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+\section{Intersection of Image with Subset of Codomain}
+Tags: Mapping Theory, Set Intersection
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $A \subseteq S$ and $B \subseteq T$.
+Then:
+:$f \sqbrk {A \cap f^{-1} \sqbrk B} = f \sqbrk A \cap B$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = f \sqbrk {A \cap f^{-1} \sqbrk B}
+      | r = \set {\map f x: x \in A \cap f^{-1} \sqbrk B}
+      | c = {{Defof|Image of Subset under Mapping}}
+}}
+{{eqn | r = \set {\map f x: x \in A \land x \in f^{-1} \sqbrk B}
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | r = \set {\map f x: x \in A \land \map f x \in B}
+      | c = {{Defof|Preimage of Subset under Mapping}}
+}}
+{{eqn | r = \set {\map f x: x \in A} \cap \set {\map f x: \map f x \in B}
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | r = f \sqbrk A \cap B
+      | c = {{Defof|Image of Subset under Mapping}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projection is Injection iff Factor is Singleton}
+Tags: Projections, Injections, Singletons
+
+\begin{theorem}
+Let $S_1, S_2, \ldots, S_n$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]].
+Let $\displaystyle S = \prod_{i \mathop = 1}^n S_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $S_1, S_2, \ldots, S_n$.
+Let $\pr_j: S \to S_j$ be the [[Definition:Projection (Mapping Theory)|$j$th projection]] on $S$.
+Then $\pr_j$ is an [[Definition:Injection|injection]] {{iff}} $S_k$ is a [[Definition:Singleton|singleton]] for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Suppose $S_k = \set {s_k}$ for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
+Let $\map {\pr_j} x = \map {\pr_j} y = z$ for $x, y \in S$.
+Then by definition of [[Definition:Projection (Mapping Theory)|$j$th projection]], $x, y \in S$ are given by:
+:$x = \tuple {s_1, s_2, \dotsc, s_{j - 1}, z, s_{j + 1}, \dotsc, s_n}$
+:$y = \tuple {s_1, s_2, \dotsc, s_{j - 1}, z, s_{j + 1}, \dotsc, s_n}$
+and so $x = y$.
+Thus $\pr_j$ is an [[Definition:Injection|injection]] by definition.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\pr_j$ be an [[Definition:Injection|injection]].
+Thus:
+:$\forall x, y \in S: \map {\pr_j} x = \map {\pr_j} y \implies x = y$
+Let $x, y \in S$ be such that:
+:$x = \tuple {x_1, x_2, \dotsc, x_{j - 1}, z, x_{j + 1}, \dotsc, x_n}$
+:$y = \tuple {y_1, y_2, \dotsc, y_{j - 1}, z, y_{j + 1}, \dotsc, y_n}$
+Then:
+:$\map {\pr_j} x = \map {\pr_j} y = z$
+For $x = y$ it is necessary that:
+:$x_1 = y_1, x_2 = y_2, \dotsc, x_{j - 1} = y_{j - 1}, x_{j + 1} = y_{j + 1}, \dotsc, x_n = y_n$
+That is:
+:$\forall k \in \set {1, 2, \dotsc, n}, k \ne j: x_k = y_k$
+That is, that $S_k$ is a [[Definition:Singleton|singleton]] for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Element under Projection}
+Tags: Projections
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Set|sets]].
+Let $A \times B$ be the [[Definition:Cartesian Product|cartesian product]] of $A$ and $B$.
+Let $\pr_1: A \times B \to A$ be the [[Definition:First Projection|first projection]] of $A \times B$.
+Let $a \in A$.
+Then:
+:$\pr_1^{-1} \sqbrk {\set a} = \set {\tuple {a, b}: b \in B}$
+that is:
+:$\pr_1^{-1} \sqbrk {\set a} = \set a \times B$
+\end{theorem}
+
+\begin{proof}
+Directly apparent from the definition of [[Definition:Cartesian Product|cartesian product]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisors of Factorial}
+Tags: Factorials, Divisors
+
+\begin{theorem}
+Let $n \in \N_{>0}$.
