_i}$ are [[Definition:Cyclic Group|cyclic groups]] of [[Definition:Prime Power|prime power]] [[Definition:Order of Structure|order]].
+Since all elements of $G$ has [[Definition:Order of Group Element|order]] not more than $2$, $a_i = 2$ for all $i$, that is:
+:$G = C_2^k$
+As $\order G = 2^k > 2$:
+:$k > 1$
+Writing each element of $G$ as $\tuple {g_1, \dots, g_k}$ where $g_i = 0$ or $1$, the swapping [[Definition:Group Automorphism|automorphism]]:
+:$\tuple {g_1, g_2, \dots, g_k} \mapsto \tuple {g_2, g_1, \dots, g_k}$
+is an [[Definition:Group Automorphism|automorphism]] that is not the [[Definition:Identity Mapping|identity automorphism]]. 
+Hence $\order {\Aut G} > 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic Subgroup is Transitive}
+Tags: Characteristic Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$.
+Let $K$ be a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $H$.
+Then $K$ is a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi: G \to G$ be a [[Definition:Group Automorphism|group automorphism]].
+Since $H$ is a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$, we have:
+:$\phi \sqbrk H = H$
+Thus, from [[Group Homomorphism Preserves Subgroups]], $\phi {\restriction_H}$, the [[Definition:Restriction of Mapping|restriction]] of $\phi$ to $H$, is an [[Definition:Group Automorphism|automorphism]] of $H$.
+Now since $K$ is a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $H$, we have that:
+:$\phi {\restriction_H} \sqbrk K = K$
+but this immediately implies that:
+:$\phi \sqbrk K = K$
+by definition of the [[Definition:Restriction of Mapping|restriction]] $\phi {\restriction_H}$.
+That is, $K$ is a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$ as well.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Orthogonal Projection}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $K$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+Let $P_K$ denote the [[Definition:Orthogonal Projection|orthogonal projection]] on $K$.
+Then $P_K$ has the following properties:
+:$(1):\qquad P_K$ is a [[Definition:Linear Transformation|linear transformation]] on $H$.
+:$(2):\qquad \forall h\in H: \left\|{P_K(h)}\right\| \le \left\|{h}\right\|$
+:$(3):\qquad P_K \circ P_K = P_K$
+:$(4):\qquad \ker P_K = K^{\perp}$ and $\operatorname{ran} P_K = K$
+{{wtd|I am not entirely happy with this setup of the page, with four results stated. Strictly speaking, each should have its own page. But I am not up to thinking of a name for each of the results separately. Probably I will put up four transcluded proof pages. Lastly, note that these four results are one theorem $I.2.7$ in Conway. --LF<br/><br/>Four transcluded subpages is a technique that's been used before for this sort of linked set of results. The fact they're all part of the same proof in Conway is incidental. -- pm}}
+\end{theorem}<|endoftext|>
+\section{Quotient Theorem for Group Homomorphisms}
+Tags: Group Homomorphisms, Quotient Groups, Quotient Theorem for Group Homomorphisms
+
+\begin{theorem}
+Let $\phi: G \to G'$ be a [[Definition:Group Homomorphism|(group) homomorphism]] between two [[Definition:Group|groups]] $G$ and $G'$.
+Then $\phi$ can be decomposed into the form:
+:$\phi = \alpha \beta \gamma$
+where:
+:$\alpha: \Img \phi \to G'$ is a [[Definition:Group Monomorphism|monomorphism]]
+:$\beta: G / \map \ker \phi \to \Img \phi$ is an [[Definition:Group Isomorphism|isomorphism]]
+:$\gamma: G \to G / \map \ker \phi$ is an [[Definition:Group Epimorphism|epimorphism]].
+\end{theorem}
+
+\begin{proof}
+=== Monomorphism ===
+The [[Definition:Mapping|mapping]] $\alpha$ is the [[Definition:Inclusion Mapping|inclusion mapping]] defined as $i: \Img \phi \to G'$:
+:$\forall x \in \Img \phi: \map i x = x$
+From [[Inclusion Mapping is Monomorphism]], it follows that $\alpha$ is a [[Definition:Group Monomorphism|monomorphism]].
+{{qed|lemma}}
+=== Isomorphism ===
+From the [[First Isomorphism Theorem for Groups]]:
+:$\Img \phi \cong G / \map \ker \phi$
+for any [[Definition:Group Homomorphism|homomorphism]] $\phi$.
+That is, the [[Definition:Image of Mapping|image]] of $\phi$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Quotient Group|quotient group]] of $G$ by the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
+Thus $\beta$ is such an [[Definition:Group Isomorphism|isomorphism]].
+{{qed|lemma}}
+=== Epimorphism ===
+The mapping $\gamma: G \to G / \map \ker \phi$ is given by:
+:$\forall a \in G: \map \gamma a = a \, \map \ker \phi$
+where $a \, \map \ker \phi$ is the [[Definition:Left Coset|left coset]] of $\map \ker \phi$ by $a$.
+This is justified by [[Kernel is Normal Subgroup of Domain]].
+By definition, each [[Definition:Left Coset|(left) coset]] of $\map \ker \phi$ is the [[Definition:Left Coset|(left) coset]] of each of its [[Definition:Element|elements]].
+So $\gamma$ is a [[Definition:Surjection|surjection]].
+$\gamma$ is shown to be a [[Definition:Group Homomorphism|homomorphism]] thus:
+{{begin-eqn}}
+{{eqn | l = \map \gamma {a b}
+      | r = \paren {a b} \, \map \ker \phi
+      | c = 
+}}
+{{eqn | r = \paren {a \, \map \ker \phi} \paren {b \, \map \ker \phi}
+      | c = {{Defof|Coset Product}}
+}}
+{{eqn | r = \map \gamma a \map \gamma b
+      | c = 
+}}
+{{end-eqn}}
+thus demonstrating the [[Definition:Morphism Property|morphism property]].
+A [[Definition:Group Homomorphism|homomorphism]] which is [[Definition:Surjection|surjective]] is an [[Definition:Group Epimorphism|epimorphism]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+This theorem can be illustrated by means of the following [[Definition:Commutative Diagram|commutative diagram]]:
+::$\begin{xy}\xymatrix@L+2mu@+1em{
+ G \ar@{-->}[r]^*{\phi}
+     \ar[d]^*{\gamma}
+&
+ G'
+\\ 
+ G / \map \ker \phi \ar[r]_*{\beta}^*{\cong}
+&
+ \Img \phi \ar[u]^*{\alpha}
+}\end{xy}$
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping is Monomorphism}
+Tags: Monomorphisms, Inclusion Mappings
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $\left({T, \circ}\right)$ be an [[Definition:Algebraic Substructure|algebraic substructure]] of $S$.
+Let $\iota: T \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $T$ to $S$.
+Then $\iota$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Inclusion Mapping is Injection|inclusion mapping is an injection]].
+Now let $x, y \in T$:
+{{begin-eqn}}
+{{eqn | l=\iota \left({x \circ y}\right)
+      | r=x \circ y
+      | c=Definition of [[Definition:Inclusion Mapping|inclusion mapping]]
+}}
+{{eqn | r=\iota \left({x}\right) \circ \iota \left({y}\right)
+      | c=Definition of [[Definition:Inclusion Mapping|inclusion mapping]]
+}}
+{{end-eqn}}
+demonstrating that $\iota$ has the [[Definition:Morphism Property|morphism property]].
+So $\iota$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also an [[Definition:Injection|injection]].
+Thus by definition, $\iota$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]].
+{{qed}}
+[[Category:Monomorphisms]]
+[[Category:Inclusion Mappings]]
+5r56ehrqxt3wyhwk48szr9ve4rktbtf
+\end{proof}<|endoftext|>
+\section{Function is Odd Iff Inverse is Odd}
+Tags: Inverse Mappings, Odd Functions
+
+\begin{theorem}
+Let $f$ be an [[Definition:Odd Function|odd]] [[Definition:Real Function|real function]] with an [[Definition:Inverse Mapping|inverse]] $f^{-1}$.
+Then $f^{-1}$ is also [[Definition:Odd Function|odd]].
+\end{theorem}
+
+\begin{proof}
+First note that we have:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = f \left({x}\right)
+      | c = 
+}}
+{{eqn | ll = \implies
+      | l = x
+      | r = f^{-1} \left({y}\right)
+      | c = [[Image of Element under Inverse Mapping]]
+}}
+{{eqn | ll = \implies
+      | l = -x
+      | r = -f^{-1} \left({y}\right)
+      | c = Multiply both sides by $-1$
+}}
+{{eqn-intertext|Then:}}
+{{eqn | l = -y 
+      | r = f \left({-x}\right)
+      | c = {{Defof|Odd Function}}
+}}
+{{eqn | ll = \implies
+      | l = f^{-1} \left({-y}\right)
+      | r = f^{-1} \circ f \left({-x}\right)
+      | c = because $f^{-1}$ is a [[Definition:Mapping|mapping]] 
+}}
+{{eqn | r = -x
+      | c = [[Composite of Bijection with Inverse is Identity Mapping]]
+}}
+{{eqn | r = -f^{-1} \left({y}\right)
+      | c = from above
+}}
+{{end-eqn}}
+The result follows by definition of an [[Definition:Odd Function|odd function]]. 
+{{qed}}
+[[Category:Inverse Mappings]]
+[[Category:Odd Functions]]
+s7y3uva37y1vsh20h9fu1t2cy1s7mfz
+\end{proof}<|endoftext|>
+\section{Integral Resulting in Arcsecant}
+Tags: Integral Calculus
+
+\begin{theorem}
+:$\displaystyle \int \frac 1 {x \sqrt{x^2 - a^2} }\ \mathrm dx = \begin{cases}
+\dfrac 1 {\left\vert{a}\right\vert} \operatorname {arcsec} \dfrac x {\left\vert{a}\right\vert} + C & : x > \left\vert{a}\right\vert \\
+-\dfrac 1 {\left\vert{a}\right\vert} \operatorname {arcsec} \dfrac x {\left\vert{a}\right\vert} + C & : x < -\left\vert{a}\right\vert
+\end{cases}$
+where $a$ is a [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac 1 {x \sqrt{x^2 - a^2} } \ \mathrm dx
+      | r = \int \frac 1 {x \sqrt{a^2 \left({\frac {x^2}{a^2} - 1}\right)} } \ \mathrm dx
+      | c = factor $a^2$ out of the [[Definition:Radicand|radicand]]
+}}
+{{eqn | r = \int \frac 1 {x \sqrt{a^2} \sqrt{\left({\frac x a}\right)^2 - 1} } \ \mathrm dx
+      | c = 
+}}
+{{eqn | r = \frac 1 {\left \vert {a} \right \vert} \int \frac 1 {x \sqrt{\left({\frac x a}\right)^2 - 1} } \ \mathrm dx 
+}}
+{{end-eqn}}
+[[Integration by Substitution|Substitute]]:
+:$\sec \theta = \dfrac x {\left\vert{a}\right\vert} \iff \left\vert{a}\right\vert \sec \theta = x$
+for $\theta \in \left({0 \,.\,.\, \dfrac \pi 2}\right) \cup \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.
+This substitution is valid for all $\dfrac x {|a|} \in \R \setminus \left({-1 \,.\,.\, 1}\right)$.
+{{MissingLinks}}
+[[Definition:By Hypothesis|By hypothesis]]:
+{{begin-eqn}}
+{{eqn | l=\left({x > \left\vert{a}\right\vert}\right)
+      | o=\lor
+      | r=\left({x < - \left\vert{a}\right\vert}\right)
+      | c=
+}}
+{{eqn | ll=\iff
+      | l=\left({\dfrac x {\left\vert{a}\right\vert} > 1}\right)
+      | o=\lor
+      | r=\left({\dfrac x {\left\vert{a}\right\vert} < -1}\right)
+      | c=
+}}
+{{end-eqn}}
+so this substitution will not change the domain of the integrand.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \left \vert{a}\right \vert \sec \theta 
+      | r = x
+      | c = from above
+}}
+{{eqn | ll = \implies
+      | l = \left \vert {a}\right \vert \sec \theta \tan \theta \frac {\mathrm d \theta}{\mathrm dx}
+      | r = 1
+      | c = differentiate [[Definition:Differentiation With Respect To|WRT]] $x$, [[Derivative of Secant Function]], [[Chain Rule for Derivatives]]
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = \int \frac 1 {x\sqrt{x^2 - a^2} } \ \mathrm dx
+      | r =  \frac 1 {\left \vert {a} \right \vert} \int \frac {\left \vert {a}\right \vert \sec \theta \tan \theta}{\left \vert {a}\right \vert \sec \theta \sqrt{\sec^2\theta - 1} } \frac {\mathrm d \theta}{\mathrm dx} \mathrm dx 
+      | c = from above
+}}
+{{eqn | r = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta}{\sqrt{\sec^2\theta - 1} }\ \mathrm d \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r =  \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta}{\sqrt{\tan^2 \theta}\ \mathrm d \theta}
+      | c = [[Sum of Squares of Sine and Cosine#Corollaries|corollary to sum of squares of sine and cosine]]
+}}
+{{eqn | r = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta
+      | c = 
+}}
+{{end-eqn}}
+By [[Shape of Tangent Function]] and the stipulated definition of $\theta$:
+:$(A): \quad \dfrac x {\left\vert{a}\right\vert} > 1 \iff \theta \in  \left({0 \,.\,.\, \dfrac \pi 2}\right)$
+and
+:$(B): \quad \dfrac x {\left\vert{a}\right\vert} < -1 \iff \theta \in \left({\dfrac \pi 2 \,.\,.\, \pi}\right)$
+If $(A)$:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta
+      | r = \frac 1 {\left \vert {a} \right \vert} \int \mathrm d \theta
+      | c = definition of [[Definition:Absolute Value|absolute value]]
+}}
+{{eqn | r = \frac 1 {\left \vert {a} \right \vert} \theta + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \frac 1 {\left \vert {a} \right \vert} \operatorname{arcsec} \frac x {\left \vert {a} \right \vert} + C
+      | c = definition of [[Definition:Arcsecant|arcsecant]]
+}}
+{{end-eqn}}
+If $(B)$:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {\left \vert {a} \right \vert} \int \frac {\tan \theta} {\left \vert \tan \theta \right \vert} \ \mathrm d \theta
+      | r = \frac 1 {\left \vert {a} \right \vert} \int -1 \ \mathrm d \theta
+      | c = definition of [[Definition:Absolute Value|absolute value]]
+}}
+{{eqn | r = -\frac 1 {\left \vert {a} \right \vert} \theta + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = -\frac 1 {\left \vert {a} \right \vert} \operatorname{arcsec} \frac x {\left \vert {a} \right \vert} + C
+      | c = definition of [[Definition:Arcsecant|arcsecant]]
+}}
+{{end-eqn}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Schreier-Zassenhaus Theorem}
+Tags: Normal Series
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $\HH_1$ and $\HH_2$ be two [[Definition:Normal Series|normal series]] for $G$.
+Then $\HH_1$ and $\HH_2$ have [[Definition:Refinement of Normal Series|refinements]] of equal [[Definition:Length of Normal Series|length]] whose [[Definition:Factor of Normal Series|factors]] are [[Definition:Group Isomorphism|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+Suppose that:
+:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
+and:
+:$(2): \quad \set e = H_0 \lhd H_1 \lhd \cdots \lhd H_{m - 1} \lhd H_m = G$
+are two [[Definition:Normal Series|normal series]] for $G$.
+Let a new series be formed:
+:$(3): \quad \set e = \hat G_0 \subseteq \hat G_1 \subseteq \cdots \subseteq \hat G_{n m - 1} \subseteq \hat G_{n m} = G$
+such that:
+:$\hat G_k = G_q \paren {G_{q + 1} \cap H_r}$
+for $k = q m + r$, where $0 \le q < n$ and $0 \le r \le m$.
+For completeness, define $\hat G_{n m} = G_n$.
+Note that, by the above construction:
+:$\hat G_{q m} = G_q \paren {G_{q + 1} \cap H_0} = G_q \paren {\set e} = G_q$
+This implies that each group in $(1)$ also appears in $(3)$.
+It needs to be demonstrated that $\hat G_k$ is [[Definition:Well-Defined Mapping|well-defined]].
+Consider the ambiguity of $k = q m + r$ in the following cases:
+:When $m \nmid k$, there is none.
+:When $m \divides k$, we have:
+:::$k = q m + m = \paren {q + 1} m$
+For $(3)$ to be well-defined, we require:
+:$G_q \paren {G_{q + 1} \cap H_m} = G_{q + 1} \paren {G_{q + 2} \cap H_0}$
+Note that the {{RHS}} is just:
+:$G_{q + 1} \paren {G_{q + 2} \cap \set e} = G_{q + 1} \paren {\set e} = G_{q + 1}$
+The {{LHS}} is:
+:$G_q \paren {G_{q + 1} \cap H_m} = G_q \paren {G_{q + 1} \cap G} = G_q G_{q + 1} = G_{q + 1}$
+Thus the construction is verified.
+It next needs to be demonstrated that:
+:$\hat G_k \lhd \hat G_{k + 1}$
+Choose the representation of $k$ in which $r < m$.
+It follows that:
+:$\hat G_{k + 1} = G_q \paren {G_{q + 1} \cap H_{r + 1} }$
+Since $H_{r + 1} \rhd H_r$:
+:$\hat G_{k + 1} = G_q \paren {G_{q + 1} \cap H_{r + 1} } \rhd G_q \paren {G_{q + 1} \cap H_r} = \hat G_k$
+That is, $\hat G_k$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\hat G_{k + 1}$.
+$(A): \quad$ But note that $(3)$ may not actually be a [[Definition:Normal Series|normal series]] as it is possible that $\hat G_k = \hat G_{k + 1}$ for some $k$.
+Having created series $(3)$, we use the same procedure to form:
+:$(4): \quad \set e = \hat H_0 \subseteq \hat H_1 \subseteq \cdots \subseteq \hat H_{n m - 1} \subseteq \hat H_{n m} = G$
+such that:
+:$\hat H_k = H_q \paren {H_{q + 1} \cap G_r}$
+for $k = q n + r$, where $0 \le q < m$ and $0 \le r \le n$.
+The same statements that were made about $(3)$ also hold about $(4)$.
+Now let $k = u m + v$ and $l = v n + u$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \frac {\hat G_{k + 1} } {\hat G_k}
+      | r = \frac {G_u \paren {G_{u + 1} \cap H_{v + 1} } } {G_u \paren {G_{u + 1} \cap H_v} }
+      | c = from above
+}}
+{{eqn | o = \cong
+      | r = \frac {H_v \paren {H_{v + 1} \cap G_{u + 1} } } {H_v \paren {H_{v + 1} \cap G_u} }
+      | c = [[Zassenhaus Lemma]]
+}}
+{{eqn | r = \frac {\hat H_{l + 1} } {\hat H_l}
+      | c = from above
+}}
+{{end-eqn}}
+So $(3)$ and $(4)$ have [[Definition:Group Isomorphism|isomorphic]] [[Definition:Factor of Normal Series|factors]].
+In $(A)$ it was remarked that some of the [[Definition:Factor of Normal Series|factors]] in these [[Definition:Group Isomorphism|isomorphic]] series may have redundant elements where $\hat G_k = \hat G_{k + 1}$ and similarly for $\hat H_l = \hat H_{l + 1}$.
+As the series consist of isomorphic elements, all we now have to do is remove these redundant elements from both (they will occur at the same place).
+Hence we end up with two [[Definition:Refinement of Normal Series|refinements]] of equal [[Definition:Length of Normal Series|length]] whose [[Definition:Factor of Normal Series|factors]] are [[Definition:Group Isomorphism|isomorphic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Trigonometric Function of Reciprocal}
+Tags: Inverse Trigonometric Functions, Reciprocals
+
+\begin{theorem}
+For all $x \in \R$ for which the expressions above are defined:
+\end{theorem}<|endoftext|>
+\section{Jordan-Hölder Theorem}
+Tags: Normal Series
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $\HH_1$ and $\HH_2$ be two [[Definition:Composition Series|composition series]] for $G$.
+Then:
+:$\HH_1$ and $\HH_2$ have the same [[Definition:Length of Normal Series|length]]
+:Corresponding [[Definition:Factor of Normal Series|factors]] of $\HH_1$ and $\HH_2$ are [[Definition:Group Isomorphism|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+By the [[Schreier-Zassenhaus Theorem]], two [[Definition:Normal Series|normal series]] have [[Definition:Refinement of Normal Series|refinements]] of equal [[Definition:Length of Normal Series|length]] whose [[Definition:Factor of Normal Series|factors]] are [[Definition:Group Isomorphism|isomorphic]].
+But from the definition of [[Definition:Composition Series|composition series]], $\HH_1$ and $\HH_2$ have no [[Definition:Proper Refinement of Normal Series|proper refinements]].
+Hence any such [[Definition:Refinement of Normal Series|refinements]] must be identical to $\HH_1$ and $\HH_2$ themselves.
+{{qed}}
+{{Namedfor|Marie Ennemond Camille Jordan|name2 = Otto Ludwig Hölder|cat = Jordan|cat2 = Hölder}}
+\end{proof}<|endoftext|>
+\section{Double Orthocomplement is Closed Linear Span}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \subseteq \HH$ be a [[Definition:Subset|subset]] of $\HH$.
+Then the following identity holds:
+:$\paren {A^\perp}^\perp = \vee A$
+Here $A^\perp$ denotes [[Definition:Orthocomplement|orthocomplementation]], and $\vee A$ denotes the [[Definition:Closed Linear Span|closed linear span]].
+\end{theorem}<|endoftext|>
+\section{Orthocomplement is Closed Linear Subspace}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \subseteq H$ be a [[Definition:Subset|subset]] of $H$.
+Then the [[Definition:Orthocomplement|orthocomplement]] $A^\perp$ of $A$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n} \subset A^\perp$ be a [[Definition:Convergent Sequence/Hilbert Space|convergent]] [[Definition:Sequence|sequence]].
+Let $x$ be the [[Definition:Limit of Sequence|limit]] of $\sequence {x_n}$.
+Then, by the definition of [[Definition:Orthocomplement|orthocomplement]]:
+:$\forall n \in \N, y \in A: \innerprod {x_n} y = 0$
+{{LinkWanted|The notation $\innerprod {x_n} y$ is to be specified}}
+Passing to the [[Definition:Limit of Sequence|limit]], from [[Inner Product is Continuous]] we have:
+:$\forall y \in A: \innerprod x y = 0$
+So:
+:$x \in A^\perp$
+This shows that $A^\perp$ is [[Definition:Closed Linear Subspace|closed]].
+{{qed}}
+[[Category:Hilbert Spaces]]
+7kd5zfnewigrdywd9swhwm13j4dw7qr
+\end{proof}<|endoftext|>
+\section{Linear Subspace Dense iff Zero Orthocomplement}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $K$ be a [[Definition:Vector Subspace|linear subspace]] of $H$.
+Then $K$ is [[Definition:Everywhere Dense|everywhere dense]] {{iff}} $K^\perp = \left({0}\right)$, where $K^\perp$ is the [[Definition:Orthocomplement|orthocomplement]] of $K$, and $\left({0}\right)$ denotes the [[Definition:Zero Subspace|zero subspace]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that $K$ is [[Definition:Everywhere Dense|everywhere dense]] and $x \in K^\perp$. Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x \in K^\perp
+}}
+{{eqn | ll = \leadsto
+      | lo = \forall y \in K :
+      | o = 
+      | r = x \perp y
+}}
+{{eqn | ll = \leadsto
+      | lo = \forall y \in H :
+      | o = 
+      | r = x \perp y
+      | c = $K$ is everywhere dense, and [[Inner Product is Continuous]]
+}}
+{{eqn | ll = \leadsto
+      | lo = 
+      | o = 
+      | r = x = 0
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume $K^\perp = 0$.
+Then by [[Double Orthocomplement is Closed Linear Span]], $\vee K=\paren{0}^\perp=H$.
+Hence $K$ is [[Definition:Everywhere Dense|everywhere dense]].
+{{qed|lemma}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Group has Composition Series}
+Tags: Normal Series
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Then $G$ has a [[Definition:Composition Series|composition series]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
+Either $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] or it does not.
+If not, then:
+:$\set e \lhd G$
+is the [[Definition:Composition Series|composition series]] for $G$.
+Otherwise, $G$ has one or more [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]].
+Of these, one or more will have a [[Definition:Maximum Element|maximum]] [[Definition:Order of Group|order]].
+Select one of these and call it $G_1$.
+Again, either $G_1$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] or it does not.
+If not, then:
+:$\set e \lhd G_1 \lhd G$
+is a [[Definition:Composition Series|composition series]] for $G$.
+By the [[Jordan-Hölder Theorem]], there can be no other [[Definition:Composition Series|composition series]] which is longer.
+As $G_1$ is a [[Definition:Proper Subgroup|proper subgroup]] of $G$:
+:$\order {G_1} < \order G$
+where $\order G$ denotes the [[Definition:Order of Group|order]] of $G$.
+Again, if $G_1$ has one or more [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]], one or more will have a [[Definition:Maximum Element|maximum]] [[Definition:Order of Group|order]].
+Select one of these and call it $G_2$.
+Thus we form a [[Definition:Normal Series|normal series]]:
+:$\set e \lhd G_2 \lhd G_1 \lhd G$
+The process can be repeated, and at each stage a [[Definition:Normal Subgroup|normal subgroup]] is added to the series of a smaller 
+[[Definition:Order of a Structure|order]] than the previous one.
+This process can not continue infinitely.
+Eventually a $G_n$ will be encountered which has no [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]].
+Thus a [[Definition:Composition Series|composition series]]:
+:$\set e \lhd G_n \cdots \lhd G_2 \lhd G_1 \lhd G$
+will be the result.
+{{qed}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Length of Subgroup Plus Length of Quotient Group}
+Tags: Normal Subgroups, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then:
+:$\map l G = \map l H + \map l {G / H}$
+where:
+:$\map l G$ denotes the [[Definition:Length of Group|length]] of $G$
+:$G / H$ denotes the [[Definition:Quotient Group|quotient group]] of $G$ by $H$.
+\end{theorem}<|endoftext|>
+\section{Length of Abelian Group}
+Tags: Abelian Groups
+
+\begin{theorem}
+Let $G$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Order of Structure|order]] is $n$.
+Let $n$ have the [[Definition:Prime Decomposition|prime decomposition]]:
+:$\displaystyle n = \prod_{i \mathop = 1}^r p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
+where $p_1 < p_2 < \cdots < p_r$ are distinct [[Definition:Prime Number|primes]] and $k_1, k_2, \ldots, k_r$ are [[Definition:Positive Integer|positive integers]].
+Then the [[Definition:Length of Group|length]] of $G$ is given by:
+:$\displaystyle \map l G = \sum_{i \mathop = 1}^r k_i = k_1 + k_2 + \cdots + k_r$
+\end{theorem}<|endoftext|>
+\section{Condition for Composition Series}
+Tags: Normal Series
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Then:
+:a [[Definition:Normal Series|normal series]] $\HH$ for $G$ is a [[Definition:Composition Series|composition series]] for $G$
+{{iff}}:
+:every [[Definition:Factor of Normal Series|factor group]] of $\HH$ is a [[Definition:Simple Group|simple group]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let:
+:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
+be a [[Definition:Normal Series|normal series]] for $G$.
+=== Necessary Condition ===
+Suppose there exists $k$ such that $G_{k + 1} / G_k$ is not a [[Definition:Simple Group|simple group]].
+Then there exists a [[Definition:Normal Subgroup|normal subgroup]] $G'$ such that:
+:$\set e \lhd G' \lhd G_{k + 1} / G_k$
+It follows that:
+:$G_k \lhd G'' \lhd G_{k + 1}$
+where:
+:$G''$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G_{k + 1}$
+and:
+:$G' = G'' / G_{k + 1}$
+{{explain|why?}}
+Thus $(1)$ has a [[Definition:Proper Refinement of Normal Series|proper refinement]] and so is not a [[Definition:Composition Series|composition series]].
+By the [[Rule of Transposition]] it follows that if [[Definition:Normal Series|normal series]] is a [[Definition:Composition Series|composition series]], every [[Definition:Factor of Normal Series|factor group]] of that normal series is a [[Definition:Simple Group|simple group]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $(1)$ is not a [[Definition:Composition Series|composition series]] for $G$.
+Then a [[Definition:Proper Refinement of Normal Series|proper refinement]] of $(1)$ can be constructed by inserting a [[Definition:Group|group]] $G''$ into the series somewhere, for example:
+:$G_k \lhd G'' \lhd G_{k + 1}$
+It follows that $G'' / G_k$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G_{k + 1} / G''$.
+{{explain|There's a specific result demonstrating that fact.}}
+Thus, by definition, $G_{k + 1} / G_k$ is not a [[Definition:Simple Group|simple group]].
+Thus it has been shown that if a [[Definition:Normal Series|normal series]] is not a [[Definition:Composition Series|composition series]], then it contains at least one [[Definition:Factor of Normal Series|factor group]] which is not a [[Definition:Simple Group|simple group]]
+By the [[Rule of Transposition]] it follows that if every [[Definition:Factor of Normal Series|factor group]] of a [[Definition:Normal Series|normal series]] is a [[Definition:Simple Group|simple group]], then that normal series is a [[Definition:Composition Series|composition series]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Factors of Composition Series for Prime Power Group}
+Tags: Normal Series, Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] such that $\order G = p^n$ where $p$ is a [[Definition:Prime Number|prime number]].
+Then $G$ has a [[Definition:Composition Series|composition series]] in which each [[Definition:Factor of Normal Series|factor group]] is [[Definition:Cyclic Group|cyclic]] of [[Definition:Order of Group|order]] $p$.
+\end{theorem}
+
+\begin{proof}
+From [[Composition Series of Group of Prime Power Order]], $G$ has a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]]:
+:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
+such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is [[Definition:Cyclic Group|cyclic]] and of [[Definition:Order of Group|order]] $p$.
+From [[Prime Group is Simple]] it follows that $G_{k + 1} / G_k$ is a [[Definition:Simple Group|simple group]] for all $k$.
+From [[Prime Group is Cyclic]] it follows that $G_{k + 1} / G_k$ is [[Definition:Cyclic Group|cyclic]].
+The result follows from [[Condition for Composition Series]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Linear Functionals}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $L$ be a [[Definition:Linear Functional|linear functional]] on $H$.
+Then the following four statements are equivalent:
+:$(1):\quad L$ is [[Definition:Continuous Mapping (Topology)|continuous]]
+:$(2):\quad L$ is [[Definition:Continuous Mapping (Topology)#Continuous at a Point|continuous]] at $\mathbf{0}_H$
+:$(3):\quad L$ is [[Definition:Continuous Mapping (Topology)#Continuous at a Point|continuous]] at some point
+:$(4):\quad \exists c > 0: \forall h \in H: \left|{Lh}\right| \le c \left\|{h}\right\|$
+\end{theorem}<|endoftext|>
+\section{Equivalence of Definitions of Closed Linear Span}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\Bbb F \in \set {\R, \C}$, and let $A \subseteq H$ be a [[Definition:Subset|subset]].
+{{TFAE|def = Closed Linear Span|view = closed linear span of $A$}}
+:$(1): \quad \displaystyle \vee A = \bigcap \Bbb M$, where $\Bbb M$ consists of all [[Definition:Closed Linear Subspace|closed linear subspaces]] $M$ of $H$ with $A \subseteq M$
+:$(2): \quad \displaystyle \vee A$ is the smallest [[Definition:Closed Linear Subspace|closed linear subspace]] $M$ of $H$ with $A \subseteq M$
+:$(3): \quad \displaystyle \vee A = \map \cl {\set {\sum_{k \mathop = 1}^n \alpha_k f_k: n \in \N_{\ge 1}, \alpha_i \in \Bbb F, f_i \in A} }$, where $\cl$ denotes [[Definition:Closure (Topology)|closure]]
+\end{theorem}
+
+\begin{proof}
+Let the proposition (1) holds; assume the closed linear subspace $M'$ contains the set $A$, then because $M' \in \Bbb M$, we have
+:$ \qquad \vee A \subseteq M' $
+The intersection of arbitrary family of subspaces is a subspace. For suppose $\mathcal C$ is a family of subspaces.
+Denotes $\bigcap \mathcal C =\{f\in H|\text{ for any }V\in \mathcal C \text{, there }f\in V \}$
+If $f\in \bigcap \mathcal C$, then for any $V\in \mathcal C $, $f\in V$ there $af\in V$ for $a\in \Bbb F$. If $f\text{ , }g\in \bigcap \mathcal C$,
+we have for any $V\in \mathcal C $, $f+g\in V$.
+Therefore, $\vee A$ is a subspace. It is closed, as intersection of arbitrary family of closed sets is closed. The choice of $M'$ is arbitrary. Hence, (2) holds.
+This is the proof (1)=>(2).
+Next if (2) holds. Since $A\subseteq \vee A$, $\vee A \in \Bbb M$. 
+$\vee A$ is the smallest one in $\Bbb M$; hence
+:$\qquad \displaystyle \vee A = \bigcap \Bbb M$
+We have established the equivalence between (1) and (2).
+Finally we come to (3):
+We prove that $\cl(\text{span}(A))$ is a subspace.
+Let $f\in \cl(\text{span}(A))$. 
+$H$, the Hilbert space, is a metric space, which satisfies the first countability axiom. For reason then one shall consider the collection of 
+open balls which all are centred at a point and all have their length of radiuses be rational.
+We then need "[[The sequence lemma|the sequence lemma]]," which is:
+:Let $A$ be a subset of a topological space $X$. If there is a sequence of points of $A$ converging to $x$, then $x\in \cl(A)$;
+:the converse holds if $X$ is first-countable.
+So there is a sequence $\{f_i\}$ in $\text{span}(A)$ that its limit is $f$. By the continuity of the function of the multiplication of numbers of
+the field $\Bbb F$ and points in $H$, the sequence $\{af_i\}$ with $a\in \Bbb F$ converges to $af$. Because all the terms of the sequence $\{af_i\}$ are points in $\text{span}(A)$, we have $af\in \cl(\text{span}(A))$.
+The proof regards the addition of two vectors is by a similar manner. If $f\in \cl(\text{span}(A))$ and $g\in \cl(\text{span}(A))$, 
+then a sequence $\{f_i\}$, converges to $f$, and a sequence $\{g_i\}$, converges to $g$, of $\text{span}(A)$ are given. By the continuity of the function of the addtion of points in $H$, $\{f_i+g_i\}$ converges to $f+g$. Because each terms of the sequence $f_i+g_i$ is in $\text{span}(A)$, we have $f+g\in \cl(\text {span}(A))$.
+Notice that in general, if $H'$ is a subspace, then $\cl(H')$ is a subspace.
+Finally we come to the proof of equivalence of (2) and (3):
+$\cl(\text{span}(A))$, the closed linear subspace, contains $\text{span}(A)$ and thus contains $A$. For any closed linear subspace $M$ which contains $A$, $\text{span}(A)\subseteq M$, since the linear span of $A$ is the smallest subspace that contains $A$. Because $M$ is closed, $\cl(\text{span}(A))\subseteq M$. $\cl(\text{span}(A))$ is the the smallest [[Definition:Closed Linear Subspace|closed linear subspace]] $M$ of $H$ with $A \subseteq M$. Because arbitrary intersection of closed sets is closed and arbitrary intersection of subspaces is a subspace, the smallestness is unique. Hence $\vee A=\cl(\text{span}(A))$.{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Norm of Linear Functional}
+Tags: Linear Transformations on Hilbert Spaces, Definition Equivalences
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $L$ be a [[Definition:Bounded Linear Functional|bounded linear functional]] on $H$.
+{{explain|We can weaken the assumption that $H$ be a Hilbert space by saying that it is just a normed space, I think. It is not strictly necessary to consider a Hilbert space.}}
+Define the following [[Definition:Norm on Bounded Linear Functional|norms]] of $L$:
+:$(1): \quad \left\|{L}\right\|_1 = \sup \left\{{\left|{Lh}\right|: \left\|{h}\right\| \le 1}\right\}$
+:$(2): \quad \left\|{L}\right\|_2 = \sup \left\{{\left|{Lh}\right|: \left\|{h}\right\| = 1}\right\}$
+:$(3): \quad \left\|{L}\right\|_3 = \displaystyle \sup \left\{{\dfrac {\left|{Lh}\right|} {\left\|{h}\right\|}: h \in H\setminus \left\{\mathbf 0 \right\}}\right\}$
+:$(4): \quad \left\|{L}\right\|_4 = \inf \left\{{c > 0: \forall h \in H: \left|{Lh}\right| \le c \left\|{h}\right\|}\right\}$
+Then:
+: $\left\|{L}\right\|_1 = \left\|{L}\right\|_2 = \left\|{L}\right\|_3 = \left\|{L}\right\|_4$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$\displaystyle\left\{{ h \in H : \left\Vert{h}\right\Vert = 1 }\right\} \subseteq \left\{{ h \in H : \left\Vert{h}\right\Vert \le 1 }\right\} \subseteq H$
+So it follows from the definition of the [[Definition:Supremum of Set|supremum]] that 
+:$\left\Vert{L}\right\Vert_2 \le \left\Vert{L}\right\Vert_1 \le \left\Vert{L}\right\Vert_3$
+Next we show that $\left\Vert{L}\right\Vert_2 = \left\Vert{L}\right\Vert_3$.
+{{begin-eqn}}
+{{eqn|l= \left\Vert{L}\right\Vert_2
+     |o= =
+     |r= \sup \left\{ {\left\Vert{Lh}\right\Vert: \left\Vert{h}\right\Vert = 1}\right\}
+}}
+{{eqn|l= 
+     |o= =
+     |r= \sup \left\{ {\left\Vert{Lh'}\right\Vert: h \in H\setminus\left\{\mathbf 0\right\},\ h' = h/\left\Vert{h}\right\Vert} \right\}
+     |c= as $\left\Vert h/\left\Vert{h}\right\Vert \right\Vert = \left\Vert{h}\right\Vert/\left\Vert{h}\right\Vert = 1$ for all $h \in H\setminus\left\{\mathbf 0 \right\}$
+}}
+{{eqn|l= 
+     |o= =
+     |r= \sup \left\{ {\left\Vert{\frac1{\left\Vert{h}\right\Vert}Lh}\right\Vert: h \in H\setminus \left\{ \mathbf 0  \right\} }\right\}
+     |c= as $L$ is [[Definition:Linear Functional|linear]]
+}}
+{{eqn|l= 
+     |o= =
+     |r= \sup \left\{ {\frac{\left\Vert{Lh}\right\Vert}{\left\Vert{h}\right\Vert}: h \in H\setminus \left\{ \mathbf 0  \right\} }\right\}
+     |c= by property $(N2)$ of a [[Definition:Norm on Vector Space|norm]]
+}}
+{{eqn|l= 
+     |o= =
+     |r= \left\Vert{L}\right\Vert_3
+}}
+{{end-eqn}}
+Therefore
+:$\left\Vert{L}\right\Vert_1 = \left\Vert{L}\right\Vert_2 = \left\Vert{L}\right\Vert_3$
+Moreover, if $\left|{Lh}\right| \le c \left\|{h}\right\|$ for all $h \in H\setminus\left\{\mathbf 0 \right\}$, then we have:
+:$\displaystyle\left\|{L}\right\|_3 = \sup \left\{{\dfrac {\left|{Lh}\right|} {\left\|{h}\right\|}: h \in H \setminus \left\{\mathbf 0 \right\}}\right\} \le c$
+Taking the [[Definition:Infimum of Set|infimum]] over all such $c$ this reads:
+:$\left\Vert{L}\right\Vert_3 \le \left\Vert{L}\right\Vert_4$
+Suppose that $c_0 := \left\Vert{L}\right\Vert_4 > \left\Vert{L}\right\Vert_3$.
+Then by the definitions of these two norms, this means that there exists $\epsilon > 0$ such that for every $h \in H\setminus\left\{\mathbf 0 \right\}$:
+:$\displaystyle\frac{|Lh|}{\Vert h\Vert} + \epsilon \le c_0$
+But this in turn implies that for every $h \in H\setminus\left\{\mathbf 0 \right\}$:
+:$|Lh| \leq c_0 \Vert h\Vert - \epsilon \Vert h\Vert = \left( c_0 - \epsilon \right) \Vert h\Vert$
+This contradicts the fact that $c_0$ is the least such number satisfying this inequality.
+Therefore:
+: $\left\Vert{L}\right\Vert_3 = \left\Vert{L}\right\Vert_4$
+and the proof is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riesz Representation Theorem (Hilbert Spaces)}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $L$ be a [[Definition:Bounded Linear Functional|bounded linear functional]] on $H$.
+Then there is a unique $h_0 \in H$ such that:
+:$\forall h \in H: L h = \innerprod h {h_0}$
+\end{theorem}
+
+\begin{proof}
+If $L \equiv 0$ identically, then $L h = 0 = \innerprod h 0$, and the theorem holds.
+Otherwise, set:
+:$M = \map \ker L = \map {L^{-1} } {\set 0}$
+Then $M$ is a subspace.
+Because $L$ is bounded, [[Continuity of Linear Functionals|it is continuous]].
+Because $\set 0$ is closed, [[Continuity Defined from Closed Sets|the continuity of $L$ implies that $M$ is closed]].
+Then we can [[Direct Sum of Subspace and Orthocomplement|decompose $H$ as a direct sum]]:
+:$H \cong M \oplus M^\perp$
+As $L \not \equiv 0$:
+:$M^\perp \ne \set 0$
+Choose a $z \in M^\perp$ with norm $1$.
+By linearity of $L$, for any $h \in H$:
+{{begin-eqn}}
+{{eqn | l = L \paren {z L h - h L z}
+      | r = L z L h - L h L z
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+So:
+:$z L h - h L z \in \ker L = M$
+Then:
+{{begin-eqn}}
+{{eqn | l = L h
+      | r = L h \innerprod z z
+      | c = as $\norm z = 1$
+}}
+{{eqn | r = \innerprod {z L h} z
+      | c = linearity in the first argument
+}}
+{{eqn | r = \innerprod {z L h - h L z + h L z} z
+      | c = adding and subtracting $h L z$ in the first argument
+}}
+{{eqn | r = \innerprod {z L h - h L z} z + \innerprod {h L z} z
+      | c = linearity in the first argument
+}}
+{{eqn | r = \innerprod {h L z} z
+      | c = $z L h - h L z \in M, z \in M^\perp$
+}}
+{{eqn | r = \innerprod h {z \paren {L z}^*}
+      | c = conjugate symmetry
+}}
+{{end-eqn}}
+Thus $L h = \innerprod h {h_0}$ for $h_0 = z (Lz)^*$.
+To show uniqueness, assume $h_0$ and $h_1$ both satisfy the above equation for all $h \in H$:
+{{begin-eqn}}
+{{eqn | l = \innerprod h {h_0}
+      | r = \innerprod h {h_1}
+}}
+{{eqn | ll= \leadsto
+      | l = \innerprod h {h_0} - \innerprod h {h_1}
+      | r = 0
+}}
+{{eqn | r = \innerprod h {h_0 - h_1}
+      | c = additivity in the second argument
+}}
+{{end-eqn}}
+The result follows from Setting $h = h_0 - h_1$ and invoking the positive definiteness of the inner product.
+{{qed}}
+{{Namedfor|Frigyes Riesz|cat = Riesz F}}
+\end{proof}<|endoftext|>
+\section{Orthonormal Subset Extends to Basis}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $S$ be an [[Definition:Orthonormal Subset|orthonormal subset]] of $H$.
+Then there exists a [[Definition:Basis (Hilbert Space)|basis]] for $H$ that contains $S$ as a subset.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+{{AoC|Zorn's Lemma}}
+\end{proof}<|endoftext|>
+\section{Gram-Schmidt Orthogonalization}
+Tags: Hilbert Spaces, Gram-Schmidt Orthogonalization
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $S = \set {h_n: n \in \N}$ be a [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Subset|subset]] of $H$. 
+Then there exists an [[Definition:Orthonormal Subset|orthonormal subset]] $E = \set {e_n: n \in \N}$ of $H$ such that:
+:$\forall k \in \N: \operatorname{span} \set {h_n: 0 \le n \le k} = \operatorname{span} \set {e_n: 0 \le n \le k}$
+where $\operatorname{span}$ denotes [[Definition:Linear Span|linear span]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+{{Namedfor|Jørgen Pedersen Gram|name2 = Erhard Schmidt|cat = Gram|cat2 = Schmidt}}
+\end{proof}<|endoftext|>
+\section{Orthogonal Projection onto Closed Linear Span}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $E = \left\{{e_1, \ldots, e_n}\right\}$ be an [[Definition:Orthonormal Subset|orthonormal subset]] of $H$.
+Let $M = \vee E$, the [[Definition:Closed Linear Span|closed linear span]] of $E$. 
+Then the [[Definition:Orthogonal Projection|orthogonal projection]] $P$ onto $M$ satisfies, $\forall h \in H$:
+:$Ph = \displaystyle \sum_{k=1}^n \left\langle{h, e_k}\right\rangle e_k$
+\end{theorem}<|endoftext|>
+\section{Bessel's Inequality}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $E = \left\{{e_n: n \in \N}\right\}$ be a [[Definition:Countable|countably infinite]] [[Definition:Orthonormal Subset|orthonormal subset]] of $H$.
+Then, for all $h \in H$, one has the inequality:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \left|{\left\langle{h, e_n}\right\rangle}\right|^2 \le \left\|{h}\right\|^2$
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+{{Namedfor|Friedrich Wilhelm Bessel|cat=Bessel}}
+\end{proof}<|endoftext|>
+\section{Between two Rational Numbers exists Irrational Number}
+Tags: Real Analysis, Between two Rational Numbers exists Irrational Number
+
+\begin{theorem}
+Let $a, b \in \Q$ where $a < b$. 
+Then:
+:$\exists \xi \in \R \setminus \Q: a < \xi < b$
+\end{theorem}
+
+\begin{proof}
+Let $d = b - a$.
+As $a, b \in \Q: a < b$ it follows from [[Rational Numbers form Ordered Integral Domain]] that $d \in \Q: d > 0$.
+From [[Square Root of 2 is Irrational]], $\sqrt 2$ is not a [[Definition:Rational Number|rational number]], so it is an [[Definition:Element|element]] of $\R \setminus \Q$.
+From [[Square Number Less than One]], for any given [[Definition:Real Number|real number]] $x$:
+:$x^2 < 1 \implies x \in \openint {-1} 1$
+Let $k = \dfrac {\sqrt 2} 2$.
+Then from [[Between two Rational Numbers exists Irrational Number/Lemma 1|Lemma 1]]:
+:$k \in \R \setminus \Q$
+As $k^2 = \dfrac 1 2$, it follows that $0 < k < 1$.
+Let $\xi = a + k d$.
+Then, since $a, d \in \Q$ and $k \in \R \setminus \Q$, it follows from [[Between two Rational Numbers exists Irrational Number/Lemma 1|Lemma 1]] and [[Between two Rational Numbers exists Irrational Number/Lemma 2|Lemma 2]] that:
+:$\xi \in \R \setminus \Q$
+$d > 0$ and $k > 0$, so:
+:$\xi = a + k d > a + 0 \cdot 0 = a$
+$k < 1$, so:
+:$\xi = a + k d < a + 1 \cdot d < a + \paren {b - a} = b$
+We thus have:
+:$\xi \in \R \setminus \Q: \xi \in \openint a b$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Between two Real Numbers exists Rational Number]], there exists $x \in \Q$ such that:
+:$a - \sqrt2 < x < b - \sqrt2$
+Since [[Square Root of 2 is Irrational]], by the [[Between two Rational Numbers exists Irrational Number/Lemma 2|Lemma]]:
+:$x + \sqrt 2$ is [[Definition:Irrational Number|irrational]].
+But:
+:$a < x + \sqrt 2 < b$
+which is what we wanted to show.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Abelian Group is Simple iff Prime}
+Tags: Simple Groups, Abelian Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Abelian Group|abelian group]].
+Then $G$ is [[Definition:Simple Group|simple]] {{iff}} $G$ is a [[Definition:Prime Group|prime group]].
+\end{theorem}
+
+\begin{proof}
+First we note that the [[Definition:Trivial Group|trivial group]] is (trivially) [[Definition:Simple Group|simple]], but not [[Definition:Prime Group|prime]] because [[One is not Prime|$1$ is not prime]].
+Hence the specification of $G$ as being [[Definition:Non-Trivial Group|non-trivial]].
+=== Necessary Condition ===
+Let $G$ be a [[Definition:Simple Group|simple]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Abelian Group|abelian group]].
+Suppose that $G$ is an [[Definition:Infinite Group|infinite group]].
+Then by [[Infinite Group has Infinite Number of Subgroups]] $G$ has an [[Definition:Infinite Set|infinite]] number of [[Definition:Distinct Elements|distinct]] [[Definition:Subgroup|subgroups]].
+But then by [[Subgroup of Abelian Group is Normal]], $G$ cannot be [[Definition:Simple Group|simple]].
+So [[Proof by Contradiction|by contradiction]] $G$ must be [[Definition:Finite Group|finite]].
+Let $n$ be the [[Definition:Order of Group|order]] of $G$.
+By definition of [[Definition:Simple Group|simple]], $G$ has no [[Definition:Non-Trivial Proper Subgroup|non-trivial proper]] [[Definition:Normal Subgroup|normal subgroups]].
+From [[Subgroup of Abelian Group is Normal]], it follows that $G$ can have no [[Definition:Non-Trivial Proper Subgroup|non-trivial proper subgroups]] at all.
+From [[Cauchy's Group Theorem]], if $p$ is a [[Definition:Prime Number|prime number]] which is a [[Definition:Divisor of Integer|divisor]] of $n$, then $G$ has a [[Definition:Subgroup|subgroup]] of order $p$.
+It follows that if $G$ is [[Definition:Simple Group|simple]], there can be no [[Definition:Prime Number|prime number]] less than $n$ which is a [[Definition:Divisor of Integer|divisor]] of $n$.
+It follows that $n$ is [[Definition:Prime Number|prime]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $G$ is a [[Definition:Prime Group|prime group]].
+By [[Prime Group is Simple]], it follows that $G$ is [[Definition:Simple Group|simple]].
+Because $G$ is [[Definition:Prime Group|prime]], its [[Definition:Order of Group|order]] is greater than $1$.
+Thus $G$ is [[Definition:Non-Trivial Group|non-trivial]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cyclic Group is Simple iff Prime}
+Tags: Simple Groups, Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
+Then $G$ is [[Definition:Simple Group|simple]] {{iff}} $G$ is a [[Definition:Prime Group|prime group]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
+From [[Cyclic Group is Abelian]] it follows that $G$ is an [[Definition:Abelian Group|abelian group]].
+The result follows from [[Abelian Group is Simple iff Prime]].
+{{qed}}
+[[Category:Simple Groups]]
+[[Category:Cyclic Groups]]
+ihiodpl0ircj9fl5b5ya623qbow3qh2
+\end{proof}<|endoftext|>
+\section{Factors of Solvable Group are Prime}
+Tags: Solvable Groups, Prime Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Solvable Group|solvable group]].
+Let $\mathcal H$ be a [[Definition:Composition Series|composition series]] of $G$.
+Then all [[Definition:Factor of Normal Series|factor groups]] of $\mathcal H$ must be [[Definition:Prime Group|prime]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Composition Series|composition series]] is a [[Definition:Normal Series|normal series]] whose [[Definition:Factor of Normal Series|factor groups]] are all [[Definition:Simple Group|simple]].
+A [[Definition:Solvable Group|solvable group]], by definition, is one which has a [[Definition:Composition Series|composition series]] whose [[Definition:Factor of Normal Series|factor groups]] are all [[Definition:Cyclic Group|cyclic]].
+From [[Cyclic Group is Simple iff Prime]], it follows that all the [[Definition:Factor of Normal Series|factor groups]] of a [[Definition:Composition Series|composition series]] of a [[Definition:Solvable Group|solvable group]] must all be [[Definition:Prime Group|prime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Subspace is Convex Set}
+Tags: Vector Subspaces
+
+\begin{theorem}
+Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a [[Definition:Vector Subspace|linear subspace]] of $V$.
+Then $L$ is a [[Definition:Convex Set (Vector Space)|convex set]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in L, t \in \closedint 0 1$ be arbitrary.
+Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that
+:$t x + \paren {1 - t} y \in L$
+Hence $L$ is [[Definition:Convex Set (Vector Space)|convex]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton is Convex Set}
+Tags: Vector Spaces, Singletons
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|vector space]] over $\R$ or $\C$, and let $v \in V$.
+Then the [[Definition:Singleton|singleton]] $S = \left\{{v}\right\}$ is a [[Definition:Convex Set (Vector Space)|convex set]].
+\end{theorem}
+
+\begin{proof}
+For any $x, y \in S$, we have $x = y = v$.
+It follows that:
+:$\forall t \in \left[{0 \,.\,.\, 1}\right]: t x + \left({1 - t}\right)y = v \in S$
+Hence $S$ is a [[Definition:Convex Set (Vector Space)|convex set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group is Solvable iff Normal Subgroup and Quotient are Solvable}
+Tags: Normal Subgroups, Quotient Groups, Solvable Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then $G$ is [[Definition:Solvable Group|solvable]] {{iff}}:
+:$(1): \quad H$ is [[Definition:Solvable Group|solvable]]
+and
+:$(2): \quad G / H$ is [[Definition:Solvable Group|solvable]]
+where $G / H$ is the [[Definition:Quotient Group|quotient group]] of $G$ by $H$.
+\end{theorem}
+
+\begin{proof}
+As $H \lhd G$ we can construct the [[Definition:Normal Series|normal series]]:
+:$(A): \quad \set e \lhd H \lhd G$
+By [[Finite Group has Composition Series]], $(A)$ can be [[Definition:Refinement of Normal Series|refined]] to a [[Definition:Composition Series|composition series]] for $G$:
+:$(B): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
+Suppose $G_k = H$.
+Then we can construct the [[Definition:Composition Series|composition series]]:
+:$(C): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_k = H$
+and:
+:$(D): \quad \set e = G_k / H \lhd G_{k + 1} / H \lhd \cdots \lhd G_n / H = G / H$
+Furthermore, by the [[Third Isomorphism Theorem]]:
+:$\dfrac {G_{i + 1} / H} {G_i / H} \cong \dfrac {G_{i + 1} } {G_i}$
+for all $k \le i \le n$.
+So each [[Definition:Factor of Normal Series|factor]] of the [[Definition:Composition Series|composition series]] for $G$ is a [[Definition:Factor of Normal Series|factor]] of either:
+:the [[Definition:Composition Series|composition series]] for $H$
+or:
+:the [[Definition:Composition Series|composition series]] for $G / H$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Abelian Group is Solvable}
+Tags: Abelian Groups, Solvable Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
+Then $G$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Second Principle of Mathematical Induction|strong induction]] on the [[Definition:Order of Structure|order]] of $G$:
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:All [[Definition:Abelian Group|abelian groups]] of [[Definition:Order of Structure|order]] $n$ and below are [[Definition:Solvable Group|solvable]].
+=== Basis for the Induction ===
+$\map P 1$ is true, as the [[Definition:Trivial Group|trivial group]] is trivially [[Definition:Solvable Group|solvable]].
+From [[Prime Group is Simple]], both $\map P 2$ and $\map P 3$ hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+It remains to be shown that, if $\map P j$ holds, for all $j$ such that $1 \le j \le k$, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ holds.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:All [[Definition:Abelian Group|abelian groups]] of [[Definition:Order of Structure|order]] $k$ and below are [[Definition:Solvable Group|solvable]].
+Then we need to show:
+:All [[Definition:Abelian Group|abelian groups]] of [[Definition:Order of Structure|order]] $k+1$ and below are [[Definition:Solvable Group|solvable]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $G$ be an [[Definition:Abelian Group|abelian group]] such that:
+:$\order G = k + 1$
+where $\order G$ denotes the [[Definition:Order of Group|order]] of $G$.
+Then from [[Positive Integer Greater than 1 has Prime Divisor]], there exists some [[Definition:Prime Number|prime]] $p$ which [[Definition:Divisor of Integer|divides]] $k + 1$.
+From [[Cauchy's Lemma (Group Theory)]], $G$ has an [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $p$.
+Since $G$ is [[Definition:Abelian Group|abelian]], this element [[Definition:Generator of Subgroup|generates]] a [[Definition:Subgroup|subgroup]] of $G$ which is [[Definition:Normal Subgroup|normal]] by [[Subgroup of Abelian Group is Normal]].
+If $p = n$ and $H = G$ the proof is finished, from [[Prime Power Group is Solvable]].
+If $p < n$, then $H$ and $G / H$ (which has [[Definition:Order of Structure|order]] $n / p$) are both [[Definition:Solvable Group|solvable]] from the [[Finite Abelian Group is Solvable#Induction Hypothesis|induction hypothesis]].
+It follows from [[Group is Solvable iff Normal Subgroup and Quotient are Solvable]] that $G$ is [[Definition:Solvable Group|solvable]].
+The result follows by the [[Second Principle of Mathematical Induction|strong principle of mathematical induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Convex Sets is Convex Set (Vector Spaces)}
+Tags: Vector Spaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|vector space]] over $\R$ or $\C$.
+Let $\mathcal C$ be a family of [[Definition:Convex Set (Vector Space)|convex subsets]] of $V$.
+Then the intersection $\displaystyle \bigcap \mathcal C$ is also a [[Definition:Convex Set (Vector Space)|convex subset]] of $V$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \displaystyle \bigcap \mathcal C$.
+Then by definition of [[Definition:Set Intersection|set intersection]], $\forall C \in \mathcal C: x, y \in C$.
+The [[Definition:Convex Set (Vector Space)|convexity]] of each $C$ yields:
+:$\forall t \in \left[{0 \,.\,.\, 1}\right]: t x + \left({1 - t}\right) y \in C$
+Therefore, these elements are also in $\displaystyle \bigcap \mathcal C$, by definition of [[Definition:Set Intersection|set intersection]].
+Hence $\displaystyle \bigcap \mathcal C$ is also [[Definition:Convex Set (Vector Space)|convex]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Power Group is Solvable}
+Tags: P-Groups, Solvable Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of a Structure|order]] is $p^n$ where $p$ is a [[Definition:Prime Number|prime number]] and $n$ is a [[Definition:Positive Integer|positive integer]].
+Then $G$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}
+
+\begin{proof}
+A direct consequence of [[Factors of Composition Series for Prime Power Group]] and the definition of [[Definition:Solvable Group|solvable group]].
+{{qed}}
+[[Category:P-Groups]]
+[[Category:Solvable Groups]]
+a3862dezf4buwaskcacaosw1234emt5
+\end{proof}<|endoftext|>
+\section{Subgroup of Solvable Group is Solvable}
+Tags: Subgroups, Solvable Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Solvable Group|solvable group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $H$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}
+
+\begin{proof}
+Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a [[Definition:Normal Series|normal series]] for $G$ with [[Definition:Abelian_Group|Abelian]] quotients.
+For every $i = 1, 2, \ldots, n$ we have:
+:$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$
+From the [[Second Isomorphism Theorem for Groups]]:
+:$\dfrac {\paren {H \cap G_i} \cap G_{i - 1} } {G_{i - 1} } \cong \dfrac {H \cap G_i} {H \cap G_{i - 1} }$
+where $\cong$ denotes [[Definition:Group Isomorphism|group isomorphism]].
+Specifically, $H \cap G_{i - 1}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $H \cap G_i$.
+We have that:
+:$\paren {H \cap G_i} \cap G_{i - 1} \subseteq G_i$
+and so from the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]]:
+:$\dfrac {\paren {H \cap G_i} \cap G_{i - 1} } {G_{i - 1} } \le G_i / G_{i - 1}$
+We have that $G_i / G_{i - 1}$ is [[Definition:Abelian Group|abelian]].
+Thus from [[Subgroup of Abelian Group is Abelian]]:
+:$\dfrac {\paren {H \cap G_i } \cap G_{i - 1} } {G_{i - 1} }$ is [[Definition:Abelian Group|abelian]].
+Hence $\dfrac {H \cap G_i} {H \cap G_{i - 1} }$ is [[Definition:Abelian Group|abelian]].
+Therefore, the series :
+:$\set e = H \cap G_0 \lhd H \cap G_1 \lhd \cdots \lhd H \cap G_n = G$
+is a [[Definition:Normal Series|normal series]] with [[Definition:Abelian Group|abelian]] [[Definition:Factor of Normal Series|factor groups]] for $H$.
+Therefore $H$ is [[Definition:Solvable Group|solvable]].
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Group with Normal Series with Solvable Factor Groups is Solvable}
+Tags: Subgroups, Solvable Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Solvable Group|solvable group]].
+Let $\HH$ be a [[Definition:Normal Series|normal series]] for $G$.
+Let all the [[Definition:Factor of Normal Series|factor groups]] of $\HH$ be [[Definition:Solvable Group|solvable]].
+Then $G$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}<|endoftext|>
+\section{Group with Order Less than 60 is Solvable}
+Tags: Solvable Groups
+
+\begin{theorem}
+Every [[Definition:Group|group]] whose [[Definition:Order of a Structure|order]] is less than $60$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}
+
+\begin{proof}
+A direct consequence of [[Burnside's Theorem]]: the smallest number with $3$ [[Definition:Distinct|distinct]] [[Definition:Prime Factor|prime factors]] is $2 \times 3 \times 5 = 60$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Solvable Groups is Solvable}
+Tags: Solvable Groups, Group Direct Products
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]] which are [[Definition:Solvable Group|solvable]].
+Then their [[Definition:Group Direct Product|(external) direct product]] $G \times H$ is also [[Definition:Solvable Group|solvable]].
+\end{theorem}<|endoftext|>
+\section{Group with One Sylow Subgroup per Prime Divisor is Solvable}
+Tags: Solvable Groups, Prime Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $n$.
+Suppose that, for each [[Definition:Prime Number|prime number]] $p$ which [[Definition:Divisor of Integer|divides]] $n$, $G$ has exactly one [[Definition:Sylow p-Subgroup|$p$-Sylow subgroup]].
+Then $G$ is [[Definition:Solvable Group|solvable]].
+\end{theorem}<|endoftext|>
+\section{Permutation Group as Effective Transformation Group}
+Tags: Group Actions
+
+\begin{theorem}
+Let $*$ be a [[Definition:Group Action|group action]] of $G$ on $X$.
+Then $G$ is a [[Definition:Permutation Group|permutation group]] {{iff}} $G$ acts [[Definition:Effective Transformation Group|acts effectively]] on $X$.
+\end{theorem}<|endoftext|>
+\section{Characterization of Bases (Hilbert Spaces)}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $E$ be an [[Definition:Orthonormal Subset|orthonormal subset]] of $H$.
+Then the following six statements are equivalent:
+:$(1): \qquad E$ is a [[Definition:Basis (Hilbert Space)|basis]] for $H$
+:$(2): \qquad h \in H, h \perp E \implies h = \mathbf 0$, where $\perp$ denotes [[Definition:Orthogonal (Hilbert Space)|orthogonality]]
+:$(3): \qquad \vee E = H$, where $\vee E$ denotes the [[Definition:Closed Linear Span|closed linear span]] of $E$
+:$(4): \qquad \forall h \in H: h = \displaystyle \sum \left\{{\left\langle{h, e}\right\rangle e: e \in E}\right\}$
+:$(5): \qquad \forall g, h \in H: \left\langle{g, h}\right\rangle = \displaystyle \sum \left\{{\left\langle{g, e}\right\rangle\left\langle{e, h}\right\rangle: e \in E}\right\}$
+:$(6): \qquad \forall h \in H: \left\Vert{h}\right\Vert^2 = \displaystyle \sum \left\{{\left\vert{\left\langle{h, e}\right\rangle}\right\vert^2: e \in E}\right\}$
+In the last three statements, $\displaystyle \sum$ denotes a [[Definition:Generalized Sum|generalized sum]].
+Statement $(6)$ is commonly known as '''Parseval's identity'''.
+\end{theorem}<|endoftext|>
+\section{Square Number Less than One}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]] such that $x^2 < 1$.
+Then:
+: $x \in \left({-1 \,.\,.\, 1}\right)$
+where $\left({-1 \,.\,.\, 1}\right)$ is the [[Definition:Open Real Interval|open interval]] $\left\{{x \in \R: -1 < x < 1}\right\}$.
+\end{theorem}
+
+\begin{proof}
+First note that from [[Square of Real Number is Non-Negative]]:
+:$x^2 \ge 0$
+From [[Ordering of Squares in Reals]]:
+:$(1): \quad x > 1 \implies x^2 > 1$
+:$(2): \quad x < 1 \implies x^2 < 1$
+From [[Identity Element of Multiplication on Numbers]]:
+:$1^2 = 1$
+so it is clear that the strict inequalities apply above.
+For clarity, therefore, the case where $x = \pm 1$ can be ignored.
+Suppose $x \notin \left({-1 \,.\,.\, 1}\right)$.
+Then either:
+:$x < -1$
+or:
+:$x > 1$
+If $x > 1$ then:
+:$x^2 > 1$
+If $x < -1$ then:
+:$-x > 1$
+From [[Square of Real Number is Non-Negative]]:
+:$\left({-x}\right)^2 = x^2$
+So again $x^2 > 1$.
+Thus:
+:$x \notin \left({-1 \,.\,.\, 1}\right) \implies x^2 > 1$
+So by the [[Rule of Transposition]]:
+:$x^2 < 1 \implies x \in \left({-1 \,.\,.\, 1}\right)$
+{{qed}}
+[[Category:Real Analysis]]
+73izy5ub74rmzmsrkm35tsizjduv10g
+\end{proof}<|endoftext|>
+\section{Subgroups of Symmetric Group Isomorphic to Product of Subgroups}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $k \in \closedint 1 n$.
+Then there are $\dbinom n k$ [[Definition:Subgroup|subgroups]] of $S_n$ which are [[Definition:Group Isomorphism|isomorphic]] to $S_k \times S_{n - k}$, where $\dbinom n k$ denotes the [[Definition:Binomial Coefficient|binomial coefficient]].
+All of these $\dbinom n k$ [[Definition:Subgroup|subgroups]] are [[Definition:Conjugate of Group Subset|conjugate]].
+\end{theorem}<|endoftext|>
+\section{Uniform Limit Theorem}
+Tags: Functional Analysis, Continuous Mappings, Metric Spaces, Named Theorems
+
+\begin{theorem}
+Let $\left({M, d_M}\right)$ and $\left({N, d_N}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\left \langle{f_n}\right \rangle$ be a [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] from $M$ to $N$ such that:
+:$(1): \quad \forall n \in \N: f_n$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous at every point of]] $M$
+:$(2): \quad  \langle f_n\rangle$ [[Definition:Uniform Convergence|converges uniformly]] to $f$
+Then:
+: $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous at every point of]] $M$.
+\end{theorem}
+
+\begin{proof}
+Let $a \in M$.
+We are given that $d_N$ is a [[Definition:Metric|metric]] on $N$.
+So:
+:$\forall x, y, z \in N: d_N \left({x, z}\right) \le d_N \left({x, y}\right) + d_N \left({y, z}\right)$
+We apply this property twice to assert that $\forall n \in \N, \forall x \in M$, we have:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = d_N \left({f \left({x}\right), f \left({a}\right)}\right)
+      | o = \le
+      | r = d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)
+}}
+{{eqn | o = +
+      | r = d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)
+}}
+{{eqn | o = +
+      | r = d_N \left({f_n \left({a}\right), f \left({a}\right)}\right)
+}}
+{{end-eqn}}
+Let $\epsilon >0$.
+Since $f_n \to f$ [[Definition:Uniform Convergence|uniformly]]:
+: $\exists \mathcal N \in \R: \forall n \ge \mathcal N: \forall x \in M$:
+{{begin-eqn}}
+{{eqn | n = 2 a
+      | l = d_N \left({f \left({a}\right), f_n \left({a}\right)}\right)
+      | o = <
+      | r = \frac \epsilon 3
+}}
+{{eqn | n = 2 b
+      | l = d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)
+      | o = <
+      | r = \frac \epsilon 3
+}}
+{{end-eqn}}
+We are given that $\forall n \in \N: f_n$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+So, for any $n\in\N$, $\exists \delta > 0: \forall x \in M$:
+{{begin-eqn}}
+{{eqn | n = 3
+      | l = d_M \left({x, a}\right)
+      | o = <
+      | r = \delta
+}}
+{{eqn | ll = \implies
+      | l = d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)
+      | o = <
+      | r = \frac \epsilon 3
+}}
+{{end-eqn}}
+By combining $(1)$, $(2 a)$, $(2 b)$ and $(3)$:
+: $\exists \delta > 0$ and $\exists n$ [[Definition:Sufficiently Large|sufficiently large]] such that $\forall x \in M$:
+{{begin-eqn}}
+{{eqn | l = d_M \left({x, a}\right)
+      | o = <
+      | r = \delta
+}}
+{{eqn | ll = \implies
+      | l = d_N \left({f \left({x}\right), f \left({a}\right)}\right)
+      | o = \le
+      | r = \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+As $a$ and $\epsilon$ are arbitrary, it follows that:
+: $\forall a \in M: \forall \epsilon > 0: \exists \delta > 0: \forall x \in M$:
+{{begin-eqn}}
+{{eqn | l = d_M \left({x, a}\right)
+      | o = <
+      | r = \delta
+}}
+{{eqn | ll = \implies
+      | l = d_N \left({f \left({x}\right), f \left({a}\right)}\right)
+      | o = \le
+      | r = \epsilon
+}}
+{{end-eqn}}
+Hence, $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous at every point of]] $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equality is Transitive}
+Tags: Logic, Equality
+
+\begin{theorem}
+:$\forall a, b, c: \paren {a = b} \land \paren {b = c} \implies a = c$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \vdash
+      | l = \map P a
+      | o = \iff
+      | r = \map P b
+      | c = [[Axiom:Leibniz's Law|Leibniz's law]]
+}}
+{{eqn | l = b
+      | r = c
+      | c = 
+}}
+{{eqn | ll= \vdash
+      | l = \map P b
+      | o = \iff
+      | r = \map P c
+      | c = [[Axiom:Leibniz's Law|Leibniz's law]]
+}}
+{{eqn | ll= \vdash
+      | l = \map P a
+      | o = \iff
+      | r = \map P c
+      | c = [[Biconditional is Transitive]]
+}}
+{{eqn | ll= \vdash
+      | l = a
+      | r = c
+      | c = [[Axiom:Leibniz's Law|Leibniz's law]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Axiom:Axioms of Equality}
+Tags: Axioms: Logic
+
+\begin{theorem}
+The '''axioms of equality''' are strictly speaking not [[Definition:Axiom|axiomatic]] at all, as they can be deduced from still more basic axioms, in particular [[Axiom:Leibniz's Law|Leibniz's law]]:
+:$x = y \dashv \vdash P \left({x}\right) \iff P \left({y}\right)$
+where $P \left({x}\right)$ and $P \left({y}\right)$ are [[Definition:Propositional Function|propositional functions]] on the elements $x$ and $y$ of the [[Definition:Universe of Discourse|universe of discourse]].
+\end{theorem}<|endoftext|>
+\section{Axiom of Pairing from Infinity and Replacement}
+Tags: Axiom of Pairing, Doubletons
+
+\begin{theorem}
+The [[Axiom:Axiom of Pairing|Axiom of Pairing]] is a consequence of:
+:the [[Axiom:Axiom of Infinity|Axiom of Infinity]]
+and
+:the [[Axiom:Axiom of Replacement|Axiom of Replacement]].
+\end{theorem}
+
+\begin{proof}
+The set $2 = \set {\O, \set \O}$ is used with the [[Axiom:Axiom of Replacement|Axiom of Replacement]] as the [[Definition:Domain of Mapping|domain]] for a [[Definition:Mapping|mapping]] whose [[Definition:Image of Mapping|image]] is $\set {A, B}$.