+Then all [[Definition:Natural Numbers|natural numbers]] less than or equal to $n$ are [[Definition:Divisor of Integer|divisors]] of $n!$:
+:$\forall k \in \left\{{1, 2, \ldots, n}\right\}: n! \equiv 0 \pmod k$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Factorial|factorial]]:
+:$n! = 1 \times 2 \times \cdots \times \left({n-1}\right) \times n$
+Thus every number less than $n$ appears as a [[Definition:Divisor of Integer|divisor]] of $n!$.
+The result follows from definition of [[Definition:Congruence (Number Theory)/Integers/Integer Multiple|congruence]].
+{{qed}}
+[[Category:Factorials]]
+[[Category:Divisors]]
+3p1tjojrrsp4gejtpfuz94uxurucs8o
+\end{proof}<|endoftext|>
+\section{Positive Difference Relation on Reals is Transitive}
+Tags: Transitive Relations
+
+\begin{theorem}
+Let $P \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Numbers|real numbers]] such that:
+:$(1): \quad 1 \in P$
+:$(2): \quad a, b \in P \implies a + b \in P$
+:$(3): \quad$ For all $x \in \R$, exactly one of these is true:
+:::$x \in P$
+:::$x = 0$
+:::$-x \in P$
+Let $Q \subseteq \R \times \R$ be the [[Definition:Relation|relation]] on $\R$ defined as:
+:$Q = \set {\tuple {a, b} \in \R:  a - b \in P}$
+Then $Q$ is a [[Definition:Transitive Relation|transitive relation]].
+\end{theorem}
+
+\begin{proof}
+Let $a - b \in P$ and $b - c \in P$.
+By condition $(2)$:
+:$\paren {a - b} + \paren {b - c} \in P$
+Simplifying:
+:$a - c \in P$
+The result follows by definition of [[Definition:Transitive Relation|transitive relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Repeated Composition of Injection is Injection}
+Tags: Injections, Composite Mappings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $f: S \to S$ be an [[Definition:Injection|injection]].
+Let the [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]]:
+:$f^0, f^1, f^2, \ldots, f^n, \ldots$
+be defined as:
+:$\forall n \in \N: \map {f^n} x = \begin {cases}
+x & : n = 0 \\
+\map f x & : n = 1 \\
+\map f {\map {f^{n - 1} } x} & : n > 1
+\end{cases}$
+Then for all $n \in \N$, $f^n$ is an [[Definition:Injection|injection]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$f^n$ is an [[Definition:Injection|injection]].
+$\map P 0$ is true, as this is the case [[Identity Mapping is Injection]].
+$\map P 1$ is true, as this is the assertion that $f$ is an [[Definition:Injection|injection]].
+=== Basis for the Induction ===
+$\map P 2$ is the assertion that $f^2$ is an [[Definition:Injection|injection]].
+{{begin-eqn}}
+{{eqn | l = \map {f^2} x
+      | r = \map f {\map f x}
+      | c = Definition of $f^n$
+}}
+{{eqn | r = \map {\paren {f \circ f} } x
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{end-eqn}}
+But $f \circ f$ is an [[Definition:Injection|injection]] by [[Composite of Injections is Injection]].
+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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$f^k$ is an [[Definition:Injection|injection]].
+Then we need to show:
+:$f^{k + 1}$ is an [[Definition:Injection|injection]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \map {f^{k + 1} } x
+      | r = \map f {\map {f^k} x}
+      | c = Definition of $f^n$
+}}
+{{eqn | r = \map {\paren {f \circ f^k} } x
+      | c = 
+}}
+{{end-eqn}}
+By the [[Repeated Composition of Injection is Injection#Induction Hypothesis|induction hypothesis]] $f^k$ is an [[Definition:Injection|injection]].
+By [[Composite of Injections is Injection]] it follows that $f^{k + 1}$ is an [[Definition:Injection|injection]].
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore for all $n \in \N$:
+:$f^n$ is an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Images of Elements under Repeated Composition of Injection form Equivalence Classes}
+Tags: Injections, Composite Mappings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $f: S \to S$ be an [[Definition:Injection|injection]].