+A suitable [[Definition:Mapping|mapping]] would be:
+:$\paren {y = \O \land z = A} \lor \paren {y = \set \O \land z = B}$
+The set $2$ is shown to exist as a member of the [[Definition:Infinite Set|infinite set]] whose existence is asserted by the [[Axiom:Axiom of Infinity|axiom of infinity]].
+{{qed}}
+[[Category:Axiom of Pairing]]
+[[Category:Doubletons]]
+pqr9ervif1gvsz3x2iy9mcbnl1fomgi
+\end{proof}<|endoftext|>
+\section{Axiom of Pairing from Powers and Replacement}
+Tags: Axiom of Pairing, Doubletons
+
+\begin{theorem}
+The [[Axiom:Axiom of Pairing|Axiom of Pairing]] is a consequence of:
+:the [[Axiom:Axiom of Empty Set|Axiom of the Empty Set]]
+and
+:the [[Axiom:Axiom of Powers|Axiom of Powers]]
+and also
+:the [[Axiom:Axiom of Replacement|Axiom of Replacement]].
+\end{theorem}
+
+\begin{proof}
+It is to be shown that $2 = \set {\O, \set \O}$ is a [[Definition:Set|set]].
+By the [[Axiom:Axiom of Empty Set|Axiom of the Empty Set]], it is seen that $\O$ is a set.
+By the [[Axiom:Axiom of Powers|Axiom of Powers]], it is seen that $\powerset \O$ is also a set.
+It is to be shown that $\powerset \O = \set \O$.
+{{refactor|The above may already exist as a theorem on {{ProofWiki}} }}
+To determine the elements of $\powerset \O$, we must find all of the sets $z$ such that:
+:$\forall x: x \in z \implies x \in \O$
+From the definition of the [[Definition:Empty Set|empty set]], the right hand side of this [[Definition:Logical Implication|expression]] is invariably false.
+So by [[Rule of Transposition]], we deduce that:
+:$\forall x: x \notin z$
+But this is the definition of the [[Definition:Empty Set|empty set]].
+So by [[Empty Set is Unique]], it follows that $\powerset \O = \set \O$.
+{{qed|lemma}}
+It is now to be shown that $\powerset {\set \O} = \set {\O, \set \O}$.
+By [[Empty Set is Subset of All Sets]], $\O \in \powerset {\set \O}$.
+By [[Set is Subset of Itself]], $\set \O \in \powerset {\set \O}$.
+And so:
+:$(1) \quad \set {\O, \set \O} \subseteq \powerset {\set \O}$
+It is now to be shown that $\set {\O, \set \O} \supseteq \powerset {\set \O}$.
+{{AimForCont}} that there exists a set $S \in \powerset {\set \O}$ such that $S \ne \O$ and $S \ne \set \O$.
+But then $\exists x \ne \O: x \in \set \O$, which is false.
+Therefore by [[Definition:Contradiction|contradiction]]:
+:$(2) \quad \powerset {\set \O} \subseteq \set {\O, \set \O}$.
+Putting $(1)$ and $(2)$ together yields:
+:$\powerset {\set \O} = \set {\O, \set \O}$
+{{qed|lemma}}
+The set $2 = \set {\O, \set \O}$ is used with the [[Axiom:Axiom of Replacement|Axiom of Replacement]] as the [[Definition:Domain of Mapping|domain]] for a [[Definition:Mapping|mapping]] whose [[Definition:Image of Mapping|image]] is $\set {A, B}$.
+A suitable [[Definition:Mapping|mapping]] would be:
+:$\paren {y = \O \land z = A} \lor \paren {y = \set \O \land z = B}$
+Usually this mapping would be written as:
+:$\map f x = \begin{cases}
+A & : x = \O \\
+B & : x = \set \O
+\end{cases}$
+{{qed}}
+[[Category:Axiom of Pairing]]
+[[Category:Doubletons]]
+0xyy1geaqee8imnyewfu01np6f4o34q
+\end{proof}<|endoftext|>
+\section{Axiom of Pairing from Axiom of Subsets}
+Tags: Axiom of Pairing, Doubletons
+
+\begin{theorem}
+Let it be supposed that there exists a [[Definition:Set|set]] which contains at least two [[Definition:Element|elements]].
+Then the [[Axiom:Axiom of Pairing|Axiom of Pairing]] is a consequence of the [[Axiom:Axiom of Specification|Axiom of Specification]].
+\end{theorem}
+
+\begin{proof}
+Let $A$ be a [[Definition:Set|set]] which contains at least two [[Definition:Element|elements]].
+Let $a$ and $b$ be any two [[Definition:Element|elements]] of $A$.
+Let $\map P x$ be the [[Definition:Propositional Function|propositional function]]:
+:$\map P x := \paren {x = a \lor x = b}$
+Then we may use the [[Axiom:Axiom of Specification|Axiom of Specification]] to define $B$ as:
+:$x \in B \iff \set {x \in A: \map P x}$
+Hence we can define:
+:$B := \set {a, b}$
+for any two $a, b \in A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Singleton Set}
+Tags: Axiomatic Set Theory, Singletons
+
+\begin{theorem}
+Let $a$ be a [[Definition:Set|set]].
+Then the [[Definition:Singleton|singleton set]] $\set a$ may be constructed such that:
+:$a \in \set a$
+\end{theorem}
+
+\begin{proof}
+Let $a$ be a [[Definition:Set|set]].
+From the [[Axiom:Axiom of Pairing|Axiom of Pairing]] the set $\set {a, a}$ may be formed.
+From the [[Axiom:Axiom of Extension|Axiom of Extension]] it follows that:
+:$\set {a, a} = \set a$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection is Subset of Union of Intersections with Complements}
+Tags: Set Union, Set Intersection, Set Complement
+
+\begin{theorem}
+Let $R, S, T$ be [[Definition:Set|sets]].
+Then:
+:$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$
+where $\overline R$ denotes the [[Definition:Set Complement|complement]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S \cap T$.
+Then by definition of [[Definition:Set Intersection|set intersection]], $x \in S \land x \in T$.
+From [[Conjunction implies Disjunction of Conjunctions with Complements]], it follows that:
+:$\paren {x \in S \land \psi} \lor \paren {x \in T \land \neg \psi}$
+where $\psi$ is any arbitrary [[Definition:Statement|statement]].
+Let $\psi$ be the [[Definition:Statement|statement]] $x \in R$.
+Thus:
+:$\paren {x \in S \land x \in T} \implies \paren {x \in S \land x \in R} \lor \paren {x \in T \land x \notin R}$
+By definition of [[Definition:Subset|subset]], [[Definition:Set Intersection|set intersection]], [[Definition:Set Union|set union]] and [[Definition:Set Complement|set complement]], it follows that:
+:$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Unions with Complements is Subset of Union}
+Tags: Set Union, Set Intersection, Set Complement
+
+\begin{theorem}
+Let $R, S, T$ be [[Definition:Set|sets]].
+Then:
+:$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \left({R \cup S}\right) \cap \left({\overline R \cup T}\right)$.
+Then by definition of [[Definition:Set Intersection|set intersection]], [[Definition:Set Union|set union]] and [[Definition:Set Complement|set complement]], we have:
+:$\left({x \in R \lor x \in S}\right) \land \left({x \notin R \lor x \in T}\right)$
+From [[Conjunction of Disjunctions with Complements implies Disjunction]] it follows that:
+:$x \in S \lor x \in T$
+By definition of [[Definition:Set Union|set union]], it follows that:
+:$x \in S \cup T$
+By definition of [[Definition:Subset|subset]] it follows that:
+:$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjunction implies Disjunction of Conjunctions with Complements}
+Tags: Conjunction, Disjunction
+
+\begin{theorem}
+:$p \land q \vdash \left({p \land r}\right) \lor \left({q \land \neg r}\right)$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|p \land q \vdash \left({p \land r}\right) \lor \left({q \land \neg r}\right)}}
+{{Premise|1|p \land q}}
+{{ExcludedMiddle|2|r \lor \neg r}}
+{{Simplification|3|1|p|1|1}}
+{{Simplification|4|1|q|1|2}}
+{{Assumption|5|r}}
+{{Conjunction|6|1, 5|p \land r|1|5}}
+{{Addition|7|1, 5|\left({p \land r}\right) \lor \left({q \land \neg r}\right)|6|1}}
+{{Assumption|8|\neg r}}
+{{Conjunction|9|1, 8|q \land \neg r|4|8}}
+{{Addition|10|1, 8|\left({p \land r}\right) \lor \left({q \land \neg r}\right)|9|2}}
+{{ProofByCases|11|1|\left({p \land r}\right) \lor \left(q \land \neg r\right)|2|5|7|8|10}}
+{{EndTableau}}
+{{qed}}
+[[Category:Conjunction]]
+[[Category:Disjunction]]
+0770tcw59m2lo3ynmatxcn9fqt8yf71
+\end{proof}<|endoftext|>
+\section{Conjunction of Disjunctions with Complements implies Disjunction}
+Tags: Conjunction, Disjunction
+
+\begin{theorem}
+:$\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q$
+\end{theorem}
+
+\begin{proof}
+{{BeginTableau|\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q}}
+{{Premise|1|\paren {p \lor r} \land \paren {q \lor \neg r} }}
+{{Simplification|2|1|p \lor r|1|1|The aim is to use [[Proof by Cases]] on this ...}}
+{{Assumption|3|p|Assume the first of the disjuncts ...}}
+{{Addition|4|3|p \lor q|3|1|... and demonstrate the conclusion}}
+{{Simplification|5|1|q \lor \neg r|1|2}}
+{{Commutation|6|1|\neg r \lor q|5|Disjunction}}
+{{SequentIntro|7|1|r \implies q|6|[[Rule of Material Implication]]}}
+{{Assumption|8|r| then assume the second of the disjuncts ...}}
+{{ModusPonens|9|1, 8|q|7|8}}
+{{Addition|10|1, 8|p \lor q|9|2|... and demonstrate the conclusion}}
+{{ProofByCases|11|1|p \lor q|2|3|4|8|10}}
+{{EndTableau}}
+{{qed}}
+[[Category:Conjunction]]
+[[Category:Disjunction]]
+8415phdouelyojbky2zc0dg3vml37w2
+\end{proof}<|endoftext|>
+\section{Intersection of Elements of Power Set}
+Tags: Intersection, Set Intersection, Power Set, Set Intersection, Empty Set
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let:
+: $\displaystyle \mathbb S = \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X$
+where $\mathcal P \left({S}\right)$ is the [[Definition:Power Set|power set]] of $S$.
+Then $\mathbb S = \varnothing$.
+\end{theorem}
+
+\begin{proof}
+By [[Intersection is Subset]]:
+:$\displaystyle \forall X \in P \left({S}\right): \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X \subseteq X$
+From [[Empty Set is Element of Power Set]]:
+: $\varnothing \in P \left({S}\right)$
+So:
+:$\displaystyle \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X \subseteq \varnothing$
+From [[Empty Set is Subset of All Sets]]:
+:$\displaystyle \varnothing \subseteq \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X$
+So by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\displaystyle \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X = \varnothing$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set of Subset}
+Tags: Power Set, Subsets
+
+\begin{theorem}
+Let $S \subseteq T$ where $S$ and $T$ are both [[Definition:Set|sets]].
+Then:
+:$\powerset S \subseteq \powerset T$
+where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = X
+      | o = \in
+      | r = \powerset S
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = X
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | ll= \leadsto
+      | l = X
+      | o = \subseteq
+      | r = T
+      | c = as $S \subseteq T$: [[Subset Relation is Transitive]]
+}}
+{{eqn | l = X
+      | o = \in
+      | r = \powerset T
+      | c = {{Defof|Power Set}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Elements of Power Set}
+Tags: Power Set, Set Union
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then:
+:$\displaystyle S = \bigcup_{X \mathop \in \powerset S} X$
+where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
+\end{theorem}
+
+\begin{proof}
+By [[Subset of Union]]:
+:$\displaystyle \forall X \in \powerset S: X \subseteq \bigcup_{X \mathop \in \powerset S} X$
+From [[Set is Subset of Itself]], $S \subseteq S$ and so $S \in \powerset S$.
+So:
+:$\displaystyle S \subseteq \bigcup_{X \mathop \in \powerset S} X$
+From [[Union is Smallest Superset]]:
+:$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
+where $\mathbb S \subseteq \powerset S$.
+But as $\powerset S \subseteq \powerset S$ from [[Set is Subset of Itself]]:
+:$\displaystyle \paren {\forall X \in \powerset S: X \subseteq S} \iff \bigcup \powerset S \subseteq S$
+The {{LHS}} is no more than the definition of the [[Definition:Power Set|power set]], making it a [[Definition:Tautology|tautological statement]], and so:
+:$\displaystyle \bigcup \powerset S \subseteq S$
+The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Ordered Pair}
+Tags: Cartesian Product, Axiomatic Set Theory
+
+\begin{theorem}
+The following definitions of an [[Definition:Ordered Pair|ordered pair]] are [[Definition:Logical Equivalence|equivalent]]:
+\end{theorem}
+
+\begin{proof}
+=== Equality of Ordered Pairs ===
+From [[Equality of Ordered Pairs]], we have that:
+:$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} } \iff a = c, b = d$
+hence showing that the [[Definition:Kuratowski Formalization of Ordered Pair|Kuratowski formalization]] fulfils the requirement for equality.
+=== Existence of Cartesian Product ===
+Let $A$ and $B$ be [[Definition:Set|sets]].
+Suppose that either $A$ or $B$ is [[Definition:Empty Set|empty]].
+Then:
+{{begin-eqn}}
+{{eqn | l = A \times B
+      | r = \set {\tuple {a, b} : a \in A \land b \in B}
+      | c = {{Defof|Cartesian Product}}
+}}
+{{eqn | r = \O
+}}
+{{end-eqn}}
+By the [[Axiom:Axiom of Empty Set|Axiom of the Empty Set]] $A \times B$ is a [[Definition:Set|set]].
+Now suppose that $A$ and $B$ are both [[Definition:Non-Empty Set|non-empty]].
+Let $a \in A$ and $b \in B$.
+By [[Singleton of Element is Subset]]:
+:$\set a \subseteq A$ and $\set b \subseteq B$
+Therefore by [[Set Union Preserves Subsets]]:
+:$\set {a, b} \subseteq A \cup B$
+$A \cup B$ is a [[Definition:Set|set]] by [[Union of Small Classes is Small]].
+Because $\set a \subseteq A \subseteq A \cup B$, it follows that:
+:$\set a, \set {a, b} \in \powerset {A \cup B}$
+where $\powerset {A \cup B}$ is the [[Definition:Power Set|power set]] of $A \cup B$.
+This a [[Definition:Set|set]] by the [[Axiom:Axiom of Powers|Axiom of Powers]].
+Thus:
+:$\set {\set a, \set {a, b} } \subseteq \powerset {A \cup B}$
+and so by definition of the [[Definition:Power Set|power set]]:
+:$\set {\set a, \set {a, b} } \in \powerset {\powerset {A \cup B} }$
+It has been shown that:
+:$\set {\tuple {a, b}: a \in A \land b \in B} \subseteq \powerset {\powerset {A \cup B} }$
+Applying the [[Axiom:Axiom of Specification|axiom of specification]] and the [[Axiom:Axiom of Extension|axiom of extension]], the unique set $A \times B$ is created which consists exactly of [[Definition:Ordered Pair|ordered pairs]] $\tuple {a, b}$ such that $a \in A$ and $b \in B$.
+Thus it has been demonstrated that if $A$ and $B$ are [[Definition:Non-Empty Set|non-empty]], then the [[Definition:Cartesian Product|cartesian product]] $A \times B$ exists and is [[Definition:Non-Empty Set|non-empty]].
+Finally, in [[Subset of Cartesian Product]] it is demonstrated that every [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]] is a [[Definition:Subset|subset]] of the [[Definition:Cartesian Product|cartesian product]] of two sets.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Cartesian Product}
+Tags: Cartesian Product, Axiomatic Set Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]].
+Then $S$ is the [[Definition:Subset|subset]] of the [[Definition:Cartesian Product|cartesian product]] of two sets.
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]].
+Let $x \in S$ such that $x = \left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\}$ as defined in [[Kuratowski Formalization of Ordered Pair]].
+Since the elements of $S$ are [[Definition:Set|sets]], we can form the [[Definition:Set Union|union]] $\mathbb S = \bigcup S$ of the sets in $S$.
+Since $x \in S$ it follows that the [[Definition:Element|elements]] of $x$ are elements of $\mathbb S$.
+Since $\left\{{a, b}\right\} \in x$ it follows that $\left\{{a, b}\right\} \in \mathbb S$.
+Now we can form the [[Definition:Set Union|union]] $\mathbb S' = \bigcup \mathbb S$ of the sets in $\mathbb S$.
+Since $\left\{{a, b}\right\} \in \mathbb S$ it follows that both $a$ and $b$ are elements of $\mathbb S' = \bigcup \bigcup S$.
+Thus from the [[Kuratowski Formalization of Ordered Pair]] we have that $S$ is a [[Definition:Subset|subset]] of some $A \times B$.
+We can at this stage take both $A$ and $B$ as being equal to $\bigcup \bigcup S$.
+Finally, the [[Axiom:Axiom of Specification|axiom of specification]] is applied to construct the sets:
+:$A = \left\{{a: \exists b: \left({a, b}\right) \in S}\right\}$
+and
+:$B = \left\{{b: \exists a: \left({a, b}\right) \in S}\right\}$
+$A$ and $B$ are seen to be the [[Definition:First Projection|first]] and [[Definition:Second Projection|second projections]] respectively of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Relation is Smallest Equivalence Relation}
+Tags: Equivalence Relations
+
+\begin{theorem}
+The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on $S$ is the [[Definition:Minimal Set|smallest]] [[Definition:Equivalence Relation|equivalence]] in $S$, in the sense that:
+:$\forall \mathcal E \subseteq S \times S: \Delta_S \subseteq \mathcal E$
+where $\mathcal E$ denotes a general [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+It is confirmed that, from [[Diagonal Relation is Equivalence]], $\Delta_S$ is an [[Definition:Equivalence Relation|equivalence relation]].
+Let $\mathcal E$ be an arbitrary [[Definition:Equivalence Relation|equivalence relation]].
+By definition, $\mathcal E$ is [[Definition:Reflexive Relation|reflexive]].
+From [[Relation Contains Diagonal Relation iff Reflexive]] it follows that as $\Delta_S \subseteq \mathcal E$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trivial Relation is Largest Equivalence Relation}
+Tags: Equivalence Relations, Trivial Relation
+
+\begin{theorem}
+The [[Definition:Trivial Relation|trivial relation]] $\mathcal T$ on $S$ is the largest [[Definition:Equivalence Relation|equivalence]] in $S$, in the sense that:
+:$\forall \mathcal E \subseteq S \times S: \mathcal E \subseteq \mathcal T$
+where $\mathcal E$ denotes a general [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Trivial Relation|trivial relation]] $\mathcal T$ on $S$ is defined as:
+:$\mathcal T = S \times S$
+It is confirmed from [[Trivial Relation is Equivalence]] that the [[Definition:Trivial Relation|trivial relation]] is in fact an [[Definition:Equivalence Relation|equivalence relation]].
+Let $\mathcal E$ be an arbitrary [[Definition:Equivalence Relation|equivalence relation]] on $S$.
+By definition of [[Definition:Relation|relation]], $\mathcal E \subseteq S \times S$ and so (trivially) $\mathcal E \subseteq \mathcal T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuously Differentiable Curve has Finite Arc Length}
+Tags: Integral Calculus, Analytic Geometry
+
+\begin{theorem}
+Let $y = \map f x$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ and [[Definition:Continuously Differentiable|continuously differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+The [[Definition:Definite Integral|definite integral]]: 
+:$s = \displaystyle \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$
+exists, and is called the '''[[Definition:Arc Length|arc length]]''' of $f$ between $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+It intuitively makes sense to define the [[Definition:Linear Measure|length]] of a [[Definition:Line Segment|line segment]] to be the distance between the two end points, as given by the [[Distance Formula]]:
+:$\sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
+Similarly, it is reasonable to assume that the actual length of the curve would be approximately equal to the sum of the lengths of each of the line segments, as shown:
+[[File:ArcLength1.png|400px]]
+{{handwaving|The intuitive arguments need to be replaced with something more rigorous, which would probably happen on another page and be linked.}}
+To calculate the sum of the length of these line segments, divide $\closedint a b$ into any number of [[Definition:Closed Real Interval|closed subintervals]] of the form $\closedint {x_{i - 1} } {x_i}$ where:
+:$a = x_0 < x_1 < \dotsb < x_{k - 1} < x_k = b$
+Define:
+:$\Delta x_i = x_i - x_{i - 1}$
+:$\Delta y_i = y_i - y_{i - 1}$
+As the length of the $i$th line segment is $\sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$, the sum of all these line segments is given by:
+:$\displaystyle \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$
+{{begin-eqn}}
+{{eqn | r = \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\frac {\Delta y_i} {\Delta x_i} {\Delta x_i} }^2}
+      | c = multiply the second term in the [[Definition:Radicand|radicand]] by $1 = \dfrac {\Delta x_i} {\Delta x_i}$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2} \Delta x_i
+      | c = factor $\Delta x_i$ out of the [[Definition:Radicand|radicand]]
+}}
+{{end-eqn}}
+Thus the approximate arc length is given by the sum:
+:$\displaystyle s \approx \sum_{i \mathop = 1}^k \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2}\Delta x_i$
+Recall that [[Definition:By Hypothesis|by hypothesis]]:
+:$f$ is [[Definition:Continuous Real Function on Interval|continuous]] on $\closedint a b$
+:$f$ is [[Definition:Differentiable on Interval|differentiable]] on $\openint a b$.
+Thus the [[Mean Value Theorem]] can be applied.
+In every [[Definition:Open Real Interval|open interval]] $\openint {x_i} {x_{i - 1} }$ there exists some $c_i$ such that:
+:$D_x \, \map f {c_i} = \dfrac {\Delta y_i} {\Delta x_i}$
+Plugging this into the above sum we have:
+:$\displaystyle s \approx \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \Delta x_i$
+[[Definition:By Hypothesis|By hypothesis]], $D_x f$ is [[Definition:Continuous Real Function on Interval|continuous]].
+As [[Square of Real Number is Non-Negative]] the [[Definition:Radicand|radicand]] is always positive.
+From [[Continuity of Root Function]] and [[Limit of Composite Function]], $\sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$ is continuous as well.
+Because [[Continuous Real Function is Darboux Integrable]], there exists a [[Definition:Definite Integral|definite integral]] that confirms the intuitive notion that there is a value that represents the exact length of the curve as the limit of the above [[Definition:Riemann Sum|Riemann sum]].
+This [[Definition:Definite Integral|definite integral]] is:
+:$\displaystyle s = \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arc Length for Parametric Equations}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $x = \map f t$ and $y = \map g t$ be [[Definition:Real Function|real functions]] of a parameter $t$. 
+Let these equations describe a [[Definition:Curve|curve]] $\mathcal C$ that is [[Definition:Continuous on Interval|continuous]] for all $t \in \closedint a b$ and [[Definition:Continuously Differentiable|continuously differentiable]] for all $t \in \openint a b$.
+Suppose that the [[Definition:Graph of Mapping|graph]] of the curve does not intersect itself for any $t \in \openint a b$.
+Then the [[Definition:Arc Length|arc length]] of $\mathcal C$ between $a$ and $b$ is given by:
+:$s = \displaystyle \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \rd t$
+for $\dfrac {\d x} {\d t} \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = s
+      | r = \int_a^b \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x
+      | c = {{Defof|Arc Length}}
+}}
+{{eqn | r = \int_a^b \sqrt {\paren {\frac {\frac {\d x} {\d t} } {\frac {\d x} {\d t} } }^2 + \paren {\frac {\frac {\d y}{\d t} } {\frac {\d x} {\d t} } }^2} \rd x
+      | c = because $\paren {\dfrac {\frac {\d x} {\d t} } {\frac {\d x}{\d t} } }^2 = 1$, and from [[Derivative of Composite Function/Corollary|corollary to chain rule]]
+}}
+{{eqn | r = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \paren {\frac 1 {\frac {\d x} {\d t} } } \rd x
+      | c = factoring $\dfrac {\d x}{\d t}$ out of the [[Definition:Radicand|radicand]]. No [[Definition:Absolute Value|absolute value]] is needed as [[Definition:Linear Measure|length]] cannot be negative.
+}}
+{{eqn | r = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \frac {\d t} {\d x} \rd x
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | r = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \rd t
+      | c = [[Integration by Substitution]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Element of Injection}
+Tags: Injections, Inverse Mappings
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be an [[Definition:Injection|injection]].
+Then:
+:$\map {f^{-1} } y = x \iff \map f x = y$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $y = \map f x$.
+From the definition of the [[Definition:Preimage of Element under Mapping|preimage of an element]]:
+:$\map {f^{-1} } y = \set {x \in S: \tuple {y, x} \in f}$
+Thus:
+:$x \in \map {f^{-1} } y$
+By definition of [[Definition:Injection/Definition 4|injection]], $\map {f^{-1} } y$ is a [[Definition:Singleton|singleton]]:
+:$\map {f^{-1} } y = \set x$
+which can be expressed as:
+:$\map {f^{-1} } y = x$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\map {f^{-1} } y = x$.
+Thus by definition of [[Definition:Direct Image Mapping of Mapping|direct image mapping]]:
+:$\map {f^\gets} {\set y} = \set x$
+Then:
+:$\map {f^\to} {\map {f^\gets} {\set y} } = \map {f^\to} {\set x}$
+So from the [[Image of Preimage under Mapping/Corollary|corollary to Image of Preimage under Mapping]]:
+:$\set y = \map {f^\to} {\set x}$
+and so by definition of [[Definition:Direct Image Mapping of Mapping|direct image mapping]]:
+:$y = \map f x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hilbert Space Separable iff Countable Dimension}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Then $H$ is a [[Definition:Separable Space|separable space]] iff it has [[Definition:Countable|countable]] [[Definition:Dimension (Hilbert Space)|dimension]].
+\end{theorem}<|endoftext|>
+\section{Net Convergence Equivalent to Absolute Convergence}
+Tags: Banach Spaces, Nets (Topology)
+
+\begin{theorem}
+Let $V$ be a [[Definition:Banach Space|Banach space]].
+Let $\left({v_n}\right)_{n \in \N}$ be a [[Definition:Sequence|sequence]] of elements in $V$.
+Then the following two statements are equivalent:
+:$(1): \qquad \displaystyle \sum_{n=1}^\infty \left\Vert{v_n}\right\Vert$ converges ([[Definition:Absolutely Convergent Series|absolute convergence]])
+:$(2): \qquad \displaystyle \sum \left\{{v_n: n \in \N}\right\}$ converges (generalised or [[Definition:Generalized Sum|net convergence]])
+\end{theorem}<|endoftext|>
+\section{Intersection of Infinite Successor Sets}
+Tags: Infinite Sets
+
+\begin{theorem}
+Let $\mathbb S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Infinite Successor Set|infinite successor sets]].
+Then $\bigcap \mathbb S$ is an [[Definition:Infinite Successor Set|infinite successor set]].
+\end{theorem}
+
+\begin{proof}
+From definition, a [[Definition:Set|set]] $X$ is an [[Definition:Infinite Successor Set|infinite successor set]] {{iff}} both the following holds:
+:$\O \in X$
+:$x \in X \implies x^+ \in X$
+For all $S \in \mathbb S$, $S$ is an [[Definition:Infinite Successor Set|infinite successor set]].
+From definition of an [[Definition:Infinite Successor Set|infinite successor set]], $\O \in S$.
+By definition of [[Definition:Set Intersection|set intersection]], $\O \in \bigcap \mathbb S$.
+Now suppose $x \in \bigcap \mathbb S$.
+By definition of [[Definition:Set Intersection|set intersection]]:
+:$\forall S \in \mathbb S: x \in S$
+Hence from definition of an [[Definition:Infinite Successor Set|infinite successor set]], $x^+ \in S$.
+By definition of [[Definition:Set Intersection|set intersection]], $x^+ \in \bigcap \mathbb S$.
+Therefore $\bigcap \mathbb S$ is an [[Definition:Infinite Successor Set|infinite successor set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Infinite Successor Set Exists}
+Tags: Infinite Sets
+
+\begin{theorem}
+There exists a [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]] $\omega$ that is a [[Definition:Subset|subset]] of every other [[Definition:Infinite Successor Set|infinite successor set]].
+\end{theorem}
+
+\begin{proof}
+From the [[Axiom:Axiom of Infinity|Axiom of Infinity]], there is a [[Definition:Set|set]] $S$ such that:
+:$\O \in S$
+:$x \in S \implies x^+ \in S$
+where $\O$ denotes the [[Definition:Empty Set|empty set]] and $x^+$ is the [[Definition:Successor Set|successor set]] of $x$.
+That is, there exists an [[Definition:Infinite Successor Set|infinite successor set]].
+Next, by the [[Axiom:Axiom of Specification|Axiom of Specification]], the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]] $\omega$:
+:$\omega := \displaystyle \bigcap \set {S' \subseteq S: \text{$S'$ is an infinite successor set} }$
+exists.
+It remains to be shown that if $T$ is an [[Definition:Infinite Successor Set|infinite successor set]], then $\omega \subseteq T$.
+By [[Intersection of Infinite Successor Sets]], $S \cap T$ is an [[Definition:Infinite Successor Set|infinite successor set]].
+Moreover, by [[Intersection is Subset]], $S \cap T$, so that:
+:$S \cap T \in \set {S' \subseteq S: \text{$S'$ is an infinite successor set} }$
+from which we conclude by definition of $\omega$ and [[Intersection is Subset/General Result|Intersection is Subset: General Result]] that:
+:$\omega \subseteq S \cap T \subseteq T$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Infinite Successor Set forms Peano Structure}
+Tags: Peano's Axioms, Minimal Infinite Successor Set
+
+\begin{theorem}
+Let $\omega$ be the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]].
+Let $\cdot^+: \omega \to \omega$ be the [[Definition:Mapping|mapping]] assigning to a [[Definition:Set|set]] its [[Definition:Successor Set|successor set]]:
+:$n^+ := n \cup \left\{{n}\right\}$
+Let $\varnothing \in \omega$ be the [[Definition:Empty Set|empty set]].
+Then $\left({\omega, \cdot^+, \varnothing}\right)$ is a [[Definition:Peano Structure|Peano structure]].
+\end{theorem}
+
+\begin{proof}
+We need to check that all of [[Axiom:Peano's Axioms|Peano's axioms]] hold for $\left({\omega, \cdot^+, \varnothing}\right)$.
+Suppose first that for $m, n \in \omega$, we have $m^+ = n^+$.
+Since $n \in n^+$ it follows that $n \in m^+$.
+Hence, either $n \in m$ or $n = m$.
+Similarly, either $m \in n$ or $m = n$.
+Now if $n \ne m$, both $m \in n$ and $n \in m$.
+By [[Element of Minimal Infinite Successor Set is Transitive Set]], it follows that $n \subseteq m$.
+As $m \in n$, this contradicts [[Finite Ordinal is not Subset of one of its Elements]].
+Hence it must be that $n = m$, and [[Axiom:Peano's Axioms|Axiom $(P3)$]] holds.
+Next, since $n \in n^+$ for all $n \in \omega$, it follows that $n^+ \ne \varnothing$.
+Hence, [[Axiom:Peano's Axioms|Axiom $(P4)$]] holds as well.
+Finally, let $S \subseteq \omega$ satisfy:
+:$\varnothing \in S$
+:$\forall n \in S: n^+ \in S$
+Then by definition, $S$ is an [[Definition:Infinite Successor Set|infinite successor set]].
+Therefore, by definition of $\omega$ as the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]]:
+:$\omega \subseteq S$
+Consequently $S = \omega$ by the definition of [[Definition:Set Equality/Definition 2|set equality]].
+Thus [[Axiom:Peano's Axioms|Axiom $(P5)$]] is seen to hold.
+That is, $\left({\omega, \cdot^+, \varnothing}\right)$ is a [[Definition:Peano Structure|Peano structure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Logarithm}
+Tags: Integral Calculus
+
+\begin{theorem}
+:$\displaystyle \int \ln x \ \mathrm d x = x \ln x - x + C$
+:$\displaystyle \int \log_a x \ \mathrm d x =  \frac 1 {\ln a}\left({x \ln x - x}\right) + C$
+for $x > 0$.
+\end{theorem}
+
+\begin{proof}
+=== Base $e$ ===
+:$\displaystyle \int \ln x \ \mathrm d x = \int 1 \cdot \ln x \ \mathrm d x$
+From [[Integration by Parts]]:
+:$\displaystyle \int fg' \, \mathrm dx = fg - \int f'g \, \mathrm dx$
+Here:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \ln x
+}}
+{{eqn | ll = \implies
+      | l = f'
+      | r = \frac 1 x
+      | c = [[Derivative of Natural Logarithm Function]]
+}}
+{{eqn | l = g'
+      | r = 1
+}}
+{{eqn | ll = \implies
+      | l = g
+      | r = x
+      | c = [[Integral of Constant]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn| l = \int \ln x \ \mathrm d x
+     | r = x \cdot \ln x -\int x \cdot \frac 1 x \ \mathrm dx
+}}
+{{eqn | r = x \ln x - x + C
+}}
+{{end-eqn}}
+{{qed}}
+=== Base $a$ ===
+Follows from [[Linear Combination of Integrals]] and [[Linear Operator on General Logarithm]].
+{{qed}}
+[[Category:Integral Calculus]]
+fzp9dlrzzftz592g530wc8klw6j0asy
+\end{proof}<|endoftext|>
+\section{Finite Ordinal is not Subset of one of its Elements}
+Tags: Finite Ordinals
+
+\begin{theorem}
+Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
+Then:
+:$\nexists x \in n: n \subseteq x$
+that is, $n$ is not a [[Definition:Subset|subset]] of one of its [[Definition:Element|elements]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be the [[Definition:Set|set]] of all those [[Definition:Finite Ordinal|finite ordinals]] $n$ which are not a [[Definition:Subset|subset]] of any of its [[Definition:Element|elements]].
+That is:
+:$n \in S \iff n \in \omega \land \forall x \in n: n \nsubseteq x$
+We know that $0 = \varnothing$ is not a [[Definition:Subset|subset]] of any of its [[Definition:Element|elements]], as $\varnothing$ by [[Definition:Empty Set|definition]] has no [[Definition:Element|elements]].
+So $0 \in S$.
+Now suppose $n \in S$.
+From [[Set is Subset of Itself]]:
+:$n \subseteq n$
+But as $n \in S$ it follows by definition of $S$ that:
+:$n \notin n$
+By definition of the [[Definition:Successor Set|successor]] of $n$, it follows that:
+:$n^+ \nsubseteq n$
+Now from [[Subset Relation is Transitive]]:
+:$n^+ \subseteq x \implies n \subseteq x$
+But since $n \in S$ it follows that:
+:$x \notin n$
+So:
+:$n^+ \nsubseteq n$
+and:
+:$\forall x \in n: n^+ \nsubseteq x$
+So $n^+$ is not a [[Definition:Subset|subset]] of any of its [[Definition:Element|elements]].
+That is:
+:$n^+ \in S$
+So by the [[Principle of Mathematical Induction]]:
+:$S = \omega$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Minimal Infinite Successor Set is Transitive Set}
+Tags: Minimal Infinite Successor Set
+
+\begin{theorem}
+Let $\omega$ be the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]].
+Let $n \in \omega$.
+Then $x \in n \implies x \subseteq n$.
+That is, every [[Definition:Element|element]] of $n$ is also a [[Definition:Subset|subset]] of it.
+In other words, each element of $\omega$ is a [[Definition:Transitive Set|transitive set]].
+\end{theorem}
+
+\begin{proof}
+Let $S \subseteq$ be the [[Definition:Set|set]] of all [[Definition:Transitive Set|transitive]] [[Definition:Element|elements]] of $\omega$.
+That is:
+:$n \in S \iff n \in \omega \land \forall x \in n: x \subseteq n$
+It is [[Definition:Vacuous Truth|vacuously true]] that $0 \in S$, as there are no $x \in 0$.
+Now suppose $n \in S$.
+If $x \in n^+$ then either $x \in n$ or $x = n$.
+In the first case:
+:$x \subseteq n$ as $n \in S$
+and so:
+:$x \subseteq n^+$
+In the second case, by definition of [[Definition:Successor Set|successor set]]:
+:$x \subseteq n^+$
+By [[Principle of Mathematical Induction for Minimal Infinite Successor Set]] it follows that:
+:$S = \omega$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Exponential Function}
+Tags: Primitives involving Exponential Function, Primitive of Exponential Function
+
+\begin{theorem}
+:$\displaystyle \int e^x \rd x = e^x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}<|endoftext|>
+\section{Natural Number Addition Commutes with Zero}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
+Then:
+:$\forall n \in \N: 0 + n = n = n + 0$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+From the [[Definition:Addition in Minimal Infinite Successor Set|definition of addition]], we have that:
+{{begin-eqn}}
+{{eqn | ll= \forall m, n \in \N:
+      | l = m + 0
+      | r = m
+      | c = 
+}}
+{{eqn | l = m + n^+
+      | r = \left({m + n}\right)^+
+      | c = 
+}}
+{{end-eqn}}
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$0 + n = n = n + 0$
+=== Basis for the Induction ===
+By definition, we have:
+:$0 + 0 = 0 = 0 + 0$
+Thus $P \left({0}\right)$ is seen to be true.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] $P \left({k}\right)$:
+:$0 + k = k = k + 0$
+Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
+:$0 + k^+ = k^+ = k^+ + 0$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = 0 + k^+
+      | r = \left({0 + k}\right)^+
+      | c = Definition of [[Definition:Addition in Minimal Infinite Successor Set|Addition]]
+}}
+{{eqn | r = \left({k + 0}\right)^+
+      | c = from the [[Natural Number Addition Commutes with Zero#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | r = k^+
+      | c = Definition of [[Definition:Addition in Minimal Infinite Successor Set|Addition]]: $k + 0 = k$
+}}
+{{end-eqn}}
+By definition:
+:$k^+ + 0 = k^+$
+So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: 0 + n = n = n + 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Number Addition Commutativity with Successor}
+Tags: Natural Numbers, Natural Number Addition Commutativity with Successor
+
+\begin{theorem}
+Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
+Then:
+:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+From the [[Definition:Addition in Minimal Infinite Successor Set|definition of addition]], we have that:
+{{begin-eqn}}
+{{eqn | ll= \forall m, n \in \N:
+      | l = m + 0
+      | r = m
+      | c = 
+}}
+{{eqn | l = m + n^+
+      | r = \left({m + n}\right)^+
+      | c = 
+}}
+{{end-eqn}}
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\forall m \in \N: m^+ + n = \left({m + n}\right)^+$
+=== Basis for the Induction ===
+By definition, we have:
+{{begin-eqn}}
+{{eqn | ll= \forall m \in \N:
+      | l = m^+ + 0
+      | r = m^+
+      | c = by definition
+}}
+{{eqn | r = \left({m + 0}\right)^+
+      | c = by definition: $m = m + 0$
+}}
+{{end-eqn}}
+Thus $P \left({0}\right)$ is seen to be true.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] $P \left({k}\right)$:
+:$\forall m \in \N: m^+ + k = \left({m + k}\right)^+$
+Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
+:$\forall m \in \N: m^+ + k^+ = \left({m + k^+}\right)^+$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = m^+ + k^+
+      | r = \left({m^+ + k}\right)^+
+      | c = by definition
+}}
+{{eqn | r = \left({\left({m + k}\right)^+}\right)^+
+      | c = from the [[Natural Number Addition Commutativity with Successor/Proof 1#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | r = \left({m + k^+}\right)^+
+      | c = by definition
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m, n \in \N: m^+ + n = \left({m + n}\right)^+$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|following axioms]]:
+{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
+Proof by [[Principle of Mathematical Induction|induction]]:
+From [[Axiom:Axiomatization of 1-Based Natural Numbers]], we have by definition that:
+{{begin-eqn}}
+{{eqn | ll= \forall m, n \in \N:
+      | l = m + 0
+      | r = m
+}}
+{{eqn | l = \left({m + n}\right)^+
+      | r = m + n^+
+}}
+{{end-eqn}}
+For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\forall m \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$
+=== Basis for the Induction ===
+When $n = 1$, we have:
+:$\left({m + 1}\right) + 1 = \left({m + 1}\right) + 1$
+which holds trivially.
+Thus $P \left({1}\right)$ is seen to be true.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] $P \left({k}\right)$:
+:$\forall m \in \N: \left({m + 1}\right) + k = \left({m + k}\right) + 1$
+Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
+:$\forall m \in \N: \left({m + 1}\right) + \left({k + 1}\right) = \left({m + \left({k + 1}\right)}\right) + 1$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left({m + \left({k + 1}\right)}\right) + 1
+      | r = \left({\left({m + k}\right) + 1}\right) + 1
+      | c = [[Natural Number Addition is Associative/Proof 3]]
+}}
+{{eqn | r = \left({\left({m + 1}\right) + k}\right) + 1
+      | c = from the [[Natural Number Addition Commutativity with Successor#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | r = \left({m + 1}\right) + \left({k + 1}\right)
+      | c = [[Natural Number Addition is Associative/Proof 3]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers are Comparable}
+Tags: Ordering on Natural Numbers, Natural Numbers are Comparable
+
+\begin{theorem}
+Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
+Let $m, n \in \N$.
+Then $m$ and $n$ are [[Definition:Comparable|comparable]] by the [[Definition:Ordering on Natural Numbers|ordering relation]] $\le$.
+That is, either:
+:$(1): \quad m \le n$
+or:
+:$(2): \quad n \le m$
+or possibly both.
+\end{theorem}
+
+\begin{proof}
+Let $\N$ be defined as the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]] $\omega$.
+By definition of the [[Definition:Ordering on Minimal Infinite Successor Set|ordering on minimal infinite successor set]]:
+:$m \le n \iff \begin {cases} m = n & \text {or} \\ m \in n & \end {cases}$
+Thus it is sufficient to prove that exactly one of the following is the case:
+:$(1): \quad m \in n$
+:$(2): \quad m = n$
+:$(3): \quad n \in m$
+The proof operates by [[Principle of Finite Induction|induction]]:
+For each $n \in \omega$, let $\map S n$ be the [[Definition:Set|set]] of all $m \in \omega$ which are [[Definition:Comparable|comparable]] with $n$.
+Let $S$ be the [[Definition:Set|set]] of all $n \in \N$ for which $\map S n = \omega$.
+Thus the proof is equivalent to demonstrating that $S = \omega$.
+=== Basis for the Induction ===
+First consider $\map S 0$.
+Clearly $0 \in \map S 0$ as $0 = 0$.
+Let $m \in \map S 0$.
+Then $m \notin 0$ as $0 = \O$.
+So either:
+:$(1): \quad 0 = m$, in which case $0 \in m^+$ by definition of [[Definition:Successor Set|successor set]]
+or:
+:$(2): \quad 0 \in m$, in which case, because $m \in m^+$ by definition of [[Definition:Successor Set|successor set]], again $0 \in m^+$.
+In all cases, $m \in \map S 0 \implies m^+ \in \map S 0$.
+So $\map S 0 = \omega$ by [[Principle of Finite Induction|induction]].
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+The [[Definition:Induction Hypothesis|induction hypothesis]] is that:
+:$\map S k = \omega$
+for some $k \in \omega$.
+It is now necessary to show that it follows that:
+:$\map S {k^+} = \omega$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+Consider the set $\map S {k^+}$, given that $\map S k = \omega$.
+From the [[Natural Numbers are Comparable/Proof using Minimal Infinite Successor Set#Basis for the Induction|basis for the induction]], we have that $k^+ \in \map S 0$.
+That is, $k^+$ is [[Definition:Comparable|comparable]] with $0$.
+So $0$ is [[Definition:Comparable|comparable]] with $k^+$ and so $0 \in \map S {k^+}$.
+Suppose $m \in \map S {k^+}$.
+Then either:
+:$(1): \quad k^+ \in m$, in which case $k^+ \in m^+$
+or:
+:$(2): \quad k^+ = m$, in which case the same applies: $k^+ \in m^+$
+or:
+:$(3): \quad m \in k^+$.
+In case $(3)$, either:
+:$(3 \text a): \quad m = k$, in which case $m^+ = k^+$
+or:
+:$(3 \text b): \quad m \in k$.
+Case $(3 \text b)$ is treated as follows.
+We have that $m^+ \in \map S k$ by the [[Natural Numbers are Comparable/Proof using Minimal Infinite Successor Set#Induction Hypothesis|induction hypothesis]].
+Therefore, either:
+:$(4 \text a): \quad k \in m^+$
+or:
+:$(4 \text b): \quad k = m^+$
+or:
+:$(4 \text c): \quad m^+ \in k$.
+$(4 \text a)$ is not compatible with $m \in k$, because either:
+:$k \in m$
+or
+:$k = m$
+and so $k \subseteq m$ which contradicts [[Finite Ordinal is not Subset of one of its Elements]].
+Both $(4 \text b)$ and $(4 \text c)$ imply that $m^+ \in n^+$.
+Hence $\map S {k^+} = \omega$.
+It follows by the [[Principle of Finite Induction]] that $S = \omega$.
+{{qed|lemma}}
+It then follows from [[Finite Ordinal is not Subset of one of its Elements]] that it is not possible for more than one of:
+:$(1): \quad m \in n$
+:$(2): \quad m = n$
+:$(3): \quad n \in m$
+to be true at the same time.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\N$ be defined as the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]] $\omega$.
+By definition of the [[Definition:Ordering on Von Neumann Construction of Natural Numbers|ordering on von Neumann construction]]:
+:$m \le n \iff m \subseteq n$
+From [[Von Neumann Construction of Natural Numbers is Minimally Inductive]], $\omega$ is [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class]] under the [[Definition:Successor Mapping|successor mapping]].
+Then from [[Minimally Inductive Class under Progressing Mapping induces Nest]], $\omega$ is a [[Definition:Nest (Class Theory)|nest]] in which:
+:$\forall m, n \in \omega: \map g x \subseteq y \lor y \subseteq x$
+The result follows from [[Class under Progressing Mapping such that Elements are Sandwiched is Nest]]:
+:$\forall m, n \in \omega: m \subseteq n \lor n \subseteq m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero is Zero Element for Natural Number Multiplication}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
+Then $0$ is a [[Definition:Zero Element|zero element]] for [[Definition:Natural Number Multiplication|multiplication]]:
+:$\forall n \in \N: 0 \times n = 0 = n \times 0$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$0 \times n = 0 = n \times 0$
+=== Basis for the Induction ===
+By definition, we have:
+:$0 \times 0 = 0 = 0 \times 0$
+Thus $\map P 0$ is seen to be true.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]] $\map P k$:
+:$0 \times k = 0 = k \times 0$
+Then we need to show that $\map P {k + 1}$ follows directly from $\map P k$:
+:$0 \times \paren {k + 1} = 0 = \paren {k + 1} \times 0$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = 0 \times \paren {k + 1}
+      | r = \paren {0 \times k} + 0
+      | c = {{Defof|Natural Number Multiplication}}
+}}
+{{eqn | r = 0 + 0
+      | c = [[Zero is Zero Element for Natural Number Multiplication#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 0
+      | c = {{Defof|Natural Number Addition}}
+}}
+{{end-eqn}}
+By definition:
+:$\paren {k + 1} \times 0 = 0$
+So $\map P k \implies \map P {k + 1}$, and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: 0 \times n = 0 = n \times 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hilbert Space Isomorphism is Equivalence Relation}
+Tags: Hilbert Spaces, Equivalence Relations
+
+\begin{theorem}
+[[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking the three defining properties of an [[Definition:Equivalence Relation|equivalence relation]] in turn:
+=== Reflexivity ===
+For any [[Definition:Hilbert Space|Hilbert space]] $H$, we have the [[Definition:Identity Mapping|identity map]] $\operatorname{id}: H \to H$.
+[[Identity Mapping is Linear]] yields $\operatorname{id}$ to be a [[Definition:Linear Mapping|linear map]].
+Also, [[Identity Mapping is Bijection]] ensures that $\operatorname{id}$ is a [[Definition:Surjection|surjection]].
+Lastly, for any $g, h \in H$, we have:
+:$\left\langle{g, h}\right\rangle = \left\langle{\operatorname{id} \left({g}\right), \operatorname{id} \left({h}\right)}\right\rangle$
+Therefore, the three defining properties for a [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] are satisfied by $\operatorname{id}$.
+Hence, any [[Definition:Hilbert Space|Hilbert space]] $H$ is isomorphic to itself.
+It follows that [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $H, K$ be [[Definition:Isomorphism (Hilbert Spaces)|isomorphic]] [[Definition:Hilbert Space|Hilbert spaces]], and let $U: H \to K$ be a [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]].
+From [[Hilbert Space Isomorphism is Bijection]] and the definition of [[Definition:Bijection/Definition 3|bijection]], it follows that $U$ has a two-sided inverse, $U^{-1}: K \to H$.
+[[Inverse of Linear Map is Linear]] yields $U^{-1}$ to be a [[Definition:Linear Mapping|linear map]] as well.
+[[Bijection iff Inverse is Bijection]] yields that $U^{-1}$ is also a [[Definition:Surjection|surjection]].
+For $k, l \in K$, the computation $\left\langle{k, l}\right\rangle_K = \left\langle{U U^{-1}k, U U^{-1}l}\right\rangle_K = \left\langle{U^{-1}k, U^{-1}l}\right\rangle_H$ shows that $U^{-1}$ preserves the inner product (where in the last step, the fact that $U$ is an isomorphism is used).
+Hence, $U^{-1}$ is an isomorphism, so $K$ and $H$ are also isomorphic.
+It follows that [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $H, K, L$ be [[Definition:Hilbert Space|Hilbert spaces]], and let $U: H \to K, V: K \to L$ be [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphisms]].
+Consider the [[Definition:Composition of Mappings|composition]] $VU : H \to L$.
+From [[Composition of Linear Maps is Linear]], $VU$ is a [[Definition:Linear Mapping|linear map]].
+[[Composite of Surjections is Surjection]] yields $VU$ to be a [[Definition:Surjection|surjection]].
+Lastly, for $g, h \in H$, we have:
+:$\left\langle{g, h}\right\rangle_H = \left\langle{Ug, Uh}\right\rangle_K = \left\langle{VUg, VUh}\right\rangle_L$
+Hence $VU$ is a [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]], and $H$ and $L$ are isomorphic.
+It follows that [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+Having verified these three conditions, it follows that [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] is an [[Definition:Equivalence Relation|equivalence relation]], 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Finite Ordinal iff Subset}
+Tags: Finite Ordinals, Subsets
+
+\begin{theorem}
+Let $m, n$ be [[Definition:Distinct Objects|distinct]] [[Definition:Finite Ordinal|finite ordinals]].
+Then:
+:$m \in n \iff m \subseteq n$
+\end{theorem}
+
+\begin{proof}
+Let $m \in n$.
+Since $n$ is an [[Definition:Ordinal|ordinal]], it is [[Definition:Transitive Set|transitive]].
+Therefore, it follows directly that $m \subseteq n$.
+Now let $m \subseteq n$.
+We have by hypothesis that $m \ne n$.
+From [[Natural Numbers are Comparable/Proof using Minimal Infinite Successor Set|Natural Numbers are Comparable: Proof using Minimal Infinite Successor Set]] it follows that either $m \in n$ or $n \in m$.
+Suppose $n \in m$.
+Then $\exists x \in m: m \subseteq x$ which contradicts [[Finite Ordinal is not Subset of one of its Elements]].
+The only option left is that $m \in n$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering on Natural Numbers is Compatible with Addition}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $m, n, k \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
+Then:
+:$m < n \iff m + k < n + k$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $k \in \N$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
+:$m < n \iff m + k < n + k$
+$\map P 0$ is true, as this just says $m + 0 = m < n = n + 0$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P j$ is true, where $j \ge 0$, then it logically follows that $\map P {j^+}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$m < n \iff m + j < n + j$
+Then we need to show:
+:$m < n \iff m + j^+ < n + j^+$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $m < n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = m
+      | o = <
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = m
+      | o = \subsetneq
+      | r = n
+      | c = [[Element of Finite Ordinal iff Subset]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = m^+
+      | o = \subset
+      | r = n
+      | c = {{Defof|Successor Set}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = m^+
+      | o = \subsetneq
+      | r = n^+
+      | c = {{Defof|Successor Set}}
+}}
+{{eqn | n = 1
+      | ll= \leadstoandfrom
+      | l = m^+
+      | o = <
+      | r = n^+
+      | c = [[Element of Finite Ordinal iff Subset]]
+}}
+{{end-eqn}}
+This gives:
+{{begin-eqn}}
+{{eqn | l = m + j
+      | o = <
+      | r = n + j
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {m + j}^+
+      | o = <
+      | r = \paren {n + j}^+
+      | c = from $(1)$ above
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = m + j^+
+      | o = <
+      | r = n + j^+
+      | c = {{Defof|Addition in Minimal Infinite Successor Set}}
+}}
+{{end-eqn}}
+So $\map P j \implies \map P {j^+}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m, n, k \in \N: m < n \iff m + k < n + k$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hilbert Space Isomorphism is Bijection}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Denote by $\left\langle{\cdot, \cdot}\right\rangle_H$ and $\left\langle{\cdot, \cdot}\right\rangle_K$ their respective [[Definition:Inner Product|inner products]].
+Let $U: H \to K$ be an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]].
+Then $U$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+As $U$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]], it is necessarily [[Definition:Surjection|surjective]].
+Suppose now that $g, h \in H$ are such that $Ug = Uh$.
+Then as $U$ is a [[Definition:Linear Mapping|linear map]], it follows that $U \left({g - h}\right) = \mathbf{0}_K$, the [[Definition:Zero Vector|zero vector]] of $K$.
+From property $(3)$ of an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]], conclude that:
+:$0 = \left\langle{\mathbf{0}_K, \mathbf{0}_K}\right\rangle_K = \left\langle{U \left({g - h}\right), U \left({g - h}\right)}\right\rangle_K = \left\langle{g - h, g - h}\right\rangle_H$
+Property $(5)$ of an [[Definition:Inner Product|inner product]] ensures us that $g - h = \mathbf{0}_H$, i.e. $g = h$.
+Hence $U$ is [[Definition:Injection|injective]].
+As $U$ is both [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]], it is a [[Definition:Bijection|bijection]].
+{{qed}}
+[[Category:Hilbert Spaces]]
+avhhbg8lwg75uz4yiuep5vqe4ari45v
+\end{proof}<|endoftext|>
+\section{Hilbert Spaces Isomorphic iff Same Dimension}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Two [[Definition:Hilbert Space|Hilbert spaces]] are [[Definition:Isomorphism (Hilbert Spaces)|isomorphic]] iff they have the same [[Definition:Dimension (Hilbert Space)|dimension]].
+\end{theorem}<|endoftext|>
+\section{Surjection that Preserves Inner Product is Linear}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]], and denote by $\left\langle{\cdot, \cdot}\right\rangle_H$ and $\left\langle{\cdot, \cdot}\right\rangle_K$ their respective [[Definition:Inner Product|inner products]].
+Let $U: H \to K$ be a [[Definition:Surjection|surjection]] such that:
+:$\forall g,h \in H: \left\langle{g, h}\right\rangle_H = \left\langle{Ug, Uh}\right\rangle_K$
+Then $U$ is a [[Definition:Linear Mapping|linear map]], and hence an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]].
+\end{theorem}<|endoftext|>
+\section{Linear Operator on General Logarithm}
+Tags: Linear Algebra, Logarithms
+
+\begin{theorem}
+Let $\phi: \R^\R \to \R^\R, y \mapsto \phi\left({y}\right)$ be a [[Definition:Linear Operator|linear operator]] on the space of functions from $\R\to\R$.
+Let $y$ be a [[Definition:Real Function|real function]] such that $\forall x \in \R$, $y\left({x}\right) > \mathbf{0}\left({x}\right) = 0$.
+Let $\log_a y$ be the [[Definition:General Logarithm|logarithm of $y$ to base $a$]].
+Then:
+:$\phi \left({\log_a y}\right) = \dfrac 1 {\ln a} \left({\phi \left({\ln y}\right)}\right)$
+where $\ln$ is the [[Definition:Natural Logarithm|natural logarithm]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \phi \left({\log_a y}\right)
+      | r = \phi \left({\frac {\ln y}{\ln a} }\right)
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = \frac 1 {\ln a} \left({\phi \left({\ln y}\right)}\right)
+      | c = definition of [[Definition:Linear Operator|linear operator]] (as $\frac 1 {\ln a}$ is [[Definition:Constant|constant]])
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Linear Algebra]]
+[[Category:Logarithms]]
+17whhp0hyp3jtcb4rjwxzv7tncm83vy
+\end{proof}<|endoftext|>
+\section{Ordering on Natural Numbers is Compatible with Multiplication}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $m, n, k \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
+Let $k \ne 0$.
+Then:
+:$m < n \iff m \times k < n \times k$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+First we note that if $k = 0$ then $m \times k = 0 = n \times k$ whatever $m$ and $n$ are, and the proposition clearly does not hold.
+So, for all $k \in \N \setminus \set 0$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
+:$m < n \iff m \times k < n \times k$
+From [[Identity Element of Natural Number Multiplication is One]]:
+:$m \times 1 = m < n = n \times 1$
+and so $\map P 1$ is true.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j^+}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$m < n \iff m \times j < n \times j$
+Then we need to show:
+:$m < n \iff m \times j^+ < n \times j^+$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $m < n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = m
+      | o = <
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = m \times j
+      | o = <
+      | r = n \times j
+      | c = [[Ordering on Natural Numbers is Compatible with Multiplication#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {m \times j} + m
+      | o = <
+      | r = \paren {n \times j} + m
+      | c = [[Ordering on Natural Numbers is Compatible with Addition]]
+}}
+{{end-eqn}}
+Similarly:
+:$m < n \iff \paren {m \times j} + n < \paren {n \times j} + n$
+But then from [[Ordering on Natural Numbers is Compatible with Addition]], we also have:
+:$m < n \iff \paren {m \times j} + m < \paren {m \times j} + n$
+So from [[Natural Number Ordering is Transitive]], we have:
+:$m < n \iff \paren {m \times j} + m < \paren {n \times j} + n$
+This gives:
+{{begin-eqn}}
+{{eqn | l = m
+      | o = <
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {m \times j} + m
+      | o = <
+      | r = \paren {n \times j} + n
+      | c = from above
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = m \times j^+
+      | o = <
+      | r = n \times j^+
+      | c = {{Defof|Natural Number Multiplication}}
+}}
+{{end-eqn}}
+So $\map P j \implies \map P {j^+}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m, n, k \in \N, k \ne 0: m \times n \iff m \times k < n \times k$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Number Ordering is Transitive}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $m, n, k \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
+Let $<$ be the [[Natural Numbers are Comparable|relation defined on $\N$]] such that:
+:$m < n \iff m \in n$
+where $\N$ is defined as the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]] $\omega$.
+Then:
+:$k < m, m < n \implies k < n$
+That is: $<$ is a [[Definition:Transitive Relation|transitive relation]].
+\end{theorem}
+
+\begin{proof}
+Let $k < m, m < n$.
+By definition it follows that $k \in m, m \in n$.
+We have from [[Element of Finite Ordinal iff Subset]] that:
+:$k \in m \iff k \subseteq m$
+:$m \in n \iff m \subseteq n$
+It follows from [[Subset Relation is Transitive]] that $k \subseteq n$.
+Hence the result.
+{{qed}}
+[[Category:Natural Numbers]]
+3h3h6rhf4et2e2b2hq753efl4aepldy
+\end{proof}<|endoftext|>
+\section{Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal}
+Tags: Finite Ordinals
+
+\begin{theorem}
+Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
+Let $x \subsetneq n$.
+Then for some [[Definition:Finite Ordinal|finite ordinal]] $m < n$:
+:$m \sim x$
+where $m \sim x$ denotes that $m$ is [[Definition:Set Equivalence|(set) equivalent]] to $x$.
+That is, every [[Definition:Proper Subset|proper subset]] of a [[Definition:Finite Ordinal|finite ordinal]] $n$ is [[Definition:Set Equivalence|equivalent]] to some finite ordinal smaller than $n$.
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all [[Definition:Finite Ordinal|finite ordinals]] $n$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$
+$P \left({0}\right)$ is [[Definition:Vacuous Truth|vacuously true]], as there are no [[Definition:Proper Subset|proper subsets]] of $0 = \varnothing$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$x \subsetneq k \implies \exists m \in \N: m < k: m \sim x$
+Then we need to show:
+:$x \subsetneq k^+ \implies \exists m \in \N: m < k^+: m \sim x$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Let $x \subsetneq k^+$.
+Then either:
+:$(1) \quad x \subsetneq k$, in which case the [[Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal#Induction Hypothesis|induction hypothesis]] applies
+or:
+:$(2) \quad x = k$, in which case the result is trivially true
+or:
+:$(3) \quad k \in x$.
+In case $(3)$, we find a [[Definition:Finite Ordinal|finite ordinal]] $j \in k$ such that $j \notin x$.
+Then we define a [[Definition:Mapping|mapping]] $f$ on $x$ as:
+:$\forall i \in x: f \left({i}\right) = \begin{cases}
+i & : i \ne k \\
+j & : i = k
+\end{cases}$
+Clearly $f$ is [[Definition:Injection|injective]] and $f$ maps $x$ into $k$.
+{{Questionable|mapping should be defined somehow differently in order to show '''explicitly''' that x is equivalent to some element of k}}
+So the [[Definition:Image of Mapping|image]] of $x$ under $f$ is either equal to $k$  or by the [[Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal#Induction Hypothesis|induction hypothesis]] equivalent to some element of $k$.
+Consequently, $x$ is always [[Definition:Set Equivalence|equivalent]] to some [[Definition:Element|element]] of $k$.
+So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hilbert Space Direct Sum is Hilbert Space}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $\left({H_i}\right)_{i \in I}$ be a [[Definition:Indexing Set|$I$-indexed family]] of [[Definition:Hilbert Space|Hilbert spaces]] over $\Bbb F \in \left\{{\R, \C}\right\}$.
+Let $H = \displaystyle \bigoplus_{i \mathop \in I} H_i$ be their [[Definition:Hilbert Space Direct Sum|Hilbert space direct sum]].
+Then $H$ is a [[Definition:Hilbert Space|Hilbert space]].
+\end{theorem}
+
+\begin{proof}
+{{Proofread}}
+{{wtd|I think some of the partial results are up on their own pages now, and thus can be replaced by a reference (or this should become the case).
+Furthermore, this is *so* long, transclusions and/or foldable templates are a must}}
+=== $H$ is a Vector Space ===
+From the definition of [[Definition:Hilbert Space Direct Sum|Hilbert space direct sum]], we see that $H$ is a [[Definition:Empty Set|nonempty]] [[Definition:Subset|subset]] of a [[Definition:Vector Space|vector space]] (namely, the [[Definition:Vector Space Direct Sum|direct sum]] of the $H_i$ as [[Definition:Vector Space|vector spaces]]).
+From the [[Two-Step Vector Subspace Test]] it follows that it is to be shown that:
+:$(1): \qquad \forall h_1, h_2 \in H: \displaystyle \sum \left\{{ \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\} < \infty$
+:$(2): \qquad \forall \lambda \in \Bbb F, h \in H: \displaystyle \sum \left\{{ \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\} < \infty$
+Considering $(1)$, have the following:
+{{begin-eqn}}
+{{eqn|l = \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}
+     |r = \left\Vert{ h_1 \left({i}\right) + h_2 \left({i}\right) }\right\Vert^2_{H_i}
+}}
+{{eqn|o = \le
+     |r = \left({ \left\Vert{h_1 \left({i}\right)}\right\Vert_{H_i} + \left\Vert{h_2 \left({i}\right)}\right\Vert_{H_i} }\right)^2
+     |c = [[Definition:Norm on Vector Space|Triangle inequality]] for $\left\Vert{\cdot}\right\Vert_{H_i}$
+}}
+{{eqn|ll= \implies
+     |l = \sum \left\{ { \left\Vert{ \left({h_1 + h_2}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\}
+     |o = \le
+     |r = \sum \left\{ { \left({ \left\Vert{h_1 \left({i}\right)}\right\Vert_{H_i} + \left\Vert{h_2 \left({i}\right)}\right\Vert_{H_i} }\right)^2: i \in I }\right\}
+     |c = [[Generalized Sum Preserves Inequality]]
+}}
+{{eqn|o = <
+     |r = \infty
+     |c = $h_1, h_2 \in H$, [[Square-Summable Indexed Sets Closed Under Addition]]
+}}
+{{end-eqn}}
+For $(2)$, observe that:
+{{begin-eqn}}
+{{eqn|l = \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}
+     |r = \left\vert{\lambda}\right\vert^2 \left\Vert{h \left({i}\right)}\right\Vert^2_{H_i}
+     |c = $\left\Vert{\cdot}\right\Vert_{H_i}$ is a [[Definition:Norm on Vector Space|norm]]
+}}
+{{eqn|ll= \implies
+     |l = \sum \left\{ { \left\Vert{ \left({\lambda h}\right) \left({i}\right) }\right\Vert^2_{H_i}: i \in I }\right\}
+     |o = \le
+     |r = \left\vert{\lambda}\right\vert^2 \sum \left\{ { \left\Vert{h \left({i}\right)}\right\Vert^2_{H_i}: i \in I }\right\}
+     |c = [[Generalized Sum is Linear]]
+}}
+{{eqn|o = <
+     |r = \infty
+     |c = As $h \in H$
+}}
+{{end-eqn}}
+Thus, by the [[Two-Step Vector Subspace Test]], $H$ is a [[Definition:Vector Space|vector space]].
+{{qed|lemma}}
+=== $\left\langle{\cdot, \cdot}\right\rangle$ is an Inner Product ===
+It suffices to check well-definedness of $\left\langle{\cdot, \cdot}\right\rangle$, and subsequently the five properties of an [[Definition:Inner Product|inner product]].
+==== Well-definedness ====
+It is necessary to verify that for $g, h \in H$, in fact $\left\langle{g, h}\right\rangle \in \Bbb F$.
+That is, it is required to show that $\left\langle{g, h}\right\rangle = \displaystyle \sum \left\{{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}$ [[Definition:Generalized Sum|converges]] in $\Bbb F$.
+[[Absolutely Convergent Generalized Sum Converges]] applies to the [[Definition:Banach Space|Banach space]] $\Bbb F$ and the [[Definition:Indexed Set|$I$-indexed]] [[Definition:Subset|subset]] $\left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}$ of $\Bbb F$.
+Hence it will suffice to show that $\displaystyle \sum \left\{{ \left\vert{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i} }\right\vert: i \in I}\right\}$ [[Definition:Generalized Sum|converges]] in $\R$.
+For brevity, denote already $\left\Vert{h}\right\Vert^2$ for the expression $\displaystyle \sum \left\{{\left\Vert{h \left({i}\right)}\right\Vert_{H_i}^2: i \in I}\right\}$.
+Define $g' \in H$ by $g' \left({i}\right) = \begin{cases}
+  g \left({i}\right) & \text{if } \left\Vert{g \left({i}\right)}\right\Vert_{H_i} \ge \left\Vert{h \left({i}\right)}\right\Vert_{H_i} \\
+  \mathbf{0}_{H_i} & \text{otherwise}
+\end{cases}$.
+Note that $\left\Vert{g'}\right\Vert^2 \le \left\Vert{g}\right\Vert^2$ by [[Generalized Sum Preserves Inequality]].
+Similarly, let $h' \in H$ be defined by $h' \left({i}\right) = \begin{cases}
+  h \left({i}\right) & \text{if } \left\Vert{h \left({i}\right)}\right\Vert_{H_i} \ge \left\Vert{g \left({i}\right)}\right\Vert_{H_i} \\
+  \mathbf{0}_{H_i} & \text{otherwise}
+\end{cases}$.
+By [[Generalized Sum Preserves Inequality]] again, have $\left\Vert{h'}\right\Vert^2 \le \left\Vert{h}\right\Vert^2$.
+More significantly, by construction of $g', h'$:
+:$(3): \qquad \left\Vert{g \left({i}\right)}\right\Vert_{H_i}, \left\Vert{h \left({i}\right)}\right\Vert_{H_i} \le \left\Vert{\left({g' + h'}\right) \left({i}\right)}\right\Vert_{H_i}$
+As $H$ is a [[Definition:Vector Space|vector space]], $g' + h' \in H$, and we can establish:
+{{begin-eqn}}
+{{eqn|l = \left\vert{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i} }\right\vert
+     |o = \le
+     |r = \left\Vert{g \left({i}\right)}\right\Vert_{H_i} \left\Vert{h \left({i}\right)}\right\Vert_{H_i}
+     |c = [[Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces|Cauchy-Bunyakovsky-Schwarz Inequality]]
+}}
+{{eqn|o = \le
+     |r = \left\Vert{\left({g' + h'}\right) \left({i}\right)}\right\Vert_{H_i}^2
+     |c = Equation $(1)$
+}}
+{{eqn|ll= \implies
+     |l = \sum \left\{ { \left\vert{ \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i} }\right\vert: i \in I}\right\}
+     |o = \le
+     |r = \sum \left\{ { \left\Vert{\left({g' + h'}\right) \left({i}\right)}\right\Vert_{H_i}^2: i \in I}\right\}
+     |c = [[Generalized Sum Preserves Inequality]]
+}}
+{{end-eqn}}
+Hence, for all $g, h \in H$, $\left\langle{g, h}\right\rangle \in \Bbb F$ by the comment on [[Generalized Sum Preserves Inequality]].
+{{qed|lemma}}
+==== Property 1: $\left\langle{g, h}\right\rangle = \overline{\left\langle{h, g}\right\rangle}$ ====
+{{begin-eqn}}
+{{eqn|l = \left\langle{g, h}\right\rangle
+     |r = \sum \left\{ { \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn|r = \sum \left\{ { \overline{\left\langle{ h \left({i}\right), g \left({i}\right) }\right\rangle_{H_i} }: i \in I}\right\}
+     |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an [[Definition:Inner Product|inner product]]
+}}
+{{eqn|r = \overline{ \sum \left\{ { \left\langle{ h \left({i}\right), g \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\} }
+     |c = [[Convergence of Generalized Sum of Complex Numbers/Corollary|Convergence of Generalized Sum of Complex Numbers: Corollary]]
+}}
+{{eqn|r = \overline{ \left\langle{h, g}\right\rangle }
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== Property 2: $\left\langle{\lambda g, h}\right\rangle = \lambda \left\langle{g, h}\right\rangle$ ====
+{{begin-eqn}}
+{{eqn|l = \left\langle{\lambda g, h}\right\rangle
+     |r = \sum \left\{ { \left\langle{\left({\lambda g}\right) \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn|r = \sum \left\{ { \lambda \left\langle{ g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an [[Definition:Inner Product|inner product]]
+}}
+{{eqn|r = \lambda \sum \left\{ { \left\langle{g \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = [[Generalized Sum is Linear]]
+}}
+{{eqn|r = \lambda \left\langle{h, g}\right\rangle
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== Property 3: $\left\langle{g_1 + g_2, h}\right\rangle = \left\langle{g_1, h}\right\rangle + \left\langle{g_2, h}\right\rangle$ ====
+{{begin-eqn}}
+{{eqn|l = \left\langle{g_1 + g_2, h}\right\rangle
+     |r = \sum \left\{ { \left\langle{\left({g_1 + g_2}\right) \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn|r = \sum \left\{ { \left\langle{g_1 \left({i}\right), h \left({i}\right)}\right\rangle_{H_i} + \left\langle{g_2 \left({i}\right), h \left({i}\right)}\right\rangle_{H_i}: i \in I}\right\}
+     |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an [[Definition:Inner Product|inner product]]
+}}
+{{eqn|r = \sum \left\{ { \left\langle{ g_1 \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\} + \sum \left\{ { \left\langle{ g_2 \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = [[Generalized Sum is Linear]]
+}}
+{{eqn|r = \left\langle{g_1, h}\right\rangle + \left\langle{g_2, h}\right\rangle
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== Property 4: $\left\langle{h, h}\right\rangle \ge 0$ ====
+{{begin-eqn}}
+{{eqn|l = \left\langle{h, h}\right\rangle
+     |r = \sum \left\{ { \left\langle{ h \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}: i \in I}\right\}
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle$
+}}
+{{eqn|o = \ge
+     |r = \sum \left\{ { 0: i \in I}\right\}
+     |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an [[Definition:Inner Product|inner product]], [[Generalized Sum Preserves Inequality]]
+}}
+{{eqn|r = 0
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== Property 5: $\left\langle{h, h}\right\rangle = 0$ [[Definition:Iff|iff]] $h = \mathbf{0}_H$ ====
+{{begin-eqn}}
+{{eqn|l = \left\langle{h, h}\right\rangle
+     |r = 0
+}}
+{{eqn|ll= \iff \forall i \in I:
+     |l = \left\langle{ h \left({i}\right), h \left({i}\right) }\right\rangle_{H_i}
+     |r = 0
+     |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an [[Definition:Inner Product|inner product]], [[Generalized Sum is Monotone]]
+}}
+{{eqn|ll= \iff \forall i \in I:
+     |l = h \left({i}\right)
+     |r = \mathbf{0}_{H_i}
+     |c = $\left\langle{\cdot, \cdot}\right\rangle_{H_i}$ is an [[Definition:Inner Product|inner product]]
+}}
+{{eqn|ll= \iff
+     |l = h
+     |r = \mathbf{0}_H
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== Conclusion ====
+$\left\langle{\cdot, \cdot}\right\rangle$ is checked to be a [[Definition:Mapping|mapping]] from $H \times H$ to $\Bbb F$, satisfying the five conditions for an [[Definition:Inner Product|inner product]].
+That is, $\left\langle{\cdot, \cdot}\right\rangle$ is an [[Definition:Inner Product|inner product]] on $H$.
+{{qed|lemma}}
+=== $H$ is complete ===
+A [[Definition:Hilbert Space|Hilbert space]] is a [[Definition:Complete Metric Space|complete]] [[Definition:Inner Product Space|inner product space]].
+Thus, it remains to verify that $H$ is [[Definition:Complete Metric Space|complete]].
+Suppose $\left({h_n}\right)_{n\in\N}$ is a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $H$.
+Let $N \in \N$ such that $n, m \ge N \implies \left\Vert{h_n - h_m}\right\Vert < \epsilon$.
+That is, $\displaystyle \sum \left\{{ \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2: i \in I}\right\} < \epsilon^2$.
+From [[Generalized Sum is Monotone]] obtain that, for all $i \in I$:
+:$\left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2 < \epsilon^2$
+It follows that $\left({h_n \left({i}\right) }\right)_{n\in\N}$ is a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $H_i$.
+$H_i$ is a [[Definition:Hilbert Space|Hilbert space]], hence [[Definition:Complete Metric Space|complete]].
+Hence there is some $h_i \in H_i$ such that $\displaystyle \lim_{n\to\infty} h_n \left({i}\right) = h_i$.
+Now let $h$ be defined by $h \left({i}\right) = h_i$; it is the only candidate for $\displaystyle \lim_{n\to\infty} h_n = h$.
+It remains to be shown that indeed $\displaystyle \lim_{n\to\infty} h_n = h$, and then that $h \in H$.
+So, for any $\epsilon > 0$, an $N \in \N$ is to be found such that for all $n \ge N$:
+:$(4): \qquad \displaystyle \sum \left\{{ \left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2: i \in I}\right\} < \epsilon^2$
+To this end, let $N \in \N$ be such that:
+:$(5): \qquad n, m \ge N \implies \left\Vert{h_n - h_m}\right\Vert^2 < \frac {\epsilon^2} 2$
+Such an $N$ exists as $\left({h_n}\right)_{n\in\N}$ is a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]].
+Now observe that, for any [[Definition:Finite|finite]] $G \subseteq I$ and $n \ge N$:
+{{begin-eqn}}
+{{eqn | ll= \forall i \in I:
+      | l = \left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2
+      | r = \lim_{m \to \infty} \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2
+      | c = Definition of $h \left({i}\right)$
+}}
+{{eqn | ll= \implies
+      | l = \sum_{i \in G} \left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2
+      | r = \sum_{i \in G} \ \lim_{m \to \infty} \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2
+}}
+{{eqn | r = \lim_{m \to \infty} \ \sum_{i \in G} \left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2
+      | c = [[Sum Rule for Sequences]]
+}}
+{{eqn | o = \le
+      | r = \lim_{m \to \infty} \ \sum \left\{ {\left\Vert{ \left({h_n - h_m}\right) \left({i}\right) }\right\Vert_{H_i}^2 : i \in I}\right\}
+      | c = [[Generalized Sum is Monotone]]
+}}
+{{eqn | r = \lim_{m \to \infty} \left\Vert{h_n - h_m}\right\Vert^2
+      | c = Definition of $\left\Vert{\cdot}\right\Vert$
+}}
+{{eqn | o = \le
+      | r = \frac {\epsilon^2} 2
+      | c = [[Upper and Lower Bounds of Sequences]]
+}}
+{{end-eqn}}
+The last inequality follows from $(5)$, as $m \ge N$ eventually when $m \to \infty$. 
+From [[Bounded Generalized Sum Converges]], it now follows that:
+:$\displaystyle \sum \left\{ {\left\Vert{ \left({h_n - h}\right) \left({i}\right) }\right\Vert_{H_i}^2 : i \in I}\right\} \le \frac {\epsilon^2} 2 < \epsilon^2$
+This precisely establishes the inequality desired in $(4)$ for $n \ge N$.
+It follows that $\displaystyle \lim_{n \to \infty} h_n = h$.
+To show that $h \in H$, it is to be shown that $\left\Vert{h}\right\Vert^2 < \infty$.
+This is done as follows:
+{{begin-eqn}}
+{{eqn | l = \left\Vert{h}\right\Vert^2
+      | r = \sum \left\{ {\left\Vert{ h \left({i}\right) }\right\Vert_{H_i}^2 : i \in I}\right\}
+      | c = Definition of $\left\Vert{\cdot}\right\Vert$
+}}
+{{eqn | o = \le
+      | r = \sum \left\{ {\left({ \left\Vert{ \left({h - h_n}\right) \left({i}\right) }\right\Vert_{H_i} + \left\Vert{ h_n \left({i}\right) }\right\Vert_{H_i} }\right)^2 : i \in I}\right\}
+      | c = [[Definition:Norm on Vector Space|Triangle Inequality]] for all $\left\Vert{\cdot}\right\Vert_{H_i}$, [[Generalized Sum Preserves Inequality]]
+}}
+{{end-eqn}}
+The latter sum converges by [[Square-Summable Indexed Sets Closed Under Addition]], yielding convergence of $\left\Vert{h}\right\Vert^2$.
+Therefore, $h \in H$.
+That is, every [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $H$ [[Definition:Convergent Sequence (Metric Space)|converges]] to a limit in $H$, hence $H$ is [[Definition:Complete Metric Space|complete]].
+By definition, $H$ is a [[Definition:Hilbert Space|Hilbert space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Morera's Theorem}
+Tags: Calculus, Complex Analysis
+
+\begin{theorem}
+Let $D$ be a [[Definition:Simply Connected Domain|simply connected domain]] in $\C$.
+Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous function]].
+Let $f$ be such that:
+:$\displaystyle \int_\gamma \map f z \rd z = 0$
+for every [[Definition:Simple Contour (Complex Plane)|simple]] [[Definition:Closed Contour (Complex Plane)|closed contour]] $\gamma$ in $D$
+Then $f$ is [[Definition:Analytic Complex Function|analytic]] on $D$.
+\end{theorem}
+
+\begin{proof}
+For a fixed $z_0 \in D$ and $z \in D$ we consider the function:
+:$\displaystyle \map F z = \int_\gamma \map f w \rd w$
+where $\gamma$ is any [[Definition:Simple Contour (Complex Plane)|(simple) contour]] starting at $z_0$ and ending at $z$.
+By [[Primitive of Function on Connected Domain]], $F$ is a [[Definition:Complex Primitive|primitive]] of $f$.
+Since $F$ is [[Definition:Analytic Complex Function|analytic]] and $F' = f$, we conclude that $f$ is [[Definition:Analytic Complex Function|analytic]] as well.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation is Antisymmetric iff Intersection with Inverse is Coreflexive}
+Tags: Symmetric Relations
+
+\begin{theorem}
+Let $\RR$ be a [[Definition:Relation|relation]] on $S$.
+Then:
+:$\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]]
+{{iff}}:
+:$\RR \cap \RR^{-1}$ is [[Definition:Coreflexive Relation|coreflexive]]
+where $\RR^{-1}$ is the [[Definition:Inverse Relation|inverse]] of $\RR$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\RR$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]].
+Let $\tuple {a, b} \in \RR \cap \RR^{-1}$.
+That means:
+:$\tuple {a, b} \in \RR$
+and
+:$\tuple {a, b} \in \RR^{-1}$
+which means, by definition of [[Definition:Inverse Relation|inverse relation]]:
+:$\tuple {b, a} \in \RR$
+But as $\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]], that means $a = b$.
+Thus:
+:$\tuple {a, b} = \tuple {a, a}$
+and so:
+:$\tuple {a, b} \in \Delta_S$
+where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]].
+Thus from the definition of [[Definition:Subset|subset]]:
+:$\RR \cap \RR^{-1} \subseteq \Delta_S$
+Hence, by definition, $\RR$ is [[Definition:Coreflexive Relation/Definition 2|coreflexive]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\RR \cap \RR^{-1}$ be [[Definition:Coreflexive Relation|coreflexive]].
+Hence by definition of [[Definition:Coreflexive Relation/Definition 2|coreflexive]]:
+:$\RR \cap \RR^{-1} \subseteq \Delta_S$
+Let $\tuple {a, b} \in \RR$ and $\tuple {b, a} \in \RR$.
+That is, by definition of [[Definition:Inverse Relation|inverse relation]]:
+:$\tuple {a, b} \in \RR$
+and
+:$\tuple {a, b} \in \RR^{-1}$
+That is:
+:$\tuple {a, b} \in \RR \cap \RR^{-1}$
+But as $\RR \cap \RR^{-1} \subseteq \Delta_S$ it follows that $a = b$.
+So by definition $\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Min Operation on Toset is Semigroup}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $\min \left({x, y}\right)$ denote the [[Definition:Min Operation|min operation]] on $x, y \in S$.
+Then $\left({S, \min}\right)$ is a [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+By the definition of the [[Definition:Min Operation|min operation]], either:
+:$\min \left({x, y}\right) = x$
+or
+:$\min \left({x, y}\right) = y$
+So $\min$ is [[Definition:Closed Algebraic Structure|closed on $S$]].
+Then we have that the [[Max and Min are Associative|min operation is associative]]:
+: $\forall x, y, z \in S: \min \left({x, \min \left({y, z}\right)}\right) = \min \left({\min \left({x, y}\right), z}\right)$
+Hence the result, by definition of [[Definition:Semigroup|semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Max Semigroup is Commutative}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]]. 
+Then the [[Max Operation on Toset is Semigroup|semigroup $\left({S, \max}\right)$]] is [[Definition:Commutative Semigroup|commutative]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$. 
+We have that the [[Max and Min are Commutative|max operation is commutative]]:
+: $\max \left({x, y}\right) = \max \left({y, x}\right)$ 
+Hence the result, by definition of [[Definition:Commutative Semigroup|commutative semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation}
+Tags: Connected Relations, Inverse Relations, Trivial Relation
+
+\begin{theorem}
+Let $\mathcal R$ be a [[Definition:Relation|relation]] on $S$.
+Then $\mathcal R$ is a [[Definition:Connected Relation|connected relation]] {{iff}}:
+:$\mathcal R \cup \mathcal R^{-1} \cup \Delta_S = S \times S$
+where $\mathcal R^{-1}$ is the [[Definition:Inverse Relation|inverse]] of $\mathcal R$ and $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\mathcal R$ be a [[Definition:Connected Relation|connected relation]].
+By definition of [[Definition:Relation|relation]]:
+: $\mathcal R \subseteq S \times S$
+: $\mathcal R^{-1} \subseteq S \times S$
+: $\Delta_S \subseteq S \times S$
+So from [[Union is Smallest Superset]] (and indeed, trivially):
+: $\mathcal R \cup \mathcal R^{-1} \cup \Delta_S \subseteq S \times S$
+Let $\left({a, b}\right) \in S \times S$.
+Suppose $a = b$.
+Then $\left({a, b}\right) \in \Delta_S$ by definition of the [[Definition:Diagonal Relation|diagonal relation]].
+Then by [[Set is Subset of Union]]:
+:$\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$
+Suppose otherwise, that is, that $a \ne b$.
+As $\mathcal R$ is [[Definition:Connected Relation|connected]], either:
+:$\left({a, b}\right) \in \mathcal R$
+or:
+:$\left({b, a}\right) \in \mathcal R$
+From the definition of [[Definition:Inverse Relation|inverse relation]], either:
+:$\left({a, b}\right) \in \mathcal R$
+or:
+:$\left({a, b}\right) \in \mathcal R^{-1}$
+That is:
+:$\left({a, b}\right) \in R \cup \mathcal R^{-1}$
+Again by [[Set is Subset of Union]]:
+:$\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$
+So, by definition of [[Definition:Subset|subset]]:
+:$S \times S \subseteq \mathcal R \cup \mathcal R^{-1} \cup \Delta_S$
+Hence, by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\mathcal R \cup \mathcal R^{-1} \cup \Delta_S = S \times S$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\mathcal R \cup \mathcal R^{-1} \cup \Delta_S = S \times S$.
+Let $\left({a, b}\right) \in S \times S$.
+Suppose $a \ne b$.
+By definition of [[Definition:Diagonal Relation|diagonal relation]]:
+:$\left({a, b}\right) \notin \Delta_S$
+So, by definition of [[Definition:Set Union|set union]]:
+:$\left({a, b}\right) \in \mathcal R$
+or:
+:$\left({a, b}\right) \in \mathcal R^{-1}$
+That is, by definition of [[Definition:Inverse Relation|inverse relation]]:
+:$\left({a, b}\right) \in \mathcal R$
+or:
+:$\left({b, a}\right) \in \mathcal R$
+So by definition $\mathcal R$ is [[Definition:Connected Relation|connected]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Triangle Inequality for Generalized Sums}
+Tags: Generalized Sums, Banach Spaces, Triangle Inequality
+
+\begin{theorem}
+Let $V$ be a [[Definition:Banach Space|Banach space]], and let $\left\Vert{\cdot}\right\Vert$ denote its [[Definition:Norm on Vector Space|norm]].
+Let $\left({v_i}\right)_{i\in I}$ be an [[Definition:Indexed Set|indexed]] [[Definition:Subset|subset]] of $V$.
+Let the [[Definition:Generalized Sum|generalized sum]] $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ [[Definition:Absolutely Convergent Generalized Sum|converge absolutely]].
+Then $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$.
+\end{theorem}
+
+\begin{proof}
+First of all, note that [[Absolutely Convergent Generalized Sum Converges]] assures us that the left expression in above equality is defined.
+Suppose that there exists an $\epsilon > 0$ such that $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\} + \epsilon$.
+This supposition is seen to be equivalent to $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$.
+{{explain|Expand below sentence to be more clear}}
+Then, by definition of a [[Definition:Generalized Sum|generalized sum]], there necessarily exists a [[Definition:Finite|finite]] [[Definition:Subset|subset]] $F$ of $I$ with:
+:$\displaystyle \left\Vert{\sum_{i \in F} v_i}\right\Vert > \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert - \epsilon > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$
+However, using the standard triangle equality on this ''finite'' sum (i.e., axiom '''N3''' for a [[Definition:Norm on Vector Space|norm]], repetitively), we also have:
+:$\displaystyle \left\Vert{\sum_{i \in F} v_i}\right\Vert \le \sum_{i \in F} \left\Vert{v_i}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$
+Here the second inequality follows from [[Generalized Sum is Monotone]].
+These two estimates constitute a contradiction, and therefore such an $\epsilon$ cannot exist.
+Hence $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$.
+{{qed}}
+[[Category:Generalized Sums]]
+[[Category:Banach Spaces]]
+[[Category:Triangle Inequality]]
+abiohhjxt7y574z92vr2w74cszc9tgy
+\end{proof}<|endoftext|>
+\section{Absolutely Convergent Generalized Sum Converges}
+Tags: Absolute Convergence, Generalized Sums, Banach Spaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Banach Space|Banach space]].
+Let $\norm {\, \cdot \,}$ denote the [[Definition:Norm on Vector Space|norm]] on $V$.
+Let $d$ denote the corresponding [[Definition:Metric Induced by Norm|induced metric]].
+Let $\family {v_i}_{i \mathop \in I}$ be an [[Definition:Indexed Set|indexed]] [[Definition:Subset|subset]] of $V$ such that the [[Definition:Generalized Sum|generalized sum]] $\displaystyle \sum_{i \mathop \in I} \set {v_i}$ [[Definition:Generalized Sum/Absolute Net Convergence|converges absolutely]].
+Then the [[Definition:Generalized Sum|generalized sum]] $\displaystyle \sum \set {v_i: i \in I}$ [[Definition:Convergent Net|converges]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds in two stages:
+:$(1): \quad$ Finding a candidate $v \in V$ where the sum might converge to
+:$(2): \quad$ Showing that the candidate is indeed sought limit.
+That $\displaystyle \sum \set {v_i: i \mathop \in I}$ [[Definition:Generalized Sum/Absolute Net Convergence|converges absolutely]] means that $\displaystyle \sum \set {\norm {v_i}: i \mathop \in I}$ [[Definition:Generalized Sum|converges]].
+Now, for all $n \in \N$, let $F_n \subseteq I$ be [[Definition:Finite Set|finite]] such that:
+:$\displaystyle \sum_{i \mathop \in G} \norm {v_i} > \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n}$
+for all [[Definition:Finite Set|finite]] $G$ with $F_n \subseteq G \subseteq I$
+It may be arranged that $n \ge m \implies F_m \subseteq F_n$ by passing over to $\displaystyle F'_n = \bigcup_{m \mathop = 1}^n F_m$ if necessary.
+Define:
+: $\displaystyle v_n = \sum_{i \mathop \in F_n} v_i$
+Next, it is to be shown that the [[Definition:Sequence|sequence]] $\sequence {v_n}_{n \mathop \in \N}$ is [[Definition:Cauchy Sequence|Cauchy]].
+So let $\epsilon > 0$, and let $N \in \N$ be such that $2^{-N} < \epsilon$.
+Then for $m \ge n \ge N$, have:
+{{begin-eqn}}
+{{eqn | l = \map d {v_m, v_n}
+      | r = \norm {\paren {\sum_{i \mathop \in F_m} v_i} - \paren {\sum_{i \mathop \in F_n} v_i} }
+      | c = {{Defof|Metric Induced by Norm}}
+}}
+{{eqn | r = \norm {\sum_{i \mathop \in F_m \setminus F_n} v_i}
+      | c = $F_n \subseteq F_m$
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
+      | c = [[Definition:Norm on Vector Space|Triangle Inequality]] for $\norm {\, \cdot \,}$
+}}
+{{end-eqn}}
+Now to estimate this last quantity, observe:
+{{begin-eqn}}
+{{eqn | l = \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n} + \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
+      | o = <
+      | r = \sum_{i \mathop \in F_n} \norm {v_i} + \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
+      | c = Defining property of $F_n$
+}}
+{{eqn | r = \sum_{i \mathop \in F_m} \norm {v_i}
+      | c = [[Union with Relative Complement]], $F_n \subseteq F_m$
+}}
+{{eqn | o = \le
+      | r = \sum \set {\norm {v_i}: i \mathop \in I}
+      | c = [[Generalized Sum is Monotone]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}
+      | o = <
+      | r = 2^{-n}
+}}
+{{end-eqn}}
+Finally, by the defining property of $N$, as $n \ge N:
+:2^{-n} < 2^{-N} < \epsilon$
+Combining all of these estimates leads to the conclusion that:
+:$\map d {v_m, v_n} < \epsilon$
+It follows that $\sequence {v_n}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+As $V$ is a [[Definition:Banach Space|Banach space]]:
+:$\displaystyle \exists v \in V: \lim_{n \mathop \to \infty} v_n = v$
+Having identified a candidate $v$ for the sum $\displaystyle \sum \set {v_i: i \in I}$ to [[Definition:Generalized Sum|converge]] to, it remains to verify that this is indeed the case.
+According to the [[Definition:Generalized Sum|definition]] of considered sum, the convergence is [[Definition:Convergent Net|convergence of a net]].
+Next, [[Metric Induces Topology]] ensures that we can limit the choice of [[Definition:Open Set (Topology)|opens]] $U$ containing $v$ to [[Definition:Open Ball|open balls]] centered at $v$.
+Now let $\epsilon > 0$.
+We want to find a [[Definition:Finite Set|finite]] $F \subseteq I$ such that:
+:$\map d {\displaystyle \sum_{i \mathop \in G} v_i, v} < \epsilon$
+for all [[Definition:Finite Set|finite]] $G$ with $F \subseteq G \subseteq I$.
+Now let $N \in \N$ such that:
+:$\forall n \ge N: \map d {v_n, v} < \dfrac \epsilon 2$
+with the $v_n$ as above.
+By taking a larger $N$ if necessary, ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well.
+Let us verify that the set $F_N$ defined above has sought properties.
+So let $G$ be [[Definition:Finite Set|finite]] with $F_N \subseteq G \subseteq I$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {\sum_{i \mathop \in G} v_i, v}
+      | r = \norm {\paren {\sum_{i \mathop \in G} v_i} - v}
+      | c = {{Defof|Metric Induced by Norm}}
+}}
+{{eqn | o = \le
+      | r = \norm {\paren {\sum_{i \mathop \in G} v_i} - \paren {\sum_{i \mathop \in F_N} v_i} } + \norm {\paren {\sum_{i \mathop \in F_N} v_i} - v}
+      | c = [[Definition:Norm on Vector Space|Triangle inequality]] for $\norm {\, \cdot \,}$
+}}
+{{eqn | o = <
+      | r = \norm {\sum_{i \mathop \in G \setminus F_N} v_i} + \frac \epsilon 2
+      | c = $F_N \subseteq G$, defining property of $N$
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop \in G \setminus F_N} \norm {v_i} + \frac \epsilon 2
+      | c = [[Definition:Norm on Vector Space|Triangle inequality]] for $\norm {\, \cdot \,}$
+}}
+{{end-eqn}}
+For the first of these terms, observe:
+{{begin-eqn}}
+{{eqn | l = \sum \set {\norm {v_i}: i \in I} - 2^{-N} + \sum_{i \mathop \in G \setminus F_N} \norm {v_i}
+      | o = <
+      | r = \sum_{i \mathop \in F_N} \norm {v_i} + \sum_{i \mathop \in G \setminus F_N} \norm {v_i}
+      | c = defining property of $F_N$
+}}
+{{eqn | r = \sum_{i \mathop \in G} \norm {v_i}
+      | c = [[Union with Relative Complement]], $F_N \subseteq G$
+}}
+{{eqn | o = \le
+      | r = \sum \set {\norm {v_i}: i \in I}
+      | c = [[Generalized Sum is Monotone]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{i \mathop \in G \setminus F_N} \norm {v_i}
+      | o = <
+      | r = 2^{-N}
+}}
+{{end-eqn}}
+Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain:
+:$\displaystyle \map d {\sum_{i \mathop \in G} v_i, v} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
+By definition of [[Definition:Convergent Net|convergence of a net]], it follows that:
+:$\displaystyle \sum \set {v_i: i \in I} = v$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generalized Sum is Monotone}
+Tags: Generalized Sums
+
+\begin{theorem}
+Let $\left({a_i}\right)_{i \in I}$ be an [[Definition:Indexed Set|$I$-indexed family]] of [[Definition:Positive Real Number|positive real numbers]].
+That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
+Then, for every [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $F$ of $I$:
+:$\displaystyle \sum_{i \mathop \in F} a_i \le \sum \left\{{a_i : i \in I}\right\}$
+provided the [[Definition:Generalized Sum|generalized sum]] on the right [[Definition:Generalized Sum/Net Convergence|converges]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Generalized Sums]]
+bflliqrb2a0u6h5gqx9lfu9zcu145dc
+\end{proof}<|endoftext|>
+\section{Convergent Generalized Sum of Positive Reals has Countably Many Non-Zero Terms}
+Tags: Generalized Sums, Real Analysis
+
+\begin{theorem}
+Let $\left({a_i}\right)_{i \mathop \in I}$ be an [[Definition:Indexed Set|$I$-indexed family]] of [[Definition:Positive Real Number|positive real numbers]].
+That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
+Suppose that $\displaystyle \sum \left\{ {a_i: i \in I}\right\}$ converges.
+Then the set $I_{>0} := \left\{ {i \in I: a_i > 0}\right\}$ is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+Denote, for $\lambda \in \R$, $I_{> \lambda} := \left\{ {i \in I: a_i > \lambda}\right\}$.
+Then as $\displaystyle \sum \left\{ {a_i: i \mathop \in I}\right\}$ converges, necessarily all of the sets $I_{> \frac 1 n}$ are [[Definition:Finite Set|finite]].
+However, we have that $\displaystyle I_{>0} = \bigcup_{n \mathop = 1}^\infty I_{> \frac 1 n}$.
+From [[Countable Union of Countable Sets is Countable]], it follows that $I_{>0}$ is [[Definition:Countable Set|countable]].
+{{qed}}
+[[Category:Generalized Sums]]
+[[Category:Real Analysis]]
+5tigvqux50ppkhvjgo0z8x8uoehpxf5
+\end{proof}<|endoftext|>
+\section{Mappings Partially Ordered by Extension}
+Tags: Mapping Theory, Order Theory
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $F$ be the set of all [[Definition:Mapping|mappings]] from $S$ to $T$.
+Let $\mathcal R \subseteq F \times F$ be the [[Definition:Relation|relation]] defined as:
+:$\left({f, g}\right) \in \mathcal R \iff \operatorname{Dom} \left({f}\right) \subseteq \operatorname{Dom} \left({g}\right) \land \forall x \in \operatorname{Dom} \left({f}\right): f \left({x}\right) = g \left({x}\right)$
+That is, $f \mathrel{\mathcal R} g$ {{iff}} $g$ is an [[Definition:Extension of Mapping|extension]] of $f$.
+Then $\mathcal R$ is an [[Definition:Ordering|ordering]] on $F$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \operatorname{Dom} \left({f}\right)$ such that $f \left({x}\right) = y$.
+Then by definition $x \in \operatorname{Dom} \left({g}\right)$ and $g \left({x}\right) = y$.
+Thus by definition of [[Definition:Subset|subset]], $f \subseteq g$.
+We have that [[Subset Relation is Ordering]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Subsets form Directed Set}
+Tags: Preorder Theory, Set Theory
+
+\begin{theorem}
+Let $I$ be a [[Definition:Set|set]].
+Denote with $\mathcal F$ the set of [[Definition:Finite|finite]] [[Definition:Subset|subsets]] of $I$.
+Let $\subseteq$ be the [[Subset Relation is Ordering|subset relation]] on $\mathcal F$.
+Then $\struct {\mathcal F, \subseteq}$ is a [[Definition:Directed Set|directed set]].
+\end{theorem}
+
+\begin{proof}
+From [[Subset Relation is Ordering]], we know that $\subseteq$ is an [[Definition:Ordering|ordering]], hence also a [[Definition:Preordering|preordering]].
+Now let $F, G \in \mathcal F$.
+From [[Set Union Preserves Subsets]], conclude that $F \cup G \subseteq I$ as $F, G \subseteq I$.
+From [[Cardinality of Set Union]], $\card {F \cup G} \le \card F + \card G$, implying that $F \cup G$ is a [[Definition:Finite|finite]] set.
+Hence $F \cup G \in \mathcal F$.
+Furthermore, $F \subseteq F \cup G$ and $G \subseteq F \cup G$.
+It follows that $\struct {\mathcal F, \subseteq}$ is a [[Definition:Directed Set|directed set]].
+{{qed}}
+[[Category:Preorder Theory]]
+[[Category:Set Theory]]
+8i9hubqhfeanlj598gtqtiraigcss6g
+\end{proof}<|endoftext|>
+\section{Strictly Precedes is Strict Ordering}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $\prec$ be the [[Definition:Relation|relation]] on $S$ defined as:
+:$a \prec b \iff (a \ne b) \land (a \preceq b)$
+That is, $a \prec b$ {{iff}} $a$ [[Definition:Strictly Precede|strictly precedes]] $b$.
+Then:
+: $a \preceq b \iff (a = b) \lor (a \prec b)$
+and $\prec$ is a [[Definition:Strict Ordering|strict ordering]] on $S$.
+\end{theorem}
+
+\begin{proof}
+We are given that $\left({S, \preceq}\right)$ is an [[Definition:Ordered Set|ordered set]].
+=== [[Definition:Antireflexive Relation|Antireflexive]] ===
+Follows immediately:
+:$\forall a \in S: a = a \implies a \not \prec a$ {{explain}}
+{{qed|lemma}}
+=== [[Definition:Transitive Relation|Transitive]] ===
+Suppose $a, b, c \in S$ such that $a \preceq b, b \preceq c$.
+$a \preceq b \land b \preceq c \implies a \preceq c$ from [[Definition:Transitive Relation|transitivity]] of $\preceq$.
+Now suppose $a \preceq b, b \preceq c, a = c$. Then $a \preceq b$ but because of the [[Definition:Antisymmetric Relation|antisymmetry]] of $\preceq$, it is not then possible for $b \preceq c$. So a condition for transitivity to be violated will not arise in this circumstance.
+If either $a = b$ or $b = c$, then $a \not \prec b$ and $b \not \prec c$ by antireflexivity of $\prec$.
+So $\prec$ is shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+=== [[Definition:Asymmetric Relation|Asymmetric]] ===
+A direct application of [[Antireflexive and Transitive Relation is Asymmetric]].
+Hence the result.
+{{qed}}
+Let $a \preceq b$.
+Either $a = b$ or $a \ne b$ by the [[Law of Excluded Middle]].
+If $a = b$ we are done.
+Suppose $a \ne b$. Then it follows that $a \prec b$.
+Thus $a \preceq b \implies (a = b) \lor (a \prec b)$.
+{{qed}}
+Now let $(a = b) \lor (a \prec b)$.
+If $a = b$ then $a \preceq b$ by the reflexivity of $\preceq$.
+If $a \prec b$ then $a \preceq b$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Smallest Element is Unique}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+If $S$ has a [[Definition:Smallest Element|smallest element]], then it can have only one.
+That is, if $a$ and $b$ are both [[Definition:Smallest Element|smallest elements]] of $S$, then $a = b$.
+\end{theorem}
+
+\begin{proof}
+Let $a$ and $b$ both be [[Definition:Smallest Element|smallest elements]] of $S$.
+Then by definition:
+:$\forall y \in S: a \preceq y$
+:$\forall y \in S: b \preceq y$
+Thus it follows that:
+:$a \preceq b$
+:$b \preceq a$
+But as $\preceq$ is an [[Definition:Ordering|ordering]], it is [[Definition:Antisymmetric Relation|antisymmetric]].
+Hence by definition of [[Definition:Antisymmetric Relation|antisymmetric]], $a = b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Element is Unique}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Ordered Set|ordered set]].
+If $S$ has a [[Definition:Greatest Element|greatest element]], then it can have only one.
+That is, if $a$ and $b$ are both [[Definition:Greatest Element|greatest elements]] of $S$, then $a = b$.
+\end{theorem}
+
+\begin{proof}
+Let $a$ and $b$ both be [[Definition:Greatest Element|greatest elements]] of $S$.
+Then by definition:
+:$\forall y \in S: y \preceq a$
+:$\forall y \in S: y \preceq b$
+Thus it follows that:
+:$b \preceq a$
+:$a \preceq b$
+But as $\preceq$ is an [[Definition:Ordering|ordering]], it is [[Definition:Antisymmetric Relation|antisymmetric]].
+Hence, by definition of [[Definition:Antisymmetric Relation|antisymmetric relation]], $a = b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Smallest Element is Lower Bound}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$.
+Let $T$ have a [[Definition:Smallest Element|smallest element]] $m \in T$.
+Then $m$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$.
+It follows by definition that $T$ is [[Definition:Bounded Below Set|bounded below]].
+\end{theorem}
+
+\begin{proof}
+Let $m \in T$ be a [[Definition:Smallest Element|smallest element]] of $T$.
+By definition:
+:$\forall y \in T: m \preceq y$
+But as $T \subseteq S$, it follows that $m \in S$.
+Hence:
+:$\exists m \in S: \forall y \in T: m \preceq y$
+Thus $T$ is [[Definition:Bounded Below Set|bounded below]] by the [[Definition:Lower Bound of Set|lower bound]] $m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Element is Upper Bound}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$.
+Let $T$ have a [[Definition:Greatest Element|greatest element]] $M \in T$.
+Then $M$ is an [[Definition:Upper Bound of Set|upper bound]] of $T$.
+It follows by definition that $T$ is [[Definition:Bounded Above Set|bounded above]].
+\end{theorem}
+
+\begin{proof}
+Let $M \in T$ be a [[Definition:Greatest Element|greatest element]] of $T$.
+By definition:
+:$\forall y \in T: y \preceq M$
+But as $T \subseteq S$, it follows that $M \in S$.
+Hence:
+:$\exists M \in S: \forall y \in T: y \preceq M$
+Thus $T$ is [[Definition:Bounded Above Set|bounded above]] by the [[Definition:Upper Bound of Set|upper bound]] $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subset with Lower Bounds}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$.
+Let $T_*$ be the [[Definition:Set|set]] of all [[Definition:Lower Bound of Set|lower bounds]] of $T$ in $S$.
+Then $T_* \cap T \ne \O$ {{iff}}:
+:$T$ has a [[Definition:Smallest Element|smallest element]] $m$
+and
+:$T_* \cap T$ is a [[Definition:Singleton|singleton]] such that $T_* \cap T = \set m$
+\end{theorem}
+
+\begin{proof}
+Suppose $T_* \cap T = \O$, where $\O$ denotes the [[Definition:Empty Set|empty set]].
+That means $T$ contains none of its [[Definition:Lower Bound of Set|lower bounds]], if indeed it has any.
+From [[Smallest Element is Lower Bound]], if $T$ had a [[Definition:Smallest Element|smallest element]], it would be a [[Definition:Lower Bound of Set|lower bound]] contained in $T$.
+It follows that $T$ can have no [[Definition:Smallest Element|smallest element]].
+Otherwise $T_* \cap T \ne \O$.
+That means $T$ contains at least one of its [[Definition:Lower Bound of Set|lower bounds]].
+Suppose $\exists a, b \in T_* \cap T$.
+From [[Intersection is Subset]] it follows that $a, b \in T$.
+Then:
+:$\forall y \in T: a \preceq y$
+:$\forall y \in T: b \preceq y$
+Thus both $a$ and $b$ fulfil the criteria for being a [[Definition:Smallest Element|smallest element]] of $T$.
+From [[Smallest Element is Unique]] it follows that $a = b$ and so $T_* \cap T$ is a [[Definition:Singleton|singleton]] containing the [[Definition:Smallest Element|smallest element]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subset with Upper Bounds}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$.
+Let $T^*$ be the [[Definition:Set|set]] of all [[Definition:Upper Bound of Set|upper bounds]] of $T$ in $S$.
+Then $T^* \cap T \ne \varnothing$ {{iff}}:
+:$T$ has a [[Definition:Greatest Element|greatest element]] $M$
+and
+:$T^* \cap T$ is a [[Definition:Singleton|singleton]] such that $T^* \cap T = \left\{{M}\right\}$
+\end{theorem}
+
+\begin{proof}
+Suppose $T^* \cap T = \varnothing$, where $\varnothing$ denotes the [[Definition:Empty Set|empty set]].
+That means $T$ contains none of its [[Definition:Upper Bound of Set|upper bounds]], if indeed it has any.
+From [[Greatest Element is Upper Bound]], if $T$ had a [[Definition:Greatest Element|greatest element]], it would be an [[Definition:Upper Bound of Set|upper bound]] contained in $T$.
+It follows that $T$ can have no [[Definition:Greatest Element|greatest element]].
+Otherwise $T^* \cap T \ne \varnothing$.
+That means $T$ contains at least one of its [[Definition:Upper Bound of Set|upper bounds]].
+Suppose $\exists a, b \in T^* \cap T$.
+From [[Intersection is Subset]] it follows that $a, b \in T$.
+Then:
+:$\forall y \in T: y \preceq a$
+:$\forall y \in T: y \preceq b$
+Thus both $a$ and $b$ fulfil the criteria for being a [[Definition:Greatest Element|greatest element]] of $T$.
+From [[Greatest Element is Unique]] it follows that $a = b$ and so $T^* \cap T$ is a [[Definition:Singleton|singleton]] containing the [[Definition:Greatest Element|greatest element]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nesbitt's Inequality}
+Tags: Algebra
+
+\begin{theorem}
+Let $a$, $b$ and $c$ be positive real numbers.
+Then:
+:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac a {b + c} + \frac b {a + c} + \frac c {a + b}
+      | o = \ge
+      | r = \dfrac 3 2
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}
+      | o = \ge
+      | r = \frac 9 2
+      | c = by adding 3
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}
+      | o = \ge
+      | r = \frac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} }
+      | c = as $\dfrac {a + b + c} {\paren {b + c} + \paren {a + c} + \paren {a + b} } = \dfrac 1 2$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\frac 1 {b + c} + \frac 1 {a + c} + \frac 1 {a + b} } 3
+      | o = \ge
+      | r = \frac 3 {\paren {b + c} + \paren {a + c} + \paren {a + b} }
+      | c = dividing by $3 \paren {a + b + c}$
+}}
+{{end-eqn}}
+These are the [[Definition:Arithmetic Mean|arithmetic mean]] and the [[Definition:Harmonic Mean|harmonic mean]] of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$.
+From [[Arithmetic Mean is Never Less than Harmonic Mean]] the last inequality is true.
+Thus [[Nesbitt's Inequality]] holds.
+{{qed}}
+{{Namedfor|A.M. Nesbitt|cat = Nesbitt, A.M.}}
+\end{proof}<|endoftext|>
+\section{Greatest Element is Maximal}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]] which has a [[Definition:Greatest Element|greatest element]].
+Let $M$ be the [[Definition:Greatest Element|greatest element]] of $\struct {S, \preceq}$.
+Then $M$ is a [[Definition:Maximal Element|maximal element]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Greatest Element|greatest element]]:
+:$\forall y \in S: y \preceq M$
+Suppose $M \preceq y$.
+As $\preceq$ is an [[Definition:Ordering|ordering]], $\preceq$ is by definition [[Definition:Antisymmetric Relation|antisymmetric]].
+Thus it follows by definition of [[Definition:Antisymmetric Relation|antisymmetry]] that $M = y$.
+That is:
+:$M \preceq y \implies M = y$
+which is precisely the definition of a [[Definition:Maximal Element|maximal element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Smallest Element is Minimal}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]] which has a [[Definition:Smallest Element|smallest element]].
+Let $m$ be the [[Definition:Smallest Element|smallest element]] of $\struct {S, \preceq}$.
+Then $m$ is a [[Definition:Minimal Element|minimal element]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\forall y \in S: m \preceq y$
+Suppose $y \preceq m$.
+As $\preceq$ is an [[Definition:Ordering|ordering]], $\preceq$ is by definition [[Definition:Antisymmetric Relation|antisymmetric]].
+Thus it follows by definition of  [[Definition:Antisymmetric Relation|antisymmetry]] that $m = y$.
+That is:
+:$y \preceq m \implies m = y$
+which is precisely the definition of a [[Definition:Minimal Element|minimal element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Element in Toset is Unique and Smallest}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $m$ be a [[Definition:Minimal Element|minimal element]] of $\struct {S, \preceq}$.
+Then:
+:$(1): \quad m$ is the [[Definition:Smallest Element|smallest element]] of $\struct {S, \preceq}$
+:$(2): \quad m$ is the only [[Definition:Minimal Element|minimal element]] of $\struct {S, \preceq}$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Minimal Element|minimal element]]:
+:$\forall y \in S: y \preceq m \implies m = y$
+As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], by definition $\preceq$ is [[Definition:Connected Relation|connected]].
+That is:
+:$\forall x, y \in S: y \preceq x \lor x \preceq y$
+It follows that:
+:$\forall y \in S: m = y \lor m \preceq y$
+But as $m = y \implies m \preceq y$ by definition of $\preceq$, it follows that:
+:$\forall y \in S: m \preceq y$
+which is precisely the definition of [[Definition:Smallest Element|smallest element]].
+Hence $(1)$ holds.
+{{qed|lemma}}
+Suppose $m_1$ and $m_2$ are both [[Definition:Minimal Element|minimal elements]] of $S$.
+By $(1)$ it follows that both are [[Definition:Smallest Element|smallest elements]].
+It follows from [[Smallest Element is Unique]] that $m_1 = m_2$.
+That is, $(2)$ holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximal Element in Toset is Unique and Greatest}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $M$ be a [[Definition:Maximal Element|maximal element]] of $\struct {S, \preceq}$.
+Then:
+:$(1): \quad M$ is the [[Definition:Greatest Element|greatest element]] of $\struct {S, \preceq}$.
+:$(2): \quad M$ is the only [[Definition:Maximal Element|maximal element]] of $\struct {S, \preceq}$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Maximal Element|maximal element]]:
+:$\forall y \in S: M \preceq y \implies M = y$
+As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], by definition $\preceq$ is a [[Definition:Connected Relation|connected]].
+That is:
+:$\forall x, y \in S: y \preceq x \lor x \preceq y$
+It follows that:
+:$\forall y \in S: M = y \lor y \preceq M$
+But as $M = y \implies y \preceq M$ by definition of $\preceq$, it follows that:
+:$\forall y \in S: y \preceq M$
+which is precisely the definition of [[Definition:Greatest Element|greatest element]].
+Hence $(1)$ holds.
+{{qed|lemma}}
+Suppose $M_1$ and $M_2$ are both [[Definition:Maximal Element|maximal elements]] of $S$.
+By $(1)$ it follows that both are [[Definition:Greatest Element|greatest elements]].
+It follows from [[Greatest Element is Unique]] that $M_1 = M_2$.
+That is, $(2)$ holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Whitney Embedding Theorem}
+Tags: Topology
+
+\begin{theorem}
+Every [[Definition:Smooth Manifold|smooth $m$-dimensional manifold]] admits a [[Definition:Smooth Embedding|smooth embedding]] into [[Definition:Euclidean Space|Euclidean space $\R^{2m+1}$]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Hassler Whitney|cat = Whitney}}
+[[Category:Topology]]
+kwaatwulh9kkwggkwwrxmp6puln8jrl
+\end{proof}<|endoftext|>
+\section{Max Semigroup is Idempotent}
+Tags: Max and Min Operations, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Then the [[Definition:Semigroup|semigroup]] $\struct {S, \max}$ is an [[Definition:Idempotent Semigroup|idempotent semigroup]].
+\end{theorem}
+
+\begin{proof}
+The fact that $\struct {S, \max}$ is a [[Definition:Semigroup|semigroup]] is demonstrated in [[Max Operation on Toset is Semigroup]].
+Then the [[Max and Min are Idempotent|max operation is idempotent]]:
+:$\forall x \in S: \max \set {x, x} = x$
+The result follows by the definition of [[Definition:Idempotent Semigroup|idempotent semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tarski's Geometry is Complete}
+Tags: Tarski's Geometry
+
+\begin{theorem}
+[[Definition:Tarski's Geometry|Tarski's geometry]] is both [[Definition:Consistent (Logic)|consistent]] and [[Definition:Complete Theory|complete]].
+\end{theorem}<|endoftext|>
+\section{Continuity of Linear Transformations}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]], and let $A: H \to K$ be a [[Definition:Linear Transformation|linear transformation]].
+Then the following four statements are equivalent:
+:$(1): \quad A$ is [[Definition:Continuous Mapping (Topology)|continuous]]
+:$(2): \quad A$ is [[Definition:Continuous at Point of Topological Space|continuous]] at $\mathbf 0_H$
+:$(3): \quad A$ is [[Definition:Continuous at Point of Topological Space|continuous]] at some point
+:$(4): \quad \exists c > 0: \forall h \in H: \norm {\map A h}_K \le c \norm h_H$
+\end{theorem}
+
+\begin{proof}
+It is clear that $(1) \implies (2) \implies (3)$ and for $(4) \implies (2)$:
+For any $\epsilon > 0$, there exists $\delta = \dfrac \epsilon c$, such that when $\norm {\mathbf 0_H - h}_H < \delta$:
+:$\norm {\map A h - \map A {\mathbf 0_H} }_K \le c \norm h_H < c\delta = \epsilon$
+Now we prove $(3) \implies (1)$:
+Let $A$ be continuous at some point $h_0$.
+For any sequence $h_n \to h$ in $H$:
+:$h_n - h + h_0 \to h_0$
+Hence:
+:$\displaystyle \lim_{n \mathop \to \infty} \map A {h_n - h + h_0} = \lim_{n \mathop \to \infty} \map A {h_n} - \map A h + \map A {h_0} = \map A {h_0}$
+We see that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map A {h_n} = \map A h$
+Thus $A$ is continuous.
+Now for the proof $(2) \implies (4)$.
+We have that $A$ is continuous at $\mathbf 0_H$.
+Hence there exists an open ball of positive real radius $a$, centred at $\mathbf 0_H$, such that its image under $A$ is included in the open ball of radius $1$, centred at $\mathbf 0_K$. 
+This is $\norm h_H < a$.
+Then:
+:$\norm {\map A h}_K < 1$
+Let $h$ be an arbitrary element in $H$.
+Let $\epsilon > 0$.
+We have:
+:$\norm a \dfrac h {\norm h_H + \epsilon}_H < a$
+Hence:
+:$\norm {\map A {a \dfrac h {\norm h_H + \epsilon} }_K} = a \dfrac {\norm {\map A h}_K} {\norm h_H + \epsilon} < 1$
+Therefore:
+:$\norm {\map A h}_K < \dfrac 1 a \norm h_H + \dfrac \epsilon a$
+Let $\epsilon \to 0$.
+Then:
+:$c = \dfrac 1 a$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Norm of Linear Transformation}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]], and let $A: H \to K$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+{{TFAE|def = Norm on Bounded Linear Transformation}}
+:$(1): \qquad \left\Vert{A}\right\Vert = \sup \left\{{\left\Vert{Ah}\right\Vert_K: \left\Vert{h}\right\Vert_H \le 1}\right\}$
+:$(2): \qquad \left\Vert{A}\right\Vert = \sup \left\{{\dfrac {\left\Vert{Ah}\right\Vert_K} {\left\Vert{h}\right\Vert_H}: h \in H, h \ne \mathbf{0}_H}\right\}$
+:$(3): \qquad \left\Vert{A}\right\Vert = \sup \left\{{\left\Vert{Ah}\right\Vert_K: \left\Vert{h}\right\Vert_H \le 1}\right\}$
+:$(4): \qquad \left\Vert{A}\right\Vert = \inf \left\{{c > 0: \forall h \in H: \left\Vert{Ah}\right\Vert_K \le c \left\Vert{h}\right\Vert_H}\right\}$
+\end{theorem}<|endoftext|>
+\section{Max Semigroup on Toset is Semilattice}
+Tags: Max and Min Operations, Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]]. 
+Then the [[Max Operation on Toset is Semigroup|max semigroup]] $\struct {S, \max}$ is a [[Definition:Semilattice|semilattice]].
+\end{theorem}
+
+\begin{proof}
+The [[Max Semigroup is Commutative]] and [[Max Semigroup is Idempotent|idempotent]]. 
+Hence the result, by definition of a [[Definition:Semilattice|semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Max yields Supremum of Parameters}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]]. 
+Let $x, y \in S$.
+Then:
+:$\max \set {x, y} = \sup \set {x, y}$
+where:
+:$\max$ denotes the [[Definition:Max Operation|max operation]]
+:$\sup$ denotes the [[Definition:Supremum of Set|supremum]].
+\end{theorem}
+
+\begin{proof}
+As $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]], all elements of $S$ are [[Definition:Comparable|comparable]] by $\preceq$.
+Therefore there are two cases to consider:
+=== Case 1: $x \preceq y$ ===
+In this case:
+:$\max \set {x, y} = y = \sup \set {x, y}$
+=== Case 2: $y \preceq x$ ===
+In this case:
+:$\max \set {x, y} = x = \sup \set {x, y}$
+In either case, the result holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mappings to Vector Space form Vector Space}
+Tags: Vector Spaces
+
+\begin{theorem}
+Let $X$ be a [[Definition:Non-Empty Set|non-empty set]].
+Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] (or [[Definition:Division Ring|division ring]]) $K$.
+Let $V^X$ denote the [[Definition:Set of All Mappings|set of all mappings]] from $X$ to $V$.
+Let $+$ denote [[Definition:Pointwise Addition of Mappings|pointwise addition]] on $V^X$.
+Let $\circ$ denote [[Definition:Pointwise Scalar Multiplication of Mappings|pointwise ($K$)-scalar multiplication]] on $V^X$.
+Then $\left({V^X, +, \circ}\right)_K$ is a [[Definition:Vector Space|vector space]] over $K$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Vector Spaces]]
+2x9le09b6zjtlrg87lf3wlrpuwn2e3y
+\end{proof}<|endoftext|>
+\section{Mappings to Algebraic Structure form Similar Algebraic Structure}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $X$ be a [[Definition:Empty Set|nonempty set]].
+Let $G$ be a [[Definition:Magma|magma]] with respect to the [[Definition:Binary Operation|binary operations]] $\circ_1, \ldots, \circ_n$ on $G$.
+Let $G^X$ be the [[Definition:Set|set]] of all [[Definition:Mapping|mappings]] from $X$ to $G$.
+Denote also by $\circ_1, \ldots, \circ_n$ the [[Definition:Binary Operation|binary operations]] defined on $G^X$ by [[Definition:Pointwise Addition of Mappings|pointwise addition]].
+{{wtd|Transcluded pages for group, monoid, abelian group, and so on, and so forth}}
+\end{theorem}<|endoftext|>
+\section{Mappings to R-Algebraic Structure form Similar R-Algebraic Structure}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $X$ be a [[Definition:Empty Set|nonempty set]], and let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\left({G, \circ}\right)_R$ be an [[Definition:R-Algebraic Structure|$R$-algebraic structure]].
+Let $G^X$ be the [[Definition:Set|set]] of all [[Definition:Mapping|mappings]] from $X$ to $G$.
+Denote also by $\circ$ the [[Definition:Binary Operation|binary operation]] defined on $G^X$ by [[Definition:Pointwise Scalar Multiplication of Mappings|pointwise ($R$)-scalar multiplication]].
+{{wtd|Transcluded pages for module, vector space, and so on, and so forth}}
+\end{theorem}<|endoftext|>
+\section{Space of Bounded Linear Transformations is Banach Space}
+Tags: Banach Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $\map B {H, K}$ denote the [[Definition:Space of Bounded Linear Transformations|space of bounded linear transformations from $H$ to $K$]].
+Let $\Bbb F \in \set {\R, \C}$ denote the [[Definition:Ground Field|ground field]] of $K$.
+Now $\map B {H, K} \subseteq K^H$, the [[Definition:Set of All Mappings|set of mappings from $H$ to $K$]].
+Therefore, $\map B {H, K}$ can be endowed with [[Definition:Pointwise Addition of Mappings|pointwise addition]] ($+$) and [[Definition:Pointwise Scalar Multiplication of Mappings|($\Bbb F$)-scalar multiplication]] ($\circ$).
+Let $\norm{\,\cdot\,}$ denote the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear transformations]].
+Then $\norm{\,\cdot\,}$ is a [[Definition:Norm on Vector Space|norm]] on $\map B {H, K}$.
+Furthermore, $B \left({H, K}\right)$ is a [[Definition:Banach Space|Banach space]] with respect to this [[Definition:Norm on Vector Space|norm]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|This will be a long one: First vector space, then normed vector space, then completeness.}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Formulations of Pasch's Axiom}
+Tags: Tarski's Geometry
+
+\begin{theorem}
+The two forms of [[Axiom:Pasch's Axiom (Tarski's Axioms)|Pasch's Axiom]] in [[Definition:Tarski's Geometry|Tarski's Geometry]] are consistent.
+That is, the expressions:
+:$(1): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B b q c \implies \mathsf B p x b \land \mathsf B q x a$
+and:
+:$(2): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B q c b \implies \mathsf B a x q \land \mathsf B b p x$
+are [[Definition:Logical Equivalence|logically equivalent]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|It is important that we identify which other axioms this equivalence is based on for foundational purposes (and possible generalisations)}}
+\end{proof}<|endoftext|>
+\section{Square-Summable Indexed Sets Closed Under Addition}
+Tags: Generalized Sums
+
+\begin{theorem}
+Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be [[Definition:Indexed Set|$I$-indexed families]] of [[Definition:Real Number|real numbers]].
+Let:
+:$\displaystyle \sum \set {a_i^2: i \in I} < \infty$
+:$\displaystyle \sum \set {b_i^2: i \in I} < \infty$
+where $\displaystyle \sum$ denotes the [[Definition:Generalized Sum|generalized sums]].
+Then:
+:$\displaystyle \sum \set {\paren {a_i + b_i}^2: i \in I} < \infty$
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Some pain in the behind with careful considerations}}
+[[Category:Generalized Sums]]
+n16s25yxvbk07lnxxvawes2qwbgebre
+\end{proof}<|endoftext|>
+\section{Generalized Sum Preserves Inequality}
+Tags: Generalized Sums
+
+\begin{theorem}
+Let $\left({a_i}\right)_{i \in I}, \left({b_i}\right)_{i \in I}$ be [[Definition:Indexed Set|$I$-indexed families]] of [[Definition:Positive Real Number|positive real numbers]].
+That is, let $a_i, b_i \in \R_{\ge 0}$ for all $i \in I$.
+Suppose that for all $i \in I$, $a_i \le b_i$.
+Furthermore, suppose that $\displaystyle \sum \left\{{ b_i: i \in I }\right\}$ [[Definition:Generalized Sum|converges]].
+Then:
+: $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$
+In particular, $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ [[Definition:Generalized Sum|converges]].
+\end{theorem}
+
+\begin{proof}
+First, it is proven that $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ [[Definition:Generalized Sum|converges]].
+Then, the inequality $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$ is well-defined, and hence can be proven.
+=== Proof of Convergence ===
+For every $n \in \N$, let $F_n \subseteq I$ be [[Definition:Finite|finite]] such that:
+:$\displaystyle \sum_{i \mathop \in G} b_i > \sum \left\{{ b_i: i \in I }\right\} - 2^{-n}$ for all [[Definition:Finite|finite]] $G$ with $F_n \subseteq G \subseteq I$
+By passing over to $\displaystyle F'_n = \bigcup_{i \mathop = 1}^n F_i$ if necessary, it may be arranged that $F_n \subseteq F_m$ for $n \le m$.
+Next, define the [[Definition:Sequence|sequence]] $\left({a_n}\right)_{n \in \N}$ by $a_n := \displaystyle \sum_{i \mathop \in F_n} a_i$.
+To show $\left({a_n}\right)_{n \in \N}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]], let $\epsilon > 0$.
+Subsequently let $N \in \N$ be such that $2^{-N} < \epsilon$, and let $m \ge n \ge N$. Then:
+{{begin-eqn}}
+{{eqn|l = d \left({a_m, a_n}\right)
+     |r = \left\vert{\left({\sum_{i \mathop \in F_m} a_i}\right) - \left({\sum_{i \mathop \in F_n} a_i}\right)}\right\vert
+     |c = Definition of [[Definition:Metric Induced by Norm|Induced Metric]]
+}}
+{{eqn|r = \left\vert{\sum_{i \mathop \in F_m \setminus F_n} a_i}\right\vert
+     |c = $F_n \subseteq F_m$
+}}
+{{eqn|o = \le
+     |r = \sum_{i \mathop \in F_m \setminus F_n} b_i
+     |c = $b_i \ge a_i \ge 0$ for all $i \in I$
+}}
+{{end-eqn}}
+Now to estimate this last quantity, observe:
+{{begin-eqn}}
+{{eqn|l = \sum \left\{ {b_i : i \in I}\right\} - 2^{-n} + \sum_{i \mathop \in F_m \setminus F_n} b_i
+     |o = <
+     |r = \sum_{i \mathop \in F_n} b_i + \sum_{i \mathop \in F_m \setminus F_n} b_i
+     |c = Definition of $F_n$
+}}
+{{eqn|r = \sum_{i \mathop \in F_m} b_i
+     |c = [[Union with Relative Complement]], $F_n \subseteq F_m$
+}}
+{{eqn|o = \le
+     |r = \sum \left\{ {b_i : i \in I}\right\}
+     |c = [[Generalized Sum is Monotone]]
+}}
+{{eqn|ll= \implies
+     |l = \sum_{i \mathop \in F_m \setminus F_n} b_i
+     |o = <
+     |r = 2^{-n}
+}}
+{{end-eqn}}
+Finally, as $n \ge N, 2^{-n} < 2^{-N} < \epsilon$ (by defining property of $N$).
+Combining all of these estimates leads to the conclusion that $d \left({a_m, a_n}\right) < \epsilon$.
+It follows that $\left({a_n}\right)_{n \in \N}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+By [[Real Number Line is Complete Metric Space]], this implies there exists an $a \in \R$ such that $\displaystyle \lim_{n \to \infty} a_n = a$.
+Having identified a candidate $a$ for the sum $\displaystyle \sum \left\{{a_i: i \in I}\right\}$ to [[Definition:Generalized Sum|converge]] to, it remains to verify that this is indeed the case.
+According to the [[Definition:Generalized Sum|definition]] of considered sum, the convergence is [[Definition:Convergent Net|convergence of a net]].
+Next, [[Metric Induces Topology]] ensures that we can limit the choice of [[Definition:Open Set (Topology)|opens]] $U$ containing $a$ to [[Definition:Open Ball|open balls]] centered at $a$.
+Now let $\epsilon > 0$.
+We want to find a [[Definition:Finite|finite]] $F \subseteq I$ such that:
+:$d \left({\displaystyle \sum_{i \mathop \in G} a_i, a}\right) < \epsilon$, for all [[Definition:Finite|finite]] $G$ with $F \subseteq G \subseteq I$
+Now let $N \in \N$ such that for all $n \ge N$, $d \left({v_n, v}\right) < \dfrac \epsilon 2$ (with the $v_n$ as above).
+By taking a larger $N$ if necessary, ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well.
+Let us verify that the set $F_N$ defined above has the sought properties.
+So let $G$ be [[Definition:Finite|finite]] with $F_N \subseteq G \subseteq I$. Then:
+{{begin-eqn}}
+{{eqn|l = d \left({\sum_{i \mathop \in G} a_i, a}\right)
+     |r = \left\vert{\left({\sum_{i \mathop \in G} a_i}\right) - a}\right\vert
+     |c = Definition of [[Definition:Metric Induced by Norm|Induced Metric]]
+}}
+{{eqn|o = \le
+     |r = \left\vert{\left({\sum_{i \mathop \in G} a_i}\right) - \left({\sum_{i \mathop \in F_N} a_i}\right)}\right\vert + \left\vert{\left({\sum_{i \mathop \in F_N} a_i}\right) - a}\right\vert
+     |c = [[Triangle Inequality]]
+}}
+{{eqn|o = <
+     |r = \left\vert{\sum_{i \mathop \in G \setminus F_N} a_i}\right\vert + \frac \epsilon 2
+     |c = $F_N \subseteq G$, defining property of $N$
+}}
+{{eqn|o = \le
+     |r = \sum_{i \mathop \in G \setminus F_N} b_i + \frac \epsilon 2
+     |c = $b_i \ge a_i \ge 0$ for all $i \in I$
+}}
+{{end-eqn}}
+For the first of these terms, observe:
+{{begin-eqn}}
+{{eqn|l = \sum \left\{ {b_i: i \in I}\right\} - 2^{-N} + \sum_{i \mathop \in G \setminus F_N} b_i
+     |o = <
+     |r = \sum_{i \mathop \in F_N} b_i + \sum_{i \mathop \in G \setminus F_N} b_i
+     |c = Defining property of $F_N$
+}}
+{{eqn|r = \sum_{i \mathop \in G} b_i
+     |c = [[Union with Relative Complement]], $F_N \subseteq G$
+}}
+{{eqn|o = \le
+     |r = \sum \left\{ {b_i: i \in I}\right\}
+     |c = [[Generalized Sum is Monotone]]
+}}
+{{eqn|ll= \implies
+     |l = \sum_{i \mathop \in G \setminus F_N} b_i
+     |o = <
+     |r = 2^{-N}
+}}
+{{end-eqn}}
+Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain:
+:$\displaystyle d \left({\sum_{i \mathop \in G} a_i, a}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
+By definition of [[Definition:Convergent Net|convergence of a net]], it follows that:
+:$\displaystyle \sum \left\{{a_i: i \in I}\right\} = a$
+{{qed|lemma}}
+=== Proof of Inequality ===
+Suppose that $\displaystyle \sum \left\{{ a_i: i \in I }\right\} > \sum \left\{{ b_i: i \in I }\right\}$.
+Then, as the sums [[Definition:Generalized Sum|converge]], there exists a [[Definition:Finite|finite]] $F \subseteq I$ such that:
+:$\displaystyle \sum_{i \mathop \in F} a_i > \sum \left\{{ a_i: i \in I }\right\} - \epsilon$
+for every $\epsilon > 0$. 
+So by picking a suitable $\epsilon$, it may be arranged that:
+{{begin-eqn}}
+{{eqn|l = \sum \left\{ { b_i: i \in I }\right\}
+     |o = <
+     |r = \sum \left\{ { a_i: i \in I }\right\} - \epsilon
+}}
+{{eqn|o = <
+     |r = \sum_{i \mathop \in F} a_i
+     |c = Defining property of $F$
+}}
+{{eqn|o = \le
+     |r = \sum_{i \mathop \in F} b_i
+     |c = $a_i \le b_i$ for all $i$
+}}
+{{eqn|o = \le
+     |r = \sum \left\{ { b_i: i \in I }\right\}
+     |c = [[Generalized Sum is Monotone]]
+}}
+{{end-eqn}}
+These inequalities together constitute a contradiction, and therefore:
+:$\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$
+{{qed}}
+[[Category:Generalized Sums]]
+rgumdbhfc82we1qtta77khjt2o7gn5n
+\end{proof}<|endoftext|>
+\section{Generalized Sum is Linear}
+Tags: Generalized Sums, Complex Numbers
+
+\begin{theorem}
+Let $\left({z_i}\right)_{i \in I}, \left({w_i}\right)_{i \in I}$ be [[Definition:Indexed Set|$I$-indexed families]] of [[Definition:Complex Number|complex numbers]].
+That is, let $z_i, w_i \in \C$ for all $i \in I$.
+Suppose that $\displaystyle \sum \left\{{ z_i: i \in I }\right\}, \sum \left\{{ w_i: i \in I }\right\}$ [[Definition:Generalized Sum|converge]] to $z, w \in \C$, respectively.
+Then:
+:$(1): \quad \displaystyle \sum \left\{{ z_i + w_i: i \in I }\right\}$ [[Definition:Generalized Sum|converges]] to $z+w$
+:$(2): \quad \forall \lambda \in \C: \displaystyle \sum \left\{{ \lambda z_i: i \in I }\right\}$ [[Definition:Generalized Sum|converges]] to $\lambda z$
+\end{theorem}
+
+\begin{proof}
+=== Proof of $(1)$ ===
+Let $\epsilon > 0$.
+To verify the [[Definition:Generalized Sum|convergence]], it is necessary to find a [[Definition:Finite|finite]] $F \subseteq I$ such that:
+:$\displaystyle d \left({\sum_{i \mathop\in G} z_i + w_i, z + w}\right) < \epsilon$ for all [[Definition:Finite|finite]] $G$ with $F \subseteq G \subseteq I$
+Now let $F_z, F_w \subseteq I$ be [[Definition:Finite|finite]] [[Definition:Subset|subsets]] of $I$ such that:
+:$\displaystyle d \left({\sum_{i \mathop\in G} z_i, z}\right) < \frac \epsilon 2$ for all [[Definition:Finite|finite]] $G$ with $F_z \subseteq G \subseteq I$
+:$\displaystyle d \left({\sum_{i \mathop\in G} w_i, w}\right) < \frac \epsilon 2$ for all [[Definition:Finite|finite]] $G$ with $F_w \subseteq G \subseteq I$
+The set $F_z \cup F_w$ will be the sought $F$. Let $G$ be [[Definition:Finite|finite]] such that $F_z \cup F_w \subseteq G \subseteq I$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = d \left({\sum_{i \mathop\in G} z_i + w_i, z + w}\right)
+      | r = \left\vert{\left({\sum_{i \mathop\in G} z_i + w_i}\right) - \left({z + w}\right)}\right\vert
+      | c = {{Defof|Metric Induced by Norm}}
+}}
+{{eqn | r = \left\vert{\left({\sum_{i \mathop\in G} z_i - z}\right) + \left({\sum_{i \mathop\in G} w_i - w}\right)}\right\vert
+}}
+{{eqn | o = \le
+      | r = \left\vert{\left({\sum_{i \mathop\in G} z_i - z}\right)}\right\vert + \left\vert{\left({\sum_{i \mathop\in G} w_i, w}\right)}\right\vert
+      | c = [[Definition:Norm on Division Ring|Triangle Inequality]] for $\left\vert{\cdot}\right\vert$
+}}
+{{eqn | r = d \left({\sum_{i \mathop\in G} z_i, z}\right) + d \left({\sum_{i \mathop\in G} w_i, w}\right)
+      | c = {{Defof|Metric Induced by Norm}}
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = $F_z, F_w \subseteq G$
+}}
+{{end-eqn}}
+From the definition of [[Definition:Generalized Sum|convergence]], conclude that $\displaystyle \sum \left\{{ z_i + w_i: i \in I }\right\} = z+w$.
+{{qed|lemma}}
+=== Proof of $(2)$ ===
+Let $\epsilon > 0$.
+To verify the [[Definition:Generalized Sum|convergence]], it is necessary to find a [[Definition:Finite|finite]] $F \subseteq I$ such that:
+:$\displaystyle d \left({\sum_{i \mathop\in G} \lambda z_i, \lambda z}\right) < \epsilon$ for all [[Definition:Finite|finite]] $G$ with $F \subseteq G \subseteq I$
+Now let $F_z \subseteq I$ be a [[Definition:Finite|finite]] [[Definition:Subset|subset]] of $I$ such that:
+:$\displaystyle d \left({\sum_{i \mathop\in G} z_i, z}\right) < \frac \epsilon {\left\vert{\lambda}\right\vert}$ for all [[Definition:Finite|finite]] $G$ with $F_z \subseteq G \subseteq I$
+The set $F_z$ will be the sought $F$. Let $G$ be [[Definition:Finite|finite]] such that $F_z \subseteq G \subseteq I$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = d \left({\sum_{i \mathop \in G} \lambda z_i, \lambda z}\right)
+      | r = \left\vert{\left({\sum_{i \mathop\in G} \lambda z_i}\right) - \lambda z}\right\vert
+      | c = {{Defof|Metric Induced by Norm}}
+}}
+{{eqn | r = \left\vert{\lambda \left({\sum_{i \mathop \in G} z_i - z}\right)}\right\vert
+}}
+{{eqn | r = \left\vert{\lambda}\right\vert \left\vert{\sum_{i \mathop \in G} z_i - z}\right\vert
+      | c = [[Definition:Norm on Division Ring|Multiplicativity]] for $\left\vert{\cdot}\right\vert$
+}}
+{{eqn | r = \left\vert{\lambda}\right\vert d \left({\sum_{i \mathop \in G} z_i, z}\right)
+      | c = {{Defof|Metric Induced by Norm}}
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = $F_z \subseteq G$
+}}
+{{end-eqn}}
+From the definition of [[Definition:Generalized Sum|convergence]], conclude that $\displaystyle \sum \left\{{ \lambda z_i: i \in I }\right\} = \lambda z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Min yields Infimum of Parameters}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]]. 
+Let $x, y \in S$.
+Then:
+: $\min \left({x, y}\right) = \inf \left({\left\{{x, y}\right\}}\right)$
+where:
+:$\min$ denotes the [[Definition:Min Operation|min operation]]
+:$\inf$ denotes the [[Definition:Infimum of Set|infimum]].
+\end{theorem}
+
+\begin{proof}
+As $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]], all elements of $S$ are [[Definition:Comparable|comparable]] by $\preceq$.
+Therefore there are two cases to consider:
+=== Case 1: $x \preceq y$ ===
+In this case:
+: $\min \left({x, y}\right) = x = \inf \left({\left\{{x, y}\right\}}\right)$
+=== Case 2: $y \preceq x$ ===
+In this case:
+: $\min \left({x, y}\right) = y = \inf \left({\left\{{x, y}\right\}}\right)$
+In either case, the result holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergence of Generalized Sum of Complex Numbers}
+Tags: Generalized Sums, Complex Numbers
+
+\begin{theorem}
+Let $\left({z_i}\right)_{i \mathop \in I}$ be an [[Definition:Indexed Set|$I$-indexed family]] of [[Definition:Complex Number|complex numbers]].
+That is, let $z_i \in \C$ for all $i \in I$.
+Denote by $\operatorname{Re} z_i$, resp. $\operatorname{Im} z_i$ the families of [[Definition:Real Part|real]], resp. [[Definition:Imaginary Part|imaginary parts]] of the family $z_i$.
+Then the following are equivalent:
+:$(1): \quad \displaystyle \sum \left\{{z_i : i \in I}\right\}$ [[Definition:Generalized Sum|converges]] to $z \in \C$
+:$(2): \quad \displaystyle \sum \left\{{\operatorname{Re} z_i : i \in I}\right\}, \sum \left\{{\operatorname{Im} z_i : i \in I}\right\}$ [[Definition:Generalized Sum|converge]] to $\operatorname{Re} z, \operatorname{Im} z \in \R$, respectively
+\end{theorem}
+
+\begin{proof}
+=== $(2)$ implies $(1)$ ===
+By [[Generalized Sum is Linear]], the stated convergences lead to:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \operatorname{Re} z + i \operatorname{Im} z
+      | c = Definition of [[Definition:Real Part|Real]] and [[Definition:Imaginary Part|Imaginary Parts]]
+}}
+{{eqn | r = \sum \left\{ {\operatorname{Re} z_i : i \in I}\right\} + i \sum \left\{ {\operatorname{Im} z_i : i \in I}\right\}
+      | c = Statement $(2)$
+}}
+{{eqn | r = \sum \left\{ {\operatorname{Re} z_j + i \operatorname{Im} z_j: j \in I}\right\}
+      | c = [[Generalized Sum is Linear]]
+}}
+{{eqn | r = \sum \left\{ {z_i : i \in I}\right\}
+      | c = Definition of [[Definition:Real Part|Real]] and [[Definition:Imaginary Part|Imaginary Parts]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $(1)$ implies $(2)$ ===
+Statement $(1)$, according to the definition of [[Definition:Generalized Sum|convergence]], amounts to the following:
+For every $\epsilon > 0$, there exists a [[Definition:Finite|finite]] $G \subseteq I$ such that:
+:For every [[Definition:Finite|finite]] $F \subseteq I$ with $G \subseteq F$:
+::$\displaystyle \left\vert{z - \sum_{i \mathop \in F} z_i}\right\vert < \epsilon$
+Now suppose that for $\epsilon > 0$, $G$ and $F$ are as above. Then observe that:
+{{begin-eqn}}
+{{eqn | l = \epsilon^2
+      | o = >
+      | r = \left\vert{z - \sum_{i \mathop \in F} z_i}\right\vert^2
+}}
+{{eqn | r = \left({\operatorname{Re} z - \sum_{i \mathop \in F} \operatorname{Re} z_i}\right)^2 + \left({\operatorname{Im} z - \sum_{i \mathop \in F} \operatorname{Im} z_i}\right)^2
+      | c = Definition of [[Definition:Modulus of Complex Number|Modulus]]
+}}
+{{end-eqn}}
+Hence, by [[Square of Real Number is Non-Negative]], both of the terms on the right hand side are smaller than $\epsilon^2$.