+Let the [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]]:
+:$f^0, f^1, f^2, \ldots, f^n, \ldots$
+be defined as:
+:$\forall n \in \N: \map {f^n} x = \begin {cases}
+x & : n = 0 \\
+\map f x & : n = 1 \\
+\map f {\map {f^{n - 1} } x} & : n > 1
+\end{cases}$
+Let $\mathcal R \subseteq S \times S$ be the [[Definition:Relation|relation]] on $S$ defined as:
+:$\mathcal R = \set {\tuple {a, b} \in S \times S: \exists k \in \Z: b = \map {f^k} a \lor \exists j \in \Z: a = \map {f^j} b}$
+Then $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+By definition, $f^0$ is the [[Definition:Identity Mapping|identity mapping]].
+So by definition of $f^0$:
+:$\forall a \in S: a = \map {f^0} a$
+That is:
+:$a \mathop {\mathcal R} a$
+So $\mathcal R$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $a \mathop {\mathcal R} b$.
+That is:
+:$b = \map {f^k} a$
+for some $k \in \Z$.
+Let $g_k$ be the [[Definition:Restriction of Mapping|restriction]] of $f^k$ to its [[Definition:Image of Mapping|image]].
+From [[Injection to Image is Bijection]], $g^k$ is a [[Definition:Bijection|bijection]].
+From [[Inverse of Bijection is Bijection]]:
+:$\map {\paren {g^k}^{-1} } b = a$
+Consider the [[Definition:Extension of Mapping|extension]] $\paren {f^k}^{-1}$ of $\paren {g^k}^{-1}$ to its [[Definition:Codomain|codomain]] which is $S$.
+From [[Inverse of Injection is Many-to-One Relation]], $\map {\paren {f^k}^{-1} } b = a$ is [[Definition:Well-Defined Mapping|well-defined]].
+By interpreting $\map {\paren {f^k}^{-1} } b = a$ as $\map {f^{-k} } b = a$, it follows that we can set $j = -k$ and so:
+:$\exists j \in \Z: \map {f^j} b = a$
+That is:
+:$b \mathop {\mathcal R} a$
+So $\mathcal R$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $a \mathop {\mathcal R} b$ and $b \mathop {\mathcal R} c$.
+That is:
+:$b = \map {f^{k_1} } a$
+:$c = \map {f^{k_2} } b$
+where $k_1, k_2 \in \Z$.
+That is:
+:$c = \map {f^{k_2} } {\map {f^{k_1} } a}$
+By [[Composition of Repeated Compositions of Injections]]:
+:$c = \map {f^{k_2} } {\map {f^{k_1} } a} = \map {f^{k_1 + k_2} } a$
+and so:
+:$a \mathop {\mathcal R} c$
+So $\mathcal R$ has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+$\mathcal R$ 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{Composition of Repeated Compositions of Injections}
+Tags: Injections, Composite Mappings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $f: S \to S$ be an [[Definition:Injection|injection]].
+Let the [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]]:
+:$f^0, f^1, f^2, \ldots, f^n, \ldots$
+be defined as:
+:$\forall n \in \N: f^n \left({x}\right) = \begin{cases}
+x & : n = 0 \\
+f \left({x}\right) & : n = 1 \\
+f \left({f^{n-1} \left({x}\right)}\right) & : n > 1
+\end{cases}$
+Then:
+:$\forall m, n \in \Z_{\ge 0}: f^n \circ f^m = f^{m + n}$
+where $f^n \circ f^m$ denotes [[Definition:Composition of Mappings|composition of mappings]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+Let $m \in \Z_{\ge 0}$ be given.
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$f^n \circ f^m = f^{m + n}$
+$P \left({0}\right)$ is true, as this is the case:
+{{begin-eqn}}
+{{eqn | l = f^0 \circ f^m
+      | r = I_S \circ f^m
+      | c = where $I_S$ is the [[Definition:Identity Mapping|identity mapping]]
+}}
+{{eqn | r = f^m
+      | c = [[Identity Mapping is Left Identity]]
+}}
+{{end-eqn}}
+=== Basis for the Induction ===
+$P \left({1}\right)$ is true, as this is the case:
+:$f^{m+1} \left({x}\right) = f \left({f^m \left({x}\right)}\right)$
+by definition.