+It follows that, taking [[Definition:Square Root|square roots]], $G$ satisfies, for any [[Definition:Finite|finite]] $F \supseteq G$:
+:$\displaystyle \left\vert{\operatorname{Re} z - \sum_{i \mathop \in F} \operatorname{Re} z_i}\right\vert < \epsilon$
+:$\displaystyle \left\vert{\operatorname{Im} z - \sum_{i \mathop \in F} \operatorname{Im} z_i}\right\vert < \epsilon$
+As $\epsilon > 0$ was arbitrary, using the definition of [[Definition:Generalized Sum|convergence]], this implies precisely that:
+:$\displaystyle \sum \left\{{\operatorname{Re} z_i : i \in I}\right\}, \sum \left\{{\operatorname{Im} z_i : i \in I}\right\}$ [[Definition:Generalized Sum|converge]] to $\operatorname{Re} z, \operatorname{Im} z \in \R$, respectively.
+Hence, $(1)$ is shown to imply $(2)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equidistance is Independent of Betweenness}
+Tags: Geometry
+
+\begin{theorem}
+Let $\mathcal G$ be a [[Definition:Formal System|formal systematic]] treatment of [[Definition:Geometry|geometry]] containing only:
+: The [[Definition:Formal Language|language]] and [[Definition:Axiom|axioms]] of [[Definition:Predicate Logic|first-order logic]], and the [[Definition:Preceding Discipline|disciplines preceding it]]
+: The [[Definition:Undefined Terms of Tarski's Geometry|undefined terms of Tarski's Geometry]] (excluding [[Definition:Equidistance|equidistance]]) 
+: Some or all of [[Definition:Tarski's Axioms|Tarski's Axioms of Geometry]].
+In $\mathcal G$, [[Definition:Equidistance|equidistance]] $\equiv$ is necessarily an [[Definition:Undefined Term|undefined term]] with respect to [[Definition:Between (Geometry)|betweenness]] $\mathsf B$.
+\end{theorem}
+
+\begin{proof}
+Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf B$.
+[[Proof by Contradiction|Seeking a contradiction]], [[Definition:Assumption|assume]] that it can.
+Call this assumption $\left({A}\right)$.
+If $\left({A}\right)$ holds, it must hold in all systems.
+Let one such system be $\left({\R^2, \mathsf B_1, \equiv_1}\right)$ where:
+:$\R^2$ is the [[Definition:Cartesian Product|cartesian product]] of the [[Definition:Real Number|set of real numbers]] with itself
+:$\mathsf B_1$ is a [[Definition:Ternary|ternary]] [[Definition:Relation|relation]] of [[Definition:Between (Geometry)|betweenness]]
+:$\equiv_1$ is a [[Definition:Quaternary|quaternary]] [[Definition:Relation|relation]] of [[Definition:Equidistance|equidistance]]
+Let $\mathcal G$ be the [[Definition:Preceding Discipline|discipline preceding the given discipline]], where $\mathcal G$ is as defined above (excluding both $\equiv$ and $\mathsf B$).
+{{questionable|if $\mathcal G$ isn't strong enough to create $\R^2$, how can we use it? There is something incorrect in my presentation, particularly since we're going to use $\cdot$ and $\le$ -- GFP.}}
+Define $\mathsf B_1$ as follows:
+{{:Definition:Between (Geometry)}}
+Define $\equiv_1$ as follows:
+{{:Definition:Equidistance}}
+Now, define the [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\phi$ on $\left({\R^2, \mathsf B_2, \equiv_2}\right)$ as:
+:$\phi: \R^2 \to \R^2$ on $\left({\R^2, \mathsf B_1, \equiv_1}\right), \left({x_1, x_2}\right) \mapsto \left({x_1, 2 x_2}\right)$
+{{finish|prove that $\phi$ is an isomorphism}}
+Now consider the system:
+:$\left({\R^2, \mathsf B_2, \equiv_2}\right)$
+where $\mathsf B_2$ and $\equiv_2$ are the relations defined as above, but on the elements in the [[Definition:Image of Mapping|images]] of $\mathsf B_1$ and $\equiv_1$, respectively.
+Observe that $\mathsf B_1$ and $\mathsf B_2$ coincide, because in:
+:$\left({x_1 - y_1}\right) \cdot \left({2 y_2 - 2 z_2}\right) = \left({2 x_2 - 2 y_2}\right) \cdot \left({y_1 - z_1}\right) \land$
+:$\left({0 \le \left({x_1 - y_1}\right) \cdot \left({y_1 - z_1}\right)}\right) \land \left({0 \le \left({2 x_2 - 2 y_2}\right) \cdot \left({2 y_2 - 2 z_2}\right)}\right)$
+we can simply factor out the $2$ and divide both sides of the equality of inequality by $2$.
+But consider the elements:
+:$p_1 = \left({0, 0}\right)$
+:$p_2 = \left({0, 1}\right)$
+:$p_3 = \left({1, 0}\right)$
+Observe that $p_1 p_2 \equiv_1 p_1 p_3$:
+:$\left({0 - 0}\right)^2 + \left({0 - 1}\right)^2 = \left({0 - 1}\right)^2 + \left({0 - 0}\right)^2$
+But $\neg \left({p_1 p_2 \equiv_2 p_1 p_3}\right)$:
+:$\left({0 - 0}\right)^2 + \left({0 - 2}\right)^2 \ne \left({0 - 1}\right)^2 + \left({0 - 0}\right)^2$
+But both $\left({\R^2, \mathsf B_1, \equiv_1}\right)$ and $\left({\R^2, \mathsf B_2, \equiv_2}\right)$ are both models of $\mathcal G$.
+{{explain|prove it}}
+Recall that if $\left({A}\right)$ holds, it must hold in all systems.
+But it does not.
+Hence $\left({A}\right)$ is false, from [[Proof by Contradiction]].
+{{qed}}
+{{proofread}}
+{{help}}
+\end{proof}<|endoftext|>
+\section{Lower and Upper Bounds for Sequences/Corollary}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}, \sequence {y_n}$ be [[Definition:Real Sequence|sequences in $\R$]].
+Let $x_n \to l, y_n \to m$ as $n \to \infty$.
+Suppose that for all $n \in \N$, $x_n \le y_n$.
+Then:
+:$l \le m$
+that is:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n \le \lim_{n \mathop \to \infty} y_n$
+This is often phrased as: ''limits preserve inequalities''.
+\end{theorem}
+
+\begin{proof}
+Consider the sequence $\sequence {z_n}=$ defined by $z_n := y_n - x_n$.
+The [[Sum Rule for Real Sequences]] grants that $z_n \to m - l$ as $n \to \infty$.
+Furthermore, the assumption that $x_n \le y_n$ for all $n \in \N$ means that $z_n \ge 0$ for all $n \in \N$.
+Applying the [[Lower and Upper Bounds for Sequences|main result]] to the sequence $\sequence {z_n}$ leads to the conclusion that $m - l \ge 0$.
+That is, $l \le m$.
+{{qed}}
+[[Category:Limits of Sequences]]
+o61xs1dfmo77y9rhq6olejsr2tjnnbe
+\end{proof}<|endoftext|>
+\section{Bounded Linear Transformation Induces Bounded Sesquilinear Form}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]] over $\Bbb F \in \left\{{\R, \C}\right\}$.
+Let $A \in B \left({H, K}\right), B \in B \left({K, H}\right)$ be [[Definition:Bounded Linear Transformation|bounded linear transformations]].
+Let $u, v: H \times K \to \Bbb F$ be defined by:
+:$u \left({h, k}\right) := \left\langle{Ah, k}\right\rangle_K$
+:$v \left({h, k}\right) := \left\langle{h, Bk}\right\rangle_H$
+Then $u$ and $v$ are [[Definition:Bounded Sesquilinear Form|bounded sesquilinear forms]].
+\end{theorem}<|endoftext|>
+\section{Classification of Bounded Sesquilinear Forms}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]] over $\Bbb F \in \left\{{\R, \C}\right\}$.
+Let $u: H \times K \to \Bbb F$ be a [[Definition:Bounded Sesquilinear Form|bounded sesquilinear form]] with bound $M$.
+Then there exist unique [[Definition:Bounded Linear Transformation|bounded linear transformations]] $A \in B \left({H, K}\right), B \in B \left({K, H}\right)$ such that:
+:$\forall h \in H, k \in K: u \left({h, k}\right) = \left\langle{Ah, k}\right\rangle_K = \left\langle{h, Bk}\right\rangle_H$
+Furthermore, $\left\Vert{A}\right\Vert, \left\Vert{B}\right\Vert \le M$.
+\end{theorem}<|endoftext|>
+\section{Isomorphism iff Inverse Equals Adjoint}
+Tags: Linear Transformations on Hilbert Spaces, Adjoints
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $U \in B \struct {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+Then the following are equivalent:
+:$(1): \quad U$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]]
+:$(2): \quad U$ is [[Definition:Invertible Mapping|invertible]] and $U^{-1} = U^*$, where $U^*$ denotes the [[Definition:Adjoint Linear Transformation|adjoint]] of $U$.
+\end{theorem}<|endoftext|>
+\section{Limit of Sequence is Limit of Real Function}
+Tags: Convergence, Sequences
+
+\begin{theorem}
+Let $\sequence {a_n}$ be a [[Definition:Real Sequence|real sequence]].
+Let $f: x \mapsto \map f x$ be a [[Definition:Real Function|real function]].
+Suppose the [[Definition:Limit at Infinity|limit]]:
+:$\displaystyle \lim_{x \mathop \to +\infty} \map f x$
+exists.
+If for every $n$ in the [[Definition:Domain of Mapping|domain]] of $\sequence {a_n}$:
+:$\map f n = a_n$
+then:
+:$\displaystyle \lim_{n \mathop \to +\infty} \ a_n = \displaystyle \lim_{x \mathop \to +\infty} \map f x$
+\end{theorem}
+
+\begin{proof}
+This is an instance of [[Limit of Function by Convergent Sequences]], as the [[Real Number Line is Complete Metric Space|reals form a metric space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Semilattice on Classical Set is Semilattice}
+Tags: Semilattices
+
+\begin{theorem}
+Let $\struct {S, \vee}$ be an [[Definition:Upper Semilattice on Classical Set|upper semilattice on a classical set]] $S$.
+Then $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]].
+\end{theorem}
+
+\begin{proof}
+To show that the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]], the following need to be verified:
+: [[Definition:Closed Algebraic Structure|Closure]]
+: [[Definition:Associative|Associativity]]
+: [[Definition:Commutative Operation|Commutativity]]
+: [[Definition:Idempotent Operation|Idempotence]]
+In order:
+=== Closure ===
+By definition of an [[Definition:Upper Semilattice on Classical Set|upper semilattice]]:
+:$\forall x, y \in S: \sup \set {x, y} \in S$
+Since $x \vee y = \sup \set {x, y} \in S$ for all $x, y \in S$:
+:$\struct {S, \vee}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== Associativity ===
+Let $x, y, z \in S$.
+By definition of $\vee$:
+{{begin-eqn}}
+{{eqn | l = \paren {x \vee y} \vee z
+      | r = \sup \set {x, y} \vee z
+}}
+{{eqn | r = \sup \set {\sup \set {x, y}, z}
+}}
+{{eqn | r = \sup \set {x, y, z}
+}}
+{{eqn | r = \sup \set {x, \sup set {y, z} }
+}}
+{{eqn | r = x \vee \sup \set {y, z}
+}}
+{{eqn | r = x \vee \paren {y \vee z}
+}}
+{{end-eqn}}
+{{MissingLinks|The supremum equalities in the middle. Result might not yet exist on PW.}}
+Hence $\struct {S, \vee}$ is [[Definition:Associative|associative]].
+{{qed|lemma}}
+=== Commutativity ===
+Let $x, y \in S$.
+By definition of $\vee$:
+{{begin-eqn}}
+{{eqn | l = x \vee y
+      | r = \sup \set {x, y}
+}}
+{{eqn | r = \sup \set {y, x}
+      | c = {{Defof|Supremum of Set}}
+}}
+{{eqn | r = y \vee x
+}}
+{{end-eqn}}
+Hence $\struct {S, \vee}$ is [[Definition:Commutative Operation|commutative]].
+{{qed|lemma}}
+=== Idempotence ===
+For all $x \in S$:
+{{begin-eqn}}
+{{eqn | l = x \vee x
+      | r = \sup \set {x, x}
+}}
+{{eqn | r = x
+}}
+{{end-eqn}}
+Hence $\vee$ is [[Definition:Idempotent Operation|idempotent]].
+{{qed|lemma}}
+Having explicitly verified all prerequisites, it follows that $\struct {S, \vee}$ is a [[Definition:Semilattice|semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equality of Cartesian Products}
+Tags: Cartesian Product, Equality
+
+\begin{theorem}
+Let $A, B, C, D$ be [[Definition:Non-Empty Set|nonempty]] [[Definition:Set|sets]].
+Then:
+: $A \times B = C \times D \iff A = C \land B = D$
+where $\times$ denotes [[Definition:Cartesian Product|cartesian product]].
+\end{theorem}
+
+\begin{proof}
+If $A = C$ and $B = D$, it is immediate that $A \times B = C \times D$.
+Now suppose that $A \times B = C \times D$.
+By definition of [[Definition:Cartesian Product|Cartesian product]]:
+{{begin-eqn}}
+{{eqn | l = x \in A, y \in B
+      | o = \leadstoandfrom
+      | r = \tuple {x, y} \in A \times B
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \tuple {x, y} \in C \times D
+      | c = {{Defof|Set Equality}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = x \in C, y \in D
+}}
+{{end-eqn}}
+It follows that $x \in A$ {{iff}} $x \in C$, and so $A = C$.
+Similarly, $y \in B$ {{iff}} $y \in D$, hence $B = D$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Adjoining is Linear}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]] over $\Bbb F \in \left\{{\R, \C}\right\}$.
+Let $A, B \in B \left({H, K}\right)$ be [[Definition:Bounded Linear Transformation|bounded linear transformations]].
+Then the operation of [[Definition:Adjoint Linear Transformation|adjoining]] $^*$ satisfies, for all $\lambda \in \Bbb F$:
+:$(1): \qquad \left({\lambda A}\right)^* = \overline \lambda A^*$
+:$(2): \qquad \left({A + B}\right)^* = A^* + B^*$
+That is, $^*: B \left({H, K}\right) \to B \left({K, H}\right)$ is a [[Definition:Linear Transformation|linear transformation]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $(1)$ ===
+Let $\lambda \in \Bbb F$, $h \in H, k \in K$. Then:
+{{begin-eqn}}
+{{eqn|l = \left\langle{\left({\lambda A}\right)h, k}\right\rangle_K
+     |r = \lambda \left\langle{Ah, k}\right\rangle_K
+     |c = Property $(2)$ of [[Definition:Inner Product|inner products]]
+}}
+{{eqn|r = \lambda \left\langle{h, A^*k}\right\rangle_H
+     |c = Definition of [[Definition:Adjoint Linear Transformation|adjoint]]
+}}
+{{eqn|r = \left\langle{h, \left({\overline \lambda A^*}\right) k}\right\rangle_H
+     |c = Properties $(1), (2)$ of [[Definition:Inner Product|inner products]]
+}}
+{{end-eqn}}
+Thus, by [[Existence and Uniqueness of Adjoint]], $\left({\lambda A}\right)^* = \overline \lambda A^*$.
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{Adjoint of Composition}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H, K, L$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $A \in \map B {K, L}, B \in \map B {H, K}$ be [[Definition:Bounded Linear Transformation|bounded linear transformations]].
+Then $\paren {A B}^* = B^* A^*$, where $^*$ denotes [[Definition:Adjoint Linear Transformation|adjoining]].
+\end{theorem}
+
+\begin{proof}
+Let $h \in H, l \in L$. Then:
+{{begin-eqn}}
+{{eqn | l = \innerprod {\paren {AB} h} l_L
+      | r = \innerprod {B h} {A^* l}_K
+      | c = {{Defof|Adjoint Linear Transformation}} of $A$
+}}
+{{eqn | r = \innerprod h {B^* A^* l}_H
+      | c = {{Defof|Adjoint Linear Transformation}} of $B$
+}}
+{{end-eqn}}
+Thus, by [[Existence and Uniqueness of Adjoint]]:
+:$\paren {A B}^* = B^* A^*$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Adjoint is Itself}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $A \in B \left({H, K}\right)$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+Then $A^{**} := \left({A^*}\right)^* = A$.
+\end{theorem}
+
+\begin{proof}
+Let $h \in H, k \in K$. Then:
+{{begin-eqn}}
+{{eqn|l = \left\langle{Ah, k}\right\rangle_K
+     |r = \left\langle{h, A^*k}\right\rangle_H
+     |c = Definition of [[Definition:Adjoint Linear Transformation|adjoint]] of $A$
+}}
+{{eqn|r = \left\langle{A^{**}h, k}\right\rangle_K
+     |c = Definition of [[Definition:Adjoint Linear Transformation|adjoint]] of $A^*$
+}}
+{{end-eqn}}
+Thus, by [[Existence and Uniqueness of Adjoint]], $A^{**} = A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Adjoining Commutes with Inverting}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H, K}\right)$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Let $A^{-1} \in B \left({K, H}\right)$ be an [[Definition:Inverse (Bounded Linear Operator)|inverse]] for $A$.
+Then the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$, $A^*$, is [[Definition:Invertible Linear Operator|invertible]].
+Furthermore, $\left({A^*}\right)^{-1} = \left({A^{-1}}\right)^*$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Inverse (Bounded Linear Operator)|inverse]], one has $AA^{-1} = I_K$, where $I_K$ is the [[Definition:Identity Operator|identity operator]] on $K$.
+Now observe from [[Adjoint of Composition]] that:
+:$I_K = I_K^* = \left({AA^{-1}}\right)^* = \left({A^{-1}}\right)^*A^*$.
+Similarly, one has:
+:$I_H = I_H^* = \left({A^{-1}A}\right)^* = A^*\left({A^{-1}}\right)^*$
+Hence, by definition of [[Definition:Inverse (Bounded Linear Operator)|inverse]], $\left({A^*}\right)^{-1} = \left({A^{-1}}\right)^*$.
+This also means that $A^*$ is invertible.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity of Points}
+Tags: Euclidean Geometry
+
+\begin{theorem}
+Two [[Definition:Point|points]] share the same [[Definition:Position|position]] {{iff}} they are the [[Definition:Equals|same]] [[Definition:Point|points]].
+\end{theorem}
+
+\begin{proof}
+Let $a$ be a [[Definition:Point|point]] with [[Definition:Position|position]] $P_1$.
+Let $b$ be a [[Definition:Point|point]] with [[Definition:Position|position]] $P_2$.
+[[Definition:By Hypothesis|By hypothesis]], $P_1 = P_2$.
+By [[Axiom:Leibniz's Law|Leibniz's Law]], two [[Definition:Object|objects]] are the same object {{iff}} they share every [[Definition:Property|property]] in common.
+By the definition of [[Definition:Point|point]], the ''only'' property possessed by a [[Definition:Point|point]] is [[Definition:Position|position]].
+We have:
+:$P_1 = P_2 \dashv \vdash a = b$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Successor of Omega}
+Tags: Transfinite Arithmetic
+
+\begin{theorem}
+:$\omega + 1 = \set {0, 1, 2, ...; \omega}$
+where $\omega$ is the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]] and $\omega + 1$ is the [[Definition:Successor Set|successor of $\omega$]].
+Note the use of the semicolon; this is the notation for [[Definition:Multipart Infinite Set|multipart infinite sets]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \omega + 1
+      | r = \omega \cup \set {\omega}
+      | c = {{Defof|Successor Set}}
+}}
+{{eqn | r = \set {0, 1, 2, \ldots} \cup \set \omega
+      | c = {{Defof|Von Neumann Construction of Natural Numbers}}
+}}
+{{eqn | r = \set {0, 1, 2, \ldots; \omega}
+      | c = {{Defof|Set Union}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Versions of Axiom of Choice}
+Tags: Axiom of Choice
+
+\begin{theorem}
+The following formulations of the [[Axiom:Axiom of Choice|Axiom of Choice]] are [[Definition:Logical Equivalence|equivalent]]:
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Versions of Axiom of Choice/Formulation 2 implies Formulation 1|Formulation 2 implies Formulation 1]] ===
+{{:Equivalence of Versions of Axiom of Choice/Formulation 2 implies Formulation 1}}{{qed|lemma}}
+=== [[Axiom:Axiom of Choice/Formulation 1|Formulation 1]] implies [[Axiom:Axiom of Choice/Formulation 2|Formulation 2]] ===
+The argument can be seen to be reversible.
+{{Handwaving}}
+{{qed|lemma}}
+=== [[Equivalence of Versions of Axiom of Choice/Formulation 1 implies Formulation 3|Formulation 1 implies Formulation 3]] ===
+{{:Equivalence of Versions of Axiom of Choice/Formulation 1 implies Formulation 3}}{{qed|lemma}}
+=== [[Equivalence of Versions of Axiom of Choice/Formulation 3 implies Formulation 1|Formulation 3 implies Formulation 1]] ===
+{{:Equivalence of Versions of Axiom of Choice/Formulation 3 implies Formulation 1}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Norm of Adjoint}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $A \in \map B {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+Then the [[Definition:Norm on Bounded Linear Transformation|norm]] of $A$ satisfies:
+:$\norm A^2 = \norm {A^*}^2 = \norm {A^* A}$
+where $A^*$ denotes the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$.
+\end{theorem}
+
+\begin{proof}
+Let $h \in H$ such that $\norm h_H \le 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {A h}_K^2
+      | r = \innerprod {A h} {A h}_K
+      | c = {{Defof|Inner Product Norm}}
+}}
+{{eqn | r = \innerprod {A^* A h} h_H
+      | c = {{Defof|Adjoint Linear Transformation}}
+}}
+{{eqn | o = \le
+      | r = \norm {A^*A h}_H \norm h_H
+      | c = [[Cauchy-Bunyakovsky-Schwarz Inequality]]
+}}
+{{eqn | o = \le
+      | r = \norm {A^* A} \norm h_H^2
+      | c = [[Submultiplicativity of Operator Norm]]
+}}
+{{eqn | o = \le
+      | r = \norm {A^* A}
+      | c = Assumption on $h$
+}}
+{{eqn | o = \le
+      | r = \norm {A^*} \norm A
+      | c = [[Submultiplicativity of Operator Norm]]
+}}
+{{end-eqn}}
+By [[Definition:Norm on Bounded Linear Transformation|definition $(1)$]] for $\norm A$, it follows that:
+:$\norm A^2 \le \norm {A^* A} \le \norm {A^*} \norm A$
+That is:
+:$\norm A \le \norm {A^*}$.
+By substituting $A^*$ for $A$, and using $A^{**} = A$ from [[Double Adjoint is Itself]], the reverse inequality is obtained.
+Hence $\norm A^2 = \norm {A^* A} = \norm {A^*}^2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Operator Self-Adjoint iff Inner Product Real}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\C$.
+Let $A \in \map B H$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Then $A$ is [[Definition:Self-Adjoint Operator|self-adjoint]] {{iff}}:
+:$\forall h \in H: \innerprod {A h} h_H \mathop \in \R$
+\end{theorem}<|endoftext|>
+\section{Norm of Self-Adjoint Operator}
+Tags: Adjoints
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H}\right)$ be a [[Definition:Self-Adjoint Operator|self-adjoint operator]].
+Then the [[Definition:Norm on Bounded Linear Transformation|norm]] of $A$ satisfies:
+:$\left\Vert{A}\right\Vert = \sup \left\{{ \left\vert{ \left\langle{Ah, h}\right\rangle_H }\right\vert: h \in H, \left\Vert{h}\right\Vert_H = 1 }\right\}$
+\end{theorem}<|endoftext|>
+\section{Operator Zero iff Inner Product Zero}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\C$.
+Let $A \in B \left({H}\right)$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Suppose that:
+:$\forall h \in H: \left\langle{Ah, h}\right\rangle_H = 0$
+Then $A$ is the [[Definition:Zero Operator|zero operator]].
+\end{theorem}<|endoftext|>
+\section{Linear Operator is Sum of Real and Imaginary Parts}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\C$.
+Let $A \in B \left({H}\right)$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Let $B$ and $C$ be the [[Definition:Real Part (Linear Operator)|real]] and [[Definition:Imaginary Part (Linear Operator)|imaginary parts]] of $A$, respectively.
+Then $A = B + iC$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn|l = B + iC
+     |r = \frac 1 2 \left({A + A^*}\right) + i \frac 1 {2i} \left({A - A^*}\right)
+     |c = Definitions of $B, C$
+}}
+{{eqn|r = \frac 1 2 A + \frac 1 2 A^* + \frac 1 2 A - \frac 1 2 A^*
+}}
+{{eqn|r = A
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Normal Operators}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H}\right)$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Then the following are equivalent:
+:$(1): \qquad AA^* = A^*A$, i.e., $A$ is [[Definition:Normal Operator|normal]]
+:$(2): \qquad \forall h \in H: \left\Vert{Ah}\right\Vert_H = \left\Vert{A^*h}\right\Vert_H$
+If $H$ is a [[Definition:Hilbert Space|Hilbert space]] over $\C$, these are also equivalent to:
+:$(3): \qquad \operatorname{Re} \left({A}\right) \operatorname{Im} \left({A}\right) = \operatorname{Im} \left({A}\right) \operatorname{Re} \left({A}\right)$, i.e., the [[Definition:Real Part (Linear Operator)|real]] and [[Definition:Imaginary Part (Linear Operator)|imaginary parts]] of $A$ [[Definition:Commute|commute]]
+\end{theorem}<|endoftext|>
+\section{Isometry iff Adjoint is Left-Inverse}
+Tags: Adjoints, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $A \in B \left({H, K}\right)$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+Then $A$ is an [[Definition:Isometry (Hilbert Spaces)|isometry]] iff:
+:$A^*A = I_H$
+where $A^*$ denotes the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$, and $I_H$ the [[Definition:Identity Operator|identity operator]] on $H$.
+\end{theorem}
+
+\begin{proof}
+Let $g, h \in H$. Then by the definition of [[Definition:Adjoint Linear Transformation|adjoint]]:
+:$\left\langle{Ag, Ah}\right\rangle_K = \left\langle{A^*Ag, h}\right\rangle_H$
+From the [[Definition:Adjoint Linear Transformation|uniqueness of the adjoint]], it follows that:
+:$\left\langle{Ag, Ah}\right\rangle_K = \left\langle{g, h}\right\rangle_H$
+holds [[Definition:Iff|iff]] $A^*A = I_H$.
+Hence the result by definition of [[Definition:Isometry (Hilbert Spaces)|isometry]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Unitary Operators}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in \map B H$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Then the following are equivalent:
+:$(1): \quad A$ is a [[Definition:Unitary Operator|unitary operator]]
+:$(2): \quad A^* A = A A^* = I$, where $A^*$ denotes the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$, and $I$ denotes the [[Definition:Identity Operator|identity operator]]
+:$(3): \quad A$ is a [[Definition:Normal Operator|normal]] [[Definition:Isometry (Hilbert Spaces)|isometry]]
+\end{theorem}<|endoftext|>
+\section{Kernel of Linear Transformation is Orthocomplement of Range of Adjoint}
+Tags: Adjoints, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $A \in B \struct {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+Then $\ker A = \paren {\Img {A^*} }^\perp$, where:
+:$A^*$ denotes the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$
+:$\ker A$ is the [[Definition:Kernel of Linear Transformation|kernel]] of $A$
+:$\Img {A^*}$ is the [[Definition:Image of Mapping|image]] of $A^*$
+:$\perp$ signifies [[Definition:Orthocomplement|orthocomplementation]]
+\end{theorem}
+
+\begin{proof}
+Let $x \in H$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in 
+      | r = \map \ker A
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = A x
+      | r = \mathbf 0_K
+      | c = {{Defof|Kernel of Linear Transformation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = A x 
+      | o = \in 
+      | r = K^\perp
+      | c = [[Linear Subspace Dense iff Zero Orthocomplement]] ($K$ is trivially dense)
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \innerprod {A x} K_K
+      | r = \set 0
+      | c = {{Defof|Orthocomplement}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \innerprod x {A^* K}_H
+      | r = \set 0
+      | c = {{Defof|Adjoint Linear Transformation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \innerprod x {\Img {A^*} }_H
+      | r = \set 0
+      | c = {{Defof|Image of Mapping}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in 
+      | r = \paren {\Img {A^*} }^\perp
+      | c = {{Defof|Orthocomplement}}
+}}
+{{end-eqn}}
+Hence by definition of [[Definition:Set Equality/Definition 1|set equality]]:
+:$\ker A = \paren {\Img {A^*} }^\perp$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Contains Mapping is Equivalent to AoC}
+Tags: 
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
+Then:
+:there exists a [[Definition:Mapping|mapping]] $f \subseteq \RR$ whose [[Definition:Domain of Mapping|domain]] is the same as the [[Definition:Preimage of Relation|preimage]] of $\RR$
+{{iff}}
+:the [[Axiom:Axiom of Choice|axiom of choice]] holds.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{AoC}}
+\end{proof}<|endoftext|>
+\section{Existence of Set with Singleton Intersections with Disjoint Collection}
+Tags: Set Intersection, Singletons, Disjoint Sets
+
+\begin{theorem}
+Let $\mathcal C$ be a [[Definition:Set of Sets|set of sets]] all of which are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+Then:
+: there exists a [[Definition:Set|set]] $A$ such that $\forall S \in \mathcal C: A \cap S$ is a [[Definition:Singleton|singleton]]
+{{iff}}
+: the [[Axiom:Axiom of Choice|axiom of choice]] holds.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+{{AoC}}
+\end{proof}<|endoftext|>
+\section{Complementary Idempotent is Idempotent}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A$ be an [[Definition:Idempotent Operator|idempotent operator]].
+Then the [[Definition:Complementary Idempotent|complementary idempotent]] $I - A$ is also [[Definition:Idempotent Operator|idempotent]].
+\end{theorem}
+
+\begin{proof}
+In $I - A$, $I$ is the [[Definition:Identity Operator|identity operator]] on $H$.
+Hence:
+{{begin-eqn}}
+{{eqn|l = \left({I - A}\right)^2
+     |r = I^2 - IA - AI + A^2
+}}
+{{eqn|r = I^2 - 2A + A^2
+     |c = $I$ is the [[Definition:Identity Operator|identity operator]]
+}}
+{{eqn|r = I - A
+     |c = $I, A$ are [[Definition:Idempotent Operator|idempotent]]
+}}
+{{end-eqn}}
+That is, $I - A$ is [[Definition:Idempotent Operator|idempotent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Range of Idempotent is Kernel of Complementary Idempotent}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A$ be an [[Definition:Idempotent Operator|idempotent operator]].
+Then $\Rng A = \map \ker {I - A}$.
+\end{theorem}<|endoftext|>
+\section{Range and Kernel of Idempotent are Algebraically Complementary}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A$ be an [[Definition:Idempotent Operator|idempotent operator]].
+Then $\ker A$ and $\Rng A$ are [[Definition:Algebraically Complementary|algebraically complementary]], that is:
+:$\ker A \cap \Rng A = \left({0}\right)$, the [[Definition:Zero Subspace|zero subspace]]
+:$\ker A + \Rng A = H$, where $+$ signifies [[Definition:Setwise Addition|setwise addition]].
+{{explain|Determine whether range means image or codomain, and replace appropriately}}
+\end{theorem}<|endoftext|>
+\section{Natural Logarithm Function is Continuous}
+Tags: Continuous Real Functions, Natural Logarithms, Natural Logarithm Function is Continuous
+
+\begin{theorem}
+The [[Definition:Natural Logarithm|natural logarithm function]] is [[Definition:Continuous Real Function|continuous]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Natural Logarithm Function is Differentiable]].
+The result follows from [[Differentiable Function is Continuous]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Bounds of Natural Logarithm]]:
+:$\dfrac 1 2 < \map \ln 2 < 1$
+Fix $x \in \R$.
+Consider $\dfrac x {\map \ln 2}$.
+From [[Rationals are Everywhere Dense in Topological Space of Reals]]:
+:$\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \size {r - \dfrac x {\map \ln 2} }
+      | o = <
+      | r = \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = \map \ln 2 \size {r - \dfrac x {\map \ln 2} }
+      | r = \size {\map \ln {2^r} - x }
+      | c = [[Natural Logarithm of Rational Power]]
+}}
+{{eqn | o = < 
+      | r = \epsilon \, \map \ln 2
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | o = < 
+      | r = \epsilon
+      | c = as $\map \ln 2 < 1$
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map \ln t - x}
+      | o = <
+      | r = \epsilon
+      | c = substituting $t = 2^r$
+}}
+{{end-eqn}}
+Thus:
+:$\forall \epsilon \in \R_{>0}: \exists t \in \R_{>0}: \size {\map \ln t - x} < \epsilon$
+Thus, the [[Definition:Image of Mapping|image]] of $\R_{>0}$ under $\ln$ is [[Definition:Everywhere Dense|everywhere dense]] in $\R$.
+From [[Monotone Real Function with Everywhere Dense Image is Continuous]], $\ln$ is continuous on $\R_{>0}$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Projections}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H}\right)$ be an [[Definition:Idempotent Operator|idempotent operator]].
+Then the following are equivalent:
+:$(1): \qquad A$ is a [[Definition:Projection (Hilbert Spaces)|projection]]
+:$(2): \qquad A$ is the [[Definition:Orthogonal Projection|orthogonal projection]] onto $\operatorname{ran} A$
+:$(3): \qquad \left\Vert{A}\right\Vert = 1$, where $\left\Vert{\cdot}\right\Vert$ is the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear operators]].
+:$(4): \qquad A$ is [[Definition:Self-Adjoint Operator|self-adjoint]]
+:$(5): \qquad A$ is [[Definition:Normal Operator|normal]]
+:$(6): \qquad \forall h \in H: \left\langle{Ah, h}\right\rangle_H \ge 0$
+\end{theorem}<|endoftext|>
+\section{Set of Linear Subspaces is Closed under Intersection}
+Tags: Vector Subspaces
+
+\begin{theorem}
+Let $\struct {V, +, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $\family {M_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Vector Subspace|subspaces]] of $V$.
+Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a [[Definition:Vector Subspace|subspace]] of $V$.
+\end{theorem}
+
+\begin{proof}
+It needs to be demonstrated that $M$ is:
+:$(1): \quad$ a [[Definition:Closed Algebraic Structure|closed algebraic structure]] under $+$
+:$(2): \quad$ [[Definition:Closed for Scalar Product|closed for scalar product]] $\circ$.
+So let $a, b \in M$.
+By definition of [[Definition:Set Intersection|intersection]], $a, b \in M_i$ for all $i \in I$.
+As the $M_i$ are [[Definition:Vector Subspace|subspaces]] of $V$, $a + b \in M_i$ for all $i \in I$.
+That is, by definition of [[Definition:Set Intersection|intersection]], $a + b \in M$.
+It follows that $M$ is [[Definition:Closed Algebraic Structure|closed]] under $+$.
+Now let $\lambda \in K, a \in M$.
+By definition of [[Definition:Set Intersection|intersection]], $a \in M_i$ for all $i \in I$.
+As the $M_i$ are [[Definition:Vector Subspace|subspaces]] of $V$, $\lambda \circ a \in M_i$ for all $i \in I$.
+That is, by definition of [[Definition:Set Intersection|intersection]], $\lambda \circ a \in M$.
+It follows that $M$ is [[Definition:Closed for Scalar Product|closed]] under $\circ$.
+Hence the result, by definition of [[Definition:Vector Subspace|subspace]].
+{{qed}}
+[[Category:Vector Subspaces]]
+n61jxciytzxpoqcg6jss1ihrwb7gbgb
+\end{proof}<|endoftext|>
+\section{Closed Linear Subspaces Closed under Intersection}
+Tags: Vector Subspaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Topological Vector Space|topological vector space]].
+Let $\family {M_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Closed Linear Subspace|closed linear subspaces]] of $V$.
+Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a [[Definition:Closed Linear Subspace|closed linear subspace]] of $V$.
+\end{theorem}
+
+\begin{proof}
+By [[Set of Linear Subspaces is Closed under Intersection]], $M$ is a [[Definition:Vector Subspace|linear subspace]] of $V$.
+By [[Topology Defined by Closed Sets]], the [[Definition:Set Intersection|intersection]] of [[Definition:Closed Set (Topology)|closed sets]] is again [[Definition:Closed Set (Topology)|closed]].
+As the $M_i$ are all [[Definition:Closed Set (Topology)|closed]], it follows that $M$ is [[Definition:Closed Set (Topology)|closed]].
+Hence $M$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $V$.
+{{qed}}
+[[Category:Vector Subspaces]]
+4qr0useq9yg5nggqucqyd0u0twbljkt
+\end{proof}<|endoftext|>
+\section{Orthogonal Difference is Closed Linear Subspace}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $M, N$ be [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
+Then the [[Definition:Orthogonal Difference|orthogonal difference]] $M \ominus N$ is also a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+\end{theorem}
+
+\begin{proof}
+By definition, $M \ominus N = M \cap N^\perp$.
+By [[Orthocomplement is Closed Linear Subspace]], $N^\perp$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+Hence the result, by [[Closed Linear Subspaces Closed under Intersection]].
+{{qed}}
+[[Category:Hilbert Spaces]]
+2lj3gg359w9cwzc1kxa4wbfnzjwhsak
+\end{proof}<|endoftext|>
+\section{Closed Linear Subspaces Closed under Setwise Addition}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $M, N$ be [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
+Then $M + N$ is also a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$, where $+$ denotes [[Definition:Setwise Addition|setwise addition]].
+\end{theorem}
+
+\begin{proof}
+By [[Linear Subspaces Closed under Setwise Addition]], $M + N$ is a [[Definition:Vector Subspace|linear subspace]] of $H$.
+Now to show that it is [[Definition:Closed Set (Topology)|closed]].
+Let $P: H \to H$ denote the [[Definition:Orthogonal Projection|orthogonal projection]] on $M$.
+Denote by $I - P$ the [[Definition:Complementary Projection|complementary projection]] of $P$.
+Define $N' := \left\{{ n - Pn: n \in N}\right\}$.
+$N'$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+{{explain|prove the above claim}}
+Observe $m + n = \left({m + Pn}\right) + \left({n - Pn}\right) \in M + N'$; hence, $M + N \subseteq M + N'$. 
+By $m + \left({n - Pn}\right) = \left({m - Pn}\right) + n \in M + N$, conclude that $M + N' \subseteq M + N$, hence equality.
+Furthermore, $N' \subseteq \operatorname{ran} \left({I - P}\right) = \operatorname{ker} P$ by [[Range of Idempotent is Kernel of Complementary Idempotent]].
+That is, $N' \subseteq M^\perp$ by [[Properties of Orthogonal Projection]], and hence $M \perp N'$.
+Denote by $P'$ the [[Definition:Orthogonal Projection|orthogonal projection]] on $N'$.
+Suppose now that $h \in M + N'$. Then:
+{{begin-eqn}}
+{{eqn | l = h
+      | r = \left({P + \left({I - P}\right)}\right)h
+}}
+{{eqn | r = \left({P + \left({I - P}\right)}\right) \left({P' + \left({I - P'}\right)}\right)h
+      | c = $I$ is the [[Definition:Identity Operator|identity operator]]
+}}
+{{eqn | r = PP'h + \left({I - P}\right)P'h + P\left({I - P'}\right)h \left({I - P}\right)\left({I - P'}\right)h
+}}
+{{eqn | r = 0 + P'h + Ph + 0
+}}
+{{end-eqn}}
+For this last equality, observe $M \perp N'$, hence $M \subseteq N'^\perp$, $N' \subseteq M^\perp$ and $\left({M + N'}\right)^\perp \subseteq M^\perp \cup N'^\perp$.
+{{finish|link to results establishing the implications of these relations for the associated projections. These results don't exist at the moment}}
+The conclusion is that every $h \in M + N'$ can be uniquely decomposed as $P'h + Ph$, with $P'h \in N', Ph \in M$. 
+So suppose there is a sequence $h_n \to h$ in $M + N'$. Then $Ph_n$ and $P'h_n$ are sequences in $M, N'$, respectively.
+As those are [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$, there are $m \in M, n \in N'$ with $Ph_n \to m, P'h_n \to n$.
+It follows that $h = m + n \in M + N'$.
+That is, $M + N' = M + N$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+{{qed}}
+[[Category:Hilbert Spaces]]
+0gbg8zzxrz1hypc60x8yl9c087m1opm
+\end{proof}<|endoftext|>
+\section{Linear Subspaces Closed under Setwise Addition}
+Tags: Vector Subspaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $M, N$ be [[Definition:Vector Subspace|linear subspaces]] of $V$.
+Then $L := M + N$ is also a [[Definition:Vector Subspace|linear subspace]] of $V$, where $+$ denotes [[Definition:Setwise Addition|setwise addition]].
+\end{theorem}
+
+\begin{proof}
+It needs to be demonstrated that $L$ is [[Definition:Closed Algebraic Structure|closed]] under $+$ and $\circ$.
+So let $m_1 + n_1, m_2 + n_2 \in L$.
+Then $\paren {m_1 + n_1} + \paren {m_2 + n_2} = \paren {m_1 + m_2} + \paren {n_1 + n_2} \in L$.
+It follows that $L$ is [[Definition:Closed Algebraic Structure|closed]] under $+$.
+Now let $\lambda \in K, m + n \in L$.
+Then $\lambda \circ \paren {m + n} = \paren {\lambda \circ m} + \paren {\lambda \circ n} \in L$.
+It follows that $L$ is [[Definition:Closed for Scalar Product |closed]] under $\circ$.
+Hence the result, by definition of [[Definition:Vector Subspace|linear subspace]].
+{{qed}}
+[[Category:Vector Subspaces]]
+fkp9zb9z6qnfaekdo3j3vmer3oxbvlp
+\end{proof}<|endoftext|>
+\section{Complementary Projection is Projection}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A$ be a [[Definition:Projection (Hilbert Spaces)|projection]].
+Then the [[Definition:Complementary Projection|complementary projection]] $I - A$ is also a [[Definition:Projection (Hilbert Spaces)|projection]].
+\end{theorem}
+
+\begin{proof}
+By [[Characterization of Projections]], $A$ is [[Definition:Self-Adjoint Operator|self-adjoint]].
+Then $\left({I - A}\right)^* = I^* - A^* = I - A$ from [[Adjoining is Linear]].
+So $I - A$ is also [[Definition:Self-Adjoint Operator|self-adjoint]].
+From [[Complementary Idempotent is Idempotent]], $I - A$ is [[Definition:Idempotent Operator|idempotent]].
+Hence, applying [[Characterization of Projections]], $I - A$ is a [[Definition:Projection (Hilbert Spaces)|projection]].
+{{qed}}
+[[Category:Linear Transformations on Hilbert Spaces]]
+0o5qci51vwo2q9gzp82zog9zl95dky6
+\end{proof}<|endoftext|>
+\section{Direct Sum of Subspace and Orthocomplement}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $M$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
+Denote by $M^\perp$ its [[Definition:Orthocomplement|orthocomplement]].
+Then the [[Definition:Hilbert Space Direct Sum|direct sum]] $M \oplus M^\perp$ is [[Definition:Isomorphism (Hilbert Spaces)|isomorphic]] to $H$.
+\end{theorem}
+
+\begin{proof}
+Assert that $U: M \oplus M^\perp \to H: \left({m, m^\perp}\right) \mapsto m + m^\perp$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]].
+According to the definition of [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]], it is sufficient to check that $U$ is [[Definition:Surjection|surjective]] and that:
+:$\left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H = \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}$
+First the [[Definition:Surjection|surjectivity]]:
+Denote by $P$ the [[Definition:Orthogonal Projection|orthogonal projection]] on $M$.
+Then for any $h \in H$, have $h = Ph + \left({h - Ph}\right)$.
+By [[Definition:Orthogonal Projection|definition]] of $P$, $Ph \in M$.
+Furthermore, by [[Properties of Orthogonal Projection]], $h - Ph \perp M$; that is, $h - Ph \in M^\perp$.
+It follows that $h = U \left({Ph, h - Ph}\right)$, showing that $U$ is a [[Definition:Surjection|surjection]].
+It remains to check that $U$ preserves the [[Definition:Inner Product|inner product]]:
+{{begin-eqn}}
+{{eqn|l = \left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H
+     |r = \left\langle{ m + m^\perp, n + n^\perp}\right\rangle_H
+}}
+{{eqn|r = \left\langle{m, n}\right\rangle_H + \left\langle{m^\perp, n}\right\rangle_H + \left\langle{m, n^\perp}\right\rangle_H + \left\langle{m^\perp, n^\perp}\right\rangle_H
+     |c = [[Definition:Inner Product|Linearity]] of $\left\langle{\cdot, \cdot}\right\rangle_H$
+}}
+{{eqn|r = \left\langle{m, n}\right\rangle_H + \left\langle{m^\perp, n^\perp}\right\rangle_H
+     |c = Defining property of [[Definition:Orthocomplement|orthocomplement]]
+}}
+{{eqn|r = \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}
+     |c = [[Definition:Hilbert Space Direct Sum|Definition]] of $\left\langle{\cdot, \cdot}\right\rangle_{M \oplus M^\perp}$
+}}
+{{end-eqn}}
+Hence $U$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]], and so $M \oplus M^\perp$ is [[Definition:Isomorphism (Hilbert Spaces)|isomorphic]] to $H$.
+{{qed}}
+[[Category:Hilbert Spaces]]
+5sfhxq1m9kc6yz967uacbnschftll3f
+\end{proof}<|endoftext|>
+\section{Characterization of Invariant Subspaces}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H}\right)$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Let $\begin{pmatrix}
+ W & X \\
+ Y & Z
+\end{pmatrix}$ be the [[Definition:Matrix Notation (Bounded Linear Operator)|matrix notation]] for $A$ with respect to $M$.
+Let $M$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$; denote by $P$ the [[Definition:Orthogonal Projection|orthogonal projection]] on $M$.
+Then the following three statements are equivalent:
+:$(1): \qquad M$ is an [[Definition:Invariant Subspace|invariant subspace]] for $A$
+:$(2): \qquad PAP = AP$
+:$(3): \qquad Y = 0$
+\end{theorem}<|endoftext|>
+\section{Characterization of Reducing Subspaces}
+Tags: Hilbert Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in \map B \HH$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Let $M$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $\HH$.
+Let $P$ denote the [[Definition:Orthogonal Projection|orthogonal projection]] on $M$.
+Let $\begin{pmatrix}
+ W & X \\
+ Y & Z
+\end{pmatrix}$ be the [[Definition:Matrix Notation (Bounded Linear Operator)|matrix notation]] for $A$ with respect to $M$.
+Then the following four statements are equivalent:
+:$(1): \quad M$ is a [[Definition:Reducing Subspace|reducing subspace]] for $A$
+:$(2): \quad P A = A P$
+:$(3): \quad X = Y = 0$
+:$(4): \quad M$ is an [[Definition:Invariant Subspace|invariant subspace]] for both $A$ and its [[Definition:Adjoint Linear Transformation|adjoint]] $A^*$
+\end{theorem}<|endoftext|>
+\section{Subset of Toset is Toset}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $T \subseteq S$.
+Then $\left({T, \preceq \restriction_T}\right)$ is also a [[Definition:Totally Ordered Set|totally ordered set]].
+In the above, $\preceq \restriction_T$ denotes the [[Definition:Restriction of Relation|restriction of $\preceq$ to $T$]].
+\end{theorem}
+
+\begin{proof}
+As $\left({S, \preceq}\right)$ is a [[Definition:Totally Ordered Set|totally ordered set]], the relation $\preceq$ is a [[Definition:Total Ordering|total ordering]], and is by definition:
+* [[Definition:Reflexive Relation|reflexive]]
+* [[Definition:Antisymmetric Relation|antisymmetric]]
+* [[Definition:Transitive Relation|transitive]]
+* [[Definition:Connected Relation|connected]]
+From [[Properties of Restriction of Relation]], a [[Definition:Restriction of Relation|restriction]] of a [[Definition:Relation|relation]] which has all those properties inherits them all.
+Thus $\preceq \restriction_T$ is also:
+* [[Definition:Reflexive Relation|reflexive]]
+* [[Definition:Antisymmetric Relation|antisymmetric]]
+* [[Definition:Transitive Relation|transitive]]
+* [[Definition:Connected Relation|connected]]
+and so is also a [[Definition:Total Ordering|total ordering]].
+Hence the result, by definition of [[Definition:Totally Ordered Set|totally ordered set]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Bretschneider's Formula}
+Tags: Areas of Quadrilaterals
+
+\begin{theorem}
+Let $ABCD$ be a general [[Definition:Quadrilateral|quadrilateral]].
+Then the area $\mathcal A$ of $ABCD$ is given by:
+:$\mathcal A = \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - a b c d \cos^2 \left({\dfrac {\alpha + \gamma} 2}\right)}$
+where:
+: $a, b, c, d$ are the [[Definition:Length (Linear Measure)|lengths]] of the [[Definition:Side of Polygon|sides]] of the [[Definition:Quadrilateral|quadrilateral]]
+: $s = \dfrac {a + b + c + d} 2$ is the [[Definition:Semiperimeter|semiperimeter]]
+: $\alpha$ and $\gamma$ are [[Definition:Opposite Angles|opposite angles]].
+\end{theorem}
+
+\begin{proof}
+: [[File:Bretschneider's Formula.png|400px]]
+Let the area of $\triangle DAB$ and $\triangle BCD$ be $\mathcal A_1$ and $\mathcal A_2$.
+From [[Area of Triangle in Terms of Two Sides and Angle]]:
+:$\mathcal A_1 = \dfrac {a b \sin \alpha} 2$ and $\mathcal A_2 = \dfrac {c d \sin \gamma} 2$
+From to the [[Axiom:Area Axioms|second axiom of area]], $\mathcal A = \mathcal A_1 + \mathcal A_2$, so:
+{{begin-eqn}}
+{{eqn | l = \mathcal A^2
+      | r = \frac 1 4 \left({a^2 b^2 \sin^2 \alpha + 2 a b c d \sin \alpha \sin \gamma + c^2 d^2 \sin^2 \gamma}\right)
+      | n = 1
+}}
+{{end-eqn}}
+The diagonal $p$ can be written in 2 ways using the [[Law of Cosines]]:
+:$p^2 = a^2 + b^2 - 2 a b \cos \alpha$
+:$p^2 = c^2 + d^2 - 2 c d \cos \gamma$
+[[Equality is Transitive|Equality is transitive]], so:
+{{begin-eqn}}
+{{eqn | l = a^2 + b^2 - 2 a b \cos \alpha
+      | r = c^2 + d^2 - 2 c d \cos \gamma
+}}
+{{eqn | l = a^2 + b^2 - c^2 - d^2
+      | r = 2 a b \cos \alpha - 2 c d \cos \gamma
+      | c = adding $2 a b \cos \alpha - c^2 - d^2$ to both sides
+}}
+{{eqn | l = \left({a^2 + b^2 - c^2 - d^2}\right)^2
+      | r = 4 a^2 b^2 \cos^2 \alpha - 8 a b c d \cos \alpha \cos \gamma + 4 c^2 d^2 \cos^2 \gamma
+      | c = squaring both sides
+}}
+{{eqn | l = 0
+      | r = \frac 1 4 \left({a^2 b^2 \cos^2 \alpha - 2 a b c d \cos \alpha \cos \gamma + c^2 d^2 \cos^2 \gamma}\right)
+      | c = algebraic manipulation
+}}
+{{eqn | ro= -
+      | r = \frac 1 {16} \left({a^2 + b^2 - c^2 - d^2}\right)^2
+      | o = 
+}}
+{{end-eqn}}
+Now add this equation to $(1)$. Then trigonometric identities can be used, as follows:
+{{begin-eqn}}
+{{eqn | l = \mathcal A^2
+      | r = \frac 1 4 \left({a^2 b^2 + c^2 d^2 - 2 a b c d \cos \left({\alpha + \gamma}\right)}\right) - \frac 1 {16} \left({a^2 + b^2 - c^2 - d^2}\right)^2
+      | c = [[Sum of Squares of Sine and Cosine]] and [[Cosine of Sum]]
+}}
+{{eqn | r = \frac 1 {16} \left({4 a^2 b^2 + 4 c^2 d^2 - \left({a^2 + b^2 - c^2 - d^2}\right)^2}\right) - \frac 1 2 a b c d \cdot \cos \left({\alpha + \gamma}\right)
+}}
+{{end-eqn}}
+By expanding the square $\left({a^2 + b^2 - c^2 - d^2}\right)^2$: 
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = \mathcal A^2
+      | r = \frac 1 {16} \left({-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2}\right) - \frac 1 2 abcd \cos \left({\alpha + \gamma}\right)
+}}
+{{end-eqn}}
+Adding and subtracting $8 a b c d$ to and from the [[Definition:Numerator|numerator]] of the first term of $(2)$:
+{{begin-eqn}}
+{{eqn | l = \mathcal A^2
+      | r = \frac 1 {16} \left({-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2 + 8 a b c d - 8 a b c d}\right) - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)
+}}
+{{end-eqn}}
+allows the product $\left({-a + b + c + d}\right) \left({a - b + c + d}\right) \left({a + b - c + d}\right) \left({a + b + c - d}\right)$ to be formed:
+{{begin-eqn}}
+{{eqn | l = \mathcal A^2
+      | r = \frac 1 {16} \left({-a + b + c + d}\right) \left({a - b + c + d}\right) \left({a + b - c + d}\right) \left({a + b + c - d}\right)
+      | c = 
+}}
+{{eqn | ro= -
+      | r = \frac 1 2 a b c d - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)
+      | o = 
+}}
+{{eqn | r = \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - \frac 1 2 a b c d - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)
+      | c = as $s = \dfrac {a + b + c + d} 2$
+}}
+{{eqn | r = \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - \frac 1 2 a b c d \left({1 + \cos \left({\alpha + \gamma}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - a b c d \cos^2 \left({\dfrac {\alpha + \gamma} 2} \right)
+      | c = [[Half Angle Formula for Cosine]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transfinite Induction/Principle 1}
+Tags: Ordinals
+
+\begin{theorem}
+Let $\On$ denote the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal|ordinals]].
+Let $A$ denote a [[Definition:Class (Class Theory)|class]].
+Suppose that:
+:For all elements $x$ of $\On$, if $x$ is a [[Definition:Subset|subset]] of $A$, then $x$ is an [[Definition:Element|element]] of $A$.
+Then $\On \subseteq A$.
+\end{theorem}
+
+\begin{proof}
+{{NotZFC}}
+{{AimForCont}} that $\neg \On \subseteq A$.
+Then:
+:$\paren {\On \setminus A} \ne \O$
+From [[Set Difference is Subset]], $\On \setminus A$ is a [[Definition:Subclass|subclass]] of the [[Definition:Ordinal|ordinals]].
+By [[Epsilon Relation is Strongly Well-Founded on Ordinal Class]], $\On \setminus A$ must have a [[Definition:Minimal Element under Relation|$\in$-minimal element]] $y$.
+By [[Element of Ordinal is Ordinal]], $y$ must be a [[Definition:Subset|subset]] of $\On$, the [[Definition:Class (Class Theory)|class]] of all [[Definition:Ordinal|ordinals]].
+However, from the fact that $y$ is a [[Definition:Minimal Element under Relation|$\in$-minimal element]] of $\On \setminus A$:
+:$\paren {\On \setminus A} \cap y = \O$
+So by its subsethood of $\On$:
+:$\paren {\On \cap y} \setminus A = \paren {y \setminus A} = \O$
+Therefore $y \subseteq A$.
+However, by the hypothesis, $y$ must also be an element of $A$.
+This contradicts the fact that $y$ is an element of $\On \setminus A$.
+Therefore $\On \subseteq A$.
+{{qed}}
+{{LEM|Reductio ad Absurdum|4}}
+\end{proof}<|endoftext|>
+\section{Limit of Functions that Agree}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]].
+Let $f$ and $g$ [[Definition:Agreement of Mappings|agree]] for all $x$ in a [[Definition:Deleted Neighborhood (Real Analysis)|deleted neighborhood]] of $c$.
+Let the [[Definition:Limit of Real Function|limit]]:
+:$\displaystyle \lim_{x \to c} \ f \left({x}\right)$
+exist.
+Then the [[Definition:Limit of Real Function|limit]]:
+:$\displaystyle \lim_{x \to c} \ g \left({x}\right)$
+also exists, and:
+:$\displaystyle \lim_{x \to c} \ f \left({x}\right) =  \lim_{x \to c} \ g \left({x}\right)$
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]]:
+:$f \left({x}\right) = g \left({x}\right)$
+for all $x$ such that:
+:$x \in \R: 0 < \left \vert{\alpha - x}\right \vert < \epsilon$
+Putting $c$ for $\alpha$ and $\delta$ for $\epsilon$, this is equivalent to:
+:$x \in \R: 0 < \left \vert{x - c}\right \vert < \delta$
+By definition, that the [[Definition:Limit of Real Function|limit]] of $f \left({x}\right)$ exists is to say:
+:$\exists L: \forall \varepsilon > 0: \exists \delta > 0: 0 < \left \vert{x - c}\right \vert < \delta \implies \left \vert{f \left({x}\right) - L}\right \vert < \varepsilon$
+For all $x$ relevant, we can substitute $f \left({x}\right)$ with $g \left({x}\right)$ and get:
+:$\exists L: \forall \varepsilon > 0: \exists \delta > 0: 0 < \left \vert{x - c}\right \vert < \delta \implies \left \vert{g \left({x}\right) - L}\right \vert < \varepsilon$
+The result follows by the definition of [[Definition:Limit of Real Function|limit]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering of Reciprocals}
+Tags: Real Numbers, Reciprocals, Inequalities, Ordering of Reciprocals
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]] such that $x, y \in \openint 0 \to$ or $x, y \in \openint \gets 0$
+Then:
+:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$
+\end{theorem}
+
+\begin{proof}
+By [[Reciprocal Function is Strictly Decreasing]], the reciprocal function is [[Definition:Strictly Decreasing Mapping|strictly decreasing]].
+By [[Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing]], the reciprocal function is a [[Definition:Dual Order Embedding|dual order embedding]].
+That is:
+:$x \le y \iff \dfrac 1 y \le \dfrac 1 x$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[Reciprocal Function is Strictly Decreasing]], the reciprocal function is [[Definition:Strictly Decreasing Mapping|strictly decreasing]].
+Thus 
+:$x \le y \implies \dfrac 1 y \le \dfrac 1 x$
+Suppose then that $\dfrac 1 y \le \dfrac 1 x$.
+If $x, y > 0$, then from [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]:
+$\dfrac 1 y, \dfrac 1 x > 0$
+Similarly, if $x, y < 0$ or $x, y > 0$, then from [[Reciprocal of Strictly Negative Real Number is Strictly Negative]]:
+:$\dfrac 1 y, \dfrac 1 x < 0$
+Thus we can apply the above to show that:
+:$\dfrac 1 {1/x} \le \dfrac 1 {1/y}$
+By [[Inverse of Multiplicative Inverse]]:
+:$x \le y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Comparison Test for Divergence}
+Tags: Convergence Tests, Named Theorems, Divergent Series
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be a [[Definition:Divergent Series|divergent series]] of [[Definition:Positive Real Number|positive real numbers]].
+Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence in $\R$]].
+Let:
+:$\forall n \in \N_{>0}: b_n \le a_n$
+Then the [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Divergent Series|diverges]].
+\end{theorem}
+
+\begin{proof}
+This is the [[Definition:Contrapositive Statement|contrapositive]] of the [[Comparison Test]]. 
+Hence the result, from the [[Rule of Transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Axiom of Choice Implies Zorn's Lemma}
+Tags: Set Theory, Axiom of Choice, Axiom of Choice Implies Zorn's Lemma
+
+\begin{theorem}
+Acceptance of the [[Axiom:Axiom of Choice|Axiom of Choice]] implies the truth of [[Zorn's Lemma]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.
+By the [[Axiom:Axiom of Choice|Axiom of Choice]], there is a [[Definition:Mapping|mapping]] $f: X \to X$ such that:
+:$\forall x \in X: x \prec \map f x$
+Let $\mathcal C$ be the [[Definition:Set|set]] of all [[Definition:Chain (Set Theory)|chains]] in $X$.
+By the premise, each element of $\mathcal C$ has an [[Definition:Upper Bound of Set|upper bound]] in $X$.
+Thus by the [[Axiom:Axiom of Choice|Axiom of Choice]], there is a [[Definition:Mapping|mapping]] $g: \mathcal C \to X$ such that for each $C \in \mathcal C$, $\map g C$ is an [[Definition:Upper Bound of Set|upper bound]] of $C$.
+Let $p$ be an arbitrary element of $X$.
+Define a [[Definition:Mapping|mapping]] $h: \operatorname {Ord} \to X$ by [[Transfinite Recursion|transfinite recursion]] thus:
+{{begin-eqn}}
+{{eqn | l = \map h 0
+      | r = p
+}}
+{{eqn | l = \map h {\alpha^+}
+      | r = \map f {\map h \alpha }
+}}
+{{eqn | l = \map h \lambda
+      | r = \map f {\map g {f \sqbrk \lambda} }
+      | c = if $\lambda$ is a [[Definition:Limit Ordinal|limit ordinal]]
+}}
+{{end-eqn}}
+{{explain|The notation $f \sqbrk \lambda$ needs to be defined.}}
+Then $h$ is [[Definition:Strictly Increasing|strictly increasing]], and thus [[Definition:Injection|injective]].
+{{explain|The above statement needs to be justified.}}
+Let $h'$ be the [[Definition:Restriction|restriction]] of $h$ to $\operatorname {On} \times \map h {\operatorname{On} }$.
+Then ${h'}^{-1}$ is a [[Definition:Surjection|surjection]] from $\map h {\operatorname{On} } \subseteq X$ onto $\operatorname{On}$.
+By the [[Axiom:Axiom of Replacement|Axiom of Replacement]], $\operatorname{On}$ is a [[Definition:Set|set]].
+By [[Burali-Forti Paradox]], this is a [[Definition:Contradiction|contradiction]].
+Thus we conclude that some element of $X$ has no [[Definition:Strict Successor|strict successor]], and is thus [[Definition:Maximal Element|maximal]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+For each $x \in X$, consider the [[Definition:Lower Closure of Element|lower closure]] $x^\preceq$:
+:$x^\preceq = \set {y \in X: y \preceq x}$
+Let $\mathbb S \subseteq \powerset X$ be the [[Definition:Image of Mapping|image]] of $\cdot^\preceq$ considered as a [[Definition:Mapping|mapping]] from $X$ to $\powerset X$, where $\powerset X$ is the [[Definition:Power Set|power set]] of $X$.
+From [[Ordering is Equivalent to Subset Relation]]:
+:$\forall x, y \in X: x^\preceq \subseteq y^\preceq \iff x \preceq y$
+Thus the task of finding a [[Definition:Maximal Element|maximal element]] of $X$ is [[Definition:Logical Equivalence|equivalent]] to finding a [[Definition:Maximal Set|maximal set]] in $\mathbb S$.
+Thus the statement of the result is equivalent to a statement about [[Definition:Chain (Set Theory)|chains]] in $\mathbb S$:
+:Let $\mathbb S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\powerset X, X \ne \O$ such that every [[Definition:Non-Empty Set|non-empty]] [[Definition:Chain (Set Theory)|chain]] in $\mathbb S$, ordered by $\subseteq$, has an [[Definition:Upper Bound of Set|upper bound]] in $\mathbb S$.
+:Then $\mathbb S$ has at least one [[Definition:Maximal Set|maximal set]].
+Let $\mathbb X$ be the [[Definition:Set|set]] of all [[Definition:Chain (Set Theory)|chains]] in $\struct {X, \preceq}$.
+Every element of $X$ is included in $\map {s^\preceq} x$ for some $x \in X$.
+$\mathbb X$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set of sets]] which are [[Definition:Ordering|ordered]] (perhaps [[Definition:Partial Ordering|partially]]) by [[Subset Relation is Ordering|subset relation]].
+If $\mathcal C$ is a [[Definition:Chain (Set Theory)|chain]] in $\mathbb X$, then:
+:$\displaystyle \bigcup_{A \mathop \in \mathcal C} A \in \mathbb X$
+Since each set in $\mathbb X$ is [[Definition:Dominate (Set Theory)|dominated]] by some set in $\mathbb S$, going from $\mathbb S$ to $\mathbb X$ can not introduce any new [[Definition:Maximal Element|maximal elements]].
+The main advantage of using $\mathbb X$ is that the chain hypothesis is in a slightly more specific form.
+Instead of saying that each chain in $\mathcal C$ has an [[Definition:Upper Bound of Set|upper bound]] in $\mathbb S$, we can explicitly state that the [[Definition:Set Union|union]] of the sets of $\mathcal C$ is an element of $\mathbb X$.
+This [[Definition:Set Union|union]] of the sets of $\mathcal C$ is clearly an [[Definition:Upper Bound of Set|upper bound]] of $\mathcal C$.
+Another advantage of $\mathbb X$ is that, from [[Subset of Toset is Toset]], it contains all the [[Definition:Subset|subsets]] of each of its [[Definition:Set|sets]].
+Thus we can enlarge non-maximal sets in $\mathbb X$ one element at a time.
+So, from now on, we need consider only this non-empty collection $\mathbb X$ of [[Definition:Subset|subsets]] of a non-empty set $X$.
+$\mathbb X$ is subject to two conditions:
+:$(1): \quad$ Every subset of each set in $\mathbb X$ is in $\mathbb X$.
+:$(2): \quad$ The union of each chain of sets in $\mathbb X$ is in $\mathbb X$.
+It follows from $(1)$ that $\O \in \mathbb X$.
+We need to show that there exists a [[Definition:Maximal Set|maximal set]] in $\mathbb X$.
+Let $f$ be a [[Definition:Choice Function|choice function]] for $X$:
+:$\forall A \in \powerset X \setminus \O: \map f A \in A$
+For each $A \in \mathbb X$, let $\hat A$ be defined as:
+:$\hat A := \set {x \in X: A \cup \set x \in \mathbb X}$
+That is, $\hat A$ consists of all the elements of $X$ which, when added to $A$, make a set which is also in $\mathbb X$.
+From its definition:
+:$\displaystyle \hat A = \bigcup_{x \mathop \in \hat A} \paren {A \cup \set x}$
+where each of $A \cup \set x$ are chains in $X$ and so elements of $\mathbb X$.
+Suppose there exists a [[Definition:Maximal Set|maximal set]] $M$ in $\mathbb X$.
+Then, by definition of [[Definition:Maximal Set|maximal]], there are no elements in $x \in X \setminus M$ which can be added to $M$ to make $M \cup \set x$ another element of $\mathbb X$.
+Thus it follows that $\hat M = M$.
+From [[Set Difference with Self is Empty Set]] it then follows that if $A$ is [[Definition:Maximal Set|maximal]], then $\hat A \setminus A = \O$.
+The [[Definition:Mapping|mapping]] $g: \mathbb X \to \mathbb X$ can now be defined as:
+:$\forall A \in \mathbb X: \map g A = \begin{cases}
+A \cup \set {\map f {\hat A \setminus A} } & : \hat A \setminus A \ne \O \\
+A & : \text{otherwise} \end{cases}$
+Thus what we now have to prove is that:
+:$\exists A \in \mathbb X: \map g A = A$
+Note that from the definition of $g$:
+:$\forall A \in \mathbb X: A \subseteq \map g A$
+Also note that $\map f {\hat A \setminus A}$ is a single element of $\hat A \setminus A$.
+Thus we obtain the ''crucial fact'' that $\map g A$ contains ''at most'' one more element than $A$.
+We (temporarily) define a '''tower''' as being a [[Definition:Subset|subset]] $\mathcal T$ of $\mathbb X$ such that:
+:$(1): \quad \O \in \mathcal T$
+:$(2): \quad A \in \mathcal T \implies \map g A \in \mathcal T$
+:$(3): \quad $ If $\mathcal C$ is a chain in $\mathcal T$, then $\displaystyle \bigcup_{A \mathop \in \mathcal C} A \in \mathcal T$
+There is of course at least one tower in $\mathbb X$, as $\mathbb X$ itself is one.
+It follows from its definition that the [[Definition:Set Intersection|intersection]] of a collection of towers is itself a tower.
+It follows in particular that if $\mathcal T_0$ is the [[Definition:Set Intersection|intersection]] of ''all'' towers in $\mathbb X$, then $\mathcal T_0$ is the ''smallest'' tower in $\mathbb X$.
+Next we demonstrate that $\mathcal T_0$ is a chain.
+We (temporarily) define a set $C \in \mathcal T_0$ as '''comparable''' if it is [[Definition:Comparable|comparable]] with every element of $\mathcal T_0$.
+That is, if $A \in \mathcal T_0$ then $C \subseteq A$ or $A \subseteq C$.
+To say that $\mathcal T_0$ is a chain means that '''all''' sets of $\mathcal T_0$ are '''comparable'''.
+There is at least one '''comparable''' set in $\mathcal T_0$, as $\O$ is one of them.
+So, suppose $C \in \mathcal T_0$ is '''comparable'''.
+Let $A \in \mathcal T_0$ such that $A \subsetneq C$.
+Consider $\map g A$.
+Because $C$ is '''comparable''', either $C \subsetneq \map g A$ or $\map g A \subseteq C$.
+In the former case $A$ is a [[Definition:Proper Subset|proper subset]] of a [[Definition:Proper Subset|proper subset]] of $\map g A$.
+This contradicts the fact that $\map g A \setminus A$ can be no more than a [[Definition:Singleton|singleton]].
+Thus if such an $A$ exists, we have that:
+:$(A): \quad \map g A \subseteq C$
+Now let $\mathcal U$ be the [[Definition:Set of Sets|set]] defined as:
+:$\mathcal U := \set {A \in \mathcal T_0: A \subseteq C \lor \map g C \subseteq A}$
+Let $\mathcal U'$ be the [[Definition:Set of Sets|set]] defined as:
+:$\mathcal U' := \set {A \in \mathcal T_0: A \subseteq \map g C \lor \map g C \subseteq A}$
+That is, $\mathcal U'$ is the set of all sets in $\mathcal T_0$ which are [[Definition:Comparable|comparable]] with $\map g C$.
+If $A \in \mathcal U$, then as $C \subseteq \map g C$, either $A \subseteq \map g C \lor \map g C \subseteq A$
+So $\mathcal U \subseteq \mathcal U'$.
+The aim now is to demonstrate that $\mathcal U$ is a '''tower'''.
+From [[Empty Set is Subset of All Sets]], $\O \subseteq C$.
+Hence condition $(1)$ is satisfied.
+Now let $A \in \mathcal U$.
+As $C$ is '''comparable''', there are three possibilities:
+:$(1'): \quad A \subsetneq C$
+Then from $(A)$ above, $\map g A \subseteq C$.