+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]]:
+:$f^k \circ f^m = f^{m + k}$
+Then we need to show:
+:$f^{k+1} \circ f^m = f^{m + k + 1}$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left({f^{k+1} \circ f^m}\right) \left({x}\right)
+      | r = \left({f \left({f^k}\right) \circ f^m}\right) \left({x}\right)
+      | c = Definition of $f^{k+1}$
+}}
+{{eqn | r = \left({\left({f \circ f^k}\right) \circ f^m}\right) \left({x}\right)
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \left({f \circ \left({f^k \circ f^m}\right)}\right) \left({x}\right)
+      | c = [[Composition of Mappings is Associative]]
+}}
+{{eqn | r = \left({f \circ f^{m+k} }\right) \left({x}\right)
+      | c = [[Composition of Repeated Compositions of Injections#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = f^{m+k+1} \left({x}\right)
+      | c = [[Composition of Repeated Compositions of Injections#Basis for the Induction|Basis for the Induction]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m, n \in \Z_{\ge 0}: f^n \circ f^m = f^{m + n}$
+{{qed}}
+[[Category:Injections]]
+[[Category:Composite Mappings]]
+l10y9a8b2nveuifhsttjiyzw45bbi2n
+\end{proof}<|endoftext|>
+\section{Derivative Function on Set of Functions induces Equivalence Relation}
+Tags: Differential Calculus, Equivalence Relations
+
+\begin{theorem}
+Let $X$ be the set of [[Definition:Real Function|real functions]] $f: \R \to \R$ which [[Definition:Continuously Differentiable|possess continuous derivatives]].
+Let $\mathcal R \subseteq X \times X$ be the [[Definition:Relation|relation]] on $X$ defined as:
+:$\mathcal R = \set {\tuple {f, g} \in X \times X: D f = D g}$
+where $D f$ denotes the [[Definition:First Derivative|first derivative]] of $f$.
+Then $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Based on the result which I can't find which says that $D f {{=}} D g \iff f {{=}} g + c$}}
+\end{proof}<|endoftext|>
+\section{Factorial Divides Product of Successive Numbers}
+Tags: Number Theory, Factorials
+
+\begin{theorem}
+Let $m, n \in \N_{\ge 1}$ be [[Definition:Natural Numbers|natural numbers]]
+Let $m^{\overline n}$ be [[Definition:Rising Factorial|$m$ to the power of $n$ rising]].
+Then:
+:$m^{\overline n} \equiv 0 \bmod n!$
+That is, the [[Definition:Factorial|factorial]] of $n$ [[Definition:Divisor of Integer|divides]] the product of $n$ successive numbers.
+\end{theorem}
+
+\begin{proof}
+{{questionable|This is not enough. It might superficially be that e.g. the factors of $2$ and $6$ coincide. Effectively this argument reduces to the theorem; it's almost circular}}
+Let $m \in \N_{\ge 1}$.
+Consider the [[Definition:Set|set]]:
+:$S = \{ {m, m + 1, m + 2, \ldots, m + n - 1 }\}$
+Note $S$ has $n$ [[Definition:Element|elements]].
+By [[Set of Successive Numbers contains Unique Multiple]]:
+:$\left\{ {m} \right\}$ contains a [[Definition:Factor (Algebra)|factor]] of $1$,
+:$\left\{ {m, m+1} \right\}$ contains a [[Definition:Factor (Algebra)|factor]] of $2$,
+and in general,
+:$\left\{ {m, m+1, \ldots, m + j - 1} \right\}$ contains a [[Definition:Factor (Algebra)|factor]] of $j$.
+It follows that $S$ contains factors of $1, 2, 3, \ldots, n$.
+Multiplying all elements of $S$ gives:
+{{begin-eqn}}
+{{eqn | l = m \left({m+1}\right)\ldots\left({m+n-1}\right)
+      | r = k_1 2 k_2 3 k_3 \ldots \left({n-1}\right)k_{n-1} n k_n
+      | c = for some $k_1, k_2, \ldots, k_n \in N$
+}}
+{{eqn | r = \left({k_1 k_2 \ldots k_{n-1}k_n}\right)\left({1 \times 2 \times \ldots \times \left({n - 1}\right) \times n}\right)
+}}
+{{eqn | ll = \implies
+      | l = m^{\overline n}
+      | r = K n!