+Therefore $\map g A \in \mathcal U$.
+:$(2'): \quad A = C$
+Then $\map g A = \map g C$ and so $\map g C \subseteq \map g A$.
+Therefore $\map g A \in \mathcal U$.
+:$(3'): \quad \map g C \subseteq A$
+Then $\map g C \subseteq \map g A$
+Therefore $\map g A \in \mathcal U$.
+Hence condition $(2)$ is satisfied.
+From the definition of $\mathcal U$, it follows immediately that the [[Definition:Set Union|union]] of a [[Definition:Chain (Set Theory)|chain]] in $\mathcal U$ is also in $\mathcal U$.
+Hence condition $(3)$ is satisfied.
+The conclusion is that $\mathcal U$ is a '''tower''' such that $\mathcal U \subseteq \mathcal T_0$.
+But as $\mathcal T_0$ is the smallest '''tower''', $\mathcal T_0 \subseteq \mathcal U$.
+It follows that $\mathcal U = \mathcal T_0$.
+Consider some '''comparable''' set $C$, then.
+From that $C$ we can form $\mathcal U$, as above.
+But as $\mathcal U = \mathcal T_0$:
+:$A \in \mathcal T_0 \implies \paren {A \subseteq C \implies A \subseteq \map g C} \lor \map g C \subseteq A$
+and so $g \left({C}\right)$ is also '''comparable'''.
+We now know that:
+:$\O$ is '''comparable'''
+:the [[Definition:Mapping|mapping]] $g$ maps '''comparable''' sets to '''comparable''' sets.
+Since the [[Definition:Set Union|union]] of a [[Definition:Chain (Set Theory)|chain]] of '''comparable''' sets is itself '''comparable''', it follows that the '''comparable''' sets all form a '''tower''' $\mathcal T_C$.
+But by the nature of $\mathcal T_0$ it follows that $\mathcal T_0 \subseteq \mathcal T_C$.
+So the elements of $\mathcal T_0$ must all be '''comparable'''.
+Since $\mathcal T_0$ is a [[Definition:Chain (Set Theory)|chain]], the [[Definition:Set Union|union]] $M$ of all the sets in $\mathcal T_0$ is itself a set in $\mathcal T_0$.
+Since the [[Definition:Set Union|union]] includes all the sets of $\mathcal T_0$, it follows that $\map g M \subseteq M$.
+Since it is always the case that $M \subseteq \map g M$, it follows that $M = \map g M$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit at Infinity of Real Identity Function}
+Tags: Unbounded Mappings, Identity Mappings
+
+\begin{theorem}
+Let $I_\R: \R \to \R$ be the [[Definition:Identity Mapping|identity function]] on $\R$.
+Then:
+:$(1): \quad \displaystyle \lim_{x \mathop \to +\infty} \map {I_\R} x = +\infty$
+:$(2): \quad \displaystyle \lim_{x \mathop \to -\infty} \map {I_\R} x = -\infty$
+\end{theorem}
+
+\begin{proof}
+We have that the [[Derivative of Identity Function]] is $1$.
+Hence, by [[Derivative of Monotone Function]], $x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Now, by the definition of [[Definition:Infinite Limit at Infinity|infinite limit at infinity]], the first assertion is:
+:$\forall M \in \R_{>0}: \exists N \in \R_{>0}: x > N \implies \map f x > M$
+For every $M$, choose $N = M$.
+The second assertion is proved similarly.
+{{qed}}
+[[Category:Unbounded Mappings]]
+[[Category:Identity Mappings]]
+1jh9mhvnfumhmsqe3h8jtjpnt84altt
+\end{proof}<|endoftext|>
+\section{Push Theorem}
+Tags: Unbounded Mappings
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function|continuous]] on the [[Definition:Open Real Interval|open interval]] $\openint a \to$, $a \in \R$, such that:
+:$\displaystyle \lim_{x \mathop \to +\infty} \map f x = +\infty$
+Let $g$ be a [[Definition:Real Function|real function]] defined on some [[Definition:Open Real Interval|open interval]] $\openint b \to$ such that, for [[Definition:Sufficiently Large|sufficiently large]] $x$:
+:$\map g x > \map f x$
+Then:
+:$\displaystyle \lim_{x \mathop \to +\infty} \map g x = +\infty$
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle \lim_{x \mathop \to +\infty} \map f x = +\infty$.
+By the definition of [[Definition:Infinite Limit at Infinity|infinite limit at infinity]], this means:
+:$\forall M_1 \in \R_{>0}: \exists N_1 \in \R_{>0}: x > N_1 \implies \map f x > M_1$
+Now, the assertion that $\map g x \to +\infty$ is:
+:$\forall M_1 \in \R_{>0}: \exists N_2 \in \R_{>0}: x > N_2 \implies \map g x > M_1$
+By the premise that $\map g x > \map f x$ for [[Definition:Sufficiently Large|sufficiently large]] $x$, there is an $N$ such that:
+:$x > N \implies \map g x > \map f x$
+Now, given $M_1$, let $N_2$ be greater than both $N_1$ and $N$.
+Then since $N_2 > N$ and $N_2 > N_1$ respectively:
+:$\map g x > \map f x > M_1$
+whence:
+:$\displaystyle \lim_{x \mathop \to +\infty} \map g x = +\infty$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Orthocomplements is Orthocomplement of Closed Linear Span}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $\family {M_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
+Then:
+:$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$
+where:
+:$\perp$ denotes [[Definition:Orthocomplement|orthocomplementation]]
+:$\vee$ denotes [[Definition:Closed Linear Span|closed linear span]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Set Equality/Definition 2|set equality]], it suffices to prove the following two [[Definition:Subset|inclusions]]:
+:$\ds \bigcap_{i \mathop \in I} M_i^\perp \subseteq \paren {\vee \set {M_i : i \in I} }^\perp$
+:$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \ds \bigcap_{i \mathop \in I} M_i^\perp$
+=== $\ds \bigcap_{i \mathop \in I} M_i^\perp$ is contained in $\paren {\vee \set {M_i : i \in I} }^\perp$ ===
+By [[Orthocomplement is Closed Linear Subspace]] and [[Closed Linear Subspaces Closed under Intersection]], both spaces considered are [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
+By [[Orthocomplement Reverses Subset]], the required containment is equivalent to:
+:$\vee \set {M_i : i \in I} \subseteq \ds \paren {\bigcap_{i \mathop \in I} M_i^\perp }^\perp$
+For $h \in \ds \bigcap_{i \mathop \in I} M_i^\perp$, by definition one has:
+:$h \perp M_i$ for all $i \in I$
+That is:
+:$M_i \perp \ds \bigcap_{i \mathop \in I} M_i^\perp$
+This is equivalent to saying that:
+:$M_i \subseteq \ds \paren {\bigcap_{i \mathop \in I} M_i^\perp }^\perp$
+Definition $(2)$ of [[Definition:Closed Linear Span|closed linear span]] now grants the desired subset relation.
+{{qed|lemma}}
+=== $\paren {\vee \set {M_i : i \in I} }^\perp$ is contained in $\ds \bigcap_{i \mathop \in I} M_i^\perp$ ===
+By definition $(2)$ of [[Definition:Closed Linear Span|closed linear span]]:
+:$\forall i \in I: M_i \subseteq \vee \set {M_i^\perp : i \in I}$
+By [[Orthocomplement Reverses Subset]], it follows that:
+:$\forall i \ni I: \paren {\vee \set {M_i : i \in I} }^\perp \subseteq M_i^\perp$
+Therefore, by definition of [[Definition:Set Intersection|set intersection]]:
+:$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \displaystyle \bigcap_{i \mathop \in I} M_i^\perp$
+{{qed|lemma}}
+Thus we have established that:
+:$h \in \paren {\vee \set {M_i : i \in I} }^\perp \iff h \in \ds \bigcap_{i \mathop \in I} M_i^\perp$
+From the definition of [[Definition:Set Equality/Definition 1|set equality]], it follows that:
+:$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zorn's Lemma Implies Axiom of Choice}
+Tags: Set Theory, Axiom of Choice
+
+\begin{theorem}
+If [[Zorn's Lemma]] is true, then so must the [[Axiom:Axiom of Choice|Axiom of Choice]] be.
+\end{theorem}
+
+\begin{proof}
+Let $X$ be a [[Definition:Set|set]].
+Let $\mathcal F$ be the [[Definition:Set|set]] of [[Definition:Mapping|mappings]] defined as:
+:$f \in \mathcal F \iff \begin{cases}
+\operatorname{Dom} \left({f}\right) \subseteq \mathcal P \left({X}\right) & \ \\
+\operatorname{Im} \left({f}\right) \subseteq X & \ \\
+\forall A \in \operatorname{Dom} \left({f}\right): f \left({A}\right) \in A & \ \end{cases}$
+Let $\preceq$ be the [[Definition:Relation|relation]] defined on $\mathcal F$ as:
+:$\forall f_1, f_2 \in \mathcal F: f_1 \preceq f_2 \iff f_2$ is an [[Definition:Extension of Mapping|extension]] of $f_1$.
+Straightforwardly, $\preceq$ is a [[Definition:Partial Ordering|partial ordering]] on $\mathcal F$.
+Suppose [[Zorn's Lemma]] holds.
+Then there exists a [[Definition:Maximal Element|maximal element]] of $\mathcal F$.
+{{explain|Check the chain condition (and nonemptiness of $\mathcal F$)}}
+We then show that if $g$ is such a [[Definition:Maximal Element|maximal element]], then:
+:$\operatorname{Dom} \left({g}\right) = \mathcal P \left({X}\right) \setminus \varnothing$
+{{proof wanted}}
+\end{proof}<|endoftext|>
+\section{Orthocomplement Reverses Subset}
+Tags: Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $H$, and let $A \subseteq B$.
+Then $B^\perp \subseteq A^\perp$, where $\perp$ signifies [[Definition:Orthocomplement|orthocomplementation]].
+\end{theorem}
+
+\begin{proof}
+Suppose $h \in B^\perp$.
+That is, for all $b \in B$: $h \perp b$.
+As $A \subseteq B$, it follows that $h \perp a$ for all $a \in A$.
+Hence, by definition of [[Definition:Orthocomplement|orthocomplement]], $h \in A^\perp$.
+Therefore, $B^\perp \subseteq A^\perp$.
+{{qed}}
+[[Category:Hilbert Spaces]]
+ci6vkbc5ru0wk49jgokohq7rlk6l14t
+\end{proof}<|endoftext|>
+\section{Limit at Infinity of x^n}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $x \mapsto x^n$, $n \in \R$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Open Interval|continuous]] on the [[Definition:Open Real Interval|open interval]] $\openint 1 {+\infty}$.
+Let $n > 0$.
+Then $x^n \to +\infty$ as $x \to +\infty$.
+\end{theorem}
+
+\begin{proof}
+From [[Upper Bound of Natural Logarithm]]:
+:$\forall n > 0: n \ln x < x^n$
+which, by [[Multiple Rule for Continuous Functions]], implies:
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to +\infty} n \ln x
+      | r = n \lim_{x \mathop \to +\infty} \ln x
+}}
+{{end-eqn}}
+From [[Logarithm Tends to Infinity]]:
+:$n \ln x \to +\infty$ as $x \to +\infty$
+The result follows from the [[Push Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monotonicity of Real Sequences}
+Tags: Sequences, Real Analysis
+
+\begin{theorem}
+Let $\sequence {a_n}: \mathbb D \to \R$ be a [[Definition:Real Sequence|real sequence]], where $\mathbb D$ is a [[Definition:Subset|subset]] of $\N$.
+Let $\Bbb X$ be a [[Definition:Real Interval|real interval]] such that $\Bbb D \subseteq \Bbb X$.
+Let $f: \Bbb X \to \R, x \mapsto \map f x$ be a [[Definition:Differentiable on Interval|differentiable]] [[Definition:Real Function|real function]].
+Suppose that for every $n \in \mathbb D$:
+:$\map f n = a_n$
+Then:
+:If $\forall x \in \Bbb X: D_x \map f x \ge 0$, $\sequence {a_n}$ is [[Definition:Increasing Sequence|increasing]]
+:If $\forall x \in \Bbb X: D_x \map f x >  0$, $\sequence {a_n}$ is [[Definition:Strictly Increasing Sequence|strictly increasing]]
+:If $\forall x \in \Bbb X: D_x \map f x \le 0$, $\sequence {a_n}$ is [[Definition:Decreasing Sequence|decreasing]]
+:If $\forall x \in \Bbb X: D_x \map f x < 0$, $\sequence {a_n}$ is [[Definition:Strictly Decreasing Sequence|strictly decreasing]]
+where $D_x$ denotes [[Definition:Differentiation|differentiation]] {{WRT|Differentiation}} $x$.
+\end{theorem}
+
+\begin{proof}
+Consider the case where $D_x \map f x \ge 0$
+Let $n \in \N$ be in the [[Definition:Domain of Mapping|domain]] of $\sequence {a_n}$.
+From [[Derivative of Monotone Function]], the sign of $D_x f$ is indicative of the [[Definition:Monotonicity|monotonicity]] of $f$.
+Because [[Differentiable Function is Continuous]] and [[Continuous Real Function is Darboux Integrable]], $D_x f$ is [[Definition:Darboux Integrable Function|integrable]].
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map f {n + 1} - \map f n
+      | r = \int_n^{n + 1} D_x \map f x \rd x \ge 0
+      | c = [[Fundamental Theorem of Calculus]]
+}}
+{{eqn | o = \ge
+      | r = 0
+      | c = [[Relative Sizes of Definite Integrals]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map f {n + 1}
+      | o = \ge
+      | r = \map f n
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall n \in \mathbb D:
+      | l = \map f n
+      | r = a_n
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall n \in \mathbb D:
+      | l = a_{n + 1}
+      | o = \ge
+      | r = a_n
+      | c = as $n \in \mathbb D$ was arbitrary
+}}
+{{end-eqn}}
+Hence the result, by the definition of [[Definition:Monotone Mapping|monotone]].
+The proofs of the other cases are similar.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hausdorff Maximal Principle is equivalent to Axiom of Choice}
+Tags: Order Theory, Axiom of Choice
+
+\begin{theorem}
+Every [[Definition:Ordered Set|ordered set]] has a [[Definition:Maximal Chain|maximal chain]] {{iff}} the [[Axiom:Axiom of Choice|axiom of choice]] holds.
+\end{theorem}
+
+\begin{proof}
+* [[Hausdorff Maximal Principle implies Well-Ordering Theorem]]
+* [[Well-Ordering Theorem is Equivalent to Axiom of Choice]]
+* [[Hausdorff Maximal Principle/Proof 3|Hausdorff Maximal Principle: Proof 3]], from whence the [[Axiom:Axiom of Choice|Axiom of Choice]] is seen to imply the [[Hausdorff Maximal Principle]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Young's Inequality for Products}
+Tags: Real Analysis, Inequalities, Young's Inequality for Products
+
+\begin{theorem}
+Let $p, q \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]] such that:
+: $\dfrac 1 p + \dfrac 1 q = 1$
+Then, for any $a, b \in \R_{\ge 0}$:
+: $a b \le \dfrac {a^p} p + \dfrac{b^q} q$
+Equality occurs {{iff}}:
+:$b = a^{p-1}$.
+\end{theorem}
+
+\begin{proof}
+The result follows directly if $a = 0$ or $b = 0$.
+{{WLOG}}, assume that $a > 0$ and $b > 0$. 
+Then:
+{{begin-eqn}}
+{{eqn | l = a b
+      | r = \map \exp {\map \ln {a b} }
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{eqn | r = \map \exp {\ln a + \ln b}
+      | c = [[Sum of Logarithms]]
+}}
+{{eqn | r = \map \exp {\frac 1 p p \ln a + \frac 1 q q \ln b}
+      | c = {{Defof|Multiplicative Identity}} and {{Defof|Multiplicative Inverse}}
+}}
+{{eqn | r = \map \exp {\frac 1 p \map \ln {a^p} + \frac 1 q \map \ln {b^q} }
+      | c = [[Logarithms of Powers]]
+}}
+{{eqn | o = \le
+      | r = \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {map \ln {b^q} }
+      | c = [[Exponential is Strictly Convex]] and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$
+}}
+{{eqn | r = \frac {a^p} p + \frac {b^q} q
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+[[File:Holder's Ineq.jpg|200px|thumb|right|The blue colored region corresponds to $\displaystyle \int_0^\alpha t^{p-1} \mathrm d t$ and the red colored region to $\displaystyle \int_0^\beta u^{q-1} \mathrm d u$.]]
+In order for $\dfrac 1 p + \dfrac 1 q = 1$ it is necessary for both $p > 1$ and $q > 1$.
+{{begin-eqn}}
+{{eqn | l = \frac 1 p + \frac 1 q
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = p + q
+      | r = p q
+      | c = multiplying both sides by $p q$
+}}
+{{eqn | ll= \implies
+      | l = p + q - p - q + 1
+      | r = p q - p - q + 1
+      | c = adding $1 - p - q$ to both sides
+}}
+{{eqn | ll= \implies
+      | l = 1
+      | r = \left({p - 1}\right) \left({q - 1}\right)
+      | c = elementary algebra
+}}
+{{eqn | ll= \implies
+      | l = \frac 1 {p - 1}
+      | r = q - 1
+      | c = 
+}}
+{{end-eqn}}
+Accordingly:
+: $u = t^{p-1} \iff t = u^{q-1}$
+Let $a, b$ be any [[Definition:Positive Real Number|positive real numbers]].
+Since $a b$ is the [[Area of Rectangle|area of the rectangle]] in the given figure, we have:
+:$\displaystyle a b \le \int_0^a t^{p-1} \ \mathrm d t + \int_0^b u^{q-1} \ \mathrm d u = \frac {a^p} p + \frac {b^q} q$
+Note that even if the [[Definition:Graph of Mapping|graph]] intersected the side of the rectangle corresponding to $t = a$, this inequality would hold.
+Also note that if either $a = 0$ or $b = 0$ then this inequality holds trivially.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Restriction of Monotone Function is Monotone}
+Tags: Analysis, Order Theory
+
+\begin{theorem}
+The [[Definition:Restriction of Mapping|restriction]] of a [[Definition:Monotone Mapping|monotone mapping]] is [[Definition:Monotone|monotone]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Restriction of Mapping|restriction]] does not introduce any new [[Definition:Mapping#Domain, Codomain, Image, Preimage|arguments]].
+Hence the result follows trivially from the definition of [[Definition:Monotone Mapping|monotone mapping]].
+{{qed}}
+{{handwaving}}
+\end{proof}<|endoftext|>
+\section{Taylor's Theorem/One Variable/Statement of Theorem}
+Tags: Taylor's Theorem
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is:
+:of [[Definition:Differentiability Class|differentiability class]] $C^n$ on the [[Definition:Closed Real Interval|closed interval]] $\closedint a x$
+and:
+:at least [[Definition:Higher Derivative|$n + 1$ times differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = \frac 1 {0!} \map f a
+      | c =
+}}
+{{eqn | o =
+      | ro= +
+      | r = \frac 1 {1!} \paren {x - a} \map {f'} a
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \frac 1 {2!} \paren {x - a}^2 \map {f''} a
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \cdots
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \frac 1 {n!} \paren {x - a}^n \map {f^{\paren n} } a
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = R_n
+      | c = 
+}}
+{{end-eqn}}
+where $R_n$ (sometimes denoted $E_n$) is known as the '''error term''' or '''remainder''', and can be presented in one of $2$ forms:
+;Lagrange Form
+:$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$
+for some $\xi \in \openint a x$.
+;Cauchy Form
+:$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi \paren {x - \xi}^n} {n!} \paren {x - a}$
+for some $\xi \in \openint a x$.
+\end{theorem}<|endoftext|>
+\section{Elements of Minimal Infinite Successor Set are Well-Ordered}
+Tags: Total Orderings, Well-Orderings
+
+\begin{theorem}
+Let $\omega$ be the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]].
+Let $a \in \omega$.
+Then $a$ is [[Definition:Well-Ordered Set|well-ordered]] by $\subseteq$.
+\end{theorem}<|endoftext|>
+\section{Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering}
+Tags: Lexicographic Order, Well-Orderings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] which is [[Definition:Well-Ordered Set|well-ordered]] by $\preceq$.
+Let $\preccurlyeq$ be the [[Definition:Lexicographic Order|lexicographic order]] on the set $T_n$ of all [[Definition:Ordered Tuple|ordered $n$-tuples]] of $S$:
+:$\left({x_1, x_2, \ldots, x_n}\right) \prec \left({y_1, y_2, \ldots, y_n}\right)$ {{iff}}:
+::$\exists k: 1 \le k \le n$ such that $\forall 1 \le j < k: x_j = y_j$ but $x_k \prec y_k$ in $S$.
+{{explain|Disentangle this.}}
+Then for a given $n \in \N_{>0}$, $\preccurlyeq$ is a [[Definition:Well-Ordering|well-ordering]] on $T_n$.
+\end{theorem}
+
+\begin{proof}
+Consider $T_n$ where $n \in \N_{>0}$.
+It is clear that $\left({T_1, \preccurlyeq}\right)$ is [[Definition:Order Isomorphism|order isomorphic]] to  $\left({S, \preceq}\right)$.
+Thus as $\preceq$ is a [[Definition:Well-Ordering|well-ordering]] on $S$, $\preccurlyeq$ is a [[Definition:Well-Ordering|well-ordering]] on $T_1$.
+Now, let us assume that $\preccurlyeq$ is a [[Definition:Well-Ordering|well-ordering]] on $T_k$ for some $k \in \N: k \ge 1$.
+Let $A$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $T_{k+1}$.
+Let $A_1$ be the set of all of the first components of the [[Definition:Ordered Tuple|ordered $n$-tuples]] that comprise $A$.
+Since $A_1$ is a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$, and $S$ is itself [[Definition:Well-Ordered Set|well-ordered]] by $\preceq$, it follows that $A_1$ contains a [[Definition:Minimal Element|minimal element]] $x$ by $\preccurlyeq$.
+Let $A_x$ be the subset of $A$ in which the first component equals $x$.
+We may consider $A_x$ to be a subset of $T_k$ where this first component $x$ has been suppressed.
+But we assumed that $T_k$ is [[Definition:Well-Ordering|well-ordered]] by $\preccurlyeq$.
+So $A_x$ contains a [[Definition:Minimal Element|minimal element]] $\left({x, x_2, x_3, \ldots, x_{k+1}}\right)$ by $\preccurlyeq$.
+This element $\left({x, x_2, x_3, \ldots, x_{k+1}}\right)$ is the [[Definition:Minimal Element|minimal element]] of $A$ by $\preccurlyeq$.
+Hence, by definition, $T_{k+1}$ is [[Definition:Well-Ordering|well-ordered]] by $\preccurlyeq$.
+The result follows by [[Principle of Mathematical Induction|induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Lexicographic Order on Well-Ordered Sets is not Well-Ordering}
+Tags: Lexicographic Order, Well-Orderings
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
+Let $\preccurlyeq$ be the [[Definition:Lexicographic Order|lexicographic order]] on the set $T$ of all [[Definition:Ordered Tuple|ordered tuples]] of $S$:
+:$\tuple {x_1, x_2, \ldots, x_m} \prec \tuple {y_1, y_2, \ldots, y_n}$ {{iff}}:
+::$\exists k: 1 \le k \le \map \min {m, n}$ such that $\forall 1 \le j < k: x_j = y_j$ but $x_k \prec y_k$ in $S$
+:or:
+::$m < n$ and $\forall 1 \le j < m: x_j = y_j$.
+Then $\preccurlyeq$ is '''not''' a [[Definition:Well-Ordering|well-ordering]] on $T$.
+\end{theorem}
+
+\begin{proof}
+Consider a set $S = \set {a, b}$ such that $a \prec b$.
+Then the set $\set {\tuple b, \tuple {a, b}, \tuple {a, a, b},  \tuple {a, a, a, b}, \ldots}$ has no [[Definition:Minimal Element|minimal element]] by $\preccurlyeq$.
+{{handwaving}}
+Thus $T$ is not [[Definition:Well-Ordering|well-ordered]] by $\preccurlyeq$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Infinite Successor Set is Well-Ordered}
+Tags: Total Orderings, Well-Orderings
+
+\begin{theorem}
+Let $\omega$ be the [[Definition:Minimal Infinite Successor Set|minimal infinite successor set]].
+Then $\omega$ is [[Definition:Well-Ordered Set|well-ordered]] by $\subseteq$.
+\end{theorem}<|endoftext|>
+\section{Transfinite Induction}
+Tags: Ordinals
+
+\begin{theorem}
+=== [[Transfinite Induction/Principle 1|Principle 1]] ===
+{{:Transfinite Induction/Principle 1}}
+=== [[Transfinite Induction/Schema 1|Schema 1]] ===
+{{:Transfinite Induction/Schema 1}}
+=== [[Transfinite Induction/Principle 2|Principle 2]] ===
+{{:Transfinite Induction/Principle 2}}
+=== [[Transfinite Induction/Schema 2|Schema 2]] ===
+{{:Transfinite Induction/Schema 2}}
+\end{theorem}<|endoftext|>
+\section{Transfinite Recursion/Theorem 1}
+Tags: Ordinals
+
+\begin{theorem}
+Let $G$ be a [[Definition:Mapping|mapping]].
+{{explain|What are the domain and range of $G$?}}
+Let $K$ be a [[Definition:Class (Class Theory)|class]] of [[Definition:Mapping|mappings]] $f$ that satisfy:
+:the [[Definition:Domain of Mapping|domain]] of $f$ is some [[Definition:Ordinal|ordinal]] $y$
+:$\forall x \in y: f \left({x}\right) = G \left({f {\restriction_x} }\right)$
+where $f {\restriction_x}$ denotes the [[Definition:Restriction of Mapping|restriction]] of $f$ to $x$.
+Let $F = \bigcup K$, the [[Definition:Union of Set of Sets|union]] of $K$.
+Then:
+:$(1): \quad F$ is a [[Definition:Mapping|mapping]] with [[Definition:Domain of Mapping|domain]] $\operatorname{On}$
+:$(2): \quad \forall x \in \operatorname{On}: F \left({x}\right) = G \left({F {\restriction_x} }\right)$
+:$(3): \quad F$ is [[Definition:Unique|unique]]. That is, if another mapping $A$ has the above two properties, then $A = F$.
+\end{theorem}
+
+\begin{proof}
+=== $K$ is a chain ===
+First, it is necessary to must show that $K$ is a [[Definition:Chain (Set Theory)|chain]] of [[Definition:Mapping|mappings]] under the [[Definition:Subset|subset]] [[Definition:Relation|relation]].
+Take any $f, g \in K$.
+Set $B$ equal to the [[Definition:Domain of Mapping|domain]] of $f$ and set $C$ equal to the [[Definition:Domain of Mapping|domain]] of $g$.
+From [[Relation between Two Ordinals]]:
+:$B \subseteq C \lor C \subseteq B$
+From [[Transfinite Recursion/Uniqueness of Transfinite Recursion|Lemma 1]]:
+: $\forall x \in B \cap C: f \left({x}\right) = g \left({x}\right)$
+So depending on whether $B \subseteq C$ or $C \subseteq B$:
+: $f \subseteq g \lor g \subseteq f$
+From [[Subset Relation is Ordering]], it follows that $\subseteq$ is a [[Definition:Total Ordering|total ordering]] on $K$.
+=== Domain of $F$ is an ordinal ===
+$K$ is a chain of [[Definition:Mapping|mappings]].
+Since the union of a chain of mappings is a mapping, $F$ is a mapping.
+Moreover, $\forall x \in B: f \left({x}\right) = F \left({x}\right)$, because $F$ is an [[Definition:Extension of Mapping|extension]] of the mapping $f$.
+The domain of $F$ is a subset of the [[Definition:Ordinal Class|ordinals]] because it is the union of a collection of mappings whose domains are themselves subsets of ordinals.
+{{begin-eqn}}
+{{eqn | l = a \in \operatorname{Dom} \left({F}\right) \land x \in a
+      | o = \implies
+      | r = \exists f: \exists \beta \in \operatorname{On}: \left({f: \beta \to \operatorname{Im} \left({f}\right) \land a \in \beta }\right)
+      | c = Definition of $F$ (above)
+}}
+{{eqn | o = \implies
+      | r = a \in \operatorname{On} \land x \in \beta
+      | c = [[Definition:Transitive Set|Transitivity]] of $\beta$ and $\operatorname{On}$
+}}
+{{eqn | o = \implies
+      | r = x \in \operatorname{Dom} \left({F}\right)
+      | c = 
+}}
+{{end-eqn}}
+From this it follows that $\operatorname{Dom} \left({F}\right)$ is an ordinal, since it is a transitive subset of $\operatorname{On}$.
+=== Domain of $F$ is $\operatorname{On}$ ===
+Assume the domain of $F$ is a set.
+Since $F$ is a mapping, the range of $F$ is a set and $F$ is a set.
+Then $\operatorname{Dom} F = \gamma$ for some ordinal $\gamma$.  Define $C$ as follows:
+:$C = F \cup \left\{{\left({\gamma, G \left({F {\restriction_\gamma}}\right)}\right)}\right\}$
+Then $C$ is a mapping with domain $\gamma^+$ and satisfies:
+: $\forall x < \gamma^+: C \left({x}\right) = G \left({F {\restriction_\gamma} }\right)$
+Therefore $C$ is a member of $K$ (note that $F {\restriction_\gamma} = C {\restriction_\gamma}$).
+:$\displaystyle C \in K \implies C \subseteq \bigcup K$
+So $C \subseteq F$.
+But if $F$ is a set, $F \subsetneq C$, a contradiction.
+Therefore, $\operatorname{Dom} \left({F}\right)$ cannot be a set and must therefore be the set of all ordinals $\operatorname{On}$.
+{{qed|lemma}}
+=== Value of $F$ is $G \left({F {\restriction_\alpha} }\right)$ ===
+:$\forall x \in \operatorname{On}: \exists f \in K: f \left({x}\right) = F \left({x}\right)$
+Since $f \in K$:
+: $f \left({x}\right) = G \left({f {\restriction_x} }\right)$
+But $\forall y < x: f \left({y}\right) = F \left({y}\right)$, so by [[Equality of Restrictions]]:
+: $f {\restriction_x} = F {\restriction_x}$
+Therefore,:
+:$F \left({\alpha}\right) = G \left({F {\restriction_\alpha} }\right)$
+{{qed|lemma}}
+=== $F$ is Unique ===
+Finally, assume there is another mapping $H$ that satisfies the first two properties.
+Then $H$ is equal to $F$ by [[Transfinite Induction/Schema 1|transfinite induction]]:
+{{begin-eqn}}
+{{eqn | l = \forall x \in y: H \left({x}\right) = F \left({x}\right)
+      | o = \implies
+      | r = \left({H \restriction y}\right) = \left({F \restriction y}\right)
+      | c = [[Equality of Restrictions]]
+}}
+{{eqn | o = \implies
+      | r = G \left({H \restriction y}\right) = G \left({F \restriction y}\right)
+      | c = by hypothesis
+}}
+{{eqn | o = \implies
+      | r = H \left({y}\right) = F \left({y}\right)
+      | c = by hypothesis
+}}
+{{end-eqn}}
+So for all ordinals:
+: $H \left({y}\right) = F \left({y}\right)$
+Thus $H = F$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Squeeze Theorem for Absolutely Convergent Series}
+Tags: Series
+
+\begin{theorem}
+Let $\displaystyle \sum \size {a_n}$ be an [[Definition:Absolutely Convergent Series|absolutely convergent series]] in $\R$.
+Suppose that:
+:$\displaystyle -\sum \size {a_n} = \sum \size {a_n}$
+Then $\displaystyle \sum a_n$ equals the above two series.
+\end{theorem}
+
+\begin{proof}
+From [[Absolutely Convergent Series is Convergent/Real Numbers|Absolutely Convergent Series is Convergent]], the [[Definition:Convergent Series|convergence]] of:
+{{refactor|We really want to be linking to [[Absolutely Convergent Series is Convergent/Real Numbers]] here, because the student who is working through basic undergraduate analysis won't know (or even care) what a Banach space is. Consequently, [[Absolutely Convergent Series is Convergent/Real Numbers]] can be given a proof which basically goes "the real numbers form a Banach space, the result follows from [[Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach]]" which gives us an in from both undergrad analysis and this more rarefied topological / functional-analytical level.}}
+:$\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}$
+implies that of:
+:$\displaystyle \sum_{n \mathop = 1}^\infty a_n$
+By [[Negative of Absolute Value]]:
+:$\displaystyle -\size {\sum_{n \mathop = 1}^j a_n} \le \sum_{n \mathop = 1}^j a_n \le \size {\sum_{n \mathop = 1}^j a_n}$
+By repeated application of [[Triangle Inequality]]:
+:$\displaystyle -\sum_{n \mathop = 1}^j \size {a_n} \le \sum_{n \mathop = 1}^j a_n \le \sum_{n \mathop = 1}^j \size {a_n}$
+[[Definition:By Hypothesis|By hypothesis]], the leftmost and rightmost terms converge as $j \to +\infty$.
+Hence the result, from [[Squeeze Theorem for Real Sequences]].
+{{qed}}
+[[Category:Series]]
+9jr4c5grr3ojim3o4cb1ep9tiqu2no3
+\end{proof}<|endoftext|>
+\section{N Choose k is not greater than n^k}
+Tags: Binomial Coefficients, N Choose k is not greater than n^k
+
+\begin{theorem}
+:$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$ 
+where $\dbinom n k$ is a [[Definition:Binomial Coefficient|binomial coefficient]].
+Equality holds when $k = 0$ and $k = 1$.
+\end{theorem}
+
+\begin{proof}
+First let $k > 1$.
+{{begin-eqn}}
+{{eqn | l = \binom n k
+      | r = \frac {n!} {k! \, \paren {n - k}!}
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{eqn | o = <
+      | r = \frac {n!} {\paren {n - k}!}
+      | c = as $k! > 0$ by definition of [[Definition:Factorial|factorial]]
+}}
+{{eqn | r = \underbrace {n \paren {n - 1} \cdots (n - k + 1)}_{k \text { factors} }
+      | c = expanding the [[Definition:Factorial|factorials]]
+}}
+{{eqn | o = <
+      | r = n^k
+      | c = for all $n, k$ here considered
+}}
+{{end-eqn}}
+For $k > 1$, the product has at least two [[Definition:Factor (Algebra)|factors]].
+Hence at least one [[Definition:Factor (Algebra)|factor]] is strictly less than $n$.
+{{qed|lemma}}
+Let $k < 0$.
+By definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
+:$\dbinom n k = 0$
+while $n^k > 0$.
+Hence the result follows for $k < 0$.
+{{qed|lemma}}
+Finally let $0 \le k \le 1$.
+From [[Binomial Coefficient with Zero]]:
+:$\dbinom n 0 = 1 = n^0$
+From [[Binomial Coefficient with One]]:
+:$\dbinom n 1 = n = n^1$
+So equality holds when $k = 0$ or $k = 1$.
+{{qed|lemma}}
+Hence the result has been shown to hold for all $k \in \Z$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We dismiss the cases where $k < 0$ by observing that in such cases $\dbinom n k = 0$ while $n^k > 0$.
+Similarly we dismiss $k = 0$: we have $\dbinom n 0 = 1 = n^0$.
+Let:
+: $N = \set {1, \ldots, n}$
+: $K = \set {1, \ldots, k}$
+From [[Cardinality of Set of Strictly Increasing Mappings]], $\dbinom n k$ is the number of [[Definition:Strictly Increasing Mapping|strictly increasing mappings]] from $K$ to $N$.
+From [[Cardinality of Set of All Mappings]], $n^k$ is the number of all [[Definition:Mapping|mappings]] from $K$ to $N$.
+For $k = 1$ there is exactly 1 [[Definition:Mapping|mapping]] from $K$ to $N$ which is trivially [[Definition:Strictly Increasing Mapping|strictly increasing]].
+Otherwise we have that $K$ has more than one [[Definition:Element|element]], and so therefore does $n$.
+The [[Definition:Mapping|mapping]] $f: K \to N: \forall s \in K: \map f s = 1$ is clearly not a [[Definition:Strictly Increasing Mapping|strictly increasing mapping]].
+So not all [[Definition:Mapping|mappings]] from $K$ to $N$ are [[Definition:Strictly Increasing Mapping|strictly increasing]].
+Hence a [[Definition:Strict Inequality|strict inequality]] holds, and so $\dbinom n k < n^k$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolutely Convergent Series is Convergent/Real Numbers}
+Tags: Absolutely Convergent Series is Convergent
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be an [[Definition:Absolutely Convergent Real Series|absolutely convergent series in $\R$]].
+Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Convergent Series|convergent]].
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Absolutely Convergent Real Series|absolute convergence]], we have that $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}$ is [[Definition:Convergent Series|convergent]].
+From the [[Comparison Test]] we have that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]] {{iff}}:
+: $\forall n \in \N_{>0}: \size {a_n} \le b_n$
+where $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ is [[Definition:Convergent Series|convergent]].
+So substituting $\size {a_n}$ for $b_n$ in the above, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Natural Numbers/Zermelo Construction}
+Tags: Definitions: Natural Numbers
+
+\begin{theorem}
+The [[Definition:Natural Numbers|natural numbers]] $\N = \set {0, 1, 2, 3, \ldots}$ can be defined as a series of [[Definition:Subset|subsets]]:
+:$0 := \O = \set {}$
+:$1 := \set 0 = \set \O$
+:$2 := \set 1 = \set {\set \O}$
+:$3 := \set 2 = \set {\set {\set \O} }$
+::::$\vdots$
+Thus the [[Definition:Natural Number|natural number]] $n$ consists of $\O$ enclosed in $n$ pairs of [[Definition:Braces|braces]].
+\end{theorem}<|endoftext|>
+\section{Set is Element of its Power Set}
+Tags: Power Set
+
+\begin{theorem}
+A [[Definition:Set|set]] is an [[Definition:Element|element]] of its [[Definition:Power Set|power set]]:
+:$S \in \powerset S$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \forall S: S
+      | o = \subseteq
+      | r = S
+      | c = [[Set is Subset of Itself]]
+}}
+{{eqn | ll= \leadsto
+      | l = \forall S: S
+      | o = \in
+      | r = \powerset S
+      | c = {{Defof|Power Set}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set of Empty Set}
+Tags: Power Set, Empty Set
+
+\begin{theorem}
+The [[Definition:Power Set|power set]] of the [[Definition:Empty Set|empty set]] $\O$ is the set $\set \O$.
+\end{theorem}
+
+\begin{proof}
+From [[Empty Set is Element of Power Set]] and [[Set is Element of its Power Set]]:
+:$\O \in \powerset \O$
+From [[Empty Set is Subset of All Sets]]:
+:$S \subseteq \O \implies S = \O$
+That is:
+:$S \in \powerset \O \implies S = \O$
+Hence the only element of $\powerset \O$ is $\O$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetric Difference is Subset of Union}
+Tags: Symmetric Difference, Set Union
+
+\begin{theorem}
+The [[Definition:Symmetric Difference|symmetric difference]] of two [[Definition:Set|sets]] is a [[Definition:Subset|subset]] of their [[Definition:Set Union|union]]:
+:$S * T \subseteq S \cup T$
+\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 | o = \subseteq
+      | r = \paren {S \cup T}
+      | c = [[Set Difference is Subset]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Projections/Binary Case}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $P, Q$ be [[Definition:Projection (Hilbert Spaces)|projections]].
+Then $P + Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]] {{iff}} $\Rng P \perp \Rng Q$.
+\end{theorem}
+
+\begin{proof}
+=== [[Definition:Necessary Condition|Necessary Condition]] ===
+Suppose $P + Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]]. Then:
+{{begin-eqn}}
+{{eqn | l = P + Q
+      | r = \paren {P + Q}^2
+      | c = $P + Q$ is an [[Definition:Idempotent Operator|idempotent]]
+}}
+{{eqn | r = P^2 + P Q + Q P + Q^2
+}}
+{{eqn | r = P + P Q + Q P + Q
+      | c = $P$ and $Q$ are [[Definition:Idempotent Operator|idempotents]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = PQ + QP
+      | r = 0
+}}
+{{end-eqn}}
+Now suppose that $h \in \Rng Q$; say $h = Q q$ for $q \in H$.
+Then it follows that $Q h = Q Q q = Q q = h$ as $Q$ is [[Definition:Idempotent Operator|idempotent]].
+It follows that:
+:$0 = P Q h + Q P h = P h + Q P h$
+From [[Characterization of Projections]], statement $(6)$, $\innerprod {Q P h} {P h}_H \ge 0$.
+Next, observe:
+:$0 = \innerprod {P h + Q p h} {P h}_H = \innerprod {P h} {P h}_H + \innerprod {Q P h} {P h}_H$
+As the second term is [[Definition:Positive|non-negative]], the first is [[Definition:Negative|non-positive]]; it follows that $P h = \mathbf 0_H$ from the definition of the [[Definition:Inner Product|inner product]].
+Hence:
+:$h \in \Ker P = \paren {\Rng P}^\perp$
+It follows that $\Rng Q \perp \Rng P$, as asserted.
+{{qed|lemma}}
+=== [[Definition:Sufficient Condition|Sufficient Condition]] ===
+Suppose that $\Rng P \perp \Rng Q$.
+Then as $P, Q$ are [[Definition:Projection (Hilbert Spaces)|projections]], have:
+:$\Rng P \subseteq \Ker Q$
+:$\Rng Q \subseteq \Ker P$
+That is, for all $h \in H$:
+:$Q P h = P Q h = \mathbf 0_H$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \paren {P + Q}^2
+      | r = P^2 + P Q + Q P = Q^2
+}}
+{{eqn | r = P + Q
+      | c = $P$ and $Q$ are [[Definition:Idempotent Operator|idempotents]]
+}}
+{{end-eqn}}
+That is, $P + Q$ is an [[Definition:Idempotent Operator|idempotent]].
+Furthermore, by [[Adjoining is Linear]], have:
+:$\paren {P + Q}^* = P^* + Q^* = P + Q$
+where the latter follows from [[Characterization of Projections]], statement $(4)$.
+This same statement implies that $P + Q$ is also a [[Definition:Projection (Hilbert Spaces)|projection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Projections/General Case}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $\left({M_i}\right)_{i \in I}$ be an [[Definition:Indexed Set|$I$-indexed set]] of [[Definition:Closed Linear Subspace|closed linear subspaces]] of $H$.
+Let $M_i$ and $M_j$ be [[Definition:Orthogonal (Hilbert Space)#Sets|orthogonal]] whenever $i \ne j$.
+Denote, for each $i \in I$, by $P_i$ the [[Definition:Orthogonal Projection|orthogonal projection]] onto $M_i$.
+Denote by $P$ the [[Definition:Orthogonal Projection|orthogonal projection]] onto the [[Definition:Closed Linear Span|closed linear span]] $\vee \left\{{M_i: i \in I}\right\}$ of the $M_i$.
+Then for all $h \in H$, $\displaystyle \sum \left\{{P_i h: i \in I}\right\} = Ph$, where $\displaystyle \sum$ denotes a [[Definition:Generalized Sum|generalized sum]].
+\end{theorem}<|endoftext|>
+\section{Product of Projections}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $P, Q$ be [[Definition:Projection (Hilbert Spaces)|projections]].
+Then the following are equivalent:
+:$(1): \quad PQ$ is a [[Definition:Projection (Hilbert Spaces)|projection]]
+:$(2): \quad PQ = QP$
+:$(3): \quad P + Q - PQ$ is a [[Definition:Projection (Hilbert Spaces)|projection]]
+{{MissingLinks|Provide proper linking to the def of addition and multiplication of operators}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by first showing that $(1)$ is equivalent to $(2)$.
+Then, these are combined and shown equivalent to $(3)$.
+=== $(1)$ implies $(2)$ ===
+Suppose $PQ$ is a [[Definition:Projection (Hilbert Spaces)|projection]]. 
+Then by [[Characterization of Projections]], statement $(4)$, one has:
+:$PQ = \left({PQ}\right)^* = Q^* P^* = QP$
+where the penultimate equality follows from [[Adjoint of Composition]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Suppose that $PQ = QP$.
+Then $\left({PQ}\right)^2 = PQPQ = P^2 Q^2 = PQ$ as $P, Q$ are [[Definition:Projection (Hilbert Spaces)|projections]].
+Hence $PQ$ is an [[Definition:Idempotent Operator|idempotent]].
+Also, note that $\left({PQ}\right)^* = Q^* P^* = QP = PQ$.
+Hence, by [[Characterization of Projections]], statement $(4)$, $PQ$ is a [[Definition:Projection (Hilbert Spaces)|projection]].
+{{qed|lemma}}
+=== $(1), (2)$ imply $(3)$ ===
+The above establishes that assuming either of $(1)$ and $(2)$ yields both to hold.
+So assuming $(1)$, $P, Q$ and $PQ$ are all [[Definition:Projection (Hilbert Spaces)|projections]], and $PQ = QP$.
+Now compute:
+{{begin-eqn}}
+{{eqn | l = \left({P + Q - PQ}\right)^2
+      | r = P^2 + Q^2 + \left({PQ}\right)^2 + PQ + QP - PPQ - PQP - QPQ - PQQ
+}}
+{{eqn | r = P + Q - PQ + PQ + PQ - PQ - PQ - PQ - PQ
+      | c = $P, Q, PQ$ [[Definition:Projection (Hilbert Spaces)|projections]], $PQ = QP$
+}}
+{{eqn | r = P + Q - PQ
+}}
+{{end-eqn}}
+It follows that $P + Q - PQ$ is an [[Definition:Idempotent Operator|idempotent]].
+Observe from [[Adjoining is Linear]] and [[Adjoint of Composition]]:
+:$\left({P + Q - PQ}\right)^* = P^* + Q^* - Q^* P^* = P + Q - QP = P + Q - PQ$
+Now applying [[Characterization of Projections]], statement $(4)$, conclude that $P + Q - PQ$ is a [[Definition:Projection (Hilbert Spaces)|projection]].
+{{qed|lemma}}
+=== $(3)$ implies $(2)$ ===
+Let $P + Q - PQ$ be a [[Definition:Projection (Hilbert Spaces)|projection]].
+Then by [[Characterization of Projections]], statement $(4)$, compute:
+{{begin-eqn}}
+{{eqn | l = P + Q - PQ
+      | r = \left({P + Q - PQ}\right)^*
+}}
+{{eqn | r = P^* + Q^* - Q^* P^*
+      | c = [[Adjoining is Linear]], [[Adjoint of Composition]]
+}}
+{{eqn | r = P + Q - QP
+}}
+{{end-eqn}}
+Hence necessarily $PQ = QP$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Projections}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $P, Q$ be [[Definition:Projection (Hilbert Spaces)|projections]].
+Then the following are equivalent:
+:$(1): \quad P - Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]]
+:$(2): \quad P Q = Q$
+:$(3): \quad Q P = Q$
+{{MissingLinks|Provide proper linking to the def of addition of operators}}
+\end{theorem}
+
+\begin{proof}
+First it is shown that $(2)$ is equivalent to $(3)$.
+Then, equivalence to $(1)$ is shown.
+=== $(2)$ iff $(3)$ ===
+Suppose $P Q = Q$.
+Then $P = P + Q - P Q$, and by [[Product of Projections]], necessarily $P Q = Q P$.
+That is, $Q P = P Q = Q$.
+The converse is obtained by swapping the rôles of $P$ and $Q$ in [[Product of Projections]].
+{{qed|lemma}}
+=== $(2), (3)$ imply $(1)$ ===
+Suppose $P Q = Q P = Q$. Then as $P, Q$ are [[Definition:Projection (Hilbert Spaces)|projections]]:
+:$\paren {P - Q}^2 = P^2 + Q^2 - P Q - Q P = P - Q$
+That is, $P - Q$ is [[Definition:Idempotent Operator|idempotent]].
+From [[Adjoining is Linear]], $\paren {P - Q}^* = P^* - Q^* = P - Q$.
+An application of [[Characterization of Projections]], statement $(4)$ shows that $P - Q$ is a [[Definition:Projection (Hilbert Spaces)|projection]].
+{{qed|lemma}}
+=== $(1)$ implies $(2)$ ===
+Let $P - Q$ be a [[Definition:Projection (Hilbert Spaces)|projection]].
+Then by [[Characterization of Projections]], statement $(6)$, one has:
+:$\innerprod {P h} h_H - \innerprod {Q h}, h_H \ge 0$
+Applying this statement on $P, Q$ also, one obtains that:
+:$P h = 0 \implies Q h = 0$
+that is:
+:$\ker P \subseteq \ker Q$
+[[Orthocomplement Reverses Subset]] and that $P, Q$ are [[Definition:Projection (Hilbert Spaces)|projections]] combine to state $\Rng Q \subseteq \Rng P$.
+So for every $h \in H$, there is a $p \in H$ with $Q h = P p$.
+It follows that:
+:$P Q h = P P p = P p = Q h$
+Hence $PQ = Q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Countable Sets of Sets}
+Tags: Set Union, Countable Sets, Union of Countable Sets of Sets
+
+\begin{theorem}
+Let $\mathcal A$ and $\mathcal B$ be [[Definition:Countable|countable]] [[Definition:Set of Sets|sets of sets]].
+Then:
+:$\left\{{A \cup B: A \in \mathcal A, B \in \mathcal B}\right\}$
+is also [[Definition:Countable|countable]].
+\end{theorem}
+
+\begin{proof}
+Since $\AA$ is [[Definition:Countable|countable]], its contents can be arranged in a [[Definition:Sequence|sequence]]:
+:$\AA = \set {A_1, A_2, \ldots}$
+Let $B \in \BB$.
+Consider the [[Definition:Sequence|sequence]] of sets:
+:$\sequence {A_1 \cup B, A_2 \cup B, \ldots}$
+We may leave out any possible repetitions, and obtain a [[Definition:Countable|countable]] set:
+:$\set {A \cup B: A \in \AA}$
+for every $B \in \BB$.
+Thus as $B$ varies over all the elements of $\BB$, we obtain the [[Definition:Countable|countable]] [[Definition:Indexed Family of Sets|family]]:
+:$\sequence {A \cup B: A \in \AA}_{\paren {B \mathop \in \BB} }$
+From [[Countable Union of Countable Sets is Countable]], their [[Definition:Set Union|union]] is [[Definition:Countable|countable]].
+{{qed}}
+{{AoC|Countable Union of Countable Sets is Countable|proof}}
+\end{proof}
+
+\begin{proof}
+Both $\mathcal A$ and $\mathcal B$ are [[Definition:Countable Set|countable]].
+Enumerate their elements by the [[Definition:Sequence|sequences]] $\sequence {A_n}_{n \mathop \in \N}$ and $\sequence {B_n}_{n \mathop \in \N}$ respectively.
+When one of $\mathcal A$ and $\mathcal B$ would be [[Definition:Finite Set|finite]], achieve such [[Definition:Sequence|sequences]] by allowing elements to occur multiple times.
+Define $\mathcal C := \set {A \cup B: A \in \mathcal A, B \in \mathcal B}$.
+For each $C \in \mathcal C$, define the set $N_C := \set {\tuple {i, j} \in \N \times \N: A_i \cup B_j = C}$.
+The definition of $\mathcal C$ ensures that $N_C \ne \o$ for all $C \in \mathcal C$.
+Consider the [[Definition:Lexicographic Ordering|lexicographic ordering]] $\preceq$ on $\N \times \N$.
+Combining the [[Well-Ordering Principle]] with [[Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering]], $\preceq$ is a [[Definition:Well-Ordering|well-ordering]] on $\N \times \N$.
+As $N_C \subseteq \N \times \N$ is [[Definition:Empty Set|nonempty]], it thus has a [[Definition:Smallest Element|smallest element]]; call it $n_C$.
+Define now the map $\phi: \mathcal C \to \N \times \N$ by $\map \phi C = n_C$.
+[[Well-Ordering Minimal Elements are Unique]] ensures that $\phi$ is well-defined.
+Now suppose that $C_1, C_2 \in \mathcal C$ satisfy:
+:$\map \phi {C_1} = \map \phi {C_2}$
+that is, $n_{C_1} = n_{C_2}$, which in turn equal some $\tuple {i, j} \in \N \times \N$.
+As $\tuple {i, j} \in N_{C_1}$ and $\tuple {i, j} \in N_{C_2}$, $C_1 = A_i \cup B_j = C_2$.
+It follows that $\phi$ is an [[Definition:Injection|injection]].
+Finally, $\N \times \N$ is [[Definition:Countable|countable]] by [[Cartesian Product of Countable Sets is Countable]].
+Let $\psi: \N \times \N \to \N$ be an [[Definition:Injection|injection]].
+Then $\psi \circ \phi: \mathcal C \to \N$ is also an [[Definition:Injection|injection]] by [[Composite of Injections is Injection]].
+Hence $\mathcal C$ is [[Definition:Countable|countable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Since $\mathcal A$ and $\mathcal B$ are [[Definition:Countable|countable]], $\mathcal A \times \mathcal B$ is countable by [[Cartesian Product of Countable Sets is Countable]].
+Thus by [[Surjection from Natural Numbers iff Countable]] there is a [[Definition:Surjection|surjection]] $f: \N \to \mathcal A \times \mathcal B$.
+Let $\mathcal C = \left\{{A \cup B: A \in \mathcal A, B \in \mathcal B}\right\}$.
+Let $g: \mathcal A \times \mathcal B \to \mathcal C$ be defined by letting $g \left({A, B}\right) = A \cup B$.
+$g$ is a [[Definition:Surjection|surjection]]:
+Let $C \in \mathcal C$.
+Then there exist $A \in \mathcal A$ and $B \in \mathcal B$ such that $C = A \cup B$.
+By the definition of [[Definition:Cartesian Product|Cartesian product]]:
+: $\left({A, B}\right) \in \mathcal A \times \mathcal B$
+Then:
+: $g \left({A, B}\right) = A \cup B = C$
+It follows that $g$ is a [[Definition:Surjection|surjection]].
+By [[Composite of Surjections is Surjection]], $g \circ f: \N \to \mathcal C$ is surjective.
+Hence $\mathcal C$ is [[Definition:Countable|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Finite Subsets of Countable Set is Countable}
+Tags: Subsets, Countable Sets, Set of Finite Subsets of Countable Set is Countable
+
+\begin{theorem}
+Let $A$ be a [[Definition:Countable Set|countable set]].
+Then the [[Definition:Set of Sets|set]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$ is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Countable Set|countable set]], there is an [[Definition:Injection|injection]] $f: A \to \N$.
+Let $\le_f$ be the [[Definition:Ordering Induced by Injection|ordering induced by $f$ on $A$]].
+By [[Injection Induces Well-Ordering]], $\le_f$ is a [[Definition:Well-Ordering|well-ordering]].
+Let $A^{\text{fin}}$ be the [[Definition:Set of Sets|set]] of [[Definition:Finite Subset|finite subsets]] of $A$.
+For $n \in \N$, denote with $A^{\left({n}\right)}$ the [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $A$ that have precisely $n$ [[Definition:Element|elements]].
+For $A' \in A^{\left({n}\right)}$, we write $A' = \left\{{a'_1, a'_2, \ldots, a'_n}\right\}$ uniquely so that $i < j$ implies $a'_i <_f a'_j$.
+Then we define $f^{\left({n}\right)}: A^{\left({n}\right)} \to \N$ by:
+:$\displaystyle f^{\left({n}\right)} \left({A'}\right) := \prod_{i \mathop = 1}^n p_i^{f \left({a'_i}\right)}$
+where $p_i$ denotes the $i$th [[Definition:Prime Number|prime]].
+By the [[Fundamental Theorem of Arithmetic]], if $f^{\left({n}\right)} \left({A'}\right) = f^{\left({n}\right)} \left({A''}\right)$, then:
+:$\left({f \left({a'_1}\right), \dotsc, f \left({a'_n}\right)}\right) = \left({f \left({a''_1}\right), \dotsc, f \left({a''_n}\right)}\right)$
+As $f$ is [[Definition:Injection|injective]], also:
+:$\left({a'_1, \dotsc, a'_n}\right) = \left({a''_1, \dotsc, a''_n}\right)$
+and in particular $A' = A''$.
+Thus $f^{\left({n}\right)}: A^{\left({n}\right)} \to \N$ is an [[Definition:Injection|injection]].
+So $A^{\left({n}\right)}$ is [[Definition:Countable Set|countable]].
+Now define $f^{\text{fin}}: A^{\text{fin}} \to \N \times \N$ by:
+:$f^{\text{fin}} \left({A'}\right) := \left({n, f^{\left({n}\right)} \left({A'}\right)}\right)$
+where $n = \left\vert{A'}\right\vert$ is the [[Definition:Cardinality|cardinality]] of $A'$.
+Suppose that:
+:$f^{\text{fin}} \left({A'}\right) = f^{\text{fin}} \left({A''}\right)$
+Then by [[Equality of Ordered Pairs]]:
+:$n = \left\vert{A'}\right\vert = \left\vert{A''}\right\vert$
+and:
+:$f^{\left({n}\right)} \left({A'}\right) = f^{\left({n}\right)} \left({A''}\right)$
+Since $f^{\left({n}\right)}$ was already shown to be an [[Definition:Injection|injection]], it follows that:
+:$A' = A''$
+Thus $f^{\text{fin}}$ is also an [[Definition:Injection|injection]].
+By [[Cartesian Product of Natural Numbers with Itself is Countable]], $\N \times \N$ is [[Definition:Countable Set|countable]].
+The result follows from [[Domain of Injection to Countable Set is Countable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By the definition of a [[Definition:Countable Set|countable set]], there exists an [[Definition:Injection|injection]] $g: A \to \N$.
+Let $\mathcal F$ denote the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$.
+Let $f: \mathcal F \to \N$ be the [[Definition:Mapping|mapping]] defined by:
+:$\displaystyle \map f F = \prod_{k \mathop \in \map g F} p_{k + 1}$
+where $p_n$ denotes the $n$th [[Definition:Prime Number|prime number]].
+We define $\map f \O = 1$, the [[Definition:Vacuous Product|vacuous product]].
+Let $F, G \in \mathcal F$, and suppose that $\map f F = \map f G$.
+By [[Expression for Integer as Product of Primes is Unique]]:
+:$\map g F = \map g G$
+By [[Subset equals Preimage of Image iff Mapping is Injection]]:
+:$F = \map {g^{-1} } {\map g F} = \map {g^{-1} } {\map g G} = G$
+That is, $f$ is an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $A^{\paren n}$ be the set of [[Definition:Subset|subsets]] of $A$ with no more than $n$ [[Definition:Element|elements]].
+Thus:
+:$A^{\paren 0} = \set \O$
+:$A^{\paren 1} = A^{\paren 0} \cup \set {\set a: a \in A}$
+and $\forall n \ge 0$:
+:$A^{\paren {n + 1} } = \set {a^{\paren n} \cup a^{\paren 1}: a^{\paren n} \in A^{\paren n} \land a^{\paren 1} \in A^{\paren 1} }$
+Let us verify by [[Principle of Mathematical Induction|induction]] that each $A^{\paren n}$ is [[Definition:Countable Set|countable]].
+Let $A$ be [[Definition:Countable Set|countable]].
+:$A^{\paren 1}$ is [[Definition:Countable Set|countable]], as its [[Definition:Cardinality|cardinality]] is $1 + \card A$.
+Suppose $A^{\paren n}$ is [[Definition:Countable Set|countable]].
+Then by [[Union of Countable Sets of Sets]], so $A^{\paren {n + 1} }$ also [[Definition:Countable Set|countable]].
+By [[Principle of Mathematical Induction|induction]], each $A^{\paren n}$ is [[Definition:Countable Set|countable]].
+Denote with $A^f$ the set of [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$.
+It is apparent that every [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] is in some $A^{\paren n}$, and so:
+:$A^f = \displaystyle \bigcup_{n \mathop \in \N} A^{\paren n}$
+The result follows from [[Countable Union of Countable Sets is Countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cantor's Theorem (Strong Version)}
+Tags: Power Set, Cantor's Theorem
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\map {\PP^n} S$ be defined [[Principle of Recursive Definition|recursively]] by:
+:$\map {\PP^n} S = \begin{cases}
+S & : n = 0 \\
+\powerset {\map {\PP^{n - 1} } S} & : n > 0
+\end{cases}$
+where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
+Then $S$ is not [[Definition:Set Equivalence|equivalent]] to $\map {\PP^n} S$ for any $n > 0$.
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^n} S$.
+=== Basis for the Induction ===
+$\map P 1$ is [[Cantor's Theorem]].
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^k} S$.
+Then we need to show:
+:There is no [[Definition:Surjection|surjection]] from $S$ onto $\map {\PP^{k + 1} } S$.
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Suppose that $\map P k$ is true.
+{{AimForCont}} that $f: S \to \map {\PP^{k + 1} } S$ is a [[Definition:Surjection|surjection]].
+Define the [[Definition:Mapping|mapping]] $g: S \to \map {\PP^k} S$ as:
+:$\displaystyle \map g x = \bigcup \map f x$
+This is actually a mapping into $\map {\PP^k} S$, as follows:
+:$\map f x \in \map {\PP^{k + 1} } S = \powerset {\map {\mathcal P^k} S}$
+By the definition of [[Definition:Power Set|power set]]:
+:$\map f x \subseteq \map {\PP^k} S$
+Thus each element of $\map f x$ is a [[Definition:Subset|subset]] of $\map {\PP^{k - 1} } S$.
+Thus by [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset]]:
+:$\displaystyle \bigcup \map f x \subseteq \map {\PP^{k - 1} } S$
+Therefore:
+:$\displaystyle \bigcup \map f x \in \map {\PP^k} S$
+That is, $\displaystyle \map g x$ is a mapping into $\map {\PP^k} S$.
+Next we have that:
+:$\forall y \in \map {\PP^k} S: \set y \in \map {\PP^{n + 1} } S$
+But [[Definition:By Hypothesis|by hypothesis]] $f$ is [[Definition:Surjection|surjective]], and so:
+:$\exists x \in S: \map f x = \set y$
+Then:
+:$\displaystyle \map g x = \bigcup \set y = y$
+As this holds for all such $y$, $g$ is [[Definition:Surjection|surjective]].
+But this contradicts the [[Cantor's Theorem (Strong Version)/Proof 2#Induction Hypothesis|induction hypothesis]].
+Thus we conclude that the theorem holds for all $n$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We temporarily introduce the notation:
+:$a^n = \begin{cases}
+a : & n = 0 \\
+\set {a^{n - 1} } : & n > 0
+\end{cases}$
+where $a \in S$.
+Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$.
+Let there be a [[Definition:Bijection|bijection]] $f: S \to \QQ^n$ where $\QQ^n \subseteq \map {\PP^n} S$.
+Then define:
+:$\AA^{n - 1} = \set {s^{n - 1} \in \map {\PP^{n - 1} } S: s^{n - 1} \ne \map f s}$
+Each $\QQ^{n - 1} \in \QQ^n$ is the [[Definition:Image of Element under Mapping|image]] of some [[Definition:Element|element]] of $S$ under $f$.
+Let $\QQ^{n - 1} = \map f x$ for some $x \in S$.
+Let $x^{n - 1} \in \QQ^{n - 1} = \map f x$.
+Then $x^{n - 1} \notin \AA^{n - 1}$, and so $\AA^{n - 1} \notin \QQ^{n - 1}$.
+On the other hand, let $x^{n - 1} \notin \QQ^{n - 1} = \map f x$.
+Then $x^{n - 1} \in \AA^{n - 1}$, and so again $ A^{n - 1} \notin \QQ^{n - 1}$.
+Thus $\AA^{n - 1} \notin Q^n$.
+Therefore $Q^n$ is a [[Definition:Proper Subset|proper subset]] of $\map {\PP^n} S$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering Induced by Injection is Ordering}
+Tags: Order Theory, Injections
+
+\begin{theorem}
+Let $\left({T, \le}\right)$ be an [[Definition:Ordered Set|ordered set]], and let $S$ be a [[Definition:Set|set]].
+Let $f: S \to T$ be an [[Definition:Injection|injection]].
+Then $\le_f$, the [[Definition:Ordering Induced by Injection|ordering induced by $f$]], is an [[Definition:Ordering|ordering]].
+\end{theorem}
+
+\begin{proof}
+To establish that $\le_f$ is an [[Definition:Ordering|ordering]] on $S$, we need to show that it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Antisymmetric Relation|antisymmetric]] and [[Definition:Transitive Relation|transitive]].
+So, checking in turn each of the criteria for an [[Definition:Ordering|ordering]]:
+=== Reflexivity ===
+For all $s \in S$, one has $f \left({s}\right) \le f \left({s}\right)$ as $\le$ is an [[Definition:Ordering|ordering]].
+Hence, by definition of $\le_f$, $s \le_f s$ for all $s \in S$.
+So $\le_f$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Antisymmetry ===
+Suppose that for some $s_1, s_2 \in S$, one has both $s_1 \le_f s_2$ and $s_2 \le_f s_1$.
+This is equivalent to $f \left({s_1}\right) \le f \left({s_2}\right)$ and $f \left({s_2}\right) \le f \left({s_1}\right)$.
+As $\le$ is an [[Definition:Ordering|ordering]], this implies $f \left({s_1}\right) = f \left({s_2}\right)$.
+That $f$ is an [[Definition:Injection|injection]] now implies that $s_1 = s_2$.
+Hence $\le_f$ is [[Definition:Antisymmetric Relation|antisymmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Suppose that for some $s_1, s_2, s_3 \in S$, one has both $s_1 \le_f s_2$ and $s_2 \le_f s_3$.
+This is equivalent to $f \left({s_1}\right) \le f \left({s_2}\right)$ and $f \left({s_2}\right) \le f \left({s_3}\right)$.
+As $\le$ is an [[Definition:Ordering|ordering]], this implies $f \left({s_1}\right) \le f \left({s_3}\right)$.
+The definition of $\le_f$ now implies that $s_1 \le_f s_3$.
+So $\le_f$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+So we have shown that $\le_f$ is an [[Definition:Ordering|ordering]] on $S$.
+{{qed}}
+[[Category:Order Theory]]
+[[Category:Injections]]
+lldi5ek9vrmgyze2lc9lv2gfmfa18yt
+\end{proof}<|endoftext|>
+\section{Injection Induces Total Ordering}
+Tags: Total Orderings, Injections
+
+\begin{theorem}
+Let $\left({T, \le}\right)$ be a [[Definition:Toset|totally ordered set]], and let $S$ be a [[Definition:Set|set]].
+Let $f: S \to T$ be an [[Definition:Injection|injection]].
+Then $\le_f$, the [[Definition:Ordering Induced by Injection|ordering induced by $f$]], is a [[Definition:Total Ordering|total ordering]].
+\end{theorem}
+
+\begin{proof}
+By [[Ordering Induced by Injection is Ordering]], $\le_f$ is an [[Definition:Ordering|ordering]].
+Let $s_1, s_2 \in S$.
+Then as $\le$ is a [[Definition:Total Ordering|total ordering]], at least one of the following holds:
+:$(1): \qquad f \left({s_1}\right) \le f \left({s_2}\right)$
+:$(2): \qquad f \left({s_2}\right) \le f \left({s_1}\right)$
+If $(1)$ holds, the [[Definition:Ordering Induced by Injection|definition]] of $\le_f$ implies that $s_1 \le s_2$.
+Similarly, if $(2)$ holds, $s_2 \le s_1$.
+This leads to the conclusion that at least one of $s_1 \le s_2$ and $s_2 \le s_1$ holds.
+Hence $\le_f$ is a [[Definition:Total Ordering|total ordering]].
+{{qed}}
+[[Category:Total Orderings]]
+[[Category:Injections]]
+2yzdr7kcr2eanxa3w2vp8budup2vozf
+\end{proof}<|endoftext|>
+\section{Injection Induces Well-Ordering}
+Tags: Well-Orderings, Injections
+
+\begin{theorem}
+Let $\left({T, \le}\right)$ be a [[Definition:Well-Ordered Set|well-ordered set]].
+Let $S$ be a [[Definition:Set|set]].
+Let $f: S \to T$ be an [[Definition:Injection|injection]].
+Then $\le_f$, the [[Definition:Ordering Induced by Injection|ordering induced by $f$]], is a [[Definition:Well-Ordering|well-ordering]].
+\end{theorem}
+
+\begin{proof}
+By [[Ordering Induced by Injection is Ordering]], $\le_f$ is an [[Definition:Ordering|ordering]].
+Let $S' \subseteq S$ be [[Definition:Non-Empty Set|non-empty]].
+Then as $\le$ is a [[Definition:Well-Ordering|well-ordering]], the set:
+:$f \left({S'}\right) = \left\{{f \left({s}\right): s \in S'}\right\}$
+has a [[Definition:Minimal Element|minimal element]], say $f \left({s_0}\right)$.
+That is, for all $s \in S', f \left({s_0}\right) \le f \left({s}\right)$.
+By [[Definition:Ordering Induced by Injection|definition]] of $\le_f$, this implies:
+:$\forall s \in S': s_0 \le_f s$
+That is, $s_0$ is a [[Definition:Minimal Element|minimal element]] of $S'$.
+Hence $\le_f$ is a [[Definition:Well-Ordering|well-ordering]], as $S'$ was arbitrary.
+{{qed}}
+[[Category:Well-Orderings]]
+[[Category:Injections]]
+7hfxhks81racj6qp7e5oupmvgi9e4rg
+\end{proof}<|endoftext|>
+\section{K-Connectivity Implies Lesser Connectivity}
+Tags: Graph Theory
+
+\begin{theorem}
+If a [[Definition:Graph|graph]] $G$ is [[Definition:K-Connected|$k$-connected]], then $G$ is $l$-connected for all $l \in \Z : 0 < l < k$.
+\end{theorem}
+
+\begin{proof}
+Suppose that $G$ is [[Definition:K-Connected|$k$-connected]]. Then
+* $\left|{V \left({G}\right) }\right| > k > l$
+* $G$ is [[Definition:Connected Graph|connected]]
+* If $W$ is a [[Definition:Vertex Cut|vertex cut]] of $G$, then $\left|{W}\right| \ge k > l$ so $\left|{W}\right| \ge l$.
+{{qed}}
+[[Category:Graph Theory]]
+rfow1tll27dgjiwxtx4rb11zjuf8xut
+\end{proof}<|endoftext|>
+\section{Hartogs' Lemma (Set Theory)}
+Tags: Set Theory, Ordinals, Hartogs' Lemma (Set Theory)
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then there exists an [[Definition:Ordinal Number|ordinal]] $\alpha$ such that there is no [[Definition:Injection|injection]] from $\alpha$ to $S$.
+\end{theorem}
+
+\begin{proof}
+Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$.
+First of all, it is to be shown that $\alpha$ is a [[Definition:Set|set]].
+To this end, define the [[Definition:Set|set]] $W$ by:
+:$W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}$
+By [[Woset is Isomorphic to Unique Ordinal]], each $w \in W$ corresponds to a unique [[Definition:Ordinal|ordinal]] $\beta_w$.
+Thus by the [[Axiom:Axiom of Replacement|Axiom of Replacement]], the following is a [[Definition:Set|set]]:
+:$\set {\beta: \exists w \in W: \beta = \beta_w}$
+It follows from [[Injection Induces Well-Ordering]] that this [[Definition:Set|set]] coincides with $\alpha$.
+In particular, then, $\alpha$ is a [[Definition:Set|set]].
+Next, to establish $\alpha$ is an [[Definition:Ordinal Number|ordinal]].
+Suppose $\beta \in \alpha$ and $\gamma < \beta$.
+Let $i: \beta \to S$ be an [[Definition:Injection|injection]], and let $\iota: \gamma \to \beta$ be the [[Definition:Inclusion Mapping|inclusion]] of $\gamma$ in $\beta$.