+      | c = for some $K \in \N$
+}}
+{{eqn | ll = \implies
+      | l = m^{\overline n}
+      | o = \equiv
+      | r = 0 \bmod n!
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Number Theory]]
+[[Category:Factorials]]
+0bxzl1xm1nz2t6qgk3dpamjhu03p06r
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient is Integer}
+Tags: Binomial Coefficients, Binomial Coefficient is Integer
+
+\begin{theorem}
+Let $\dbinom n k$ be a [[Definition:Binomial Coefficient|binomial coefficient]].
+Then $\dbinom n k$ is an [[Definition:Integer|integer]].
+\end{theorem}
+
+\begin{proof}
+If it is not the case that $0 \le k \le n$, then the result holds trivially.
+So let $0 \le k \le n$.
+By the definition of [[Definition:Binomial Coefficient|binomial coefficients]]:
+{{begin-eqn}}
+{{eqn | l = \binom n k
+      | r = \frac {n!} {k!\left({n - k}\right)!}
+}}
+{{eqn | r = \frac {n \left({n - 1}\right) \left({n - 2}\right) \ldots \left({n - k + 1}\right)} {k!}
+}}
+{{end-eqn}}
+The [[Definition:Numerator|numerator]] is a [[Definition:Integer Multiplication|product]] of $k$ successive [[Definition:Integer|integers]].
+From [[Factorial Divides Product of Successive Numbers]], $k!$ [[Definition:Divisor of Integer|divides]] it.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The result follows by [[Pascal's Rule]] and [[Integer Addition is Closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence Classes induced by Derivative Function on Set of Functions}
+Tags: Differential Calculus, Equivalence Classes
+
+\begin{theorem}
+Let $X$ be the set of [[Definition:Real Function|real functions]] $f: \R \to \R$ which [[Definition:Continuously Differentiable|possess continuous derivatives]].
+Let $\mathcal R \subseteq X \times X$ be the [[Definition:Equivalence Relation|equivalence relation]] on $X$ defined as:
+:$\mathcal R = \set {\tuple {f, g} \in X \times X: D f = D g}$
+where $D f$ denotes the [[Definition:First Derivative|first derivative]] of $f$.
+Then the [[Definition:Equivalence Class|equivalence classes]] of $\mathcal R$ are defined as:
+:$\map {\eqclass f {\mathcal R} } x = \set {g \in X: \exists c \in \R: \forall x \in \R: \map g x = \map f x + c}$
+That is, it consists of the [[Definition:Set|set]] of all [[Definition:Real Function|real functions]] $f \in X$ which differ by a [[Definition:Real Number|real]] [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Derivative Function on Set of Functions induces Equivalence Relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Mappings which map to Same Element induces Equivalence Relation}
+Tags: Mapping Theory, Equivalence Relations
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Set|sets]].
+Let $E$ be the [[Definition:Set|set]] of all [[Definition:Mapping|mappings]] from $X$ to $Y$.
+Let $b \in X$.
+Let $\mathcal R \subseteq E \times E$ be the [[Definition:Relation|relation]] on $E$ defined as:
+:$\mathcal R := \set {\tuple {f, g} \in \mathcal R: \map f b = \map g b}$
+Then $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+:$\forall f \in E: \map f b = \map f b$
+That is:
+:$\forall f \in E: \tuple {f, f} \in \mathcal R$
+Thus $\mathcal R$ is seen to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+{{begin-eqn}}
+{{eqn | l = \tuple {f, g}
+      | o = \in
+      | r = \mathcal R
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map g b
+      | r = \map f b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map g b
+      | r = \map f b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {g, f}
+      | o = \in
+      | r = \mathcal R
+      | c = 
+}}
+{{end-eqn}}
+Thus $\mathcal R$ is seen to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $f, g, h \in E$.