+Then by [[Composite of Injections is Injection]], $i \circ \iota: \gamma \to S$ is an [[Definition:Injection|injection]].
+Hence $\gamma \in \alpha$, and therefore $\alpha$ is an [[Definition:Ordinal Number|ordinal]].
+Finally, by [[Ordinal is not Element of Itself]], $\alpha \notin \alpha$.
+That is to say, there is no [[Definition:Injection|injection]] $\alpha \to S$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $W$ be the [[Definition:Set|set]] of all [[Definition:Well-Ordering|well-orderings]] on [[Definition:Subset|subsets]] of $S$.
+By [[Woset is Isomorphic to Unique Ordinal]], there exists a [[Definition:Mapping|mapping]] $F: W \to \On$ defined by letting $\map F s$ be the [[Definition:Ordinal|ordinal]] which is isomorphic to $s$.
+{{explain|"isomorphism" between ordinals}}
+By [[Mapping from Set to Ordinal Class is Bounded Above]], $F \sqbrk W$ has an [[Definition:Upper Bound of Set|upper bound]] $\alpha_0$.
+Then if $\alpha$ is any ordinal strictly greater than $\alpha_0$, then $\alpha \notin F \sqbrk {W}$.
+[[Proof by Contradiction|Seeking a contradiction]], suppose that there is an [[Definition:Injection|injection]] $g: \alpha \to S$.
+Then by [[Injection to Image is Bijection]], there is a [[Definition:Bijection|bijection]] from $\alpha$ onto $g \sqbrk \alpha$.
+But this induces a well-ordering on $g \sqbrk \alpha \subseteq S$ which is isomorphic to $\alpha$, contradicting the fact that $\alpha \notin F \sqbrk W$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bourbaki-Witt Fixed Point Theorem}
+Tags: Set Theory, Fixed Point Theorems
+
+\begin{theorem}
+Let $\struct {X, \le}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Chain Complete Poset|chain complete poset]] (that is, an [[Definition:Ordered Set|ordered set]] in which every [[Definition:Chain (Set Theory)|chain]] has a [[Definition:Supremum of Set|supremum]]).
+Let $f: X \to X$ be an [[Definition:Inflationary Mapping|inflationary mapping]], that is, so that $\map f x \ge x$.
+Then for every $x \in X$ there exists $y \in X$ where $y \ge x$ such that $\map f y = y$.
+\end{theorem}
+
+\begin{proof}
+Let $\gamma$ be the [[Definition:Hartogs Number|Hartogs number]] of $X$.
+Suppose that $x \in X$.
+Define $g : \gamma \to X$ by [[Principle of Transfinite Recursion|transfinite recursion]] as follows: 
+:$\map g 0 = x$
+:$\map g {\alpha + 1} = \map f {\map g \alpha}$
+:$\map g \alpha = \sup \set {\map g \beta: \beta < \alpha}$ when $\alpha$ is a [[Definition:Limit Ordinal|limit ordinal]].
+That $f$ is [[Definition:Inflationary Mapping|inflationary]] guarantees both that:
+:$(1) \quad \set {\map g \beta: \beta < \alpha}$ is a [[Definition:Chain (Set Theory)|chain]] for each $\alpha < \gamma$
+:$(2) \quad g$ is [[Definition:Increasing Mapping|increasing]].
+{{explain|I'm not sure whether further explanation is needed as to why the above should be so, or whether I'm just slow.}}
+If $g$ is [[Definition:Strictly Increasing Mapping|strictly increasing]], then $g$ is [[Definition:Strictly Monotone Mapping|strictly monotone]] and, by [[Strictly Monotone Mapping with Totally Ordered Domain is Injective]], an [[Definition:Injection|injection]].
+But by definition of [[Definition:Hartogs Number|Hartogs number]], $\gamma$ is the least [[Definition:Ordinal|ordinal]] such that there is no [[Definition:Injection|injection]] from $\gamma$ to $X$.
+Therefore there must be some $\alpha < \beta < \gamma$ with $\map g \alpha = \map g \beta$.
+Then:
+:$\alpha + 1 \le \beta$
+so:
+:$\map g \alpha \le \map g {\alpha + 1} \le \map g \beta = \map g \alpha$
+whence:
+:$\map f {\map g \alpha} = \map g \alpha$
+Because $x = \map g 0$:
+:$\exists y \in \map g \gamma \subseteq \set {y \in X : x \le y}: \map f y = y$
+{{qed}}
+{{Namedfor|Nicolas Bourbaki|name2 = Ernst Witt|cat = Bourbaki|cat2 = Witt}}
+\end{proof}<|endoftext|>
+\section{Convergent Sequence with Finite Number of Terms Deleted is Convergent}
+Tags: Series, Sequences, Convergence Tests
+
+\begin{theorem}
+Let $\left({X, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\left\langle{x_k}\right\rangle$ be a [[Definition:Sequence|sequence in $X$]].
+Let $\left\langle{x_k}\right\rangle$ be [[Definition:Convergent Sequence (Metric Space)|convergent]].
+Let a [[Definition:Finite Set|finite]] number of [[Definition:Term (Algebra)|terms]] be deleted from $\left \langle {x_k} \right \rangle$.
+Then the resulting [[Definition:Subsequence|subsequence]] is [[Definition:Convergent Sequence (Metric Space)|convergent]].
+\end{theorem}
+
+\begin{proof}
+Suppose the sequence $\left \langle {x_k} \right \rangle$ converges to $x \in X$.
+That is for every $\epsilon > 0$ there is some index $N$ such that $d \left({x_n, x}\right) < \epsilon$ for all $n \ge N$.
+The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the original sequence.
+For the second part note that also adding finitely many terms to a convergent sequence will still result in a convergent sequence.
+This implies that removing finitely many terms from a divergent sequence will still result in a divergent sequence (if it were convergent then the original sequence must already have been convergent).
+{{handwaving|Very little more is being done here than repeat the exposition.}}
+{{refactor}}
+{{MissingLinks}}
+\end{proof}<|endoftext|>
+\section{Cauchy Condensation Test}
+Tags: Series, Convergence Tests, Analytic Geometry
+
+\begin{theorem}
+Let $\sequence {a_n}: n \mapsto \map a n$ be a [[Definition:Decreasing Sequence|decreasing sequence]] of [[Definition:Strictly Positive|strictly positive]] terms in $\R$ which [[Definition:Convergent Sequence|converges]] with a [[Definition:Limit of Sequence (Number Field)|limit]] of zero.
+That is, for every $n$ in the [[Definition:Domain (Set Theory)/Mapping|domain]] of $\sequence {a_n}$: $a_n > 0$, $a_n \ge a_{n + 1}$, and $a_n \to 0$ as $n \to +\infty$.
+Then the [[Definition:Series|series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty a_n$ 
+[[Definition:Convergent Series|converges]] {{iff}} the [[Definition:Condensed Series|condensed series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$
+[[Definition:Convergent Series|converges]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+We will first show that:
+:if the [[Definition:Condensed Series|condensed series]] $\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ [[Definition:Convergent Series|converges]]
+then:
+:$\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]] as well.
+Assume $\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ [[Definition:Convergent Series|converges]].
+Consider the [[Definition:Graph of Mapping|graph]] of $\sequence {a_n}$ and the [[Definition:Partial Sum|partial sums]] of $\sum 2^n \map a {2^n}$:
+:[[File:Cauchycondensation1.png|700px]]
+The dotted black line represents the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
+The $n$th [[Definition:Rectangle|rectangle]] has:
+:[[Definition:Base of Parallelogram|Base]] equal to $2^n$
+:[[Definition:Altitude of Parallelogram|Altitude]] equal to $\map a {2^n}$
+The series of partial sums of the [[Area of Rectangle|areas of the rectangles]] are:
+:$\displaystyle \sum_{n \mathop = 1}^N 2^n \map a {2^n}$
+which is precisely the defined condensed series.
+The series $\sum a_n$ can be viewed as the sum of the areas of rectangles with width $1$ and height $a_n$. 
+The diagram suggests that the partial sums of $\sum a_n$ are not greater than the condensed partial sums:
+:$\displaystyle \sum_{n \mathop = 1}^N a_n \le \sum_{n \mathop = 1}^N 2^n \map a {2^n}$
+To formalize this claim, observe that because $\sequence {a_n}$ is decreasing:
+:$a_1 + \underbrace {a_2 + a_3}_{\le 2 a_2} + \underbrace {a_4 + a_5 + a_6 + a_7}_{\le 4 a_4} + \cdots + a_N \le a_1 + 2 a_2 + 4 a_4 + \cdots + 2^N \map a {2^N}$
+As $n \to +\infty$, the {{RHS}} converges [[Definition:By Hypothesis|by hypothesis]].
+Hence $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ also converges, by the [[Comparison Test]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+We will show that if $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges, the [[Definition:Condensed Series|condensed series]] $\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges as well.
+Assume $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges.
+From the [[Combination Theorem for Sequences]], $2\displaystyle \sum_{n \mathop = 1}^\infty a_n = \displaystyle \sum_{n \mathop = 1}^\infty 2 a_n$ converges as well.
+Consider, then, the [[Definition:Graph of Mapping|graph]] of $\sequence {a_n}$ and of the [[Definition:Partial Sum|partial sums]] of what will be shown to equal $\displaystyle \frac 1 2 \sum 2^n \map a {2^n}$:
+:[[File:Cauchycondensation2.png|700px]]
+The dotted black line represents the sequence $\sequence {a_n}$.
+The $n$th [[Definition:Rectangle|rectangle]] has:
+:[[Definition:Base of Parallelogram|Base]] equal to $2^n$
+:[[Definition:Altitude of Parallelogram|Altitude]] equal to $\map a {2 \cdot 2^n}$
+The series of partial sums of the [[Area of Rectangle|areas of the rectangles]] are:
+{{begin-eqn}}
+{{eqn | o =
+      | r = \sum_{n \mathop = 1}^N 2^n \map a {2 \cdot 2^n}
+}}
+{{eqn | r = \sum_{n \mathop = 1}^N 2^n \map a {2^{n + 1} }
+}}
+{{eqn | r = \sum_{n - 1 \mathop = 1}^N 2^{n - 1} \map a {2^n}
+      | c = replacing $n$ with $n - 1$
+}}
+{{eqn | r = \frac 1 2 \sum_{n \mathop = 2}^N 2^n \map a {2^n}
+}}
+{{end-eqn}}
+From [[Deletion of Terms from a Sequence]], it is justified to examine only the behavior of $n$ for $n \ge 2$.
+The series $\sum a_n$ can be viewed as the sum of the area of rectangles with height $a_n$ and width $1$.
+The diagram suggests that the partial sums of $\sum a_n$ satisfy the following inequality:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 2}^N a_n
+      | o = \ge
+      | r = \frac 1 2 \sum_{n \mathop = 2}^N 2^n \map a {2^n}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \displaystyle \sum_{n \mathop = 2}^N 2 a_n
+      | o = \ge
+      | r = \sum_{n \mathop = 2}^N 2^n \map a {2^n}
+      | c = 
+}}
+{{end-eqn}}
+To formalize this claim, observe that because $\sequence {a_n}$ is decreasing: 
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 2}^N 2a_n
+      | r = a_2 + \underbrace {a_2 + a_3}_{\ge 2 a_4} + \underbrace {a_3 + 2 a_4 + a_5}_{\ge 4 a_8} + \underbrace {a_5 + 2 a_6 + 2 a_7 + 2 a_8 + a_9}_{\ge 8 a_{16} } + \cdots + 2 a_N
+}}
+{{eqn | o = \ge 
+      | r = a_2 + 2 a_4 + 4 a_8 + 8 a_{16} + \cdots + 2^n \map a {2^n}
+}}
+{{end-eqn}}
+[[Definition:By Hypothesis|By hypothesis]], the {{LHS}} converges as $n \to +\infty$.
+Hence the condensed series converges as well, by the [[Comparison Test]].
+{{qed}}
+{{Namedfor|Augustin Louis Cauchy|cat = Cauchy}}
+\end{proof}<|endoftext|>
+\section{Between Two Sets Exists Injection or Surjection}
+Tags: Injections, Surjections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Then either or both of the following cases hold:
+:$(1):$ There exists a [[Definition:Mapping|mapping]] $f: S \to T$ such that $f$ is an [[Definition:Injection|injection]]
+:$(2):$ There exists a [[Definition:Mapping|mapping]] $f: S \to T$ such that $f$ is a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+From [[Zermelo's Theorem (Set Theory)|Zermelo's Theorem]], exactly one of the following cases holds:
+:$(1): \quad S \le T$, that is, $T$ [[Definition:Dominate (Set Theory)|dominates]] $S$
+:$(2): \quad S \equiv T$, that is, $T$ is [[Definition:Set Equivalence|equivalent]] to $S$
+:$(3): \quad S \ge T$, that is, $S$ [[Definition:Dominate (Set Theory)|dominates]] $T$.
+Suppose $(1)$ holds, that is, $T$ [[Definition:Dominate (Set Theory)|dominates]] $S$.
+By definition, there exists an [[Definition:Injection|injection]] from $S$ to $T$.
+Hence in this case the result holds.
+{{qed|lemma}}
+Suppose $(2)$ holds, that is, $T$ is [[Definition:Set Equivalence|equivalent]] to $S$.
+By definition, there then exists a [[Definition:Bijection|bijection]] from $S$ to $T$.
+A [[Definition:Bijection|bijection]] is, by definition, a [[Definition:Mapping|mapping]] which is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
+Hence in this case the result holds.
+{{qed|lemma}}
+Suppose $(3)$ holds, that is, $S$ [[Definition:Dominate (Set Theory)|dominates]] $T$.
+By definition, there exists an [[Definition:Injection|injection]] from $S$ to $T$.
+By [[Injection has Surjective Left Inverse Mapping]], $f$ has a [[Definition:Left Inverse Mapping|left inverse]] $g: T \to S$ which is [[Definition:Surjection|surjective]].
+Hence in this case the result holds.
+{{qed|lemma}}
+All cases have been shown to lead to the same conclusion, that is, that there exists either an [[Definition:Injection|injection]] or a [[Definition:Surjection|surjection]] or both between $S$ and $T$.
+Hence the result, by [[Proof by Cases]].
+{{qed}}
+{{AoC|Zermelo's Theorem (Set Theory)}}
+[[Category:Injections]]
+[[Category:Surjections]]
+7i5scj9fe0as9hfjtfpi0sdn2csd23b
+\end{proof}<|endoftext|>
+\section{Injection from Set to Power Set}
+Tags: Injections, Power Set
+
+\begin{theorem}
+For every set $S$, there exists an [[Definition:Injection|injection]] from $S$ to its [[Definition:Power Set|power set]] $\mathcal P \left({S}\right)$.
+\end{theorem}
+
+\begin{proof}
+If $S = \varnothing$, the [[Definition:Empty Mapping|empty mapping]] suffices, as it is [[Definition:Vacuous Truth|vacuously]] an [[Definition:Injection|injection]].
+Let $f: S \to \mathcal P \left({S}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall s \in S: f \left({s}\right) = \left\{{s}\right\}$
+Let $s, t \in S$ such that $f \left({s}\right) = f \left({t}\right)$.
+Then $\left\{{s}\right\} = \left\{{t}\right\}$.
+By definition of [[Definition:Set Equality|set equality]] it follows directly that $s = t$.
+Hence $f$ is the required [[Definition:Injection|injection]], by definition.
+{{qed}}
+[[Category:Injections]]
+[[Category:Power Set]]
+tp08qw91z590k5pnjljyrv6uxsnrtn1
+\end{proof}<|endoftext|>
+\section{Minimum Degree is at Least Connectivity}
+Tags: Graph Theory
+
+\begin{theorem}
+Let $G = \left({V, E}\right)$ be a [[Definition:Simple Graph|simple graph]].
+Then:
+:$\delta \left({G}\right) \geq \kappa \left({G}\right)$
+That is, the [[Definition:Minimum Degree|minimum degree]] of $G$ is at least its [[Definition:Connectivity|connectivity]].
+\end{theorem}
+
+\begin{proof}
+Pick a [[Definition:Vertex (Graph Theory)|vertex]] $v \in G$ with $\deg_G \left({v}\right) = \delta \left({G}\right)$, that is, a vertex with minimum degree.
+Recall that $\Gamma_G \left({v}\right)$ is the [[Definition:Neighborhood (Graph Theory)|neighborhood]] of $v$ in $G$.
+Suppose that $V = \Gamma_G \left({v}\right) \cup \left \{{v}\right\}$.
+That is, that $v$ is adjacent to all other vertices of $G$.
+Then:
+:$\left\vert{V}\right\vert = \left\vert{\Gamma_G \left({v}\right)}\right\vert + 1 = \delta \left({G}\right) + 1$
+By definition, $\kappa \left({G}\right) < \left\vert{V}\right\vert$, so $\kappa \left({G}\right) \leq \delta \left({G}\right)$.
+Otherwise, there is at least one vertex of $G$ not adjacent to $v$.
+So, [[Definition:Vertex Deletion|deleting]] $\Gamma_G \left({v}\right)$ from $G$ leaves $v$ [[Definition:Isolated Vertex|isolated]].
+Thus $\Gamma_G\left({v}\right)$ is a [[Definition:Vertex Cut|vertex cut]] of size $\delta \left({G}\right)$. 
+Since $G$ has a vertex cut of size $\delta \left({G}\right)$, it follows that $\kappa \left({G}\right) \leq \delta \left({G}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homeomorphism Relation is Equivalence}
+Tags: Homeomorphisms, Equivalence Relations
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Let $T_1 \sim T_2$ denote that $T_1$ and $T_2$ are [[Definition:Homeomorphism (Topological Spaces)|homeomorphic]].
+The relation $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+Let $T$ be a [[Definition:Topological Space|topological space]].
+From [[Identity Mapping is Homeomorphism]], the [[Definition:Identity Mapping|identity mapping]] $I_T: T \to T$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+So $T \sim T$, and $\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]] such that $T_1 \sim T_2$.
+By definition, there exists a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]] $f: T_1 \to T_2$.
+From [[Inverse of Homeomorphism is Homeomorphism]] it follows that $f^{-1}: T_2 \to T_1$ is also a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+So $T_2 \sim T_1$, and $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $T_1, T_2, T_3$ be [[Definition:Topological Space|topological spaces]] such that $T_1 \sim T_2$ and $T_2 \sim T_3$.
+By definition, there exist [[Definition:Homeomorphism (Topological Spaces)|homeomorphisms]] $f: T_1 \to T_2$ and $g: T_2 \to T_3$.
+From [[Composite of Homeomorphisms is Homeomorphism]] it follows that $g \circ f: T_1 \to T_3$ is also a [[Definition:Homeomorphism|homeomorphism]].
+So $T_1 \sim T_3$, and $\sim$ 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{Identity Mapping is Homeomorphism}
+Tags: Homeomorphisms, Identity Mappings
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+The [[Definition:Identity Mapping|identity mapping]] $I_T: T \to T$ defined as:
+:$\forall x \in T: I_T \left({x}\right) = x$
+is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+We have [[Identity Mapping is Bijection]].
+We also have [[Identity Mapping is Continuous]].
+Hence, by definition, $I_T$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+{{qed}}
+[[Category:Homeomorphisms]]
+[[Category:Identity Mappings]]
+5s2tqa0d7guhqxk08qpyx4u63jif5wa
+\end{proof}<|endoftext|>
+\section{Composite of Homeomorphisms is Homeomorphism}
+Tags: Homeomorphisms
+
+\begin{theorem}
+Let $T_1, T_2, T_3$ be [[Definition:Topological Space|topological spaces]]. 
+Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be [[Definition:Homeomorphism (Topological Spaces)|homeomorphisms]].
+Then $g \circ f: T_1 \to T_3$ is also a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], $f$ and $g$ are both [[Definition:Bijection|bijections]].
+From [[Composite of Bijections is Bijection]] it follows that $g \circ f$ is also a [[Definition:Bijection|bijection]].
+Similarly, also by definition of [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]], $f$ and $g$ are both [[Definition:Everywhere Continuous Mapping (Topology)|continuous mappings]].
+From [[Composite of Continuous Mappings is Continuous]] it follows that $g \circ f$ is also a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+Hence the result, from definition of [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+{{qed}}
+[[Category:Homeomorphisms]]
+6wti8nffp909u0kceehmavp8m0shjq1
+\end{proof}<|endoftext|>
+\section{Identity Mapping is Continuous}
+Tags: Continuous Mappings, Identity Mappings
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+The [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ defined as:
+:$\forall x \in S: \map {I_S} x = x$
+is a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$.
+We have [[Identity Mapping is Bijection]].
+So $I_S^{-1}$ is [[Definition:Well-Defined Mapping|well-defined]] and:
+:$\forall x \in U: \map {I_S^{-1} } x = x$
+Thus $I_S^{-1} \sqbrk U = U \in \tau$.
+Hence, by definition of [[Definition:Continuous Mapping (Topology)|continuous mapping]], $I_S$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Clopen Sets in Finite Complement Topology}
+Tags: Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then the only [[Definition:Clopen Set|clopen]] sets of $T$ are $S$ and $\O$.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be [[Definition:Open Set (Topology)|open]] in $T$.
+Then by definition of [[Definition:Finite Complement Topology|finite complement topology]], $S \setminus U$ is [[Definition:Finite Set|finite]].
+By definition of [[Definition:Open Set (Topology)|open set]], $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]].
+As $S$ is [[Definition:Infinite Set|infinite]], it follows that $U$ must also be [[Definition:Infinite Set|infinite]].
+Thus unless $U = S$, $S \setminus U$ can not be [[Definition:Open Set (Topology)|open]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kuratowski's Lemma}
+Tags: Set Theory, Axiom of Choice
+
+\begin{theorem}
+Let $\struct {S, \preceq}, S \ne \O$ be a [[Definition:Empty Set|non-empty]] [[Definition:Ordered Set|ordered set]]. 
+Then every [[Definition:Chain (Set Theory)|chain]] in $S$ is the [[Definition:Subset|subset]] of some [[Definition:Maximal Chain|maximal chain]].
+\end{theorem}<|endoftext|>
+\section{Included Set Topology is Topology}
+Tags: Included Set Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau_H}\right)$ be an [[Definition:Included Set Space|included set space]].
+Then $\tau_H$ is a [[Definition:Topology|topology]] on $S$, and $T$ is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+We have [[Definition:Included Set Space|by definition]] that $\varnothing \in \tau_H$, and as $H \subseteq S$ we have that $S \in \tau_H$.
+Now let $U_1, U_2 \in \tau_H$.
+By definition $H \subseteq U_1 $ and $H \subseteq U_2$.
+From [[Intersection is Largest Subset]] it follows that $H \subseteq \left({U_1 \cap U_2}\right)$.
+So $U_1 \cap U_2 \in \tau_H$.
+Now let $\mathcal U \subseteq \tau_H$.
+We have [[Definition:Included Set Space|by definition]] that $\forall U \in \mathcal U: H \subseteq U$.
+Hence from [[Subset of Union]] $H \subseteq \bigcup \mathcal U$.
+So all the properties are fulfilled for $\tau_H$ to be a [[Definition:Topology|topology]] on $S$.
+{{qed}}
+[[Category:Included Set Topology]]
+tcccaqqmlgqp9o29glyh206padtt3m3
+\end{proof}<|endoftext|>
+\section{Space of Compact Linear Transformations is Banach Space}
+Tags: Banach Spaces, Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]], and let $B_0 \left({H, K}\right)$ be the [[Definition:Space of Compact Linear Transformations|space of compact linear transformations from $H$ to $K$]].
+Let $\Bbb F \in \left\{{\R, \C}\right\}$ be the [[Definition:Ground Field|ground field]] of $K$.
+Now $B_0 \left({H, K}\right) \subseteq K^H$, the [[Definition:Set of All Mappings|set of mappings from $H$ to $K$]].
+Therefore, $B_0 \left({H, K}\right)$ can be endowed with [[Definition:Pointwise Addition of Mappings|pointwise addition]] ($+$) and [[Definition:Pointwise Scalar Multiplication of Mappings|($\Bbb F$)-scalar multiplication]] ($\circ$).
+Let $\left\Vert{\cdot}\right\Vert$ denote the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear operators]].
+Then $\left\Vert{\cdot}\right\Vert$ is a [[Definition:Norm on Vector Space|norm]] on $B_0 \left({H, K}\right)$.
+Furthermore, $B_0 \left({H, K}\right)$ is a [[Definition:Banach Space|Banach space]] with respect to this norm.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Subspace of $B(H,K)$; completeness is nontrivial.}}
+\end{proof}<|endoftext|>
+\section{Compact Linear Transformation is Bounded}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $T \in B_0 \left({H, K}\right)$ be a [[Definition:Compact Linear Transformation|compact linear transformation]].
+Then $T$ is also a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
+That is, $B_0 \left({H, K}\right) \subseteq B \left({H, K}\right)$.
+\end{theorem}<|endoftext|>
+\section{Compact Linear Transformations Composed with Bounded Linear Operator}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $T \in B_0 \left({H, K}\right)$ be a [[Definition:Compact Linear Transformation|compact linear transformation]].
+Let $A \in B \left({H}\right), B \in B \left({K}\right)$ be [[Definition:Bounded Linear Operator|bounded linear operators]].
+Then the [[Definition:Composition of Mappings|compositions]] $TA$ and $BT$ are also [[Definition:Compact Linear Transformation|compact linear transformations]].
+\end{theorem}<|endoftext|>
+\section{Expansion of Included Set Topology}
+Tags: Included Set Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $A_1 \subseteq S$ and $A_2 \subseteq S$.
+Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be [[Definition:Included Set Space|included set spaces]] on $S$.
+Then:
+:$(1): \quad T_1 \ge T_2 \iff A_1 \subseteq A_2$
+:$(2): \quad T_1 > T_2 \iff A_1 \subsetneq A_2$
+where:
+:$T_1 \ge T_2$ denotes that $T_1$ is [[Definition:Finer Topology|finer]] than $T_2$
+:$T_1 > T_2$ denotes that $T_1$ is [[Definition:Strictly Finer Topology|strictly finer]] than $T_2$.
+\end{theorem}
+
+\begin{proof}
+Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be [[Definition:Included Set Space|included set spaces]] on $S$.
+=== Necessary Condition ===
+Let $T_1 \ge T_2$.
+We have that $A_2 \in \tau_{A_2}$ by definition of [[Definition:Included Set Topology|included set topology]].
+But by definition of [[Definition:Finer Topology|finer topology]], $A_2 \in \tau_{A_1}$.
+So by definition of [[Definition:Included Set Topology|included set topology]], $A_1 \subseteq A_2$.
+Now suppose that $T_1 > T_2$.
+That is, $T_1 \ge T_2$ and $T_1 \ne T_2$.
+It follows that $\exists X \in \tau_{A_1}: X \notin \tau_{A_2}$.
+Then  $A_1 \subseteq X \subsetneq A_2$ from which it follows that $A_1 \subsetneq A_2$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $A_1 \subseteq A_2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = U
+      | o = \in
+      | r = \tau_{A_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_2
+      | o = \subseteq
+      | r = U
+      | c = {{Defof|Included Set Topology}}
+}}
+{{eqn | ll= \leadsto
+      | l = A_1
+      | o = \subseteq
+      | r = U
+      | c = [[Subset Relation is Transitive]]
+}}
+{{eqn | ll= \leadsto
+      | l = U
+      | o = \in
+      | r = \tau_{A_1}
+      | c = {{Defof|Included Set Topology}}
+}}
+{{end-eqn}}
+So $T_1 \ge T_2$ by definition of [[Definition:Finer Topology|finer topology]].
+Now suppose that $A_1 \subsetneq A_2$.
+Then $A_2 \not \subseteq A_1$ and so $A_1 \notin \tau_{A_2}$.
+Thus $\tau_{A_1} \not \subseteq \tau_{A_2}$ and so $T_1 > T_2$.
+{{qed|lemma}}
+Both [[Definition:Necessary Condition|necessity]] and [[Definition:Sufficient Condition|sufficiency]] have been proved, hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformation Compact iff Adjoint Compact}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $T: H \to K$ be a [[Definition:Linear Transformation|linear transformation]].
+Then $T$ is [[Definition:Compact Linear Transformation|compact]] iff its [[Definition:Adjoint Linear Transformation|adjoint]] $T^*$ is.
+\end{theorem}<|endoftext|>
+\section{Finite Rank Operators Dense in Compact Linear Transformations}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Then:
+:$\map {B_{00} } {H, K}$ is [[Definition:Everywhere Dense|everywhere dense]] in $\map {B_0} {H, K}$
+where:
+:$\map {B_{00} } {H, K}$ is the [[Definition:Space of Continuous Finite Rank Operators|space of continuous finite rank operators from $H$ to $K$]]
+:$\map {B_0} {H, K}$ is the [[Definition:Space of Compact Linear Transformations|space of compact linear transformations from $H$ to $K$]].
+That is, for every $T \in \map {B_0} {H, K}$, there is a [[Definition:Sequence|sequence]] $\sequence {T_n}_{n \mathop \in \N}$ in $\map {B_{00} } {H, K}$ such that:
+:$\ds \lim_{n \mathop \to \infty} \norm {T_n - T} = 0$
+where $\norm {\, \cdot \,}$ denotes the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear transformations]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Proceeds by sequence definition of limit point}}
+\end{proof}<|endoftext|>
+\section{Inverse of Matrix Product}
+Tags: Inverse Matrices, Conventional Matrix Multiplication
+
+\begin{theorem}
+Let $\mathbf {A, B}$ be [[Definition:Square Matrix|square matrices of order $n$]]
+Let $\mathbf I$ be the $n \times n$ [[Definition:Unit Matrix|unit matrix]].
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Invertible Matrix|invertible]].
+Then the [[Definition:Matrix Product (Conventional)|matrix product]] $\mathbf {AB}$ is also [[Definition:Invertible Matrix|invertible]], and:
+:$\paren {\mathbf A \mathbf B}^{-1} = \mathbf B^{-1} \mathbf A^{-1}$
+\end{theorem}
+
+\begin{proof}
+We are given that $\mathbf A$ and $\mathbf B$ are [[Definition:Invertible Matrix|invertible]].
+From [[Product of Matrices is Invertible iff Matrices are Invertible]], $\mathbf A \mathbf B$ is also [[Definition:Invertible Matrix|invertible]].
+By the definition of [[Definition:Inverse Matrix|inverse matrix]]:
+:$\mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$
+and 
+:$\mathbf B \mathbf B^{-1} = \mathbf B^{-1} \mathbf B = \mathbf I$
+Now, observe that:
+{{begin-eqn}}
+{{eqn | l = \paren {\mathbf A \mathbf B} \paren {\mathbf B^{-1} \mathbf A^{-1} }
+      | r = \paren {\mathbf A \paren {\mathbf B \mathbf B^{-1} } } \mathbf A^{-1}
+      | c = [[Matrix Multiplication is Associative]]
+}}
+{{eqn | r = \paren {\mathbf A \mathbf I} \mathbf A^{-1}
+}}
+{{eqn | r = \mathbf A \mathbf A^{-1}
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \mathbf I
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \paren {\mathbf B^{-1} \mathbf A^{-1} } \paren {\mathbf A \mathbf B}
+      | r = \paren {\mathbf B^{-1} \paren {\mathbf A^{-1} \mathbf A} } \mathbf B
+      | c = [[Matrix Multiplication is Associative]]
+}}
+{{eqn | r = \paren {\mathbf B^{-1} \mathbf I} \mathbf B
+}}
+{{eqn | r = \mathbf B^{-1} \mathbf B
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \mathbf I
+}}
+{{end-eqn}}
+The result follows from the definition of [[Definition:Inverse Matrix|inverse]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Range of Compact Linear Transformation is Separable}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $T \in \map {B_0} {H, K}$ be a [[Definition:Compact Linear Transformation|compact linear transformation]].
+Then $\map \cl {\Rng T}$ is [[Definition:Separable Space|separable]].
+\end{theorem}<|endoftext|>
+\section{Elementary Row Operations as Matrix Multiplications}
+Tags: Conventional Matrix Multiplication, Elementary Row Operations, Elementary Matrices
+
+\begin{theorem}
+Let $e$ be an [[Definition:Elementary Row Operation|elementary row operation]].
+Let $\mathbf E$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] of [[Definition:Order of Square Matrix|order]] $m$ defined as:
+:$\mathbf E = e \paren {\mathbf I}$
+where $\mathbf I$ is the [[Definition:Unit Matrix|unit matrix]].
+Then for every $m \times n$ [[Definition:Matrix|matrix]] $\mathbf A$:
+:$e \paren {\mathbf A} = \mathbf {E A}$
+where $\mathbf {E A}$ denotes the [[Definition:Matrix Product (Conventional)|conventional matrix product]].
+\end{theorem}
+
+\begin{proof}
+Let $s, t \in \closedint 1 m$ such that $s \ne t$.
+=== Case $1$ ===
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] $r_s \to \lambda r_s$:
+:$E_{ik} = \begin{cases}
+\delta_{ik} & : i \ne s \\
+\lambda \delta_{ik} & : i = s
+\end{cases}$
+where $\delta$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {E A}_{i j}
+      | r = \sum_{k \mathop = 1}^m E_{i k} A_{k j}
+      | c = 
+}}
+{{eqn | r = \begin{cases} A_{i j} & : i \ne r \\ \lambda A_{i j} & : i = r \end{cases}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf {E A}
+      | r = e \paren {\mathbf A}
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case $2$ ===
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] $r_s \to r_s + \lambda r_t$:
+:$E_{i k} = \begin{cases}
+\delta_{i k} & : i \ne s \\
+\delta_{s k} + \lambda \delta_{t k} & : i = s
+\end{cases}$
+where $\delta$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {E A}_{i j}
+      | r = \sum_{k \mathop = 1}^m E_{i k} A_{k j}
+      | c = 
+}}
+{{eqn | r = \begin {cases} A_{i j} & : i \ne s \\ A_{i j} + \lambda A_{t j} & : i = s \end {cases}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf {E A}
+      | r = e \paren {\mathbf A}
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case $3$ ===
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] $r_s \leftrightarrow r_t$:
+By [[Exchange of Rows as Sequence of Other Elementary Row Operations]], this [[Definition:Elementary Row Operation|elementary row operation]] can be expressed as:
+:$e_1 e_2 e_3 e_4 \paren {\mathbf A} = e \paren {\mathbf A}$
+where the $e_i$ are [[Definition:Elementary Row Operation|elementary row operation]] of the other two types.
+For each $e_i$, let $\mathbf E_i = e_i \paren {\mathbf I}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = e \paren {\mathbf A}
+      | r = e_1 e_2 e_3 e_4 \paren {\mathbf A}
+      | c = Definition of $e$
+}}
+{{eqn | r = \mathbf E_1 \mathbf E_2 \mathbf E_3 \mathbf E_4 \mathbf A
+      | c = Cases $1$ and $2$
+}}
+{{eqn | r = \mathbf E_1 \mathbf E_2 \mathbf E_3 e_4 \paren {\mathbf I} \mathbf A
+}}
+{{eqn | r = \mathbf E_1 \mathbf E_2 e_3 e_4 \paren {\mathbf I} \mathbf A
+}}
+{{eqn | r = \mathbf E_1 e_2 e_3 e_4 \paren {\mathbf I} \mathbf A
+}}
+{{eqn | r = e_1 e_2 e_3 e_4 \paren {\mathbf I} \mathbf A
+}}
+{{eqn | r = e \paren {\mathbf I} \mathbf A
+}}
+{{eqn | r = \mathbf {E A}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Included Set Topology on Finite Intersection}
+Tags: Included Set Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] on a [[Definition:Set|set]] $S$.
+Let $A_1, A_2, \ldots, A_n$ be a [[Definition:Finite Set|finite set]] of [[Definition:Subset|subsets]] of $S$:
+:$\forall i \in \closedint 1 n: A_i \subseteq S$
+Let $\forall i \in \closedint 1 n: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the [[Definition:Included Set Space|included set spaces]] on $S$ by $A_i$.
+Let:
+:$\forall i \in \closedint 1 n: \map T {A_i} \le T$
+where $\map T {A_i} \le T$ denotes that $\map T {A_i}$ is [[Definition:Coarser Topology|coarser]] than $T$.
+Then:
+:$\displaystyle \map T {\bigcap A_i} \le T$
+where $\displaystyle \map T {\bigcap A_i}$ is the [[Definition:Included Set Space|included set space]] on $S$ by $\displaystyle \bigcap_{i \mathop = 1}^n A_i$.
+\end{theorem}
+
+\begin{proof}
+For ease of notation, define:
+:$A := \displaystyle \bigcap_{i \mathop = 1}^n A_i$
+and let $\tau_A$ denote the [[Definition:Included Set Topology|included set topology]] on $S$ by $A$.
+Let $U \in \tau_A$ be [[Definition:Empty Set|nonempty]].
+Then by definition, $A \subseteq U$.
+Hence there is a [[Definition:Subset|subset]] $Z \subseteq S$ of $S$, such that $U = A \cup Z$; that is:
+:$U = \displaystyle \paren {\bigcap_{i \mathop = 1}^n A_i} \cup Z = \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$
+where the latter equality follows from [[Union Distributes over Intersection]].
+For every $i \in \closedint 1 n$, one has $A_i \cup Z \in \tau_{A_i}$ by definition of [[Definition:Included Set Topology|included set topology]].
+Further, by assumption, $\map T {A_i} \le T$, that is $\tau_{A_i} \subseteq \tau$.
+It follows that $A_i \cup Z \in \tau$ for all $i \in \closedint 1 n$.
+Hence $U = \displaystyle \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$ is in $\tau$, because $\tau$ is a [[Definition:Topology|topology]].
+This comes down to:
+:$\tau_A \subseteq \tau$
+Hence by definition of [[Definition:Coarser Topology|coarser topology]]:
+:$\map T {\bigcap A_i} \le T$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transformation of Unit Matrix into Inverse}
+Tags: Unit Matrices, Inverse Matrices, Elementary Row Operations
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix of order $n$]] of the [[Definition:Matrix Space|matrix space]] $\map {\MM_\R} n$.
+Let $\mathbf I$ be the [[Definition:Unit Matrix|unit matrix]] of order $n$.
+Suppose there exists a [[Definition:Sequence|sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] that reduces $\mathbf A$ to $\mathbf I$.
+Then $\mathbf A$ is [[Definition:Invertible Matrix|invertible]].
+Futhermore, the same sequence, when performed on $\mathbf I$, results in the [[Definition:Inverse Element|inverse]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$.
+We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.
+By repeated application of [[Elementary Row Operations as Matrix Multiplications]], we can write this assertion as:
+{{begin-eqn}}
+{{eqn | l = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf A
+      | r = \mathbf I
+}}
+{{end-eqn}}
+From [[Elementary Row Matrix is Invertible]]:
+:$\mathbf E_1, \dotsc, \mathbf E_t \in \GL {n, \R}$
+{{MissingLinks|$\GL {n, \R}$, and explain the significance of this. It's General Linear Group, clearly.}}
+We can multiply on the left both sides of this equation by:
+{{begin-eqn}}
+{{eqn | l = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf  A 
+      | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf {II} \cdots \mathbf {IIA} 
+      | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I
+      | c = definition of inverse
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A
+      | r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t
+      | c = definition of identity
+}}
+{{eqn | ll= \leadsto
+      | l = \breve {\mathbf A}
+      | r = \breve {\breve {\mathbf E} }_t \breve {\breve {\mathbf E} }_{t - 1} \cdots \breve {\breve {\mathbf E} }_2 \breve {\breve {\mathbf E} }_1
+      | c = [[Inverse of Matrix Product]], [[Axiom:Leibniz's Law|Leibniz's Law]]
+}}
+{{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1
+      | c = [[Inverse of Group Inverse]]
+}}
+{{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf I
+      | c = {{Defof|Unit Matrix}}
+}}
+{{end-eqn}}
+By repeated application of [[Elementary Row Operations as Matrix Multiplications]], each $\mathbf E_n$ on the right hand side corresponds to an elementary row operation.
+Hence the result.
+{{qed}}
+{{proofread}}
+[[Category:Unit Matrices]]
+[[Category:Inverse Matrices]]
+[[Category:Elementary Row Operations]]
+fqmfgxd8fey29torcvzebsnm83fzq1f
+\end{proof}<|endoftext|>
+\section{Included Set Topology on Union}
+Tags: Included Set Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] on a [[Definition:Set|set]] $S$.
+Let $\family {A_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Subsets|family of subsets]] of $S$ indexed by the [[Definition:Indexing Set|indexing set]] $I$:
+:$\forall i \in I: A_i \subseteq S$
+Let $\forall i \in I: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the [[Definition:Included Set Space|included set spaces]] on $S$ by $A_i$.
+Let:
+:$\forall i \in I: \map T {A_i} \ge T$
+where $\map T {A_i} \ge T$ denotes that $\map T {A_i}$ is [[Definition:Finer Topology|finer]] than $T$.
+Then:
+:$\map T {\bigcup A_i} \ge T$
+where $\map T {\bigcup A_i}$ is the [[Definition:Included Set Space|included set space]] on $S$ by $\displaystyle \bigcup_{i \mathop \in I} A_i$.
+\end{theorem}
+
+\begin{proof}
+For ease of notation, define:
+:$A := \displaystyle \bigcup_{i \mathop \in I} A_i$
+and let $\tau_A$ denote the [[Definition:Included Set Topology|included set topology]] on $S$ by $A$.
+Let $U \in \tau$ be [[Definition:Non-Empty Set|nonempty]].
+As $\map T {A_i}$ is [[Definition:Finer Topology|finer]] than $T$ it follows by definition that:
+:$\forall i \in I: \tau \subseteq \tau_{A_i}$
+Thus:
+:$\forall i \in I: U \in \tau_{A_i}$
+Hence for all $i$ there is a [[Definition:Subset|subset]] $Z_i \subseteq S$ of $S$, such that $U = A_i \cup Z_i$; that is:
+:$\displaystyle U = \bigcup_{i \mathop \in I} \paren {A_i \cup Z_i} = \paren {\bigcup_{i \mathop \in I} A_i} \cup \paren {\bigcup_{i \mathop \in I} Z_i}$
+where the latter equality follows from [[Union is Associative|associativity]] and [[Union is Commutative|commutativity]] of [[Definition:Set Union|set union]].
+That is:
+:$U = A \cup Z$
+where:
+:$\displaystyle Z = \paren {\bigcup_{i \mathop \in I} Z_i}$
+Hence $U \in \tau_A$. by definition of the [[Definition:Included Set Topology|included set topology]].
+This comes down to $\tau \subseteq \tau_A$, and hence $\map T {\bigcup A_i} \ge T$, by definition of a [[Definition:Finer Topology|finer topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kuratowski's Free Set Theorem}
+Tags: 
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]].
+Let $X$ be an [[Definition:Infinite Set|infinite set]].
+Let $\left[{X}\right]^n$ denote the [[Definition:Set of Sets|set]] of $n$-element [[Definition:Subset|subsets]] of $X$ whose [[Definition:Cardinality|cardinality]] is $n$.
+Let $\left[{X}\right]^{< \aleph_0}$ denote the [[Definition:Set of Sets|set]] of [[Definition:Finite Set|finite]][[Definition:Subset|subsets]] of $X$.
+Then:
+: the [[Definition:Cardinality|cardinality]] of $X$ is greater than or equal to $\aleph_n$
+{{iff}}
+: for every [[Definition:Mapping|mapping]] $f: \left[{X}\right]^n \to \left[{X}\right]^{<\aleph_0}$ there exists an $(n + 1)$-element [[Definition:Free Subset|free subset]] of $X$ with respect to $f$.
+That is, a [[Definition:Subset|subset]] $Y$ of $X$ with $n+1$ elements such that $y \notin f \left({Y \setminus \left\{{y}\right\} }\right)$ for all $y \in Y$.
+\end{theorem}<|endoftext|>
+\section{Diagonalizable Operator Bounded iff Value Set Bounded}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A: H \to H$ be a [[Definition:Diagonalizable Operator|diagonalizable operator]].
+Let $\left({\alpha_e}\right)_{e \in E}$ be the [[Definition:Diagonalizable Operator#Value Set|value set]] of $A$, with respect to a suitable [[Definition:Basis (Hilbert Space)|basis]] $E$ for $H$.
+Then $A$ is [[Definition:Bounded Linear Operator|bounded]] iff $\left({\alpha_e}\right)_{e \in E}$ is, i.e., iff:
+:$\exists M \in \R: \forall e \in E: \left\vert{\alpha_e}\right\vert \le M$
+\end{theorem}<|endoftext|>
+\section{Diagonalizable Operator Compact iff Value Set Converges to Zero}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] of [[Definition:Countable|countable]] [[Definition:Dimension (Hilbert Space)|dimension]].
+Let $A: H \to H$ be a [[Definition:Diagonalizable Operator|diagonalizable operator]].
+Let $\left({\alpha_n}\right)_{n \in \N}$ be the [[Definition:Diagonalizable Operator#Value Set|value set]] of $A$, with respect to a suitable [[Definition:Basis (Hilbert Space)|basis]] $E = \left({e_n}\right)_{n \in \N}$ for $H$.
+Then $A$ is [[Definition:Compact Linear Operator|compact]] iff:
+:$\displaystyle \lim_{n \to \infty} \alpha_n = 0$
+\end{theorem}<|endoftext|>
+\section{Nonzero Eigenvalue of Compact Operator has Finite Dimensional Eigenspace}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $T \in B_0 \left({H}\right)$ be a [[Definition:Compact Operator|compact operator]].
+Let $\lambda \in \sigma_p \left({T}\right), \lambda \ne 0$ be a nonzero [[Definition:Eigenvalue|eigenvalue]] of $T$.
+Then the [[Definition:Eigenspace|eigenspace]] for $\lambda$ has [[Definition:Finite|finite]] [[Definition:Dimension (Hilbert Space)|dimension]].
+\end{theorem}<|endoftext|>
+\section{Condition for Nonzero Eigenvalue of Compact Operator}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\Bbb F \in \set {\R, \C}$.
+Let $T \in \map {B_0} H$ be a [[Definition:Compact Operator|compact operator]].
+Let $\lambda \in \Bbb F, \lambda \ne 0$ be a nonzero [[Definition:Scalar (Vector Space)|scalar]].
+Suppose that the following holds:
+:$\inf \set {\norm {\paren {T - \lambda I} h}_H: \norm h_H = 1} = 0$
+Then $\lambda \in \map {\sigma_p} T$, that is, $\lambda$ is an [[Definition:Eigenvalue|eigenvalue]] for $T$.
+\end{theorem}<|endoftext|>
+\section{Finite Rank Operator is Compact}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H, K$ be [[Definition: Hilbert Space|Hilbert spaces]].
+Let $T \in \map {B_{00} } {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded]] [[Definition:Finite Rank Operator|finite rank operator]].
+Then $T \in \map {B_0} {H, K}$, that is, $T$ is [[Definition:Compact Linear Transformation|compact]].
+\end{theorem}
+
+\begin{proof}
+Since $T$ is bounded, it maps bounded sets to bounded sets. Because its range is finite dimensional, the closure of an image of a bounded set is compact. Therefore, $T$ is a compact linear operator.
+\end{proof}<|endoftext|>
+\section{Adjoint of Finite Rank Operator}
+Tags: Linear Transformations on Hilbert Spaces, Adjoints
+
+\begin{theorem}
+Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
+Let $T \in \map {B_{00} } {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded]] [[Definition:Finite Rank Operator|finite rank operator]].
+Then:
+:$T^* \in \map {B_{00} } {K, H}$
+that is, the [[Definition:Adjoint Linear Transformation|adjoint]] of $T$ is also a [[Definition:Bounded Linear Transformation|bounded]] [[Definition:Finite Rank Operator|finite rank operator]].
+\end{theorem}<|endoftext|>
+\section{Compact Idempotent is of Finite Rank}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $T \in \map {B_0} H$ be a [[Definition:Compact Linear Operator|compact linear operator]], and let $T$ be [[Definition:Idempotent Operator|idempotent]].
+Then:
+:$T \in \map {B_{00} } H$
+that is, $T$ is a [[Definition:Bounded Linear Transformation|bounded]] [[Definition:Finite Rank Operator|finite rank operator]].
+\end{theorem}<|endoftext|>
+\section{Characterization of Finite Rank Operators}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $T \in B_{00} \left({H}\right)$ be a [[Definition:Bounded Linear Operator|bounded]] [[Definition:Finite Rank Operator|finite rank operator]].
+Let $n = \operatorname{dim} \left({\operatorname{ran} T}\right)$ be the [[Definition:Rank|rank]] of $T$.
+Then there are [[Definition:Orthonormal|orthonormal]] [[Definition:Vector (Linear Algebra)|vectors]] $e_1, \ldots, e_n$ and [[Definition:Vector (Linear Algebra)|vectors]] $g_1, \ldots, g_n$ of $H$ such that:
+:$\forall h \in H: Th = \displaystyle \sum_{i=1}^n \left\langle{h, e_i}\right\rangle_H g_i$
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Induction}}
+\end{proof}<|endoftext|>
+\section{Compact Operator on Hilbert Space Direct Sum}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $\sequence {H_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Hilbert Space|Hilbert spaces]].
+Denote by $H = \displaystyle \bigoplus_{n \mathop = 1}^\infty H_n$ their [[Definition:Hilbert Space Direct Sum|Hilbert space direct sum]].
+For each $n \in \N$, let $T_n \in \map B {H_n}$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Suppose that one has $\displaystyle \sup_{n \mathop \in \N} \norm {T_n} < \infty$, where $\norm {\, \cdot \, }$ signifies the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear operators]].
+Define $T \in \map B H$ by: 
+:$\forall h = \sequence {h_n}_{n \mathop \in \N}: T h = \sequence {T_n h_n}_{n \mathop \in \N} \in H$
+(That $T$ is indeed [[Definition:Bounded Linear Operator|bounded]] follows from [[Bounded Linear Operator on Hilbert Space Direct Sum]].)
+Then $T$ is [[Definition:Compact Linear Operator|compact]] {{iff}} the following conditions hold:
+: For each $n \in \N$, $T_n$ is [[Definition:Compact Linear Operator|compact]]
+: $\displaystyle \lim_{n \mathop \to \infty} \norm {T_n} = 0$
+\end{theorem}<|endoftext|>
+\section{Bounded Linear Operator on Hilbert Space Direct Sum}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $\left({H_n}\right)_{n \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Hilbert Space|Hilbert spaces]].
+Denote by $H = \displaystyle \bigoplus_{n \mathop = 1}^\infty H_n$ their [[Definition:Hilbert Space Direct Sum|Hilbert space direct sum]].
+For each $n \in \N$, let $T_n \in B \left({H_n}\right)$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Suppose that one has $\displaystyle \sup_{n \mathop \in \N} \, \left\Vert{T_n}\right\Vert < \infty$, where $\left\Vert{\cdot}\right\Vert$ signifies the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear operators]].
+Define $T: H \to H$ by: 
+:$\forall h = \left({h_n}\right)_{n \in \N}: T h = \left({T_n h_n}\right)_{n \in \N} \in H$
+Then $T \in B \left({H}\right)$ is a [[Definition:Bounded Linear Operator|bounded linear operator]].
+\end{theorem}
+
+\begin{proof}
+Let $c = \displaystyle \sup_{n \mathop \in \N} \, \left\Vert{T_n}\right\Vert$.
+By assumption, $c < \infty$.
+Let $h = \left({h_n}\right)_{n \in \N} \in H$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn|l = \left\Vert{ T h }\right\Vert_H^2
+     |r = \left\Vert{ \left({T_n h_n}\right)_{n \in \N} }\right\Vert_H^2
+     |c = Definition of $T$
+}}
+{{eqn|r = \sum_{n \mathop = 1}^\infty \left\Vert{ T_n h_n }\right\Vert_{H_n}^2
+     |c = Definition of $\left\Vert{\cdot}\right\Vert_H$
+}}
+{{eqn|o = \le
+     |r = \sum_{n \mathop = 1}^\infty \left\Vert{ T_n }\right\Vert^2 \left\Vert{ h_n }\right\Vert_{H_n}^2
+}}
+{{eqn|o = \le
+     |r = \sum_{n \mathop = 1}^\infty c^2 \left\Vert{ h_n }\right\Vert_{H_n}^2
+}}
+{{eqn|r = c^2 \sum_{n \mathop = 1}^\infty \left\Vert{ h_n }\right\Vert_{H_n}^2
+}}
+{{eqn|r = c^2 \left\Vert{ h }\right\Vert_H^2
+}}
+{{end-eqn}}
+{{MissingLinks|results for some steps}}
+In summary:
+:$\left\Vert{ T h }\right\Vert_H^2 \le c^2 \left\Vert{ h }\right\Vert_H^2$
+It follows that $T$ is [[Definition:Bounded Linear Transformation|bounded]].
+{{qed}}
+[[Category:Linear Transformations on Hilbert Spaces]]
+t68mo69inf7su8yo7n236b3e1laax81
+\end{proof}<|endoftext|>
+\section{Compact Self-Adjoint Operator has Countable Point Spectrum}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $T \in B_0 \left({H}\right)$ be a [[Definition:Compact Linear Operator|compact]], [[Definition:Self-Adjoint Operator|self-adjoint operator]].
+Then its [[Definition:Point Spectrum|point spectrum]] $\sigma_p \left({T}\right)$ is [[Definition:Countable|countable]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Use the next result, the spectral theorem}}
+\end{proof}<|endoftext|>
+\section{Spectral Theorem for Compact Self-Adjoint Operators}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $T \in B_0 \left({H}\right)$ be a [[Definition:Compact Linear Operator|compact]], [[Definition:Self-Adjoint Operator|self-adjoint operator]].
+Then there exists a (possibly [[Definition:Finite Sequence|finite]]) [[Definition:Sequence|sequence]] $\left({\lambda_n}\right)$ of distinct nonzero [[Definition:Eigenvalue|eigenvalues]] of $T$ such that:
+* $P_n P_m = P_m P_n = 0$ if $n \ne m$
+* $\displaystyle \lim_{k \to \infty} \left\Vert{T - \sum_{n=1}^k \lambda_n P_n}\right\Vert = 0$, that is, $T = \displaystyle \sum_{n=1}^\infty \lambda_n P_n$
+where $P_n$ is the [[Definition:Orthogonal Projection|orthogonal projection]] onto the [[Definition:Eigenspace|eigenspace]] of $\lambda_n$, and $\left\Vert{\cdot}\right\Vert$ denotes the [[Definition:Norm on Bounded Linear Transformation|norm on bounded linear operators]].
+{{refactor|Split corollaries to subpages}}
+=== Corollary 1 ===
+There exists a (possibly [[Definition:Finite Sequence|finite]]) [[Definition:Sequence|sequence]] $\left({\mu_n}\right)$ of [[Definition:Real Number|real numbers]] and a [[Definition:Basis (Hilbert Space)|basis]] $E = \left({e_n}\right)$ for $\left({\operatorname{ker} T}\right)^\perp$ such that:
+:$\forall h \in H: Th = \displaystyle \sum_{n=1}^\infty \left\langle{h, e_n}\right\rangle_H \mu_n e_n$
+\end{theorem}<|endoftext|>
+\section{Eigenspace for Normal Operator is Reducing Subspace}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\Bbb F \in \left\{{\R, \C}\right\}$.
+Let $A \in B \left({H}\right)$ be a [[Definition:Normal Operator|normal operator]].
+Let $\lambda \in \Bbb F$.
+Then $\ker \left({A - \lambda}\right)$ is a [[Definition:Reducing Subspace|reducing subspace]] for $A$.
+Here $\ker$ denotes [[Definition:Kernel of Linear Transformation|kernel]].
+\end{theorem}
+
+\begin{proof}
+Since $A$ is [[Definition:Normal Operator|normal]], we have by [[Kernel of Linear Transformation is Orthocomplement of Range of Adjoint/Corollary|Kernel of Linear Transformation is Orthocomplement of Range of Adjoint: Corollary]] that:
+:$\ker A = \left({\operatorname{ran} A}\right)^\perp$
+and in particular, that:
+:$\ker A \subseteq \left({\operatorname{ran} A}\right)^\perp$
+Now, by [[Orthocomplement of Subset of Orthocomplement is Superset]]:
+:$\operatorname{ran} A \subseteq \left({\ker A}\right)^\perp$
+Applying this to the [[Definition:Normal Operator|normal operator]] $A - \lambda$, we find:
+:$\operatorname{ran} \left({A - \lambda}\right) \subseteq \left({\ker \left({A - \lambda}\right)}\right)^\perp$
+{{MissingLinks|Link to result proving $A - \lambda$ normal (not up presently)}}
+We are now set up to prove that $\ker \left({A - \lambda}\right)$ is a [[Definition:Reducing Subspace|reducing subspace]] for $A$.
+Let $x \in \ker \left({A - \lambda}\right)$.
+Then:
+{{begin-eqn}}
+{{eqn|l = A x
+     |r = \lambda x + \left({A - \lambda}\right) x
+}}
+{{eqn|r = \lambda x
+     |c = Definition of [[Definition:Kernel of Linear Transformation|kernel]], $x \in \ker \left({A - \lambda}\right)$
+}}
+{{end-eqn}}
+Therefore, $A \ker \left({A - \lambda}\right) \subseteq \ker \left({A - \lambda}\right)$; that is to say, $\ker \left({A - \lambda}\right)$ is [[Definition:Invariant Subspace|$A$-invariant]].
+Now, let $x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$.
+Observe that:
+:$A x = \lambda x + \left({A - \lambda}\right) x$
+Now $\left({A - \lambda}\right) x \in \operatorname{ran} \left({A - \lambda}\right)$, and by our derivation above, this means that:
+:$\left({A - \lambda}\right) x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$
+In conclusion, since $\left({\ker \left({A - \lambda}\right)}\right)^\perp$ is a [[Definition:Linear Subspace|linear subspace]] of $H$, it follows that:
+:$\lambda x + \left({A - \lambda}\right) x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$
+as desired.
+Hence both $\ker \left({A - \lambda}\right)$ and $\left({\ker \left({A - \lambda}\right)}\right)^\perp$ have been shown to be [[Definition:Invariant Subspace|$A$-invariant subspaces]] of $H$.
+That is, $\ker \left({A - \lambda}\right)$ is a [[Definition:Reducing Subspace|reducing subspace]] for $A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Eigenvalues of Normal Operator have Orthogonal Eigenspaces}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H}\right)$ be a [[Definition:Normal Operator|normal operator]].
+Let $\lambda, \mu$ be distinct [[Definition:Eigenvalue|eigenvalues]] of $A$.
+Then:
+:$\ker \left({A - \lambda}\right) \perp \ker \left({A - \mu}\right)$
+where:
+: $\ker$ denotes [[Definition:Kernel of Linear Transformation|kernel]]
+: $\perp$ denotes [[Definition:Orthogonal (Hilbert Space)#Sets|orthogonality]].
+\end{theorem}
+
+\begin{proof}
+{{improve}}
+Let $\mathcal V$ be an [[Definition:Inner Product Space|inner product space]].
+{{explain|The theorem calls for a [[Definition:Hilbert Space|Hilbert space]]. What is actually required?}}
+Let $T: \mathcal V \to \mathcal V$ be a [[Definition:Normal Operator|normal linear operator]].
+Requisite knowledge: $T^*$ is the [[Definition:Adjoint Linear Transformation|adjoint]] of $T$ and is defined by the fact that for any $u, w \in \mathcal V$, we have
+:$\left\langle{T u, w}\right\rangle = \left\langle{T^* w}\right\rangle$
+It is important to note the existence and uniqueness of adjoint operators.
+{{explain|Link to the required proofs and/or definitions as given above.}}
+'''Claim''': We know that for $v \in \mathcal V$:
+:$T v = \lambda v \iff T^* v = \overline \lambda v$
+This is true because for all [[Definition:Normal Operator|normal operators]], by definition:
+:$T^* T = T T*$
+and so:
+{{begin-eqn}}
+{{eqn | l = \left\Vert{T v}\right\Vert^2
+      | r = \left\langle{T v, T v}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\langle{T^* T v, v}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\langle{T T^* v, v}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\langle{T^* v, T^*v}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\Vert{T^* v}\right\Vert^2
+      | c = 
+}}
+{{end-eqn}}
+Since for [[Definition:Normal Operator|normal]] $T$, $\left({T - \lambda I}\right)$ is [[Definition:Normal Operator|normal]], we have:
+{{explain|Link to a page demonstrating the above}}
+{{begin-eqn}}
+{{eqn | l = T v
+      | r = \lambda v
+      | c = 
+}}
+{{eqn | ll= \iff 
+      | l = \left\Vert{\left({T - \lambda I}\right) v}\right\Vert
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \iff 
+      | l = \left\Vert{\left({T - \lambda I}\right)^* v}\right\Vert
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \iff 
+      | l = \left\Vert{T^* v - \overline \lambda v}\right\Vert
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \iff 
+      | l = T^* v
+      | r = \overline \lambda v
+      | c = 
+}}
+{{end-eqn}}
+Now, if $T v_1 = \lambda_1 v_1$ and $T v_2 = \lambda_2 v_2$, where $\lambda_1 \ne \lambda_2$ and $v_1, v_2$ are eigenvectors (i.e. $v_1, v_2 \ne \vec 0$), we have:
+{{begin-eqn}}
+{{eqn | l = \lambda_1 \left\langle{v_1, v_2}\right\rangle
+      | r = \left\langle{\lambda_1 v_1, v_2}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\langle{T v_1, v_2}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\langle{v_1, T^* v_2}\right\rangle
+      | c = 
+}}
+{{eqn | r = \left\langle{v_1, \overline{\lambda_2} v_2}\right\rangle
+      | c = 
+}}
+{{eqn | r = \lambda_2 \left\langle{v_1, v_2}\right\rangle
+      | c = 
+}}
+{{end-eqn}}
+Since $\lambda_1 \ne \lambda_2$, this is only possible if $\left\langle{v_1, v_2}\right\rangle = 0$, which means the [[Definition:Eigenvector|eigenvectors]] of our [[Definition:Normal Operator|normal operator]] are [[Definition:Orthogonal (Hilbert Space)#Sets|orthogonal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Eigenvalues of Self-Adjoint Operator are Real}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A \in B \left({H}\right)$ be a [[Definition:Self-Adjoint Operator|self-adjoint operator]].
+Then all [[Definition:Eigenvalue|eigenvalues]] of $A$ are [[Definition:Real Number|real]].
+\end{theorem}
+
+\begin{proof}
+Let $\lambda$ be an [[Definition:Eigenvalue|eigenvalue]] of $A$.
+Let $v \in H$ be an [[Definition:Eigenvector|eigenvector]] for $\lambda$; i.e. $Av = \lambda v$.
+Now compute:
+{{begin-eqn}}
+{{eqn|l = \lambda \left\langle{v, v}\right\rangle
+     |r = \left\langle{\lambda v, v}\right\rangle
+     |c = Property $(2)$ of an [[Definition:Inner Product|inner product]]
+}}
+{{eqn|r = \left\langle{Av, v}\right\rangle
+     |c = $v$ is an [[Definition:Eigenvector|eigenvector]]
+}}
+{{eqn|r = \left\langle{v, Av}\right\rangle
+     |c = $A$ is [[Definition:Self-Adjoint Operator|self-adjoint]]
+}}
+{{eqn|r = \overline{\left\langle{Av, v}\right\rangle}
+     |c = Property $(1)$ of an [[Definition:Inner Product|inner product]]
+}}
+{{eqn|r = \overline{\left\langle{\lambda v, v}\right\rangle}
+     |c = $v$ is an [[Definition:Eigenvector|eigenvector]]
+}}
+{{eqn|r = \bar \lambda \left\langle{v, v}\right\rangle
+     |c = Properties $(2)$, $(4)$ of an [[Definition:Inner Product|inner product]] 
+}}
+{{end-eqn}}
+Now $v$, being an [[Definition:Eigenvector|eigenvector]], is [[Definition:Zero Vector|non-zero]].
+By property $(5)$ of an [[Definition:Inner Product|inner product]], this implies $\left\langle{v, v}\right\rangle \ne 0$.
+Thence, dividing out $\left\langle{v, v}\right\rangle$, obtain $\lambda = \bar \lambda$.
+From [[Complex Number equals Conjugate iff Wholly Real]], $\lambda \in \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Operator Diagonalizable iff Basis of Eigenvectors}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A:H \to H$ be a [[Definition:Linear Operator|linear operator]] on $H$.
+Then $A$ is [[Definition:Diagonalizable Operator|diagonalizable]] iff there exists a [[Definition:Basis (Hilbert Space)|basis]] $E$ of $H$, consisting of [[Definition:Eigenvector|eigenvectors]] for $A$.
+\end{theorem}<|endoftext|>
+\section{Existence of Hartogs Number}
+Tags: Ordinals, Existence of Hartogs Number
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then $S$ has a [[Definition:Hartogs Number|Hartogs number]].
+\end{theorem}
+
+\begin{proof}
+From [[Hartogs' Lemma (Set Theory)|Hartogs' lemma]] there exists an [[Definition:Ordinal|ordinal]] $\alpha$ such that there is no [[Definition:Injection|injection]] from $\alpha$ to $S$.
+We also have that [[Ordinals are Well-Ordered]].
+It follows from the definition of [[Definition:Well-Ordering|well-ordering]] that there exists a [[Definition:Smallest Element|smallest]] such [[Definition:Ordinal|ordinal]].
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows immediately from [[Cardinal Equal to Collection of All Dominated Ordinals]].
+The collection of all [[Definition:Dominate (Set Theory)|dominated]] [[Definition:Ordinal|ordinals]]:
+:$\set {y \in \On: y \preccurlyeq S}$
+is the [[Definition:Hartogs Number|Hartogs number]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse not always Unique for Non-Associative Operation}
+Tags: Inverse Elements
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $\circ$ be a [[Definition:Associative|non-associative]] operation.
+Then for any $x \in S$, it is possible for $x$ to have more than one [[Definition:Inverse Element|inverse element]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Consider the [[Definition:Algebraic Structure|algebraic structure]] $\left({S, \circ}\right)$ consisting of:
+:The [[Definition:Set|set]] $S = \left\{{a, b, e}\right\}$
+:The [[Definition:Binary Operation|binary operation]] $\circ$
+whose [[Definition:Cayley Table|Cayley table]] is given as follows:
+:$\begin{array}{c|cccc}
+\circ & e & a & b \\
+\hline
+e & e & a & b \\
+a & a & e & e \\
+b & b & e & e \\
+\end{array}$
+By inspection, we see that $e$ is the [[Definition:Identity Element|identity element]] of $\left({S, \circ}\right)$.
+We also note that:
+{{begin-eqn}}
+{{eqn | l = \left({a \circ a}\right) \circ b
+      | r = e \circ b
+      | c = 
+}}
+{{eqn | r = b
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = a \circ \left({a \circ b}\right)
+      | r = a \circ e
+      | c = 
+}}
+{{eqn | r = a
+      | c = 
+}}
+{{end-eqn}}
+and so $\circ$ is not [[Definition:Associative|associative]].
+Note further that:
+{{begin-eqn}}
+{{eqn | l = a \circ b
+      | r = e
+      | c = 
+}}
+{{eqn | r = b \circ a
+      | c = 
+}}
+{{end-eqn}}
+and also:
+{{begin-eqn}}
+{{eqn | l = a \circ a
+      | r = e
+      | c = 
+}}
+{{end-eqn}}
+So both $a$ and $b$ are [[Definition:Inverse Element|inverses]] of $a$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left and Right Inverse Mappings Implies Bijection}
+Tags: Bijections, Inverse Mappings, Left and Right Inverse Mappings Implies Bijection
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $f$ be such that:
+: $\exists g_1: T \to S: g_1 \circ f = I_S$
+: $\exists g_2: T \to S: f \circ g_2 = I_T$
+where both $g_1$ and $g_2$ are [[Definition:Mapping|mappings]].
+Then $f$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+Suppose:
+: $\exists g_1: T \to S: g_1 \circ f = I_S$
+: $\exists g_2: T \to S: f \circ g_2 = I_T$
+From [[Injection iff Left Inverse]], it follows that $f$ is an [[Definition:Injection|injection]].
+From [[Surjection iff Right Inverse]], it follows that $f$ is a [[Definition:Surjection|surjection]].
+So $f$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]] and, by definition, therefore also a [[Definition:Bijection|bijection]].
+\end{proof}
+
+\begin{proof}
+Suppose:
+: $\exists g_1: T \to S: g_1 \circ f = I_S$
+: $\exists g_2: T \to S: f \circ g_2 = I_T$
+We have that the [[Identity Mapping is Bijection]], thus $I_S$ and $I_T$ are both [[Definition:Bijection|bijections]].
+From [[Injection if Composite is Injection]], if $I_S = g_1 \circ f$ is an [[Definition:Injection|injection]], then so is $f$.
+From [[Surjection if Composite is Surjection]], if $I_T = f \circ g_2$ is a [[Definition:Surjection|surjection]], then so is $f$.
+So $f$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]] and, by definition, therefore also a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left and Right Inverses of Mapping are Inverse Mapping}
+Tags: Bijections, Inverse Mappings, Left and Right Inverses of Mapping are Inverse Mapping
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]] such that:
+:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
+:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
+Then:
+: $g_1 = g_2 = f^{-1}$
+where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = g_2
+      | r = I_S \circ g_2
+      | c = {{Defof|Identity Mapping}}
+}}
+{{eqn | r = \paren {g_1 \circ f} \circ g_2
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = g_1 \circ \paren {f \circ g_2}
+      | c = [[Composition of Mappings is Associative]]
+}}
+{{eqn | r = g_1 \circ I_T
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = g_1
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+From [[Left and Right Inverse Mappings Implies Bijection]] it follows that $f$ is a [[Definition:Bijection|bijection]].
+It follows from [[Composite of Bijection with Inverse is Identity Mapping]] that $g_1 = g_2 = f^{-1}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Left and Right Inverse Mappings Implies Bijection]], we have that $f$ is a [[Definition:Bijection|bijection]].
+Let $y \in T$ and let $x = f^{-1} \left({y}\right)$.
+Such an $x$ exists because $f$ is [[Definition:Surjection|surjective]], and [[Definition:Unique|unique]] within $S$ as $f$ is [[Definition:Injection|injective]].
+Then $y = f \left({x}\right)$ and so:
+{{begin-eqn}}
+{{eqn | l = f^{-1} \left({y}\right)
+      | r = x
+      | c = 
+}}
+{{eqn | r = I_S \left({x}\right)
+      | c = $I_S$ is the [[Definition:Identity Mapping|identity mapping]] on $S$
+}}
+{{eqn | r = g_1 \circ f \left({x}\right)
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = g_1 \left({y}\right)
+      | c = from above: $y = f \left({x}\right)$
+}}
+{{end-eqn}}
+So $f^{-1} = g_1$.
+Also:
+{{begin-eqn}}
+{{eqn | l = f \left({x}\right)
+      | r = y
+      | c = 
+}}
+{{eqn | r = I_T \left({y}\right)
+      | c = $I_T$ is the [[Definition:Identity Mapping|identity mapping]] on $T$
+}}
+{{eqn | r = f \circ g_2 \left({y}\right)
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \implies
+      | l = f^{-1} \left({y}\right)
+      | r = x
+      | c = as $f$ is [[Definition:Injection|injective]]
+}}
+{{eqn | r = g_2 \left({y}\right)
+      | c = 
+}}
+{{end-eqn}}
+So $f^{-1} = g_2$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Because [[Composition of Mappings is Associative]], brackets do not need to be used.
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = g_1 \circ f
+      | r = I_S
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = g_1 \circ f \circ g_2
+      | r = I_S \circ g_2
+      | c = 
+}}
+{{eqn | r = g_2
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = f \circ g_2
+      | r = I_T
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = g_1 \circ f \circ g_2
+      | r = g_1 \circ I_T
+      | c = 
+}}
+{{eqn | r = g_1
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+Thus $g_1 = g_2$.
+Now suppose there exists $g_3: T \to S: g_3 \circ f = I_S$.
+By the same argument as above, $g_3 = g_2$.
+This means that $g_1 (= g_3)$ is the only [[Definition:Left Inverse Mapping|left inverse]] of $f$.
+Similarly, suppose there exists $g_4: T \to S: f \circ g_4 = I_T$.
+By the same argument as above, $g_4 = g_1$.
+This means that $g_2 (= g_4)$ is the only [[Definition:Right Inverse Mapping|right inverse]] of $f$.
+So $g_1 = g_2 = g_3 = g_4$ are all the same.
+By [[Composite of Bijection with Inverse is Identity Mapping]], it follows that this unique mapping is the [[Definition:Inverse Mapping|inverse]] $f^{-1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Operator Commuting with Diagonalizable Operator}
+Tags: Linear Transformations on Hilbert Spaces
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
+Let $A = \displaystyle \sum_{i \mathop \in I} \alpha_i P_i$ be a [[Definition:Diagonalizable Operator|diagonalizable operator]] on $H$.
+Let $B \in \map B H$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
+Then the following are equivalent:
+:$(1): \quad A B = B A$
+:$(2): \quad$ For all $i \in I$, $\Rng {P_i}$ is a [[Definition:Reducing Subspace|reducing subspace]] for $B$
+where $\Rng {P_i}$ denotes [[Definition:Range of Relation|range]].
+\end{theorem}<|endoftext|>
+\section{Composition of Mappings is Composition of Relations}
+Tags: Composite Mappings, Composite Relations
+
+\begin{theorem}
+Let $S_1, S_2, S_3$ be [[Definition:Set|sets]].
+Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be [[Definition:Mapping|mappings]] such that the [[Definition:Domain of Mapping|domain]] of $f_2$ is the same set as the [[Definition:Codomain of Mapping|codomain]] of $f_1$.
+Let $f_2 \circ f_1$ be the [[Definition:Composition of Mappings|composite of $f_1$ and $f_2$]].
+Let $f_1$ and $f_2$ be considered as [[Definition:Relation|relations]] on $S_1 \times S_2$ and $S_2 \times S_3$ respectively.
+Then $f_2 \circ f_1$ defined as the [[Definition:Composition of Relations|composition of relations]] coincides with the definition of $f_2 \circ f_1$ as the [[Definition:Composition of Mappings|composition of mappings]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$f_2 \circ f_1 = \left\{{\left({x, z}\right) \in S_1 \times S_3: \exists y \in S_2: \left({x, y}\right) \in f_1 \land \left({y, z}\right) \in f_2}\right\}$
+Let $\mathcal R_1 \subseteq S_1 \times T_1$ and $\mathcal R_2 \subseteq S_2 \times T_2$ be [[Definition:Relation|relations]].
+The [[Definition:Composition of Relations|composite of $\mathcal R_1$ and $\mathcal R_2$]] is defined as:
+:$\mathcal R_2 \circ \mathcal R_1 := \left\{{\left({x, z}\right) \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \left({x, y}\right) \in \mathcal R_1 \land \left({y, z}\right) \in \mathcal R_2}\right\}$
+When $T_1 = S_2$ and $T_2 = S_3$, this becomes:
+:$\mathcal R_2 \circ \mathcal R_1 := \left\{{\left({x, z}\right) \in S_1 \times T_3: \exists y \in S_2: \left({x, y}\right) \in \mathcal R_1 \land \left({y, z}\right) \in \mathcal R_2}\right\}$
+The definitions can be seen to be identical.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequence in Exponential Sequence}
+Tags: Exponential Function, Sequences, Null Sequence in Exponential Sequence
+
+\begin{theorem}
+Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a [[Definition:Complex Sequence|sequence of complex numbers]] such that:
+:$\displaystyle \lim_{n \mathop \to +\infty}a_n = 0$
+Then:
+:$\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$
+\end{theorem}<|endoftext|>
+\section{Identity Mapping is Right Identity}
+Tags: Identity Mappings
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then:
+:$f \circ I_S = f$
+where $I_S$ is the [[Definition:Identity Mapping|identity mapping]] on $S$, and $\circ$ signifies [[Definition:Composition of Mappings|composition of mappings]].
+\end{theorem}
+
+\begin{proof}
+=== Equality of Codomains ===
+The [[Definition:Codomain of Mapping|codomains]] of $f$ and $f \circ I_S$ are both equal to $T$ from [[Codomain of Composite Relation]].
+{{qed|lemma}}
+=== Equality of Domains ===
+From [[Domain of Composite Relation]]:
+:$\Dom {f \circ I_S} = \Dom {I_S}$
+But from the definition of the [[Definition:Identity Mapping|identity mapping]]:
+:$\Dom {I_S} = \Img {I_S} = S$
+{{qed|lemma}}
+=== Equality of Mappings ===
+The [[Definition:Composition of Mappings|composite of $I_S$ and $f$]] is defined as:
+:$f \circ I_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in I_S \land \tuple {y, z} \in f}$
+But by definition of the [[Definition:Identity Mapping|identity mapping]] on $S$, we have that:
+:$\tuple {x, y} \in I_S \implies x = y$
+Hence:
+:$f \circ I_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y}\ \in I_S \land \tuple {y, z} \in f}$
+But as $\forall y \in S: \tuple {y, y} \in I_S$, this means:
+:$f \circ I_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in f}$
+That is:
+:$f \circ I_S = f$
+{{qed|lemma}}
+Hence the result, by [[Equality of Mappings]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, a [[Definition:Mapping|mapping]] is also a [[Definition:Relation|relation]].
+Also by definition, the [[Definition:Identity Mapping|identity mapping]] is the same as the [[Definition:Diagonal Relation|diagonal relation]].
+Thus [[Diagonal Relation is Right Identity]] can be applied directly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Relation is Left Identity}
+Tags: Relation Theory, Identity Mappings
+
+\begin{theorem}
+Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
+Then:
+:$\Delta_T \circ \RR = \RR$
+where $\Delta_T$ is the [[Definition:Diagonal Relation|diagonal relation]] on $T$, and $\circ$ signifies [[Definition:Composition of Relations|composition of relations]].
+\end{theorem}
+
+\begin{proof}
+We use the [[Equality of Relations|definition of relation equality]], as follows:
+=== Equality of Domains ===
+The [[Definition:Domain of Relation|domains]] of $\RR$ and $\Delta_T \circ \RR$ are equal from [[Domain of Composite Relation]]:
+:$\Dom {\Delta_T \circ \RR} = \Dom \RR$
+=== Equality of Codomains ===
+From [[Codomain of Composite Relation]], the [[Definition:Codomain of Relation|codomains]] of $\Delta_T \circ \RR$ and $\Delta_T$ are both equal to $T$.
+But from the definition of the [[Definition:Diagonal Relation|diagonal relation]], the [[Definition:Codomain of Relation|codomain]] of $\Delta_T$ is $\Dom {\Delta_T} = T$.
+=== Equality of Relations ===
+The [[Definition:Composition of Relations|composite of $\RR$ and $\Delta_T$]] is defined as:
+:$\Delta_T \circ \RR = \set {\tuple {x, z} \in S \times T: \exists y \in T: \tuple {x, y} \in \RR \land \tuple {y, z} \in \Delta_T}$
+But by definition of the [[Definition:Diagonal Relation|diagonal relation]] on $T$, we have that:
+:$\tuple {y, z} \in \Delta_T \implies y = z$
+Hence:
+:$\Delta_T \circ \RR = \set {\tuple {x, y} \in S \times T: \exists y \in T: \tuple {x, y} \in \RR \land \tuple {y, y} \in \Delta_T}$
+But as $\forall y \in T: \tuple {y, y} \in \Delta_T$, this means:
+:$\Delta_T \circ \RR = \set {\tuple {x, y} \in S \times T: \tuple {x, y} \in \RR}$
+That is:
+:$\Delta_T \circ \RR = \RR$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Associativity}
+Tags: Isomorphism Preserves Associativity, Isomorphisms
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Then $\circ$ is [[Definition:Associative|associative]] {{iff}} $*$ is [[Definition:Associative|associative]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ is [[Definition:Associative|associative]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism|isomorphism]].
+As an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is [[Definition:Surjection|surjective]], it follows that:
+:$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$
+So:
+{{begin-eqn}}
+{{eqn | l = \paren {u * v} * w
+      | r = \paren {\map \phi x * \map \phi y} * \map \phi z
+      | c = as $\phi$ is a [[Definition:Surjection|Surjection]]
+}}
+{{eqn | r = \map \phi {x \circ y} * \map \phi z
+      | c = {{Defof|Morphism Property}}
+}}
+{{eqn | r = \map \phi {\paren {x \circ y} \circ z}
+      | c = {{Defof|Morphism Property}}
+}}
+{{eqn | r = \map \phi {x \circ \paren {y \circ z} }
+      | c = [[Definition:Associative|Associativity of $\circ$]]
+}}
+{{eqn | r = \map \phi x * \map \phi {y \circ z}
+      | c = {{Defof|Morphism Property}}
+}}
+{{eqn | r = \map \phi x * \paren {\map \phi y * \map \phi z}
+      | c = {{Defof|Morphism Property}}
+}}
+{{eqn | r = u * \paren {v * w}
+      | c = by definition as above
+}}
+{{end-eqn}}
+As $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], it follows from [[Inverse of Algebraic Structure Isomorphism is Isomorphism]] that $\phi^{-1}$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Thus the result for $\phi$ can be applied to $\phi^{-1}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Bijection|bijection]].
+By definition, an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Surjection|surjection]].
+That is, an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] which is also an [[Definition:Injection|injection]].
+Thus [[Epimorphism Preserves Associativity]] can be applied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Commutativity}
+Tags: Isomorphisms, Isomorphism Preserves Commutativity
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Then $\circ$ is [[Definition:Commutative Operation|commutative]] {{iff}} $*$ is [[Definition:Commutative Operation|commutative]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ is [[Definition:Commutative Operation|commutative]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+As an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is [[Definition:Surjection|surjective]], it follows that:
+:$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$
+So:
+{{begin-eqn}}
+{{eqn | l = u * v
+      | r = \map \phi x * \map \phi y
+      | c = [[Definition:Surjection|as $\phi$ is a surjection]]
+}}
+{{eqn | r = \map \phi {x \circ y}
+      | c = [[Definition:Morphism Property|Morphism Property]]
+}}
+{{eqn | r = \map \phi {y \circ x}
+      | c = [[Definition:Commutative Operation|Commutativity of $\circ$]]
+}}
+{{eqn | r = \map \phi y * \map \phi x
+      | c = [[Definition:Morphism Property|Morphism Property]]
+}}
+{{eqn | r = v * u
+      | c = by definition as above
+}}
+{{end-eqn}}
+As $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], it follows from [[Inverse of Algebraic Structure Isomorphism is Isomorphism]] that $\phi^{-1}$ is also a [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Thus the result for $\phi$ can be applied to $\phi^{-1}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Bijection|bijection]].
+By definition, an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Surjection|surjection]].
+That is, an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] which is also an [[Definition:Injection|injection]].
+Thus [[Epimorphism Preserves Commutativity]] can be applied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Inverses}
+Tags: Isomorphisms, Isomorphism Preserves Inverses
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Let $\left({S, \circ}\right)$ have an [[Definition:Identity Element|identity]] $e_S$.
+Then $x^{-1}$ is an [[Definition:Inverse Element|inverse]] of $x$ for $\circ$ [[Definition:Iff|iff]] $\phi \left({x^{-1}}\right)$ is an [[Definition:Inverse Element|inverse]] of $\phi \left({x}\right)$ for $*$.
+That is, [[Definition:Iff|iff]]:
+: $\phi \left({x^{-1}}\right) = \left({\phi \left({x}\right)}\right)^{-1}$
+\end{theorem}
+
+\begin{proof}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
+From [[Epimorphism Preserves Identity]], it follows that $\left({T, *}\right)$ also has an identity, which is $\phi \left({e_S}\right)$.
+Let $y$ be an [[Definition:Inverse Element|inverse]] of $x$ in $\left({S, \circ}\right)$. 
+Then:
+{{begin-eqn}}
+{{eqn | l = \phi \left({x}\right) * \phi \left({y}\right)
+      | r = \phi \left({x \circ y}\right)
+      | c = 
+}}
+{{eqn | r = \phi \left({e_S}\right)
+      | c = 
+}}
+{{eqn | r = \phi \left({y \circ x}\right)
+      | c = 
+}}
+{{eqn | r = \phi \left({y}\right) * \phi \left({x}\right)
+      | c = 
+}}
+{{end-eqn}}
+So $\phi \left({y}\right)$ is an [[Definition:Inverse Element|inverse]] of $\phi \left({x}\right)$ in $\left({T, *}\right)$.
+As $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], it follows from [[Inverse of Algebraic Structure Isomorphism is Isomorphism]] that $\phi^{-1}$ is also a [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Thus the result for $\phi \left({x}\right)$ can be applied to $\phi^{-1} \left({\phi \left({x}\right)}\right)$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Bijection|bijection]].
+By definition, an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Surjection|surjection]].
+That is, an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] which is also an [[Definition:Injection|injection]].
+Thus [[Epimorphism Preserves Inverses]] can be applied.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+An [[Definition:Group Isomorphism|group isomorphism]] is by definition a [[Definition:Group Epimorphism|group epimorphism]].
+The result follows from [[Epimorphism Preserves Inverses]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $g \in G$.
+{{begin-eqn}}
+{{eqn | l = \phi \left({g}\right) * \phi \left({g^{-1} }\right)
+      | r = \phi \left({g \circ g^{-1} }\right)
+      | c = {{Defof|Group Isomorphism}}
+}}
+{{eqn | r = \phi \left({e_G}\right)
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | r = e_H
+      | c = [[Group Isomorphism Preserves Identity]]
+}}
+{{end-eqn}}
+It follows from [[Inverse in Group is Unique]] that $\phi \left({g^{-1} }\right)$ is the [[Definition:Unique|unique]] [[Definition:Inverse Element|inverse element]] of $\phi \left({g}\right)$ in $\left({H, *}\right)$.
+That is:
+: $\forall g \in G: \phi \left({g^{-1}}\right) = \left({\phi \left({g}\right)}\right)^{-1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Identity}
+Tags: Isomorphisms, Isomorphism Preserves Identity
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Then $\circ$ has an [[Definition:Identity Element|identity]] $e_S$ {{iff}} $\phi \left({e_S}\right)$ is the [[Definition:Identity Element|identity]] for $*$.
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Group Isomorphism|group isomorphism]] is by definition a [[Definition:Group Epimorphism|group epimorphism]].
+The result follows from [[Epimorphism Preserves Identity]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \phi {e_G}
+      | r = \map \phi {e_G \circ e_G}
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
+      | c = {{Defof|Group Isomorphism}}
+}}
+{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
+      | c = {{Defof|Group Isomorphism}}
+}}
+{{end-eqn}}
+It follows from [[Identity is only Idempotent Element in Group]] that $\map \phi {e_G}$ is the [[Definition:Identity Element|identity]] of $H$.
+That is:
+:$\map \phi {e_G} = e_H$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
+Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.
+The result follows directly from the [[Definition:Morphism Property|morphism property]] of $\circ$ under $\phi$:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x \circ e_S}
+      | r = \map \phi x
+      | c =
+}}
+{{eqn | r = \map \phi {e_S \circ x}
+      | c =
+}}
+{{eqn | l = \map \phi x * \map \phi {e_S}
+      | r = \map \phi x
+      | c = 
+}}
+{{eqn | r = \map \phi {e_S} * \map \phi x
+      | c = 
+}}
+{{end-eqn}}
+As $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], it follows from [[Inverse of Algebraic Structure Isomorphism is Isomorphism]] that $\phi^{-1}$ is also a [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Thus the result for $\map \phi {e_S}$ can be applied to $\map {\phi^{-1} } {\map \phi {e_S} }$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Bijection|bijection]].
+By definition, an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Surjection|surjection]].
+That is, an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] which is also an [[Definition:Injection|injection]].
+Thus [[Epimorphism Preserves Identity]] can be applied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Natural Numbers under Multiplication form Commutative Semigroup}
+Tags: Natural Numbers, Examples of Semigroups
+
+\begin{theorem}
+Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without [[Definition:Zero (Number)|zero]], that is, $\N_{>0} = \N \setminus \set 0$.
+The [[Definition:Algebraic Structure|structure]] $\struct {\N_{>0}, \times}$ forms an [[Definition:Infinite Set|infinite]] [[Definition:Commutative Semigroup|commutative semigroup]].
+\end{theorem}
+
+\begin{proof}
+=== Closure ===
+From [[Natural Numbers have No Proper Zero Divisors]]:
+:$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$
+It follows that:
+:$\forall m, n \in \N_{>0}: m \times n \ne 0$
+and so:
+:$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$
+That is, $\struct {\N_{>0}, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== Associativity ===
+[[Natural Number Multiplication is Associative]].
+{{qed|lemma}}
+=== Commutativity ===
+	
+[[Natural Number Multiplication is Commutative]].
+{{qed|lemma}}
+=== Infinite ===
+We have that the [[Natural Numbers are Infinite]].
+Then we have that [[Infinite if Injection from Natural Numbers]].
+The mapping $s: \N \to \N_{>0}: \map s n = n + 1$ is such an [[Definition:Injection|injection]].
+Hence $\N_{>0}$ is [[Definition:Infinite Set|infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Groups}
+Tags: Isomorphisms, Group Theory, Isomorphism Preserves Groups
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+If $\left({S, \circ}\right)$ is a [[Definition:Group|group]], then so is $\left({T, *}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Isomorphism Preserves Semigroups]], if $\left({S, \circ}\right)$ is a [[Definition:Semigroup|semigroup]] then so is $\left({T, *}\right)$.
+From [[Isomorphism Preserves Identity]], if $\left({S, \circ}\right)$ has an [[Definition:Identity Element|identity]] $e_S$, then $\phi \left({e_S}\right)$ is the [[Definition:Identity Element|identity]] for $*$.
+From [[Isomorphism Preserves Inverses]], if $x^{-1}$ is an [[Definition:Inverse Element|inverse]] of $x$ for $\circ$, then $\phi \left({x^{-1}}\right)$ is an [[Definition:Inverse Element|inverse]] of $\phi \left({x}\right)$ for $*$.
+The result follows from the definition of [[Definition:Group|group]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
+The result follows as a direct [[Definition:Corollary|corollary]] of [[Epimorphism Preserves Groups]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Semigroups}
+Tags: Isomorphisms, Semigroups, Isomorphism Preserves Semigroups
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: S \to T$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+If $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], then so is $\struct {T, *}$.
+\end{theorem}
+
+\begin{proof}
+If $\left({S, \circ}\right)$ is a [[Definition:Semigroup|semigroup]], then by definition it is [[Definition:Closed Algebraic Structure|closed]].
+From [[Morphism Property Preserves Closure]], $\left({T, *}\right)$ is therefore also [[Definition:Closed Algebraic Structure|closed]].
+If $\left({S, \circ}\right)$ is a [[Definition:Semigroup|semigroup]], then by definition $\circ$ is [[Definition:Associative|associative]].
+From [[Isomorphism Preserves Associativity]], $*$ is therefore also [[Definition:Associative|associative]].
+So $\left({T, *}\right)$ is closed, and $*$ is associative, and therefore [[Definition:Semigroup|by definition]], $\left({T, *}\right)$ is a [[Definition:Semigroup|semigroup]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
+The result follows as a direct [[Definition:Corollary|corollary]] of [[Epimorphism Preserves Semigroups]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left Cancellable iff Left Regular Representation Injective}
+Tags: Regular Representations
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Then $a \in S$ is [[Definition:Left Cancellable Element|left cancellable]] {{iff}} the [[Definition:Left Regular Representation|left regular representation]] $\map {\lambda_a} x$ is [[Definition:Injection|injective]].
+\end{theorem}
+
+\begin{proof}
+Suppose $a \in S$ is [[Definition:Left Cancellable Element|left cancellable]].
+Then:
+:$a \circ x = a \circ y \implies x = y$
+From the definition of the [[Definition:Left Regular Representation|left regular representation]]:
+:$\map {\lambda_a} x = a \circ x$
+Thus:
+:$\map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
+and so the [[Definition:Left Regular Representation|left regular representation]] is [[Definition:Injection|injective]].
+{{qed|lemma}}
+Suppose $\map {\lambda_a} x$ is [[Definition:Injection|injective]].
+Then:
+:$\map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
+From the definition of the [[Definition:Left Regular Representation|left regular representation]]:
+:$\map {\lambda_a} x = a \circ x$
+Thus:
+: $a \circ x = a \circ y \implies x = y$
+and so $a$ is [[Definition:Left Cancellable Element|left cancellable]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Right Cancellable iff Right Regular Representation Injective}
+Tags: Regular Representations
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Then $a \in S$ is [[Definition:Right Cancellable Element|right cancellable]] {{iff}} the [[Definition:Right Regular Representation|right regular representation]] $\map {\rho_a} x$ is [[Definition:Injection|injective]].
+\end{theorem}
+
+\begin{proof}
+Suppose $a \in S$ is [[Definition:Right Cancellable Element|right cancellable]].
+Then:
+:$x \circ a = y \circ a \implies x = y$
+From the definition of the [[Definition:Right Regular Representation|right regular representation]]:
+:$\map {\rho_a} x = x \circ a$
+Thus:
+:$\map {\rho_a} x = \map {\rho_a} y \implies x = y$
+and so the [[Definition:Right Regular Representation|right regular representation]] is [[Definition:Injection|injective]].
+{{qed|lemma}}
+Suppose $\map {\rho_a} x$ is [[Definition:Injection|injective]].
+Then:
+:$\map {\rho_a} x = \map {\rho_a} y \implies x = y$
+From the definition of the [[Definition:Right Regular Representation|right regular representation]]:
+:$\map {\rho_a} x = x \circ a$
+Thus:
+: $x \circ a = y \circ a \implies x = y$
+and so $a$ is [[Definition:Right Cancellable Element|right cancellable]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Cancellability}
+Tags: Isomorphisms, Cancellability
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Then:
+: $a \in S$ is [[Definition:Cancellable Element|cancellable]] in $\left({S, \circ}\right)$ [[Definition:Iff|iff]] $\phi \left({a}\right) \in T$ is [[Definition:Cancellable Element|cancellable]] in $\left({T, *}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $a$ is [[Definition:Cancellable Element|cancellable]].
+From [[Isomorphism Preserves Left Cancellability]] and [[Isomorphism Preserves Right Cancellability]]:
+: $a \in S$ is [[Definition:Left Cancellable Element|left cancellable]] in $\left({S, \circ}\right)$ [[Definition:Iff|iff]] $\phi \left({a}\right) \in T$ is [[Definition:Left Cancellable Element|left cancellable]] in $\left({T, *}\right)$
+and
+: $a \in S$ is [[Definition:Right Cancellable Element|right cancellable]] in $\left({S, \circ}\right)$ [[Definition:Iff|iff]] $\phi \left({a}\right) \in T$ is [[Definition:Right Cancellable Element|right cancellable]] in $\left({T, *}\right)$.
+It follows that:
+: $a \in S$ is [[Definition:Cancellable Element|cancellable]] in $\left({S, \circ}\right)$
+[[Definition:Iff|iff]]
+: $\phi \left({a}\right) \in T$ is [[Definition:Cancellable Element|cancellable]] in $\left({T, *}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Left Cancellability}
+Tags: Isomorphisms, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Then:
+:$a \in S$ is [[Definition:Left Cancellable Element|left cancellable]] in $\struct {S, \circ}$
+{{iff}}:
+:$\map \phi a \in T$ is [[Definition:Left Cancellable Element|left cancellable]] in $\struct {T, *}$.
+\end{theorem}
+
+\begin{proof}
+Let $a$ be [[Definition:Left Cancellable Element|left cancellable]] in $\struct {S, \circ}$.
+Let $x \in S$ and $y \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi a * \map \phi x
+      | r = \map \phi a * \map \phi y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi {a \circ x}
+      | r = \map \phi {a \circ y}
+      | c = [[Definition:Morphism Property|Morphism Property]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi x
+      | r = \map \phi y
+      | c = as $a$ is [[Definition:Left Cancellable Element|left cancellable]]
+}}
+{{end-eqn}}
+That is, $\map \phi a$ is [[Definition:Left Cancellable Element|left cancellable]] in $\struct {T, *}$.
+As $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], then so is $\phi^{-1}$.
+So the same proof works in reverse in exactly the same way.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism Preserves Right Cancellability}
+Tags: Isomorphisms, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Then:
+:$a \in S$ is [[Definition:Right Cancellable Element|right cancellable]] in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is [[Definition:Right Cancellable Element|right cancellable]] in $\struct {T, *}$.
+\end{theorem}
+
+\begin{proof}
+Let $a$ be [[Definition:Right Cancellable Element|right cancellable]] in $\struct {S, \circ}$.
+Let $x \in S$ and $y \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi x * \map \phi a
+      | r = \map \phi y * \map \phi a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \map \phi {x \circ a}
+      | r = \map \phi {y \circ a}
+      | c = [[Definition:Morphism Property|Morphism property]]
+}}
+{{eqn | ll= \implies
+      | l = \map \phi x
+      | r = \map \phi y
+      | c = as $a$ is [[Definition:Right Cancellable Element|right cancellable]]
+}}
+{{end-eqn}}
+That is, $\map \phi a$ is [[Definition:Right Cancellable Element|right cancellable]] in $T$.
+As $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], then so is $\phi^{-1}$.
+So the same proof works in reverse in exactly the same way.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differentiation of Power Series}
+Tags: Differential Calculus, Power Series
+
+\begin{theorem}
+Let $\xi \in \R$ be a [[Definition:Real Number|real number]].
+Let $\left \langle {a_n} \right \rangle$ be a [[Definition:Sequence|sequence in $\R$]].
+Let $\displaystyle \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m$ be the [[Definition:Power Series|power series]] in $x$ about the point $\xi$.
+Then within the [[Definition:Interval of Convergence|interval of convergence]]:
+:$\displaystyle \frac {\mathrm d^n} {\mathrm d x^n} \sum_{m \mathop \ge 0} a_m \left({x - \xi}\right)^m = \sum_{m \mathop \ge n} a_m m^{\underline n} \left({x - \xi}\right)^{m - n}$
+where $m^{\underline n}$ denotes the [[Definition:Falling Factorial|falling factorial]].
+\end{theorem}
+
+\begin{proof}
+First we can make the substitution $z = x - \xi$ and convert the expression into:
+:$\displaystyle \frac {\mathrm d^n} {\mathrm d x^n} \sum_{m \mathop \ge 0} a_m z^m$
+We then use [[Nth Derivative of Mth Power]]:
+:$\displaystyle \frac {\mathrm d^n} {\mathrm d z^n} z^m = \begin{cases}
+m^{\underline n} z^{m - n} & : n \le m \\
+0 & : n > m
+\end{cases}$
+[[Definition:By Hypothesis|By hypothesis]] $x$ is within the [[Definition:Interval of Convergence|interval of convergence]].
+It follows from [[Abel's Theorem]] that:
+:$\displaystyle \frac {\mathrm d^n}{\mathrm dz^n} \sum_{m \mathop \ge 0} a_m z^m = \sum_{m \mathop \ge n} a_m m^{\underline n} z^{m - n}$
+Then from [[Derivative of Identity Function]] and others, we have:
+: $\displaystyle \frac {\mathrm d} {\mathrm d x} \left({x - \xi}\right) = 1$
+The result follows from the [[Chain Rule for Derivatives]].
+{{qed}}
+{{Handwaving|An explanation needed for how [[Abel's Theorem]] allows term-by-term differentiation of the infinite power series.}}
+[[Category:Differential Calculus]]
+[[Category:Power Series]]
+8xgvqg2hwbch8uw5h77lw0hmg688v5d
+\end{proof}<|endoftext|>
+\section{General Associativity Theorem/Formulation 1}
+Tags: General Associativity Theorem
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
+Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Element|elements]] of $S$.
+Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] of [[Definition:Natural Numbers|natural numbers]] such that $r_0 = p$ and $r_s = p+n$.
+Suppose:
+:$\displaystyle \forall k \in \closedint 1 s: b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} {a_j}$
+Then:
+:$\displaystyle \forall n \in \N_{>0}: \prod_{k \mathop = 1}^s {b_k} = \prod_{k \mathop = p \mathop + 1}^{p \mathop + n} {a_k}$
+That is:
+:$\displaystyle \forall n \in \N_{>0}: \prod_{k \mathop = 1}^s \paren {a_{r_{k - 1} + 1} \circ a_{r_{k - 1} + 2} \circ \ldots \circ a_{r_k} } = a_{p + 1} \circ \ldots \circ a_{p + n}$
+\end{theorem}
+
+\begin{proof}
+The proof will proceed by the [[Principle of Mathematical Induction]] on $\N$.
+Let $T$ be the [[Definition:Set|set]] of all $n \in \N_{>0}$ such that:
+:$(1): \quad$ for every [[Definition:Sequence|sequence]] $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ of [[Definition:Element|elements]] of $S$
+and:
+:$(2): \quad$ for every [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of [[Definition:Natural Numbers|natural numbers]] such that $r_0 = p$ and $r_s = p + n$:
+the statement:
+:$\displaystyle b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} a_j$
+holds.
+=== Basis for the Induction ===
+Let $n = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r_s
+      | r = r_0 + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = s
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \prod_{k \mathop = 1}^s b_k
+      | r = b_1
+      | c = 
+}}
+{{eqn | r = a_{p + 1}
+      | c = 
+}}
+{{eqn | r = \prod_{k \mathop = p + 1}^{p \mathop + n} a_k
+      | c = 
+}}
+{{end-eqn}}
+So $1 \in T$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+It is to be shown that, if $m \in T$ where $m \ge 1$, then it follows that $m + 1 \in T$.
+This is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$(1): \quad$ for every [[Definition:Sequence|sequence]] $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m}$ of [[Definition:Element|elements]] of $S$
+and:
+:$(2): \quad$ for every [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of [[Definition:Natural Numbers|natural numbers]] such that $r_0 = p$ and $r_s = p + m$:
+the statement:
+:$\displaystyle b_k = \prod_{j \mathop = r_{k-1} \mathop + 1}^{r_k} a_j$
+holds.
+It is to be demonstrated that it follows that:
+:$(1): \quad$ for every [[Definition:Sequence|sequence]] $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m + 1}$ of [[Definition:Element|elements]] of $S$
+and:
+:$(2): \quad$ for every [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of [[Definition:Natural Numbers|natural numbers]] such that $r_0 = p$ and $r_s = p + m + 1$:
+the statement:
+:$\displaystyle b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} a_j$
+holds.
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m + 1}$ be a [[Definition:Sequence|sequence]] of [[Definition:Element|elements]] of $S$.
+Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] of [[Definition:Natural Numbers|natural numbers]] such that $r_0 = p$ and $r_s = p + m + 1$.
+Then $r_{s - 1} \le p + m$.
+There are two cases:
+:$(1): \quad r_{s - 1} = p + m$
+:$(2): \quad r_{s - 1} < p + m$
+First, suppose:
+:$r_{s - 1} = p + m$
+Then:
+:$b_s = a_{p + m + 1}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = a_{p + 1} \circ \ldots \circ a_{p + m}
+      | r = b_1 \circ \ldots \circ b_{s-1}
+      | c = $m \in T$ by [[General Associativity Theorem/Formulation 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | ll=\leadsto
+      | l = a_{p + 1} \circ \ldots \circ a_{p + m + 1}
+      | r = \paren {a_{p + 1} \circ \ldots \circ a_{p + m} } \circ a_{p + m + 1}
+      | c = {{Defof|Composite (Abstract Algebra)}}
+}}
+{{eqn | r = \paren {b_1 \circ \ldots \circ b_{s - 1} } \circ b_s
+      | c = 
+}}
+{{eqn | r = b_1 \circ \ldots \circ b_s
+      | c = {{Defof|Composite (Abstract Algebra)}}
+}}
+{{end-eqn}}
+Secondly, suppose:
+:$r_{s - 1} < p + m$
+Let $b\,'_s = a_{r_{s - 1} + 1} \circ \ldots \circ a_{r_s + 1}$.
+Then by definition of [[Definition:Composite (Abstract Algebra)|composite]]:
+:$b_s = b\,'_s \circ a_{p + m + 1}$
+Now by the [[General Associativity Theorem/Formulation 1#Induction Hypothesis|Induction Hypothesis]]:
+:$m \in T \implies a_{p + 1} \circ \ldots \circ a_{p + m} = b_1 \circ \ldots \circ b_{s - 1} \circ b\,'_s$
+Thus:
+{{begin-eqn}}
+{{eqn | l = b_1 \circ \ldots \circ b_s
+      | r = \paren {b_1 \circ \ldots \circ b_{s - 1} } \circ \paren {b\,'_s \circ a_{p + m + 1} }
+      | c = 
+}}
+{{eqn | r = \paren {\paren {b_1 \circ \ldots \circ b_{s - 1} } \circ b\,'_s} \circ a_{p + m + 1}
+      | c = 
+}}
+{{eqn | r = \paren {b_1 \circ \ldots \circ b_{s - 1} \circ b\,'_s} \circ a_{p + m + 1}
+      | c = 
+}}
+{{eqn | r = \paren {a_{p + 1} \circ \ldots \circ a_{p + m} } \circ a_{p + m + 1}
+      | c = 
+}}
+{{eqn | r = a_{p + 1} \circ \ldots \circ a_{p + m} \circ a_{p + m + 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus in both cases $m + 1 \in T$.
+So by the [[Principle of Mathematical Induction]]:
+:$T = \N_{>0}$
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{General Associativity Theorem/Formulation 2}
+Tags: General Associativity Theorem
+
+\begin{theorem}
+Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
+Let $\circ$ be an [[Definition:Associative Operation|associative operation]] on $S$.
+Let the [[Definition:Set|set]] $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
+:$\map {P_1} {a_1} = \set {a_1}$
+:$\map {P_2} {a_1, a_2} = \set {a_1 \circ a_2}$
+:$\map {P_n} {a_1, a_2, \ldots, a_n} = \set {x \circ y: x \in \map {P_r} {a_1, a_2, \ldots, a_r} \land y \in \map {P_s} {a_{r + 1}, a_{r + 2}, \ldots, a_{r + s} }, n = r + s}$
+Then $\map {P_n} {a_1, a_2, \ldots, a_n}$ consists of a unique entity which we can denote $a_1 \circ a_2 \circ \ldots \circ a_n$.
+\end{theorem}<|endoftext|>
+\section{Monoid of Self-Inverse Elements is Abelian Group}
+Tags: Monoids, Abelian Groups
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Monoid|monoid]] such that:
+:$\forall x \in S: x \circ x = e$
+where $e$ is the [[Definition:Identity Element|identity element]] of $\left({S, \circ}\right)$.
+Then $\left({S, \circ}\right)$ is an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From [[Equivalence of Definitions of Self-Inverse]], $x \circ x = e \implies x = x^{-1}$.
+From [[Inverse in Monoid is Unique]], it follows that every element of $\left({S, \circ}\right)$ has a unique [[Definition:Inverse Element|inverse]].
+So by definition, $\left({S, \circ}\right)$ is a [[Definition:Group|group]].
+From [[All Elements Self-Inverse then Abelian]], it follows that $\left({S, \circ}\right)$ is an [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product is Left Identity Therefore Left Cancellable}
+Tags: Semigroups, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
+Let $e_L \in S$  be a [[Definition:Left Identity|left identity]] of $S$.
+Let $a \in S$ such that:
+:$\exists b \in S: b \circ a = e_L$
+Then $a$ is [[Definition:Left Cancellable Element|left cancellable]] in $\struct {S, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \circ x
+      | r = a \circ y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = b \circ \paren {a \circ x}
+      | r = b \circ \paren {a \circ y}
+      | c =
+}}
+{{eqn | ll= \implies
+      | l = \paren {b \circ a} \circ x
+      | r = \paren {b \circ a} \circ y
+      | c = as $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], $\circ$ is [[Definition:Associative|associative]]
+}}
+{{eqn | ll= \implies
+      | l = e_L \circ x
+      | r = e_L \circ y
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = y
+      | c = Definition of [[Definition:Left Identity|Left Identity]]
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Left Cancellable Element|left cancellable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product is Right Identity Therefore Right Cancellable}
+Tags: Semigroups, Cancellability
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Semigroup|semigroup]].
+Let $e_R \in S$  be a [[Definition:Right Identity|right identity]] of $S$.
+Let $a \in S$ such that:
+:$\exists b \in S: a \circ b = e_R$
+Then $a$ is [[Definition:Right Cancellable Element|right cancellable]] in $\left({S, \circ}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \circ a
+      | r = y \circ a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({x \circ a}\right) \circ b
+      | r = \left({y \circ a}\right) \circ b
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({x \circ a}\right) \circ b
+      | r = \left({y \circ a}\right) \circ b
+      | c = as $\left({S, \circ}\right)$ is a [[Definition:Semigroup|semigroup]], $\circ$ is [[Definition:Associative|associative]]
+}}
+{{eqn | ll= \implies
+      | l = x \circ e_R
+      | r = y \circ e_R
+      | c = [[Definition:By Hypothesis|By hypothesis]]
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = y
+      | c = Definition of [[Definition:Right Identity|Right Identity]]
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Right Cancellable Element|right cancellable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiples Greater than Positive Integer Closed under Addition}
+Tags: Integer Addition, Algebraic Closure
+
+\begin{theorem}
+Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
+Let $p \in \Z: p \ge 0$ be a [[Definition:Positive Integer|positive integer]].
+Let $S \subseteq n \Z$  be defined as:
+:$S := \set {x \in n \Z: x > p}$
+that is, the [[Definition:Set|set]] of [[Definition:Set of Integer Multiples|integer multiples of $n$]] greater than $p$.
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, +}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Addition|addition]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+From [[Integer Multiples Closed under Addition]], $x + y \in n \Z$.
+As $x, y > p$ we have that:
+:$\exists r \in \Z_{>0}: x = p + r$
+:$\exists s \in \Z_{>0}: y = p + s$
+Thus it follows that:
+{{begin-eqn}}
+{{eqn | l = x + y
+      | r = \paren {p + r} + \paren {p + s}
+      | c = 
+}}
+{{eqn | r = 2 p + r + s
+      | c = 
+}}
+{{eqn | o = >
+      | r = p
+      | c = as $r, s > 0$
+}}
+{{end-eqn}}
+So $x + y > p$ and $x + y \in n \Z$.
+Hence by definition $x + y \in S$, and so $S$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Addition|addition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiples Greater than Positive Integer Closed under Multiplication}
+Tags: Integer Multiplication, Algebraic Closure
+
+\begin{theorem}
+Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
+Let $p \in \Z: p \ge 0$ be a [[Definition:Positive Integer|positive integer]].
+Let $S \subseteq n \Z$  be defined as:
+:$S := \left\{{x \in n \Z: x > p}\right\}$
+that is, the [[Definition:Set|set]] of [[Definition:Set of Integer Multiples|integer multiples of $n$]] greater than $p$.
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \times}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+From [[Integer Multiples Closed under Multiplication]], $x y \in n \Z$.
+As $x, y > p$ we have that:
+:$\exists r \in \Z_{>0}: x = p + r$
+:$\exists s \in \Z_{>0}: y = p + s$
+Thus it follows that:
+{{begin-eqn}}
+{{eqn | l = x + y
+      | r = \paren {p + r} \paren {p + s}
+      | c = 
+}}
+{{eqn | r = p^2 + p \paren {r + s} + r s
+      | c = 
+}}
+{{eqn | o = >
+      | r = p
+      | c = as $r, s > 0$
+}}
+{{end-eqn}}
+So $x y > p$ and $x y \in n \Z$.
+Hence by definition $x y \in S$, and so $S$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Multiplication|multiplication]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Rational Numbers Closed under Multiplication}
+Tags: Rational Multiplication, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Zero (Algebra)|non-zero]] [[Definition:Rational Number|rational numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Rational Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+Recall that [[Rational Numbers form Field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]].
+By definition of a [[Definition:Field (Abstract Algebra)|field]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Group|group]].
+Thus, by definition, $\times$ is [[Definition:Closed Algebraic Structure|closed]] in $\struct {\Q_{\ne 0}, \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Real Numbers Closed under Multiplication}
+Tags: Real Multiplication, Algebraic Closure, Non-Zero Real Numbers Closed under Multiplication
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Real Number|real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]]:
+:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$
+\end{theorem}
+
+\begin{proof}
+Recall that [[Real Numbers form Field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]].
+By definition of a [[Definition:Field (Abstract Algebra)|field]], the [[Definition:Algebraic Structure|algebraic structure]] $\left({\R_{\ne 0}, \times}\right)$ is a [[Definition:Group|group]].
+Thus, by definition, $\times$ is [[Definition:Closed Algebraic Structure|closed]] in $\left({\R_{\ne 0}, \times}\right)$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x \times y = 0$.
+{{WLOG}}, suppose that $x \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \times y
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 x \times \paren {x \times y}
+      | r = \frac 1 x \times 0
+      | c = as $x \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\frac 1 x \times x} \times y
+      | r = \frac 1 x \times 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1 \times y
+      | r = \frac 1 x \times 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 4$: Inverse]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \frac 1 x \times 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 3$: Identity]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = 0
+      | c = [[Real Zero is Zero Element]]
+}}
+{{end-eqn}}
+Thus:
+:$x \times y = 0, x \ne 0 \implies y = 0$
+[[Definition:Mutatis Mutandis|Mutatis mutandis]]
+:$x \times y = 0, y \ne 0 \implies x = 0$
+and so:
+:$x \times y = 0 \implies y = 0 \lor x = 0$
+So:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x \times y = 0
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \paren {x = 0} \lor \paren {y = 0}
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \neg \paren {x \ne 0 \land y \ne 0}
+      | c = [[De Morgan's Laws (Logic)/Disjunction|De Morgan's Laws: Disjunction]]
+}}
+{{end-eqn}}
+The result follows by the [[Rule of Transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Complex Numbers Closed under Multiplication}
+Tags: Complex Multiplication, Algebraic Closure, Non-Zero Complex Numbers Closed under Multiplication
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Zero (Algebra)|non-zero]] [[Definition:Complex Number|complex numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+Recall that [[Complex Numbers form Field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]].
+By definition of a [[Definition:Field (Abstract Algebra)|field]], the [[Definition:Algebraic Structure|algebraic structure]] $\left({\C_{\ne 0}, \times}\right)$ is a [[Definition:Group|group]].
+Thus, by definition, $\times$ is [[Definition:Closed Algebraic Structure|closed]] in $\left({\C_{\ne 0}, \times}\right)$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $z_1, z_2 \in \C_{\ne 0}$.
+Then by definition of [[Definition:Complex Number/Definition 1|complex number]]:
+:$z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$
+for some $x_1, y_1, x_2, y_2 \in \R$ such that:
+:$x_1 \ne 0$ or $y_1 \ne 0$
+:$x_2 \ne 0$ or $y_2 \ne 0$
+Expressing $z_1$ and $z_2$ in [[Definition:Exponential Form of Complex Number|exponential form]] (although [[Definition:Polar Form of Complex Number|polar form]] is equally adequate):
+:$z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$
+for some $r_1, r_2, \theta_1, \theta_2 \in \R$.
+Then by [[Product of Complex Numbers in Polar Form]]:
+:$z_1 \times z_2 = \paren {r_1 \times r_2} e^{i \paren {\theta_1 + \theta_2} }$
+By definition of [[Definition:Exponential Form of Complex Number|exponential form]]:
+:$r_1 = \sqrt {x_1^2 + y_1^2}$
+:$r_2 = \sqrt {x_2^2 + y_2^2}$
+Thus $r_1 > 0$ and $r_2 > 0$.
+Hence $r_1 \times r_2 > 0$ and so $z_1 \times z_2 \ne 0$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Equivalently this is to say:
+:$z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$
+Let $z_1 z_2 = 0$.
+{{begin-eqn}}
+{{eqn | l = z_1
+      | r = \tuple {x_1, y_1}
+      | c = {{Defof|Complex Number|index = 2}}: for some $x_1, y_1 \in \R$
+}}
+{{eqn | l = z_2
+      | r = \tuple {x_2, y_2}
+      | c = {{Defof|Complex Number|index = 2}}: for some $x_2, y_2 \in \R$
+}}
+{{eqn | ll= \leadsto
+      | l = z_1 z_2
+      | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2}
+      | c = Definition of [[Definition:Complex Number/Definition 2/Multiplication|Complex Multiplication]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = x_1 x_2
+      | r = y_1 y_2
+      | c = 
+}}
+{{eqn | n = 2
+      | lo= \land
+      | l = x_1 y_2
+      | r = -y_1 x_2
+      | c = as $z_1 z_2 = 0$
+}}
+{{end-eqn}}
+{{WLOG}}, let $\tuple {x_2, y_2} \ne \tuple {0, 0}$.
+Suppose also that $\tuple {x_1, y_1} \ne \tuple {0, 0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | r = \frac {y_1 y_2} {x_2}
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y_1 y_2} {x_2} y_2
+      | r = -y_1 x_2
+      | c = substituting in $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = y_2^2
+      | r = - x_2^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_2 = y_2
+      | r = 0
+      | c = as both $x_2$ and $y_2$ are [[Definition:Real Number|real]]
+}}
+{{end-eqn}}
+From this [[Proof by Contradiction|contradiction]] it follows that:
+:$\tuple {x_1, y_1} = \tuple {0, 0}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Modulo Numbers Closed under Multiplication then Modulo is Prime}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $\left({\Z_m, +_m, \times_m}\right)$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]] for $m > 1$.
+Let $\Z'_m$ be the [[Definition:Set|set]] of [[Definition:Zero (Algebra)|non-zero]] [[Definition:Integers Modulo m|integers modulo $m$]].
+Let $\left({\Z_m, \times_m}\right)$ be [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Modulo Multiplication|modulo multiplication]].
+Then $m$ is [[Definition:Prime Number|prime]].
+\end{theorem}
+
+\begin{proof}
+Suppose $m$ is not [[Definition:Prime Number|prime]].
+Then $m = r s$ for some $r, s \in \Z: 1 < r < m, 1 < s < m$.
+So $r, s \in \Z'_m$.
+But:
+:$r \times_m s \equiv 0 \pmod m$
+and so $r \times_m s \notin \Z'_m$.
+So if $m$ is not [[Definition:Prime Number|prime]], $\left({\Z_m, \times_m}\right)$ is not [[Definition:Closed Algebraic Structure|closed]].
+The result follows from the [[Rule of Transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Magma is Submagma of Itself}
+Tags: Magmas
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Then $\left({S, \circ}\right)$ is a [[Definition:Submagma|submagma]] of itself.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Magma|magma]] is an [[Definition:Algebraic Structure|algebraic structure]] $\left({S, \circ}\right)$ where $\circ$ is [[Definition:Closed Algebraic Structure|closed]].
+That is:
+:$\forall x, y \in S: x \circ y \in S$
+From [[Set is Subset of Itself]], $S \subseteq S$.
+By definition, $\left({T, \circ}\right)$ is a [[Definition:Submagma|submagma]] of $S$ if:
+:$\forall x, y \in T: x \circ y \in T$
+It follows from the above that $\left({S, \circ}\right)$ is, by definition, a [[Definition:Submagma|submagma]] of itself.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Submagma of Magma}
+Tags: Magmas
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Magma|magma]].
+Then:
+: $\left({\varnothing, \circ}\right)$ is a [[Definition:Submagma|submagma]] of $\left({S, \circ}\right)$
+where $\varnothing$ is the [[Definition:Empty Set|empty set]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Magma|magma]] is an [[Definition:Algebraic Structure|algebraic structure]] $\left({S, \circ}\right)$ where $\circ$ is [[Definition:Closed Algebraic Structure|closed]].
+That is:
+:$\forall x, y \in S: x \circ y \in S$
+By definition, $\left({T, \circ}\right)$ is a [[Definition:Submagma|submagma]] of $S$ if:
+:$\forall x, y \in T: x \circ y \in T$
+But:
+:$\not \exists x, y \in \varnothing: x \circ y \notin \varnothing$
+it follows [[Definition:Vacuous Truth|vacuously]] that:
+:$\forall x, y \in \varnothing: x \circ y \in \varnothing$
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Max Operation on Toset is Semigroup}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $\max \left({x, y}\right)$ denote the [[Definition:Max Operation|max operation]] on $x, y \in S$.
+Then $\left({S, \max}\right)$ is a [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+By the definition of the [[Definition:Max Operation|max operation]], either:
+:$\max \left({x, y}\right) = x$
+or
+:$\max \left({x, y}\right) = y$
+So $\max$ is [[Definition:Closed Algebraic Structure|closed on $S$]].
+Then we have that the [[Max and Min are Associative|max operation is associative]]:
+: $\forall x, y, z \in S: \max \left({x, \max \left({y, z}\right)}\right) = \max \left({\max \left({x, y}\right), z}\right)$
+Hence the result, by definition of [[Definition:Semigroup|semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Min Semigroup on Toset is Semilattice}
+Tags: Max and Min Operations, Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]]. 
+Then the [[Min Operation on Toset is Semigroup|min semigroup]] $\struct {S, \min}$ is a [[Definition:Semilattice|semilattice]].
+\end{theorem}
+
+\begin{proof}
+The [[Min Semigroup is Commutative]] and [[Min Semigroup is Idempotent|idempotent]]. 
+Hence the result, by definition of a [[Definition:Semilattice|semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Min Semigroup is Commutative}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]]. 
+Then the [[Min Operation on Toset is Semigroup|semigroup $\left({S, \min}\right)$]] is [[Definition:Commutative Semigroup|commutative]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$. 
+We have that the [[Max and Min are Commutative|min operation is commutative]]:
+: $\min \left({x, y}\right) = \min \left({y, x}\right)$ 
+Hence the result, by definition of [[Definition:Commutative Semigroup|commutative semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Min Semigroup is Idempotent}
+Tags: Max and Min Operations, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Then the [[Definition:Semigroup|semigroup]] $\struct {S, \min}$ is an [[Definition:Idempotent Semigroup|idempotent semigroup]].
+\end{theorem}
+
+\begin{proof}
+The fact that $\struct {S, \min}$ is a [[Definition:Semigroup|semigroup]] is demonstrated in [[Min Operation on Toset is Semigroup]].
+Then the [[Max and Min are Idempotent|min operation is idempotent]]:
+:$\forall x \in S: \min \set {x, x} = x$
+The result follows by the definition of [[Definition:Idempotent Semigroup|idempotent semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Max Operation on Woset is Monoid}
+Tags: Max and Min Operations, Well-Orderings, Examples of Monoids
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
+Let $\max \set {x, y}$ denote the [[Definition:Max Operation|max operation]] on $x, y \in S$.
+Then $\struct {S, \max}$ is a [[Definition:Monoid|monoid]].
+Its [[Definition:Identity Element|identity element]] is the [[Definition:Smallest Element|smallest element]] of $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Well-Ordering is Total Ordering]], we have that $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]].
+Thus, by [[Max Operation on Toset is Semigroup]], $\struct {S, \max}$ is a [[Definition:Semigroup|semigroup]].
+By definition, a [[Definition:Well-Ordered Set|well-ordered set]] has a [[Definition:Smallest Element|smallest element]].
+So:
+:$\exists m \in S: \forall x \in S: m \preceq x$
+where $m$ is that [[Definition:Smallest Element|smallest element]].
+It follows by definition of the [[Definition:Max Operation|max operation]] that:
+:$\forall x \in S: \max \set {m, x} = x$
+Thus $m$ is the [[Definition:Identity Element|identity element]] of $\struct {S, \max}$.
+Hence the result, by definition of [[Definition:Monoid|monoid]].
+{{qed}}
+[[Category:Max and Min Operations]]
+[[Category:Well-Orderings]]
+[[Category:Examples of Monoids]]
+9k93zgqq1r9n7kp28j7h8ndq6a8hhdv
+\end{proof}<|endoftext|>
+\section{Max Operation on Natural Numbers forms Monoid}
+Tags: Natural Numbers, Examples of Monoids, Max and Min Operations
+
+\begin{theorem}
+Let $\left({\N, \max}\right)$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed from the [[Definition:Natural Numbers|natural numbers]] $\N$ and the [[Definition:Max Operation|max operation]].
+Then $\left({\N, \max}\right)$ is a [[Definition:Monoid|monoid]].
+Its [[Definition:Identity Element|identity element]] is the [[Definition:Zero|zero]].
+\end{theorem}
+
+\begin{proof}
+By the [[Well-Ordering Principle]], $\N$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+The result follows from [[Max Operation on Woset is Monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Modulus Function is Continuous}
+Tags: Complex Modulus, Continuous Functions
+
+\begin{theorem}
+Let $z_0 \in \C$ be a [[Definition:Complex Number|complex number]].
+Then the [[Definition:Modulus of Complex Number|complex modulus function]] is [[Definition:Continuous Complex Function|continuous]] at $z_0$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon>0$.
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]] satisfying $\left\vert{z - z_0}\right\vert < \epsilon$.
+By the [[Reverse Triangle Inequality/Real and Complex Fields|Reverse Triangle Inequality]]:
+: $\left\vert{ \left\vert{z}\right\vert - \left\vert{z_0}\right\vert }\right\vert \le \left\vert{z - z_0}\right\vert < \epsilon$
+Hence the result, by the [[Definition:Continuous Complex Function#Epsilon-Delta Definition|$\epsilon$-$\delta$ definition of continuity]] (taking $\delta = \epsilon$).
+{{qed}}
+[[Category:Complex Modulus]]
+[[Category:Continuous Functions]]
+c5b85ilzan02knmwpgg6iyhjvy62lag
+\end{proof}<|endoftext|>
+\section{Subset of Natural Numbers under Max Operation is Monoid}
+Tags: Natural Numbers, Examples of Monoids, Max and Min Operations
+
+\begin{theorem}
+Let $S \subseteq \N$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] $\N$.
+Let $\left({S, \max}\right)$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed from $S$ and the [[Definition:Max Operation|max operation]].
+Then $\left({S, \max}\right)$ is a [[Definition:Monoid|monoid]].
+Its [[Definition:Identity Element|identity element]] is the [[Definition:Smallest Element|smallest element]] of $S$.
+\end{theorem}
+
+\begin{proof}
+By the [[Well-Ordering Principle]], $\N$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+By definition, every [[Definition:Subset|subset]] of a [[Definition:Well-Ordered Set|well-ordered set]] is also [[Definition:Well-Ordered Set|well-ordered]].
+Thus $S$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+The result follows from [[Max Operation on Woset is Monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Space is Banach Space}
+Tags: Euclidean Space, Banach Spaces, Euclidean Space is Banach Space
+
+\begin{theorem}
+Let $m$ be a [[Definition:Positive Integer|positive integer]].
+Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Norm|Euclidean norm]], forms a [[Definition:Banach Space|Banach space]] over $\R$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Euclidean Space|Euclidean space]] $\R^m$ is a [[Definition:Vector Space|vector space]] over $\R$.
+That the [[Definition:Norm Axioms (Vector Space)|norm axioms]] are satisfied is proven in [[Euclidean Space is Normed Space]].
+Then we have [[Euclidean Space is Complete Metric Space]].
+The result follows by the definition of a [[Definition:Banach Space|Banach space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, [[Definition:Euclidean Norm|Euclidean norm]] is the same as [[Definition:P-Norm|$p$-norm]] with $p = 2$.
+Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m}}}_{n \mathop \in \N} $ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\R^m$.
+Let $k \in \N_{> 0} : k \le m$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {x_n^{\paren k} - x_m^{\paren k} }
+      | r = \paren {\paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}
+}}
+{{eqn | o = \le
+      | r = \paren {\sum_{k \mathop = 0}^m \paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}
+}}
+{{eqn | r = \norm { {\bf x}_n - {\bf x}_m}_2
+}}
+{{end-eqn}}
+Hence, $\sequence {x_n^{\paren k} }_{n \mathop \in \N}$ is a [[Definition:Real Cauchy Sequence|Cauchy sequence]] in $\R$.
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n^{\paren k} = L^{\paren k}$
+Let $\mathbf L = \tuple {L^{\paren 1}, \ldots, L^{\paren m}} \in \R^m$
+We have that for all $n > N$:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x_n - \mathbf L}
+      | r = \paren {\sum_{k \mathop = 1}^m \size {x^{\paren k}_n - x^{\paren k} }^2}^{\frac 1 2}
+}}
+{{eqn | o = \le
+      | r = \paren {\sum_{k \mathop = 1}^m \frac {\epsilon^2} m}^{\frac 1 2}
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges to $\mathbf L$ in $\struct {\R^m, \norm {\, \cdot \,}_2}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Plane is Complete Metric Space}
+Tags: Complete Metric Spaces
+
+\begin{theorem}
+The [[Definition:Complex Number|complex plane]], along with the [[Definition:Metric Induced by Norm|metric induced]] by the norm given by the [[Definition:Modulus of Complex Number|complex modulus]], forms a [[Definition:Complete Metric Space|complete metric space]].
+\end{theorem}
+
+\begin{proof}
+Let $z = x + iy$ be a [[Definition:Complex Number|complex number]], where $x, y \in \R$.
+Now, we can identify the complex number $z$ with the [[Definition:Ordered Pair|ordered pair]] $\left( x, y \right) \in \R^2$.
+The norm on $\C$ given by the [[Definition:Modulus of Complex Number|complex modulus]] is then identical to the [[Definition:Euclidean Norm|Euclidean norm]] on $\R^2$.
+Therefore, metric on $\C$ [[Definition:Metric Induced by Norm|induced]] by the norm given by the [[Definition:Modulus of Complex Number|complex modulus]] is identical to the [[Definition:Euclidean Metric on Real Number Plane|Euclidean metric]] on $\R^2$ induced by the [[Definition:Euclidean Norm|Euclidean norm]].
+By [[Euclidean Space is Complete Metric Space]], $\C$ is a complete metric space.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Space is Complete Metric Space}
+Tags: Euclidean Space, Complete Metric Spaces
+
+\begin{theorem}
+Let $m$ be a [[Definition:Positive Integer|positive integer]].
+Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]], forms a [[Definition:Complete Metric Space|complete metric space]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {\tuple {x_{n,1}, x_{n,2}, \ldots, x_{n,m} } }_{n \mathop \in \N} $ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\R^m$.
+Let $\epsilon > 0$.
+Then $\dfrac \epsilon m > 0$.
+We have that [[Real Number Line is Complete Metric Space]].
+Therefore:
+:$\exists y_1, y_2, \ldots, y_m \in \R$ and $N_1, N_2, \ldots, N_m \in \N$ (depending on $\epsilon$)
+such that:
+:$\forall k \in \N: 1 \le k \le m: \forall n_k > N_k: \sequence {x_{n,k} – y_k}  < \dfrac \epsilon m$
+From [[Euclidean Space is Normed Space]]:
+:$\displaystyle \norm {\tuple {x_{n,1}, x_{n,2}, \ldots, x_{n,m} } - \tuple {y_1, y_2, \ldots, y_m} } \le \sum_{k \mathop = 1}^m \size {x_{n, k} – y_k} < \epsilon$
+Hence the [[Definition:Euclidean Space|Euclidean space]] is a [[Definition:Complete Metric Space|complete metric space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Plane is Banach Space}
+Tags: Banach Spaces
+
+\begin{theorem}
+The [[Definition:Complex Plane|complex plane]], along with the [[Definition:Modulus of Complex Number|complex modulus]], forms a [[Definition:Banach Space|Banach space]] over $\C$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Complex Plane|complex plane]] $\C$ is a [[Definition:Vector Space|vector space]] over $\C$.
+That the [[Definition:Norm Axioms (Vector Space)|norm axioms]] are satisfied is proven in [[Complex Modulus is Norm]].
+Then we have [[Complex Plane is Complete Metric Space]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Number Space is not Complete Metric Space}
+Tags: Rational Number Space
+
+\begin{theorem}
+Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]] under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]] $d$.
+Then $\struct {\Q, \tau_d}$ is not a [[Definition:Complete Metric Space|complete metric space]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+First note that [[Rational Numbers form Metric Space]].
+It remains to be shown that $\struct {\Q, \tau_d}$ is not [[Definition:Complete Metric Space|complete]].
+Consider the [[Definition:Sequence|sequence]] $\sequence {a_n}$ given by:
+:$a_n := \dfrac {f_{n + 1} } {f_n}$
+where $\sequence {f_n}$ is the sequence of [[Definition:Fibonacci Numbers|Fibonacci numbers]].
+By [[Ratio of Consecutive Fibonacci Numbers]]:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = \phi := \dfrac {1 + \sqrt 5} 2$
+But [[Square Root of Prime is Irrational]].
+That means $\dfrac {1 + \sqrt 5} 2$ is likewise [[Definition:Irrational Number|irrational]].
+Thus $\sequence {a_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]] of [[Definition:Rational Number|rational numbers]] which [[Definition:Convergent Sequence (Metric Space)|converges]] to a number which is not in $\Q$.
+The result follows by definition of [[Definition:Complete Metric Space|complete metric space]].
+{{qed}}
+In fact, any [[Definition:Sequence|sequence]] of rational numbers that converges to an [[Definition:Irrational Number|irrational number]] (in the metric space $\R$) is a [[Definition:Cauchy Sequence|Cauchy sequence]] that does not converge in $\Q$.
+\end{proof}<|endoftext|>
+\section{Limit of Subsequence equals Limit of Sequence/Real Numbers}
+Tags: Convergence, Subsequences, Real Analysis
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $l \in \R$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
+Let $\sequence {x_{n_r} }$ be a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+Then:
+:$\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+As $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows that:
+:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
+Now let $R = N$.
+Then from [[Strictly Increasing Sequence of Natural Numbers]]:
+:$\forall r > R: n_r \ge r$
+Thus $n_r > N$ and so:
+:$\size {x_n - l} < \epsilon$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance Function of Metric Space is Continuous}
+Tags: Continuous Mappings, Metric Spaces, Product Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\tau_A$ be the [[Definition:Topology Induced by Metric|topology on $A$ induced by $d$]].
+Let $\struct {A \times A, \tau}$ be the [[Definition:Product Space (Topology)|product space]] of $\struct {A, \tau_A}$ and itself.
+Then the [[Definition:Distance Function|distance function]] $d: A \times A \to \R$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+\end{theorem}
+
+\begin{proof}
+Let $d_\infty: \paren{A \times A}\times \paren{A \times A} \to \R$ be the [[Definition:Metric|metric]] on $A \times A$ defined by:
+:$\map {d_\infty} {\tuple{x, y}, \tuple{x', y'} } = \max \set{\map d {x, x'}, \map d {y, y'} }$
+By [[P-Product Metric Induces Product Topology]], $\tau$ is the [[Definition:Topology Induced by Metric|topology on $A \times A$ induced by $d_\infty$]].
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $\tuple {x_0, y_0} \in A \times A$.
+Suppose that $\tuple {x, y} \in A \times A$ and $\map {d_\infty} {\tuple{x, y}, \tuple{x_0, y_0} } < \dfrac 1 2 \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\map d {x, y} - \map d {x_0, y_0} }
+      | o = \le
+      | r = \size {\map d {x, y} - \map d {x_0, y} } + \size {\map d {x_0, y} - \map d {x_0, y_0} }
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \map d {x, x_0} + \map d {y, y_0}
+      | c = [[Reverse Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = by the definition of $d_\infty$
+}}
+{{end-eqn}}
+The result follows from the definition of a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+{{qed}}
+[[Category:Continuous Mappings]]
+[[Category:Metric Spaces]]
+[[Category:Product Spaces]]
+6bu7lenh5xonjrmuql24zb8x2595eq4
+\end{proof}<|endoftext|>
+\section{Sequence Converges to Within Half Limit/Real Numbers}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\sequence {x_n}$ be [[Definition:Convergent Real Sequence|convergent]] to the [[Definition:Limit of Real Sequence|limit]] $l$.
+That is, let $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.
+Suppose $l > 0$.
+Then:
+: $\exists N: \forall n > N: x_n > \dfrac l 2$
+Similarly, suppose $l < 0$.
+Then:
+: $\exists N: \forall n > N: x_n < \dfrac l 2$
+\end{theorem}
+
+\begin{proof}
+Suppose $l > 0$.
+From the definition of [[Definition:Convergent Real Sequence|convergence to a limit]]:
+:$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$
+That is, $l - \epsilon < x_n < l + \epsilon$.
+As this is true for ''all'' $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.
+Thus:
+:$\exists N: \forall n > N: x_n > \dfrac l 2$
+as required.
+Now suppose $l < 0$.
+By a similar argument:
+:$\forall \epsilon > 0: \exists N: \forall n > N: l - \epsilon < x_n < l + \epsilon$
+Thus it is also true for $\epsilon = -\dfrac l 2$ for some value of $N$.
+Thus:
+:$\exists N: \forall n > N: x_n < \dfrac l 2$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence Converges to Within Half Limit/Complex Numbers}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {z_n}$ be a [[Definition:Complex Sequence|sequence in $\C$]].
+Let $\sequence {z_n}$ be [[Definition:Convergent Sequence (Analysis)|convergent]] to the [[Definition:Limit of Sequence (Number Field)|limit]] $l$.
+That is, let $\displaystyle \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.
+Then:
+: $\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$
+\end{theorem}
+
+\begin{proof}
+Suppose $l > 0$.
+Let us choose $N$ such that:
+:$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$
+Then:
+{{begin-eqn}}
+{{eqn | l = \cmod {z_n - l}
+      | o = <
+      | r = \frac {\cmod l} 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cmod l - \cmod {z_n}
+      | o = \le
+      | r = \cmod {z_n - l}
+      | c = [[Reverse Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \frac {\cmod l} 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cmod {z_n}
+      | o = >
+      | r = \cmod l - \frac {\cmod l} 2
+      | c = 
+}}
+{{eqn | r = \frac {\cmod l} 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Limits of Sequences]]
+3ttfog1p4psqtbv9iperj3dyw9e0edy
+\end{proof}<|endoftext|>
+\section{Number to Reciprocal Power is Decreasing}
+Tags: Real Analysis, Reciprocals
+
+\begin{theorem}
+The [[Definition:Real Sequence|real sequence]] $\sequence {n^{1/n} }$ is [[Definition:Decreasing Real Sequence|decreasing]] for $n \ge 3$.
+\end{theorem}
+
+\begin{proof}
+We want to show that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {n + 1}^{1 / \paren {n + 1} }
+      | o = \le
+      | r = n^{1/n}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {n + 1}^n
+      | o = \le
+      | r = n^{n + 1}
+      | c = raising both sides to the power of $n \paren {n + 1}$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {n \paren {1 + \frac 1 n} }^n
+      | o = \le
+      | r = n^{n + 1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {1 + \frac 1 n}^n
+      | o = \le
+      | r = \frac {n^{n + 1} } {n^n} = n
+      | c = 
+}}
+{{end-eqn}}
+But from [[One Plus Reciprocal to the Nth]]:
+:$\paren {1 + \dfrac 1 n}^n < 3$
+Thus the reversible chain of implication can be invoked and we see that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$ when $n \ge 3$.
+So $\sequence {n^{1 / n} }$ is [[Definition:Decreasing Real Sequence|decreasing]] for $n \ge 3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Diagonal Matrix}
+Tags: Diagonal Matrices, Inverse Matrices
+
+\begin{theorem}
+Let:
+:$\mathbf D = \begin{bmatrix}
+a_{11} & 0 & \cdots & 0 \\
+0 & a_{22} & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & a_{nn} \\
+\end{bmatrix}$
+be an $n \times n$ [[Definition:Diagonal Matrix|diagonal matrix]].
+Then its [[Definition:Inverse Matrix|inverse]] is given by:
+: $\mathbf D^{-1} = \begin{bmatrix}
+\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\
+0 & \dfrac 1 {a_{22}} & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\
+\end{bmatrix}$
+provided that none of the [[Definition:Diagonal Element|diagonal elements]] are zero.
+If any of the diagonal elements are zero, $\mathbf D$ is not [[Definition:Invertible Matrix|invertible]]. 
+{{expand|If any diagonal element isn't a [[Definition:Unit of Ring|unit]]}}
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, consider the [[Definition:Right Inverse Element|right inverse]] of $\mathbf D$.
+Suppose none of the diagonal elements are zero. 
+Then by the definition of inverse, our assertion is that the [[Definition:Matrix Product (Conventional)|matrix product]] of the two matrices in question is the [[Definition:Unit Matrix|unit matrix of order $n$]].
+Now, observe that:
+{{begin-eqn}}
+{{eqn | l = \begin{bmatrix}
+a_{11} & 0 & \cdots & 0 \\
+0 & a_{22} & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & a_{nn} \\
+\end{bmatrix}
+\begin{bmatrix}
+\frac 1 {a_{11} } & 0 & \cdots & 0 \\
+0 & \frac 1 {a_{22} } & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & \frac 1 {a_{nn} } \\
+\end{bmatrix}
+      | r = \begin{bmatrix}
+\frac{a_{11} }{a_{11} } & 0 & \cdots & 0 \\
+0 & \frac{a_{22} }{a_{22} } & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & \frac{a_{nn} }{a_{nn} } \\
+\end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix}
+1 & 0 & \cdots & 0 \\
+0 & 1 & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & 1 \\
+\end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf I
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Now suppose one of the [[Definition:Diagonal Element|diagonal elements]] is zero.
+Then $\map \det {\mathbf D} = 0$, where $\det$ indicates the [[Definition:Determinant of Matrix|determinant]] of $\mathbf D$.
+From [[Determinant of Inverse Matrix]] it would follow that:
+:$\map \det {\mathbf D^{-1} } = \dfrac 1 {\map \det {\mathbf D} }$
+But this equation has no solution, and so $\mathbf D$ cannot admit an [[Definition:Inverse Matrix|inverse]].
+{{qed}}
+[[Category:Diagonal Matrices]]
+[[Category:Inverse Matrices]]
+gh1d57u7f1tiwpqjlan4hfxa3atfd4x
+\end{proof}<|endoftext|>
+\section{Subspace of Complete Metric Space is Closed iff Complete}
+Tags: Metric Subspaces, Complete Metric Spaces
+
+\begin{theorem}
+Let $\struct {M, d}$ be a [[Definition:Complete Metric Space|complete metric space]].
+Let $\struct {S, d}$ be a [[Definition:Metric Subspace|subspace]] of $\struct {M, d}$.
+Then $S$ is [[Definition:Closed Set (Metric Space)|closed]] {{iff}} $S$ is [[Definition:Complete Metric Space|complete]].
+\end{theorem}
+
+\begin{proof}
+This will be proved by demonstrating the [[Definition:Contrapositive Statement|contrapositive]]:
+:$S$ is not [[Definition:Complete Metric Space|complete]] {{iff}} $S$ is not [[Definition:Closed Set (Metric Space)|closed]].
+=== Necessary Condition ===
+Suppose that $S$ is not [[Definition:Complete Metric Space|complete]].
+Then there exists a [[Definition:Cauchy Sequence in Metric Space|Cauchy sequence]] $\sequence {x_n}$ in $S$ such that the [[Definition:Limit of Sequence (Metric Space)|limit]] $\displaystyle x = \lim_{n \mathop \to \infty} x_n$, which exists in the [[Definition:Complete Metric Space|complete metric space]] $M$, is not a member of $S$.
+For all $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \ge N$:
+:$\map d {x, x_n} < \epsilon$
+Hence $M \setminus S$ is not [[Definition:Open Set (Metric Space)|open]].
+Therefore, $S$ is not [[Definition:Closed Set (Metric Space)|closed]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $S$ is not [[Definition:Closed Set (Topology)|closed]].
+Then $M \setminus S$ is not [[Definition:Open Set (Metric Space)|open]].
+Therefore, there exists a $x \in M \setminus S$ such that for all $\epsilon > 0$, there exists a $y \in S$ such that $\map d {x, y} < \epsilon$.
+So there exists a [[Definition:Sequence|sequence]] $\sequence {y_n}$ in $S$ such that for all $n \in \N$:
+:$\map d {x, y_n} < \dfrac 1 n$
+Now, we show that $\sequence {y_n}$ is a [[Definition:Cauchy Sequence in Metric Space|Cauchy sequence]].
+Let $N \in \N$ be such that for all $n \ge N$:
+:$\map d {x, y_n} < \dfrac \epsilon 2$
+Let $m, n \ge N$.
+Then, by the [[Definition:Triangle Inequality|triangle inequality]]:
+:$\map d {y_m, y_n} \le \map d {x, y_m} + \map d {x, y_n} < \epsilon$
+Hence $\sequence {y_n}$ is a [[Definition:Cauchy Sequence in Metric Space|Cauchy sequence]].
+Because $\struct {M, d}$ is a [[Definition:Complete Metric Space|complete metric space]] by assumption, the [[Definition:Limit of Sequence (Metric Space)|limit]] $\displaystyle \lim_{n \mathop \to \infty} y_n$ exists and is in $M$.
+Denote this [[Definition:Limit of Sequence (Metric Space)|limit]] by $y$.
+By the definition of $\sequence {y_n}$:
+:$\displaystyle \lim_{n \mathop \to \infty} \map d {x, y_n} = 0$
+From [[Distance Function of Metric Space is Continuous]] and [[Continuity of Composite Mapping]]:
+:$\map d {x, y} = 0$
+{{explain|Which two mappings are being composed here?}}
+By definition of a [[Definition:Metric|metric]], this implies that $x = y$.
+Since $y \notin S$, $S$ is not [[Definition:Complete Metric Space|complete]].
+{{qed}}
+[[Category:Metric Subspaces]]
+[[Category:Complete Metric Spaces]]
+tsfu0pbjo82jwtsxn4b3m4i6dzxdk7w
+\end{proof}<|endoftext|>
+\section{Set of Chains is Chain Complete for Inclusion}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $\left({P, \leq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $\operatorname{ch} \left({P}\right)$ be the [[Definition:Set of Sets|set]] of [[Definition:Chain (Set Theory)|chains]] in $P$.
+Let $\mathcal C$ be a [[Definition:Chain (Set Theory)|chain]] in $\operatorname{ch} \left({P}\right)$ with respect to the [[Subset Relation is Ordering|subset ordering]].
+Then the [[Definition:Set Union|union]] $\bigcup \mathcal C$ of $\mathcal C$ is also a [[Definition:Chain (Set Theory)|chain]] in $P$.
+\end{theorem}
+
+\begin{proof}
+Let $D = \bigcup \mathcal C$. 
+Observe that any $C \in \mathcal C$ is a [[Definition:Chain (Set Theory)|chain]] in $P$, hence $C \subseteq P$.
+Therefore, $D \subseteq P$, by [[Set Union Preserves Subsets]].
+Now suppose that $a, b \in D$.
+By definition of [[Definition:Set Union|union]], there exist $A, B \in \mathcal C$ such that $a \in A$ and $b \in B$. 
+Since $\mathcal C$ is a [[Definition:Chain (Set Theory)|chain]] in $\left({\operatorname{ch} \left({P}\right), \subseteq}\right)$, either $A \subseteq B$ or $B \subseteq A$.
+Hence either $a \in B$ or $b \in A$.
+Both $A$ and $B$ are [[Definition:Chain (Set Theory)|chains]] in $P$, and thence $a \leq b$ or $b \leq a$.
+Therefore, $D$ is [[Definition:Totally Ordered Set|totally ordered]] by $\le$.
+That is, it is a [[Definition:Chain (Set Theory)|chain]] in $P$.
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Order Theory]]
+bqwyzf61gost9w8p44l40rgcb9flhyx
+\end{proof}<|endoftext|>
+\section{Jensen's Inequality (Real Analysis)}
+Tags: Real Analysis, Inequalities, Jensen's Inequality (Real Analysis)
+
+\begin{theorem}
+Let $I$ be a [[Definition:Real Interval|real interval]].
+Let $\phi: I \to \R$ be a [[Definition:Convex Real Function|convex function]].
+Let $x_1, x_2, \ldots, x_n \in I$.
+Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be [[Definition:Real Number|real numbers]], at least one of which is non-zero.
+Then:
+:$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
+For $\phi$ [[Definition:Strictly Convex Real Function|strictly convex]], equality holds {{iff}} $x_1 = x_2 = \cdots = x_n$.
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|mathematical induction]] on $n$.
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
+$\map P 1$ is true, as this just says:
+:$\displaystyle \map \phi {\frac {\lambda_1 x_1} {\lambda_1} } \le \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}$
+{{begin-eqn}}
+{{eqn | l = \map \phi {\frac {\lambda_1 x_1} {\lambda_1} }
+      | r = \map \phi {x_1}
+      | c = 
+}}
+{{eqn | r = \frac {\lambda_1 \map \phi {x_1} } {\lambda_1}
+      | c = 
+}}
+{{end-eqn}}
+trivially.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^r \lambda_k x_k} {\sum_{k \mathop = 1}^r \lambda_k} } \le \frac {\sum_{k \mathop = 1}^r \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^r \lambda_k}$
+where $\lambda_1, \lambda_2, \ldots, \lambda_r \ge 0$ are [[Definition:Real Number|real numbers]], at least one of which is non-zero.
+Then we need to show:
+:$\displaystyle \map \phi {\frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k x_k} {\sum_{k \mathop = 1}^{r + 1} \lambda_k} } \le \frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^{r + 1} \lambda_k}$
+where $\lambda_1, \lambda_2, \ldots, \lambda_{r + 1} \ge 0$ are [[Definition:Real Number|real numbers]], at least one of which is non-zero.
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{WLOG}}, assume that $\lambda_r$ and $\lambda_{r + 1}$ are non-zero.
+Otherwise, the statement reduces to the [[Jensen's Inequality (Real Analysis)#Induction Hypothesis|induction hypothesis]].
+Define:
+:$y := \dfrac {\lambda_r x_r + \lambda_{r + 1} x_{r + 1} } {\lambda_r + \lambda_{r + 1} }$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {\frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k x_k} {\sum_{k \mathop = 1}^{r + 1} \lambda_k} }
+      | r = \map \phi {\frac {\sum_{k \mathop = 1}^{r - 1} \lambda_k x_k + \paren {\lambda_r + \lambda_{r + 1} } y} {\sum_{k \mathop = 1}^{r - 1} \lambda_k + \paren {\lambda_r + \lambda_{r + 1} } } }
+}}
+{{eqn | o = \le
+      | r = \frac {\sum_{k \mathop = 1}^{r - 1} \lambda_k \map \phi {x_k} + \paren {\lambda_r + \lambda_{r + 1} } \map \phi y} {\sum_{k \mathop = 1}^{r - 1} \lambda_k + \paren {\lambda_r + \lambda_{r + 1} } }
+      | c = by the [[Jensen's Inequality (Real Analysis)#Induction Hypothesis|induction hypothesis]]
+}}
+{{eqn | o = \le
+      | r = \frac {\sum_{k \mathop = 1}^{r + 1} \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^{r + 1} \lambda_k}
+      | c = {{Defof|Convex Real Function}}
+}}
+{{end-eqn}}
+Note that in the case that $\phi$ is [[Definition:Strictly Convex Real Function|strictly convex]], equality holds {{iff}} $x_1 = x_2 = \cdots = x_r = x_{r + 1}$.
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \N_{>0}: \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
+{{qed}}
+{{Namedfor|Johan Ludwig William Valdemar Jensen|cat = Jensen}}
+[[Category:Real Analysis]]
+[[Category:Inequalities]]
+[[Category:Jensen's Inequality (Real Analysis)]]
+l6ou3qzrwkkf4ri0es47nvpdnz3z14p
+\end{proof}<|endoftext|>
+\section{Inequality of Hölder Means}
+Tags: Real Analysis, Inequalities, Measures of Central Tendency
+
+\begin{theorem}
+Let $p, q \in \R_{\ne 0}$ be non-zero [[Definition:Real Number|real numbers]] with $p < q$.
+Let $x_1, x_2, \ldots, x_n \ge 0$ be [[Definition:Real Number|real numbers]].
+If $p < 0$, then we require that $x_1, x_2, \ldots, x_n > 0$.
+Then the [[Definition:Hölder Mean|Hölder mean]] satisfies the inequality:
+: $M_p \left({x_1, x_2, \ldots, x_n}\right) \le M_q \left({x_1, x_2, \ldots, x_n}\right)$
+Equality holds if and only if $x_1 = x_2 = \cdots = x_n$.
+\end{theorem}
+
+\begin{proof}
+For [[Definition:Real Number|real]] $p \ne 0$, the [[Definition:Hölder Mean|Hölder mean]] is defined as:
+:$\displaystyle M_p \left({x_1, x_2, \ldots, x_n}\right) = \left({\frac 1 n \sum_{k \mathop = 1}^n x_k^p}\right)^{1/p}$
+whenever the above expression is defined.
+Consider the [[Definition:Real Function|function]]:
+: $\phi : \R_{\ge 0} \to \R_{\ge 0}$ defined by $\phi \left({x}\right) = x^{q/p}$
+{{begin-eqn}}
+{{eqn| l=D_x \left({x^{q/p} }\right)
+     | r=\dfrac{q}{p} x^{q/p-1}
+     | c=[[Power Rule for Derivatives|power rule]]
+}}
+{{end-eqn}}
+From [[Real Function is Strictly Convex iff Derivative is Strictly Increasing]] that $\phi$ is [[Definition:Strictly Convex Real Function|strictly convex]] if $q > 0$
+Similarly, from [[Real Function is Strictly Concave iff Derivative is Strictly Decreasing]] and [[Definition:Strictly Concave Real Function|strictly concave]] if $q < 0$.
+{{explain|The above needs to be amplified, as it is far from clear why.}}
+Now apply [[Jensen's Inequality (Real Analysis)|Jensen's inequality]] to $x_1^p, x_2^p, \ldots, x_n^p$.
+For $q > 0$, this gives:
+:$\displaystyle \left({ \frac 1 n \sum_{k \mathop = 1}^n x_k^p }\right)^{q/p} \le \frac 1 n \sum_{k \mathop = 1}^n x_k^q$
+For $q < 0$, the reverse inequality holds.
+Also by [[Jensen's Inequality (Real Analysis)|Jensen's inequality]], equality holds if and only if $x_1 = x_2 = \cdots = x_n$.
+In either case, the result follows.
+{{qed}}
+[[Category:Real Analysis]]
+[[Category:Inequalities]]
+[[Category:Measures of Central Tendency]]
+ssitrtujj0bf4izv2svida7swzg7uw0
+\end{proof}<|endoftext|>
+\section{Real Linear Subspace Contains Zero Vector}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $\mathbb W \subseteq \R^n$ such that $\mathbb W$ is a [[Definition:Vector Subspace|linear subspace of $\R^n$]].
+Then $\mathbb W$ contains the [[Definition:Zero Vector|zero vector]]:
+:$\mathbf 0_{n \times 1} = \begin{bmatrix}
+0 \\ 0 \\ \vdots \\ 0
+\end{bmatrix} \in \mathbb W$
+\end{theorem}
+
+\begin{proof}
+This is a consequence of [[Vector Subspace of Real Vector Space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Subspace is Subspace}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|vector space]] over $K$ with [[Definition:Zero Vector|zero vector]] $\mathbf 0$.
+The [[Definition:Zero Subspace|zero subspace]] $\left\{{\mathbf 0}\right\}$ is a [[Definition:Vector Subspace|subspace]] of $V$.
+\end{theorem}
+
+\begin{proof}
+We use the [[Two-Step Vector Subspace Test]].
+$\left\{{\mathbf 0}\right\}$ is [[Definition:Non-Empty Set|not empty]], because it contains $\mathbf{0}$.
+$\left\{{\mathbf 0}\right\}$ is [[Definition:Closure (Abstract Algebra)|closed]] under $+$ because:
+:$\forall \mathbf{x},\mathbf{y} \in \left\{{\mathbf 0}\right\}, \mathbf{x} + \mathbf{y} = \mathbf{0} + \mathbf{0} = \mathbf{0} \in \left\{{\mathbf 0}\right\}$
+$\left\{{\mathbf 0}\right\}$ is [[Definition:Closure (Abstract Algebra)|closed]] under multiplication because:
+:$\forall \lambda \in K, \mathbf{x}\in \left\{{\mathbf 0}\right\}: \lambda\mathbf{x} = \lambda\mathbf{0} = \mathbf{0}\in \left\{{\mathbf 0}\right\}$
+Hence the result, from the [[Two-Step Vector Subspace Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cartesian Product of Natural Numbers with Itself is Countable}
+Tags: Set Theory, Cartesian Product, Countable Sets
+
+\begin{theorem}
+The [[Definition:Cartesian Product|Cartesian product]] $\N \times \N$ of the set of [[Definition:Natural Numbers|natural numbers]] $\N$ with itself is [[Definition:Countable|countable]].
+\end{theorem}
+
+\begin{proof}
+This is simply a special case of [[Cartesian Product of Countable Sets is Countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minkowski's Inequality for Sums}
+Tags: Algebra, Analysis, Inequalities
+
+\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 $p \in \R$, $p \ne 0$ 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 $p > 1$, then:
+:$\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/p} \le \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p}$
+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}
+=== [[Minkowski's Inequality for Sums/Index 2|Proof for $p = 2$]] ===
+$p = 2$ is an easily proved special case:
+{{:Minkowski's Inequality for Sums/Index 2}}
+{{qed|lemma}}
+=== [[Minkowski's Inequality for Sums/Index Greater than 1|Proof for $p > 1$]] ===
+{{:Minkowski's Inequality for Sums/Index Greater than 1}}
+{{qed|lemma}}
+=== [[Minkowski's Inequality for Sums/Index Less than 1|Proof for $p < 1$, $p \ne 0$]] ===
+{{:Minkowski's Inequality for Sums/Index Less than 1}}
+{{qed}}
+{{namedfor|Hermann Minkowski|cat=Minkowski}}
+[[Category:Algebra]]
+[[Category:Analysis]]
+[[Category:Inequalities]]
+602bzuaclyhirrgrph9d48773r6rc2m
+\end{proof}<|endoftext|>
+\section{Minkowski's Inequality for Integrals}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f, g$ be integrable functions in $X \subseteq \R^n$ with respect to the volume element $dV$.
+{{explain|Riemann integrable or Lebesgue integrable?}}
+:$(1):\quad$ Let $p > 1$. Then:
+:: $\displaystyle \paren {\int_X \size {f + g}^p \rd V}^{1/p} \le \paren {\int_X \size f^p \rd V}^{1/p} + \paren {\int_X \size g^p \rd V}^{1/p}$
+:$(2):\quad$ Let $p < 1, p \ne 0$. Then:
+:: $\displaystyle \paren {\int_X \size {f + g}^p \rd V}^{1/p} \ge \paren {\int_X \size f^p \rd V}^{1/p} + \paren {\int_X \size g^p \rd V}^{1/p}$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Hermann Minkowski|cat = Minkowski}}
+\end{proof}<|endoftext|>
+\section{Characteristic Function Determined by 1-Fiber}
+Tags: Characteristic Functions
+
+\begin{theorem}
+Let $A \subseteq S$.
+Let $f:S \to \left\{{0, 1}\right\}$ be a [[Definition:Mapping|mapping]].
+Denote by $\chi_A$ the [[Definition:Characteristic Function of Set|characteristic function]] on $A$.
+Then the following are equivalent:
+:$(1):\quad f = \chi_A$
+:$(2):\quad \forall s \in S: f \left({s}\right) = 1 \iff s \in A$
+Using the notion of a [[Definition:Fiber (Mapping)|fiber]], $(2)$ may also be expressed as:
+:$(2'):\quad f^{-1} \left({1}\right) = A$
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Follows directly from the definition of [[Definition:Characteristic Function of Set|characteristic function]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $s \in S$.
+Suppose that $s \in A$.
+Then by assumption, $f \left({s}\right) = 1$.
+Also, by definition of [[Definition:Characteristic Function of Set|characteristic function]], $\chi_A \left({s}\right) = 1$.
+Next, suppose that $s \notin A$.
+Then $f \left({s}\right) \ne 1$ by assumption.
+As $f \left({s}\right) \in \left\{{0, 1}\right\}$, it follows that $f \left({s}\right) = 0$.
+Again, by definition of [[Definition:Characteristic Function of Set|characteristic function]], also have $\chi_A \left({s}\right) = 0$.
+Hence, for all $s \in S$, have $f \left({s}\right) = \chi_A \left({s}\right)$.
+By [[Equality of Mappings]], it follows that $f = \chi_A$.
+{{qed}}
+[[Category:Characteristic Functions]]
+07z7f5u39h69xemwks9f4667lhgeey0
+\end{proof}<|endoftext|>
+\section{Characteristic Function of Intersection}
+Tags: Characteristic Functions, Set Intersection
+
+\begin{theorem}
+Let $A, B \subseteq S$.
+Let $\chi_{A \cap B}$ be the [[Definition:Characteristic Function of Set|characteristic function]] of their [[Definition:Set Intersection|intersection]] $A \cap B$.
+\end{theorem}<|endoftext|>
+\section{Characteristic Function of Union}
+Tags: Characteristic Functions, Set Union, Characteristic Function of Union
+
+\begin{theorem}
+Let $A, B \subseteq S$.
+Let $\chi_{A \mathop \cup B}$ be the [[Definition:Characteristic Function of Set|characteristic function]] of their [[Definition:Set Union|union]] $A \cup B$.
+\end{theorem}<|endoftext|>
+\section{Characteristic Function of Set Difference}
+Tags: Characteristic Functions
+
+\begin{theorem}
+Let $A, B \subseteq S$.
+Then:
+:$\chi_{A \mathop \setminus B} = \chi_A - \chi_{A \cap B}$ 
+where:
+:$A \setminus B$ denotes [[Definition:Set Difference|set difference]]
+:$\chi$ denotes [[Definition:Characteristic Function of Set|characteristic function]].
+\end{theorem}
+
+\begin{proof}
+Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$.
+Then by definition of [[Definition:Characteristic Function of Set|characteristic function]]:
+:$s \in A \setminus B$
+That is, by definition of [[Definition:Set Difference|set difference]], $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$.
+Hence:
+:$\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$
+Conclude that:
+:$\map {\chi_{A \mathop \setminus B} } s = 1 \implies \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
+Now conversely suppose $\map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$.
+Considering that $\map {\chi_{A \cap B} } s \ge 0$, it must be that $\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$.
+Hence, $s \in A$, and $s \notin A \cap B$.
+By definition of [[Definition:Set Intersection|set intersection]], this means $s \notin B$.
+Hence $s \in A \setminus B$ by definition of [[Definition:Set Difference|set difference]].
+Thus:
+:$\map {\chi_{A \mathop \setminus B} } s = 1$
+Hence, it has been established that:
+:$\map {\chi_{A \mathop \setminus B} } s = 1 \iff \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
+which, by application of [[Characteristic Function Determined by 1-Fiber]], gives the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic Function of Symmetric Difference}
+Tags: Characteristic Functions
+
+\begin{theorem}
+Let $A, B \subseteq S$.
+Then:
+:$\chi_{A * B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$ 
+where:
+: $\chi$ denotes [[Definition:Characteristic Function of Set|characteristic function]]
+: $*$ denotes [[Definition:Symmetric Difference|symmetric difference]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Symmetric Difference/Definition 2|symmetric difference]]:
+: $A * B = \left({A \cup B}\right) \setminus \left({A \cap B}\right)$
+Thus:
+:$\chi_{A * B} = \chi_{A \cup B} - \chi_{\left({A \cup B}\right) \cap \left({A \cap B}\right)}$
+by [[Characteristic Function of Set Difference]].
+But by [[Intersection is Subset of Union]] and [[Intersection with Subset is Subset]]:
+: $\left({A \cup B}\right) \cap \left({A \cap B}\right) = A \cap B$
+Hence it follows that:
+:$\chi_{A * B} = \chi_{A \cup B} - \chi_{A \cap B}$
+which by [[Characteristic Function of Union/Variant 2|Characteristic Function of Union: Variant 2]] becomes:
+:$\chi_{A * B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Subspace of Real Vector Space}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $\R^n$ be a [[Definition:Real Vector Space|real vector space]].
+Let $\mathbb W \subseteq \R^n$.
+Then $\mathbb W$ is a [[Definition:Vector Subspace|linear subspace]] of $\R^n$ {{iff}}:
+:$(1): \quad \mathbf 0 \in \mathbb W$, where $\mathbf 0$ is the [[Definition:Zero Vector|zero vector]] with $n$ entries
+:$(2): \quad \mathbb W$ is [[Definition:Closure (Abstract Algebra)|closed]] under [[Definition:Vector Sum|vector addition]]
+:$(3): \quad \mathbb W$ is [[Definition:Closure (Abstract Algebra)|closed]] under [[Definition:Scalar Multiplication on Vector Space|scalar multiplication]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+=== Proof of Corollary ===
+Suppose $\mathbf 0 \in \mathbb W$.
+Then $\mathbb W$ contains an [[Definition:Element|element]] and is non-empty.
+Suppose $\mathbb W$ contains an element $\mathbf x \in \R^n$.
+Then, by criterion $(3)$:
+:$0 \mathbf x \in \mathbb W$
+where $0$ is the [[Definition:Zero Scalar|zero scalar]].
+But $0 \mathbf x = \mathbf 0$ from [[Vector Scaled by Zero is Zero Vector]], so $\mathbf 0 \in \mathbb W$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reverse Young's Inequality for Products}
+Tags: Analysis
+
+\begin{theorem}
+Let $p, q \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]] satisfying:
+:$\displaystyle \frac 1 p - \frac 1 q = 1$
+Let $a \in \R_{\ge 0}$ be a [[Definition:Positive Real Number|positive real number]].
+Let $b \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Then:
+:$\displaystyle a b \ge \frac {a^p} p - \frac {b^{-q}} q$
+\end{theorem}
+
+\begin{proof}
+Define:
+{{begin-eqn}}
+{{eqn|n = 1
+     |l = u
+     |r = \frac 1 p
+}}
+{{eqn|l = v
+     |r = \frac q p
+}}
+{{eqn|l = x
+     |r = \left({a b}\right)^p
+}}
+{{eqn|l = y
+     |r = b^{-p}
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn|l = \frac 1 u + \frac 1 v
+     |r = p + \frac p q
+}}
+{{eqn|r = p \left({1 + \frac 1 q}\right)
+}}
+{{eqn|r = p \left({\frac 1 p}\right)
+     |c = because $\dfrac 1 p - \dfrac 1 q = 1$, [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn|r = 1
+}}
+{{end-eqn}}
+Thus [[Young's Inequality for Products]] can be applied:
+{{begin-eqn}}
+{{eqn|l = x y
+     |o = \le
+     |r = \frac {x^u} u + \frac {y^v} v
+}}
+{{eqn|ll = \implies
+     |l = \left({a b}\right)^p b^{-p}
+     |o = \le
+     |r = \frac {\left({\left({a b}\right)^p}\right)^{1/p} } {1/p} + \frac {\left({b^{-p} }\right)^{q/p} } {q/p}
+     |c = substituting for $u, v, x, y$ from $(1)$ above
+}}
+{{eqn|ll = \implies
+     |l = a^p
+     |o = \le
+     |r = pab + p \frac { b^{-q} } q
+     |c = algebraic simplification
+}}
+{{eqn|ll = \implies
+     |l = \frac {a^p} p
+     |o = \le
+     |r = ab + \frac { b^{-q} } q
+}}
+{{eqn|ll = \implies
+     |l = ab
+     |o = \ge
+     |r = \frac {a^p} p - \frac { b^{-q} } q
+}}
+{{end-eqn}}
+{{qed}}
+{{namedfor|William Henry Young|cat=Young}}
+[[Category:Analysis]]
+47nxl9wd0fqa22sq8bin4oka1gbbbll
+\end{proof}<|endoftext|>
+\section{Reverse Hölder's Inequality for Sums}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $p, q \in \R_{>0}$ be [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]] such that $\dfrac 1 p - \dfrac 1 q = 1$.
+Suppose that the [[Definition:Sequence|sequences]] $\mathbf x = \sequence {x_n}$ and $\mathbf y = \sequence {y_n}$ in [[Definition:Complex Number|$\C$]] (or [[Definition:Real Number|$\R$]]) are such that the [[Definition:Series|series]]
+:$\displaystyle \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p}$
+and
+:$\displaystyle \paren {\sum_{n \mathop = 1}^\infty \size {y_n}^{-q} }^{-1/q}$
+are [[Definition:Convergent Series|convergent]].
+Denote these values by $\norm {\mathbf x}_p$ and $\norm {\mathbf y}_{-q}$, respectively.
+Here, the notation $\norm {\mathbf x}_p$ does not denote a [[Definition:Norm on Vector Space|norm]], but is instead just a convenient notation similar to that of the [[Definition:P-Norm|$p$-norm]], which is only defined when $p \ge 1$.
+Let $\norm {\mathbf {x y} }_1$ denote the $1$-norm of $\mathbf {x y}$, if $\mathbf {x y}$ is in the [[Definition:Lebesgue Space|Lebesgue space]] $\ell^1$.
+Then $\norm {\mathbf {x y} }_1 \ge \norm {\mathbf x}_p \norm {\mathbf y}_{-q}$.
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, assume that $\mathbf x$ and $\mathbf y$ are non-zero.
+Let:
+:$\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$
+and:
+:$\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_{-q} }$
+Then:
+:$\displaystyle \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p} = 1$
+Similarly:
+:$\norm {\mathbf v}_{-q} = 1$
+It then suffices to prove that:
+:$\norm {\mathbf {u v} }_1 = \dfrac {\norm {\mathbf {x y} }_1} {\norm {\mathbf x}_p norm {\mathbf y}_{-q} } \ge 1$
+By [[Reverse Young's Inequality for Products]]:
+:$\size {u_n v_n} \ge \dfrac 1 p \size {u_n}^p - \dfrac 1 q \size {v_n}^{-q}$
+Summing over all $n \in \N$ gives:
+:$\norm {\mathbf {u v} }_1 \ge \dfrac 1 p \norm {\mathbf u}_p - \dfrac 1 q \norm {\mathbf v}_{-q} = 1$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Vector Space is Vector Space}
+Tags: Examples of Vector Spaces, Real Numbers, Analytic Geometry
+
+\begin{theorem}
+Let $\R$ be the set of [[Definition:Real Number|real numbers]].
+Then the [[Definition:Real Vector Space|real vector space $\R^n$]] is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+=== Construction of Real Vector Space ===
+From the definition, a [[Definition:Vector Space|vector space]] is a [[Definition:Unitary Module|unitary module]] whose [[Definition:Scalar Ring of Unitary Module|scalar ring]] is a [[Definition:Field (Abstract Algebra)|field]].
+In order to call attention to the precise scope of the operators, let [[Definition:Real Addition|real addition]] and [[Definition:Real Multiplication|real multiplication]] be expressed as $+_\R$ and $\times_\R$ respectively.
+Then we can express the [[Definition:Field of Real Numbers|field of real numbers]] as $\struct {\R, +_\R, \times_\R}$.
+From [[Real Numbers under Addition form Abelian Group]], $\struct {\R, +}$ is a [[Definition:Group|group]].
+Again, in order to call attention to the precise scope of the operator, let [[Definition:Real Addition|real addition]] be expressed on $\struct {\R, +}$ as $+_G$.
+That is, the [[Definition:Group|group]] under consideration is $\struct {\R, +_G}$.
+Consider the [[Definition:Cartesian Product|cartesian product]]:
+:$\displaystyle \R^n = \prod_{i \mathop = 1}^n \struct {\R, +_G} = \underbrace {\struct {\R, +_G} \times \cdots \times \struct {\R, +_G} }_{n \text{ copies} }$
+Let:
+:$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
+:$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$
+be arbitrary elements of $\R^n$.
+Let $\lambda$ be an arbitrary element of $\R$.
+Let $+$ be the [[Definition:Binary Operation|binary operation]] defined on $\R^n$ as:
+:$\mathbf a + \mathbf b = \tuple {a_1 +_G b_1, a_2 +_G b_2, \ldots, a_n +_G b_n}$
+Also let $\cdot$ be the [[Definition:Binary Operation|binary operation]] defined on $\R \times \R^n$ as:
+:$\lambda \cdot \mathbf a = \tuple {\lambda \times_\R a_1, \lambda \times_\R a_2, \ldots, \lambda \times_\R a_n}$
+In this context, $\lambda \times_\R a_j$ is defined as [[Definition:Real Multiplication|real multiplication]], as is appropriate (both $\lambda$ and $a_j$ are [[Definition:Real Number|real numbers]]).
+With this set of definitions, the structure $\struct {\R^n, +, \cdot}$ is a [[Definition:Vector Space|vector space]], as is shown in [[Real Vector Space is Vector Space#Proof of Real Vector Space|Proof of Real Vector Space]] below.
+=== Proof of Real Vector Space ===
+In order to show that $\struct {\R^n, +, \cdot}$ is a [[Definition:Vector Space|vector space]], we need to show that:
+$\forall \mathbf x, \mathbf y \in \R^n, \forall \lambda, \mu \in \R$:
+: $(1): \quad \lambda \cdot \paren {\mathbf x + \mathbf y} = \paren {\lambda \cdot \mathbf x} + \paren {\lambda \cdot \mathbf y}$
+: $(2): \quad \paren {\lambda +_\R \mu} \cdot x = \paren {\lambda \cdot \mathbf x} + \paren {\mu \cdot \mathbf x}$
+: $(3): \quad \paren {\lambda \times_\R \mu} \cdot x = \lambda \cdot \paren {\mu \cdot \mathbf x}$
+: $(4): \quad \forall \mathbf x \in \R^n: 1 \cdot \mathbf x = \mathbf x$.
+where $1$ in this context means the [[Definition:One|real number one]].
+From [[External Direct Product of Groups is Group]], we have that $\struct {\R^n, +}$ is a [[Definition:Group|group]] in its own right.
+Let:
+:$\mathbf x = \tuple {x_1, x_2, \ldots, x_n}$
+:$\mathbf y = \tuple {y_1, y_2, \ldots, y_n}$
+Checking the criteria in turn:
+$(1): \quad \lambda \cdot \paren {\mathbf x + \mathbf y} = \paren {\lambda \cdot \mathbf x} + \paren {\lambda \cdot \mathbf y}$:
+{{begin-eqn}}
+{{eqn | l = \lambda \cdot \paren {\mathbf x + \mathbf y}
+      | r = \tuple {\lambda \times_\R \paren {x_1 +_G y_1}, \lambda \times_\R \paren {x_2 +_G y_2}, \ldots, \lambda \times_\R \paren {x_n +_G y_n} }
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \tuple {\paren {\lambda \times_\R x_1 +_G \lambda \times_\R y_1}, \paren {\lambda \times_\R x_2 +_G \lambda \times_\R y_2}, \ldots, \paren {\lambda \times_\R x_n +_G \lambda \times_\R y_n} }
+      | c = [[Real Multiplication Distributes over Addition|$\times_\R$ distributes over $+_G$]]
+}}
+{{eqn | r = \tuple {\lambda \times_\R x_1, \lambda \times_\R x_2, \ldots, \lambda \times_\R x_n} + \tuple {\lambda \times_\R y_1, \lambda \times_\R y_2, \ldots, \lambda \times_\R y_n}
+      | c = Definition of $+$ over $\R^n$
+}}
+{{eqn | r = \lambda \cdot \tuple {x_1, x_2, \ldots, x_n} + \lambda \cdot \tuple {y_1, y_2, \ldots, y_n}
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \lambda \cdot \mathbf x + \lambda \cdot \mathbf y
+      | c = Definition of $\mathbf x$ and $\mathbf y$
+}}
+{{end-eqn}}
+So $(1)$ has been shown to hold.
+$(2): \quad \paren {\lambda +_\R \mu} \cdot x = \paren {\lambda \cdot \mathbf x} + \paren {\mu \cdot \mathbf x}$:
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda +_\R \mu} \cdot x
+      | r = \tuple {\paren {\lambda +_\R \mu} \times_\R x_1, \paren {\lambda +_\R \mu} \times_\R x_2, \ldots, \paren {\lambda +_\R \mu} \times_\R x_n}
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \tuple {\paren {\lambda \times_\R x_1 +_G \mu \times_\R x_1}, \paren {\lambda \times_\R x_2 +_G \mu \times_\R x_2}, \ldots, \paren {\lambda \times_\R x_n +_G \mu \times_\R x_n} }
+      | c = [[Real Multiplication Distributes over Addition|$\times_\R$ distributes over $+_\R$]], and $+_\R$ is the same operation as $+_G$
+}}
+{{eqn | r = \tuple {\lambda \times_\R x_1, \lambda \times_\R x_2, \ldots, \lambda \times_\R x_n} + \tuple {\mu \times_\R x_1, \mu \times_\R x_2, \ldots, \mu \times_\R x_n}
+      | c = Definition of $+$ over $\R^n$
+}}
+{{eqn | r = \lambda \cdot \tuple {x_1, x_2, \ldots, x_n} + \mu \cdot \tuple {x_1, x_2, \ldots, x_n}
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \lambda \cdot \mathbf x + \mu \cdot \mathbf x
+      | c = Definition of $\mathbf x$ and $\mathbf y$
+}}
+{{end-eqn}}
+So $(2)$ has been shown to hold.
+$(3): \quad \paren {\lambda \times_\R \mu} \cdot x = \lambda \cdot \paren {\mu \cdot \mathbf x}$:
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda \times_\R \mu} \cdot x
+      | r = \tuple {\paren {\lambda \times_\R \mu} \times_\R x_1, \paren {\lambda \times_\R \mu} \times_\R x_2, \ldots, \paren {\lambda \times_\R \mu} \times_\R x_n}
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \tuple {\lambda \times_\R \paren {\mu \times_\R x_1}, \lambda \times_\R \paren {\mu \times_\R x_2}, \ldots, \lambda \times_\R \paren {\mu \times_\R x_n} }
+      | c = [[Real Multiplication is Associative]]
+}}
+{{eqn | r = \lambda \cdot \tuple {\mu \times_\R x_1, \mu \times_\R x_2, \ldots, \mu \times_\R x_n}
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \lambda \cdot \paren {\mu \cdot \tuple {x_1, x_2, \ldots, x_n} }
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \lambda \cdot \paren {\mu \cdot \mathbf x}
+      | c = Definition of $\mathbf x$
+}}
+{{end-eqn}}
+So $(3)$ has been shown to hold.
+$(4): \quad \forall \mathbf x \in \R^n: 1 \cdot \mathbf x = \mathbf x$:
+{{begin-eqn}}
+{{eqn | l = 1 \cdot \mathbf x = \mathbf x
+      | r = \tuple {1 \times_\R x_1, 1 \times_\R x_2, \ldots, 1 \times_\R x_n}
+      | c = Definition of $\cdot$ over $\R \times \R^n$
+}}
+{{eqn | r = \tuple {x_1, x_2, \ldots, x_n}
+      | c = [[Real Multiplication Identity is One]]
+}}
+{{eqn | r = \mathbf x
+      | c = Definition of $\mathbf x$
+}}
+{{end-eqn}}
+So $(4)$ has been shown to hold.
+So the [[Definition:Module on Cartesian Product|$\R$-module $\R^n$]] is a [[Definition:Vector Space|vector space]], as we were to prove.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetric Difference is Subset of Union of Symmetric Differences}
+Tags: Union, Set Union, Symmetric Difference, Set Union
+
+\begin{theorem}
+Let $R, S, T$ be [[Definition:Set|sets]].
+Then:
+:$R * S \subseteq \left({R * T}\right) \cup \left({S * T}\right)$
+where $R * S$ denotes the [[Definition:Symmetric Difference|symmetric difference]] between $R$ and $S$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Symmetric Difference|symmetric difference]], we have:
+:$R * S = \left({R \setminus S}\right) \cup \left({S \setminus R}\right)$
+Then from [[Set Difference is Subset of Union of Differences]], we have:
+: $R \setminus S \subseteq \left({R \setminus T}\right) \cup \left({T \setminus S}\right)$
+: $S \setminus R \subseteq \left({S \setminus T}\right) \cup \left({T \setminus R}\right)$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \left({R \setminus S}\right) \cup \left({S \setminus R}\right)
+      | o = \subseteq
+      | r = \left({R \setminus T}\right) \cup \left({T \setminus S}\right) \cup \left({S \setminus T}\right) \cup \left({T \setminus R}\right)
+      | c = [[Set Union Preserves Subsets]]
+}}
+{{eqn | r = \left({R \setminus T}\right) \cup \left({T \setminus R}\right) \cup \left({S \setminus T}\right) \cup \left({T \setminus S}\right)
+      | c = [[Union is Commutative]]
+}}
+{{eqn | r = \left({R * T}\right) \cup \left({S * T}\right)
+      | c = {{Defof|Symmetric Difference}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Symmetric Difference]]
+[[Category:Set Union]]
+c9zji7vn26npz9naldzlpdrxwyyft32
+\end{proof}<|endoftext|>
+\section{Sigma-Algebra Contains Empty Set}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
+Then $\varnothing \in \Sigma$.
+\end{theorem}
+
+\begin{proof}
+Axiom $(1)$ of a [[Definition:Sigma-Algebra|$\sigma$-algebra]] grants $X \in \Sigma$.
+By axiom $(2)$ and [[Set Difference with Self is Empty Set]], it follows that $\varnothing = X \setminus X \in \Sigma$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sigma-Algebra Closed under Union}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
+Let $A, B \in \Sigma$ be [[Definition:Measurable Set|measurable sets]].
+Then $A \cup B \in \Sigma$, where $\cup$ denotes [[Definition:Set Union|set union]].
+\end{theorem}
+
+\begin{proof}
+Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \varnothing$.
+Then by [[Sigma-Algebra Contains Empty Set]], axiom $(3)$ of a [[Definition:Sigma-Algebra|$\sigma$-algebra]] applies.
+Hence $\displaystyle \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sigma-Algebra Closed under Countable Intersection}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
+Suppose that $\left({E_n}\right)_{n \in \N} \in \Sigma$ is a [[Definition:Indexed Set|collection]] of [[Definition:Measurable Set|measurable sets]].
+Then $\displaystyle \bigcap_{n \mathop \in \N} E_n \in \Sigma$, where $\displaystyle \bigcap$ denotes [[Definition:Set Intersection|set intersection]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | ll= \forall n \in \N:
+      | l = E_n \in \Sigma
+      | o = \implies
+      | r = X \setminus E_n \in \Sigma
+      | c = Axiom $(2)$ for [[Definition:Sigma-Algebra/Definition 1|$\sigma$-algebras]]
+}}
+{{eqn | o = \implies
+      | r = \bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) \in \Sigma
+      | c = Axiom $(3)$ for [[Definition:Sigma-Algebra/Definition 1|$\sigma$-algebras]]
+}}
+{{eqn | o = \implies
+      | r = X \setminus \left({\bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) }\right) \in \Sigma
+      | c = Axiom $(2)$ for [[Definition:Sigma-Algebra/Definition 1|$\sigma$-algebras]]
+}}
+{{end-eqn}}
+From [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Intersection|De Morgan's laws: Complement of Intersection]]:
+:$\displaystyle \bigcup_{n \mathop \in \N} \left({X \setminus E_n}\right) = X \setminus \left({\bigcap_{n \mathop \in \N} E_n}\right)$
+Also, by [[Set Difference with Set Difference]] and [[Set Union Preserves Subsets]]:
+:$\displaystyle X \setminus \left({X \setminus \left({\bigcap_{n \mathop \in \N} E_n}\right) }\right) = \bigcap_{n \mathop \in \N} E_n$
+Combining the previous equalities, it follows that:
+: $\displaystyle \bigcap_{n \mathop \in \N} E_n \in \Sigma$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sigma-Algebra of Countable Sets}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\Sigma$ be the collection of [[Definition:Countable Set|countable]] and [[Definition:Co-Countable Set|co-countable]] [[Definition:Subset|subsets]] of $X$.
+Then $\Sigma$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+Let us verify the axioms of a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn.
+=== Axiom $(1)$ ===
+By [[Relative Complement with Self is Empty Set]]:
+: $\complement_X \left({X}\right) = \varnothing$
+In particular, it is [[Definition:Countable Set|countable]].
+Hence $X$ is [[Definition:Co-Countable Set|co-countable]], and so $X \in \Sigma$.
+{{qed|lemma}}
+=== Axiom $(2)$ ===
+The [[Definition:Relative Complement|relative complement]] of a [[Definition:Countable Set|countable set]] is by definition [[Definition:Co-Countable Set|co-countable]].
+The converse holds by [[Relative Complement of Relative Complement]].
+Hence:
+: $E \in \Sigma \implies \complement_X \left({E}\right) \in \Sigma$
+{{qed|lemma}}
+=== Axiom $(3)$ ===
+Let $\left({E_n}\right)_{n \mathop \in \N} \in \Sigma$ be a [[Definition:Indexed Set|collection]].
+Suppose that all the $E_n$ are [[Definition:Countable Set|countable]].
+Then by [[Countable Union of Countable Sets is Countable]], $\displaystyle \bigcup_{n \mathop \in \N} E_n$ is also [[Definition:Countable Set|countable]], hence in $\Sigma$.
+Now suppose that at least one $E_n$ is only [[Definition:Co-Countable Set|co-countable]].
+Then by [[Set Complement inverts Subsets]] and [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]]:
+:$\displaystyle \complement_X \left({\bigcup_{n \mathop \in \N} E_n}\right) \subseteq \complement_X \left({E_n}\right)$
+By definition of [[Definition:Co-Countable Set|co-countable]], the latter is [[Definition:Countable Set|countable]].
+Thus, by [[Subset of Countably Infinite Set is Countable]], it follows that $\displaystyle \bigcup_{n \mathop \in \N} E_n$ is [[Definition:Co-Countable Set|co-countable]], hence in $\Sigma$.
+Hence the result, from [[Proof by Cases]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trace Sigma-Algebra is Sigma-Algebra}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
+Let $E \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
+Then the [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]] $\Sigma_E$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $E$.
+\end{theorem}
+
+\begin{proof}
+Verifying the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn:
+=== Axiom $(1)$ ===
+Have $E = X \cap E$ from [[Intersection with Subset is Subset]].
+Hence, as $X \in \Sigma$ by axiom $(1)$, $E \in \Sigma_E$ by definition of [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]].
+{{qed|lemma}}
+=== Axiom $(2)$ ===
+Suppose that $S \in \Sigma_E$.
+Then there is some $A \in \Sigma$ such that $S = E \cap A$, by definition of [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]].
+Hence $X \setminus A \in \Sigma$ by axiom $(2)$ for [[Definition:Sigma-Algebra|$\sigma$-algebras]].
+So $E \cap \left({X \setminus A}\right)$ in $\Sigma_E$.
+By [[Intersection with Set Difference is Set Difference with Intersection]], this equals $\left({X \cap E}\right) \setminus A = E \setminus A$.
+Now $E \setminus A = E \setminus S$ by [[Set Difference with Intersection is Difference]].
+Hence $E \setminus S \in \Sigma_E$.
+{{qed|lemma}}
+=== Axiom $(3)$ ===
+Let $\left({S_i}\right)_{i \in \N}$ be a [[Definition:Sequence|sequence]] in $\Sigma_E$.
+For each $i \in \N$, let $A_i \in \Sigma$ such that $S_i = E \cap A_i$.
+Then observe:
+{{begin-eqn}}
+{{eqn|l = \bigcup_{i \mathop \in \N} S_i
+     |r = \bigcup_{i \mathop \in \N} E \cap A_i
+}}
+{{eqn|r = E \cap \bigcup_{i \mathop \in \N} A_i
+     |c = [[Intersection Distributes over Union|Intersection Distributes over Union: General Result]]
+}}
+{{end-eqn}}
+As $\Sigma$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]], $\displaystyle \bigcup_{i \mathop \in \N} A_i \in \Sigma$.
+Hence $\displaystyle \bigcup_{i \mathop \in \N} S_i \in \Sigma_E$ by definition of [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pre-Image Sigma-Algebra on Domain is Sigma-Algebra}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X, X'$ be [[Definition:Set|sets]], and let $f: X \to X'$ be a [[Definition:Mapping|mapping]].
+Let $\Sigma'$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X'$.
+Then $f^{-1} \left({\Sigma'}\right)$, the [[Definition:Pre-Image Sigma-Algebra on Domain|pre-image $\sigma$-algebra]] on the [[Definition:Domain of Mapping|domain]] of $f$, is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
+\end{theorem}
+
+\begin{proof}
+Verify the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn:
+=== Axiom $(1)$ ===
+As $f$ is a [[Definition:Mapping|mapping]], it is immediate that $f^{-1} \left({X'}\right) = X$.
+Also $X' \in \Sigma'$ as $\Sigma'$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
+Hence $X \in f^{-1} \left({\Sigma'}\right)$.
+{{qed|lemma}}
+=== Axiom $(2)$ ===
+Let $S = f^{-1} \left({E'}\right) \in f^{-1} \left({\Sigma'}\right)$. 
+By [[Preimage of Set Difference under Mapping]] and $f^{-1} \left({X'}\right) = X$, have:
+:$X \setminus S = f^{-1} \left({X'}\right) \setminus f^{-1} \left({E'}\right) = f^{-1} \left({X' \setminus E'}\right)$
+As $X' \setminus E' \in \Sigma'$, this means $X \setminus S \in f^{-1} \left({\Sigma'}\right)$.
+{{qed|lemma}}
+=== Axiom $(3)$ ===
+Let $\left({S_i}\right)_{i \in \N}$ be a [[Definition:Sequence|sequence]] in $f^{-1} \left({\Sigma'}\right)$.
+Let, for all $i \in \N$, $E'_i \in \Sigma'$ such that $S_i = f^{-1} \left({E'}\right)$.
+By [[Preimage of Union under Mapping/General Result|Preimage of Union under Mapping: General Result]], have:
+:$\displaystyle \bigcup_{i \mathop \in \N} S_i = \bigcup_{i \mathop \in \N} f^{-1} \left({E'_i}\right) = f^{-1} \left({\bigcup_{i \mathop \in \N} E'_i}\right)$
+Now $\displaystyle \bigcup_{i \mathop \in \N} E'_i \in \Sigma'$ as $\Sigma'$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
+Hence $\displaystyle \bigcup_{i \mathop \in \N} S_i \in f^{-1} \left({\Sigma'}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Sigma-Algebras}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\family {\Sigma_i}_{i \mathop \in I}$ be a [[Definition:Indexed Set|indexed set]] of [[Definition:Sigma-Algebra|$\sigma$-algebras]] on $X$.
+Then $\Sigma := \displaystyle \bigcap_{i \mathop \in I} \Sigma_i$ is also a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
+Here, $\displaystyle \bigcap$ denotes [[Definition:Set Intersection|set intersection]].
+\end{theorem}
+
+\begin{proof}
+Verify the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn:
+=== Axiom $(1)$ ===
+As all the $\Sigma_i$ are [[Definition:Sigma-Algebra|$\sigma$-algebras]], $X \in \Sigma_i$ for all $i \in I$.
+Hence $X \in \Sigma$ by definition of [[Definition:Set Intersection|intersection]].
+{{qed|lemma}}
+=== Axiom $(2)$ ===
+Let $E \in \Sigma$. 
+Then for all $i \in I$, $E \in \Sigma_i$, by definition of [[Definition:Set Intersection|intersection]].
+As all the $\Sigma_i$ are [[Definition:Sigma-Algebra|$\sigma$-algebras]], also, for all $i \in I$:
+:$X \setminus E \in \Sigma_i$
+Hence $X \setminus E \in \Sigma$ by definition of [[Definition:Set Intersection|intersection]].
+{{qed|lemma}}
+=== Axiom $(3)$ ===
+Let $\sequence {E_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $\Sigma$.
+Then it is also a [[Definition:Sequence|sequence]] in every $\Sigma_i$.
+Hence, for all $i \in I$, as the $\Sigma_i$ are all [[Definition:Sigma-Algebra|$\sigma$-algebras]]:
+:$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \Sigma_i$
+By definition of [[Definition:Set Intersection|intersection]], it follows that:
+:$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \Sigma$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generated Sigma-Algebra Preserves Subset}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\mathcal F, \mathcal G \subseteq \mathcal P \left({X}\right)$ be collections of [[Definition:Subset|subsets]] of $X$.
+Suppose that $\mathcal F \subseteq \mathcal G$.
+Then $\sigma \left({\mathcal F}\right) \subseteq \sigma \left({\mathcal G}\right)$, where $\sigma \left({\mathcal G}\right)$ denotes the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\mathcal G$]]
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Sigma-Algebra Generated by Collection of Subsets|definition]] of $\sigma \left({\mathcal G}\right)$, $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$.
+It follows that also $\mathcal F \subseteq \sigma \left({\mathcal G}\right)$.
+Hence, by [[Definition:Sigma-Algebra Generated by Collection of Subsets|definition]] of $\sigma \left({\mathcal F}\right)$, $\sigma \left({\mathcal F}\right) \subseteq \sigma \left({\mathcal G}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Euclidean Borel Sigma-Algebra}
+Tags: Sigma-Algebras, Characterization of Euclidean Borel Sigma-Algebra
+
+\begin{theorem}
+Let $\mathcal{O}^n$, $\mathcal{C}^n$ and $\mathcal{K}^n$ be the collections of [[Definition:Open Set (Topology)|open]], [[Definition:Closed Set (Topology)|closed]] and [[Definition:Compact (Real Analysis)|compact]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\left({\R^n, \tau}\right)$, respectively.
+Let $\mathcal{J}_{ho}^n$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$.
+Let $\mathcal{J}^n_{ho, \text{rat}}$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$ with [[Definition:Rational Number|rational]] endpoints.
+Then the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] $\mathcal B \left({\R^n}\right)$ satisfies:
+:$\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{C}^n}\right) = \sigma \left({\mathcal{K}^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right) = \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$
+where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], $\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{O}^n}\right)$.
+The rest of the proof will be split in proving the following equalities:
+:$(1): \quad \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{C}^n}\right)$
+:$(2): \quad \sigma \left({\mathcal{C}^n}\right) = \sigma \left({\mathcal{K}^n}\right)$
+:$(3): \quad \sigma \left({\mathcal{O}^n}\right) = \sigma \left({\mathcal{J}_{ho}^n}\right)$
+:$(4): \quad \sigma \left({\mathcal{J}_{ho}^n}\right) = \sigma \left({\mathcal{J}^n_{ho, \text{rat}}}\right)$
+=== [[Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed|Proof of $(1)$]] ===
+{{:Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed}}
+=== [[Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact|Proof of $(2)$]] ===
+{{:Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact}}
+=== [[Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle|Proof of $(3)$]] ===
+{{:Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle}}
+=== [[Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle|Proof of $(4)$]] ===
+{{:Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle}}
+\end{proof}<|endoftext|>
+\section{Generated Sigma-Algebra Preserves Finiteness}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]], and let $A_1, \ldots, A_n \subseteq X$ be [[Definition:Subset|subsets]] of $X$.
+Then $\sigma \left({\left\{{A_1, \ldots, A_n}\right\}}\right)$ is a [[Definition:Finite Set|finite set]], where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+Proceed by [[Principle of Mathematical Induction|induction]] on $n$, that is, on the number of [[Definition:Sigma-Algebra Generated by Collection of Subsets#Generator|generators]].
+=== Basis for the Induction ===
+The case $n = 0$ is verified as follows:
+:$\sigma \left({\varnothing}\right) = \left\{{\varnothing, X}\right\}$
+from [[Sigma-Algebra Generated by Empty Set]].
+Thus $\sigma \left({\varnothing}\right)$ is [[Definition:Finite|finite]].
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $n \in \N$.
+Assume that $\sigma \left({\left\{{A_1, \ldots, A_n}\right\}}\right)$ is [[Definition:Finite|finite]].
+This is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+For brevity, write $\Sigma_n$ for $\sigma \left({A_1, \ldots, A_n}\right)$.
+Observe that:
+:$\left\{{A_1, \ldots, A_{n+1}}\right\} \subseteq \Sigma_n \cup \left\{{A_{n+1}}\right\} \subseteq \Sigma_{n+1}$
+Hence [[Condition on Equality of Generated Sigma-Algebras]] applies to yield:
+:$\Sigma_{n+1} = \sigma \left({\Sigma_n \cup \left\{{A_{n+1}}\right\}}\right)$
+From [[Sigma-Algebra Extended by Single Set]], it follows that:
+:$\sigma \left({\Sigma_n \cup \left\{{A_{n+1}}\right\}}\right) = \left\{{\left({E_1 \cap A_{n+1}}\right) \cup \left({E_2 \cap A_{n+1}^c}\right): E_1, E_2 \in \Sigma_n}\right\}$
+Thus there is a [[Definition:Surjection|surjection]] $f: \Sigma_n \times \Sigma_n \to \Sigma_{n+1}$, defined by:
+:$f \left({E_1, E_2}\right) := \left({E_1 \cap A_{n+1}}\right) \cup \left({E_2 \cap A_{n+1}^c}\right)$
+By [[Cardinality of Surjection]], it now follows that:
+:$\left\vert{\Sigma_{n+1}}\right\vert \le \left\vert{\Sigma_n \times \Sigma_n}\right\vert$
+and the latter equals $\left\vert{\Sigma_n}\right\vert^2$ by [[Cardinality of Cartesian Product]].
+By the induction hypothesis, this is necessarily [[Definition:Finite|finite]].
+Therefore, $\Sigma_{n+1}$ is also [[Definition:Finite|finite]].
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Invertible Mappings forms Symmetric Group}
+Tags: Symmetric Groups, Permutation Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\struct {S^S, \circ}$ be the [[Set of all Self-Maps is Monoid|monoid]] of all the [[Definition:Self-Map|mappings from $S$ to itself]].
+Let $\mathcal G$ be the [[Definition:Set|set]] of all [[Definition:Invertible Mapping|invertible mappings]] from $S$ to $S$.
+Then $\struct {\mathcal G, \circ}$ is the [[Definition:Symmetric Group|symmetric group]] on $S$.
+\end{theorem}
+
+\begin{proof}
+We have [[Set of all Self-Maps is Monoid]].
+By [[Inverse of Permutation is Permutation]], if $f$ is a [[Definition:Permutation|permutation]] of $S$, then so is its [[Definition:Inverse of Mapping|inverse]] $f^{-1}$.
+By [[Bijection iff Inverse is Bijection]], it follows that all the [[Definition:Invertible Element|invertible elements]] of $S^S$ are exactly the [[Definition:Permutation|permutations]] on $S$.
+The result follows from [[Invertible Elements of Monoid form Subgroup of Cancellable Elements]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Units of Field}
+Tags: Field Theory
+
+\begin{theorem}
+Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
+Then $k^\times = k \setminus \set 0$
+\end{theorem}
+
+\begin{proof}
+$0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$.
+Thus $0 \notin k^\times$.
+Consider now $0 \ne a \in k$.
+From the axioms of [[Definition:Field (Abstract Algebra)|fields]] it follows that
+there exists $a^{-1}$.
+Thus $a \in k^\times$.
+{{qed}}
+[[Category:Field Theory]]
+i7zr3adupmhtbqdhthmn6w0x117qiuw
+\end{proof}<|endoftext|>
+\section{Automorphism Group is Subgroup of Symmetric Group}
+Tags: Examples of Groups, Automorphism Groups, Group Automorphisms
+
+\begin{theorem}
+Let $\struct {S, *}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $\Aut S$ be the [[Definition:Automorphism Group of Group|automorphism group]] of $\struct {S, *}$.
+Then $\Aut S$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Symmetric Group|symmetric group ]] $\struct {\Gamma \paren S, \circ}$ on $S$.
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Automorphism (Abstract Algebra)|automorphism]] is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\phi: S \to S$ from an [[Definition:Algebraic Structure|algebraic structure]] $S$ to itself.
+The [[Identity Mapping is Automorphism]], so $\Aut S$ is not [[Definition:Empty Set|empty]].
+The [[Definition:Composition of Mappings|composite]] of [[Definition:Isomorphism (Abstract Algebra)|isomorphisms]] is itself an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], as demonstrated on [[Isomorphism is Equivalence Relation]].
+So:
+:$\phi_1, \phi_2 \in \Aut S \implies \phi_1 \circ \phi_2 \in \Aut S$
+demonstrating [[Definition:Closed Algebraic Structure|closure]].
+If $\phi \in \Aut S$, then $\phi$ is [[Definition:Bijection|bijective]] and an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Hence from [[Inverse of Algebraic Structure Isomorphism is Isomorphism]], $\phi^{-1}$ is also [[Definition:Bijection|bijective]] and an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+So $\phi^{-1} \in \Aut S$.
+The result follows by the [[Two-Step Subgroup Test]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Differentiation of Vector-Valued Function Componentwise}
+Tags: Differential Calculus, Vector Calculus
+
+\begin{theorem}
+Let:
+:$\mathbf r: t \mapsto \left\langle{r_1 \left({t}\right), r_2 \left({t}\right), \ldots, r_n \left({t}\right)}\right\rangle$ 
+be a [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued function]].
+The [[Definition:Derivative of Vector-Valued Function|derivative]] of a [[Definition:Vector-Valued Function|vector-valued function]] can be calculated by differentiating each of its [[Definition:Vector-Valued Function/Component Function|component functions]]:
+:$\dfrac {\d \mathbf r \left({t}\right)} {\d t} = \left\langle{D_t r_1 \left({t}\right), D_t r_2 \left({t}\right), \ldots, D_t r_n \left({t}\right)}\right\rangle$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {\d \mathbf r \left({t}\right)} {\d t}
+      | r = \lim_{\Delta t \mathop \to 0} \frac {\mathbf r \left({t + \Delta t}\right) - \mathbf r \left({t}\right)} {\Delta t}
+      | c = {{Defof|Derivative of Vector-Valued Function}}
+}}
+{{eqn | r =\begin{bmatrix}
+\lim_{\Delta t \mathop \to 0} \dfrac {r_1 \left({t + \Delta t}\right) - r_1 \left({t}\right)} {\Delta t} \\ \lim_{\Delta t \mathop \to 0} \dfrac {r_2 \left({t + \Delta t}\right) - r_2 \left({t}\right)} {\Delta t} \\ \vdots \\ \lim_{\Delta t \mathop \to 0} \dfrac {r_n \left({t + \Delta t}\right) - r_n \left({t}\right)} {\Delta t}
+\end{bmatrix}
+      | c = {{Defof|Limit of Vector-Valued Function}}
+}}
+{{eqn | r = \begin{bmatrix} D_t r_1 \left({t}\right) \\D_t r_2 \left({t}\right) \\ \vdots \\ D_t r_n \left({t}\right) \end{bmatrix}
+      | c = {{Defof|Derivative of Real Function}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Differential Calculus]]
+[[Category:Vector Calculus]]
+da87vru0642rff4qw28k638djpgnpcs
+\end{proof}<|endoftext|>
+\section{Derivative of Dot Product of Vector-Valued Functions}
+Tags: Differential Calculus, Vector Calculus, Dot Product
+
+\begin{theorem}
+Let:
+: $\mathbf r: x \mapsto \left\langle{r_1 \left({x}\right), r_2 \left({x}\right), \ldots, r_n \left({x}\right)}\right\rangle$
+: $\mathbf q: x \mapsto \left\langle{q_1 \left({x}\right), q_2 \left({x}\right), \ldots, q_n \left({x}\right)}\right\rangle$
+be [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued functions]].
+The [[Definition:Derivative of Vector-Valued Function|derivative]] of their [[Definition:Dot Product|dot product]] is given by:
+:$\dfrac \d {\d x} \left({\mathbf r \left({x}\right) \cdot \mathbf q \left({x}\right)}\right) = \mathbf r' \left({x}\right) \cdot \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \cdot \mathbf q' \left({x}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac \d {\d x} \left({\mathbf r \left({x}\right) \cdot \mathbf q \left({x}\right) }\right)
+      | r = \frac \d {\d x} \left({\sum_{i \mathop = 1}^n r_i \left({x}\right) q_i \left({x}\right)}\right)
+      | c = {{Defof|Dot Product}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \frac \d {\d x} \left({r_i \left({x}\right) q_i \left({x}\right)}\right)
+      | c = [[Sum Rule for Derivatives/General Result|Sum Rule for Derivatives]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \left({ {r_i}' \left({x}\right) q_i \left({x}\right) + r_i \left({x}\right) {q_i}' \left({x}\right)}\right)
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n {r_i}' \left({x}\right) q_i \left({x}\right) + \sum_{i \mathop = 1}^n r_i \left({x}\right) {q_i}' \left({x}\right) 
+      | c = [[Summation is Linear]]
+}}
+{{eqn | r = \mathbf r'\left({x}\right) \cdot \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \cdot \mathbf q' \left({x}\right)
+      | c = {{Defof|Dot Product}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Borel Sigma-Algebra of Subset is Trace Sigma-Algebra}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $\left({X, \tau}\right)$ be a [[Definition:Topological Space|topological space]], and let $A \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
+Let $\tau_A$ be the [[Definition:Subspace Topology|subspace topology]] on $A$.
+Then the following equality of [[Definition:Sigma-Algebra|$\sigma$-algebras]] on $A$ holds:
+:$\mathcal B \left({A, \tau_A}\right) = \mathcal B \left({X, \tau}\right)_A$
+where $\mathcal B$ signifies [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], and $\mathcal B \left({X, \tau}\right)_A$ signifies [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], it holds that:
+:$\mathcal B \left({X, \tau}\right) = \sigma \left({\tau}\right)$
+and also, by definition of [[Definition:Subspace Topology|subspace topology]]:
+:$\tau_A = A \cap \tau = \left\{{A \cap U: U \in \tau}\right\}$
+Thus, it follows that:
+:$\mathcal B \left({A, \tau_A}\right) = \sigma \left({A \cap \tau}\right)$
+Thereby, the desired equality:
+:$\mathcal B \left({A, \tau_A}\right) = \mathcal B \left({X, \tau}\right)_A$
+follows directly from applying [[Trace Sigma-Algebra of Generated Sigma-Algebra]] with $\mathcal G = \tau$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Vector Cross Product of Vector-Valued Functions}
+Tags: Differential Calculus, Vector Calculus, Vector Cross Product
+
+\begin{theorem}
+Let:
+:$\mathbf r: t \mapsto \begin {bmatrix} x \\ y \\ z \end{bmatrix}$
+:$\mathbf q: t \mapsto \begin {bmatrix} \chi \\ \gamma \\ \zeta \end{bmatrix}$
+be [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued functions]], where:
+:$x, y, z, \chi, \gamma, \zeta$
+are ([[Definition:Image of Mapping|images]] of) [[Definition:Differentiable Real Function|differentiable]] [[Definition:Real Function|real functions]].
+The [[Definition:Derivative of Vector-Valued Function|derivative]] of the [[Definition:Vector Cross Product|vector cross product]] of $\mathbf r$ and $\mathbf q$ is given by:
+:$D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right) = \mathbf r' \left({x}\right) \times \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \times \mathbf q'\left({x}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf r \left({t}\right) \times \mathbf q \left({t}\right)
+      | r = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \times \begin{bmatrix} \chi \\ \gamma \\ \zeta \end{bmatrix}
+}}
+{{eqn | r = \begin{bmatrix} y \zeta - z \gamma \\ z \chi - x \zeta \\ x\gamma -y \chi \end{bmatrix}
+      | c = {{Defof|Vector Cross Product}}
+}}
+{{eqn | ll= \leadsto
+      | l = D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right)
+      | r = \begin {bmatrix} D_t\left({y \zeta - z \gamma}\right) \\ D_t\left({z \chi - x \zeta}\right) \\ D_t\left({x\gamma -y \chi}\right) \end{bmatrix} 
+      | c = [[Differentiation of Vector-Valued Function Componentwise]]
+}}
+{{eqn | r = \begin {bmatrix} \dfrac{\d y} {\d t}\zeta + y \dfrac {\d \zeta} {\d t} - \dfrac {\d z} {\d t} \gamma - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi + z \dfrac {\d \chi} {\d t} - \dfrac {\d x} {\d t} \zeta - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma + x \dfrac {\d \gamma} {\d t} - \dfrac {\d y} {\d t} \chi - y \dfrac {\d \chi} {\d t} \end{bmatrix}
+      | c = [[Product Rule for Derivatives]], [[Linear Combination of Derivatives]]
+}}
+{{eqn | l = \mathbf r' \left({x}\right) \times \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \times \mathbf q'\left({x}\right)
+      | r = \begin {bmatrix} \dfrac {\d x} {\d t} \\ \dfrac {\d y} {\d t} \\ \dfrac {\d z} {\d t} \end {bmatrix} \times \begin {bmatrix} \chi \\ \gamma \\ \zeta \end {bmatrix} + \begin {bmatrix} x \\ y \\ z \end {bmatrix} \times \begin {bmatrix} \dfrac {\d \chi} {\d t} \\ \dfrac {\d \gamma} {\d t} \\ \dfrac {\d \zeta} {\d t} \end{bmatrix}
+}}
+{{eqn | r = \begin {bmatrix} \dfrac {\d y} {\d t} \zeta - \dfrac {\d z} {\d t} \gamma \\ \dfrac {\d z} {\d t} \chi - \dfrac {\d x} {\d t} \zeta \\ \dfrac {\d x} {\d t} \gamma - \dfrac {\d y} {\d t} \chi \end {bmatrix} + \begin {bmatrix} y \dfrac {\d \zeta} {\d t} - z \dfrac {\d \gamma} {\d t} \\ z \dfrac {\d \chi} {\d t} - x \dfrac {\d \zeta} {\d t} \\ x \dfrac {\d \gamma} {\d t} - y \dfrac {\d \chi} {\d t} \end {bmatrix}
+      | c = {{Defof|Vector Cross Product}}
+}}
+{{eqn | r = \begin {bmatrix} \dfrac {\d y} {\d t} \zeta - \dfrac {\d z} {\d t} \gamma + y \dfrac {\d \zeta} {\d t} - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi - \dfrac {\d x} {\d t} \zeta + z \dfrac {\d \chi} {\d t} - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma - \dfrac {\d y} {\d t} \chi + x \dfrac {\d \gamma} {\d t} - y \dfrac {\d \chi} {\d t} \end {bmatrix}
+}}
+{{eqn | r = \begin {bmatrix} \dfrac {\d y} {\d t} \zeta + y \dfrac {\d \zeta} {\d t} - \dfrac {\d z} {\d t} \gamma - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi + z \dfrac {\d \chi} {\d t} - \dfrac {\d x} {\d t} \zeta - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma + x \dfrac {\d \gamma} {\d t} - \dfrac {\d y} {\d t} \chi - y \dfrac {\d \chi} {\d t} \end {bmatrix}
+}}
+{{eqn | r = D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right)
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Monotone Class Generated by Collection of Subsets}
+Tags: Monotone Classes
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\mathcal G \subseteq \mathcal P \left({X}\right)$ be a collection of [[Definition:Subset|subsets]] of $X$.
+Then $\mathfrak m \left({\mathcal G}\right)$, the [[Definition:Monotone Class Generated by Collection of Subsets|monotone class generated by $\mathcal G$]], exists and is unique.
+\end{theorem}
+
+\begin{proof}
+=== Existence ===
+By [[Power Set is Monotone Class]], there is at least one [[Definition:Monotone Class|monotone class]] containing $\mathcal G$.
+Now let $\Bbb M$ be the collection of [[Definition:Monotone Class|monotone classes]] containing $\mathcal G$:
+:$\Bbb M := \left\{{\mathfrak{m}': \mathcal G \subseteq \mathfrak{m}', \text{$\mathfrak{m}'$ is a monotone class}}\right\}$
+By [[Intersection of Monotone Classes is Monotone Class]], $\mathfrak m := \bigcap \Bbb M$ is a [[Definition:Monotone Class|monotone class]].
+Also, by [[Set Intersection Preserves Subsets/Families of Sets|Set Intersection Preserves Subsets]]:
+: $\mathcal G \subseteq \mathfrak m$
+Now let $\mathfrak{m}'$ be a [[Definition:Monotone Class|monotone class]] containing $\mathcal G$.
+Then by construction of $\mathfrak m$, and [[Intersection is Subset/General Result|Intersection is Subset: General Result]]:
+: $\mathfrak m \subseteq \mathfrak{m}'$
+{{qed|lemma}}
+=== Uniqueness ===
+Suppose $\mathfrak{m}_1$ and $\mathfrak{m}_2$ are both [[Definition:Monotone Class Generated by Collection of Subsets|monotone classes generated by $\mathcal G$]].
+Then property $(2)$ for these [[Definition:Monotone Class|monotone classes]] implies both $\mathfrak{m}_1 \subseteq \mathfrak{m}_2$ and $\mathfrak{m}_2 \subseteq \mathfrak{m}_1$.
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+: $\mathfrak{m}_1 = \mathfrak{m}_2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Product of Real Function and Vector-Valued Function}
+Tags: Differential Calculus, Vector Calculus
+
+\begin{theorem}
+Let:
+:$\mathbf r:x \mapsto \mathbf z$
+be a [[Definition:Differentiable Vector-Valued Function|differentiable]] [[Definition:Vector-Valued Function|vector-valued function]], where:
+:$\mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix}$
+such that:
+:$z_1, z_2, \cdots, z_n$
+are ([[Definition:Image of Mapping|images]] of) [[Definition:Differentiable Real Function|differentiable]] [[Definition:Real Function|real functions]].
+Let:
+:$f: x \mapsto y$
+be a [[Definition:Differentiable Real Function|differentiable]] [[Definition:Real Function|real function]].
+Then:
+:$D_x \left({y \, \mathbf z}\right) = \dfrac {\d y} {\d x} \mathbf z + y \dfrac {\d \mathbf z} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = D_x \left({y \, \mathbf{z} }\right)
+      | r = D_x \left({\begin{bmatrix} y \ z_1 \\ y \ z_2 \\ \vdots \\ y \ z_n \end{bmatrix} }\right)
+}}
+{{eqn | r = \begin{bmatrix} D_x \left({y \ z_1}\right) \\ D_x \left({y \ z_2 }\right) \\ \vdots \\ D_x \left({y \ z_n }\right) \end{bmatrix}
+      | c = [[Differentiation of Vector-Valued Function Componentwise]]
+}}
+{{eqn | r = \begin{bmatrix} \dfrac {\d y} {\d x} z_1 + y \dfrac {\d z_1} {\d x} \\ \dfrac {\d y} {\d x}z_2 + y \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d y} {\d x} z_n + y \dfrac {\d z_n} {\d x} \end{bmatrix}
+      | c = [[Product Rule for Derivatives|Product Rule for Derivatives of Real Functions]]
+}}
+{{eqn | r = \begin{bmatrix} \dfrac {\d y} {\d x} z_1 \\ \dfrac {\d y} {\d x} z_2 \\ \vdots \\ \dfrac {\d y} {\d x} z_n \end{bmatrix} + \begin{bmatrix} y \dfrac {\d z_1} {\d x} \\ y \dfrac {\d z_2} {\d x} \\ \vdots \\ y \dfrac {\d z_n} {\d x} \end{bmatrix}
+}}
+{{eqn | r = \dfrac {\d y} {\d x} \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} + y \begin{bmatrix} \dfrac {\d z_1} {\d x} \\ \dfrac {\d z_2} {\d x} \\ \vdots \\ \dfrac {\d z_n} {\d x} \end{bmatrix}
+}}
+{{eqn | r = \frac {\d y} {\d x} \mathbf z + y \frac {\d \mathbf z} {\d x}
+      | c = [[Differentiation of Vector-Valued Function Componentwise]]
+}}
+{{end-eqn}}
+{{qed}}
+{{proofread}}
+\end{proof}
\ No newline at end of file