+{{begin-eqn}}
+{{eqn | l = \tuple {f, g}
+      | o = \in
+      | r = \mathcal R
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \tuple {g, h}
+      | o = \in
+      | r = \mathcal R
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map f b
+      | r = \map g b
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \map g b
+      | r = \map h b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map f b
+      | r = \map h b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {f, h}
+      | o = \in
+      | r = \mathcal R
+      | c = 
+}}
+{{end-eqn}}
+Thus $\mathcal R$ is seen to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+$\mathcal R$ 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{Renaming Mapping from Set of Mappings on Single Element}
+Tags: Mapping Theory, Equivalence Relations
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Set|sets]].
+Let $E$ be the [[Definition:Set|set]] of all [[Definition:Mapping|mappings]] from $X$ to $Y$.
+Let $b \in X$.
+Let $\mathcal R \subseteq E \times E$ be the [[Definition:Relation|relation]] on $E$ defined as:
+:$\mathcal R := \set {\tuple {f, g} \in \mathcal R: \map f b = \map g b}$
+Let $e_b: E / \mathcal R \to Y$ be the [[Definition:Renaming Mapping|renaming mapping]] induced by $\mathcal R$.
+Then $e_b$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+This is an instance of [[Renaming Mapping is Bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Inverse Mapping}
+Tags: Inverse Mappings
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+{{TFAE|def = Inverse Mapping}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $f^{-1}: T \to S$ be an [[Definition:Inverse Mapping/Definition 1|inverse mapping of $f: S \to T$ by definition 1]].
+From [[Mapping is Injection and Surjection iff Inverse is Mapping]] it follows that $f^{-1}$ is a [[Definition:Bijection|bijection]].
+By [[Composite of Bijection with Inverse is Identity Mapping]]:
+: $f^{-1} \circ f = I_S$
+: $f \circ f^{-1} = I_T$
+Thus $f^{-1}: T \to S$ is an [[Definition:Inverse Mapping/Definition 2|inverse mapping of $f: S \to T$ by definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $g: T \to S$ be an [[Definition:Inverse Mapping/Definition 2|inverse mapping of $f: S \to T$ by definition 2]].
+From [[Left and Right Inverse Mappings Implies Bijection]] it follows that both $f$ and $g$ are [[Definition:Bijection|bijections]].
+From [[Mapping is Injection and Surjection iff Inverse is Mapping]] it follows that $g = f^{-1}$.
+Thus $g: T \to S$ is an [[Definition:Inverse Mapping/Definition 1|inverse mapping of $f: S \to T$ by definition 1]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Mapping is Bijection}
+Tags: Bijections, Inverse Mappings
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ and $g: T \to S$ be [[Definition:Inverse Mapping|inverse mappings]] of each other.
+Then $f$ and $g$ are [[Definition:Bijection|bijections]].
+\end{theorem}
+
+\begin{proof}
+From [[Inverse is Mapping implies Mapping is Injection and Surjection]]:
+:$f$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
+Again from [[Inverse is Mapping implies Mapping is Injection and Surjection]]:
+:$g$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
+The result follows by definition of [[Definition:Bijection/Definition 1|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping is Injection and Surjection iff Inverse is Mapping/Proof 2}
+Tags: Mapping is Injection and Surjection iff Inverse is Mapping
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then:
+: $f: S \to T$ can be defined as a [[Definition:Bijection/Definition 1|bijection]] in the sense that:
+::$(1): \quad f$ is an [[Definition:Injection|injection]]
+::$(2): \quad f$ is a [[Definition:Surjection|surjection]].
+{{iff}}:
+:the [[Definition:Inverse of Mapping|inverse]] $f^{-1}$ of $f$ is such that:
+::[[Definition:Universal Quantifier|for each]] $y \in T$, the [[Definition:Preimage of Element under Mapping|preimage]] $\map {f^{-1} } y$ has [[Definition:Unique|exactly one]] [[Definition:Element|element]].
+:That is, such that $f^{-1} \subseteq T \times S$ is itself a [[Definition:Mapping|mapping]].
+\end{theorem}
+
+\begin{proof}
+=== [[Inverse of Injective and Surjective Mapping is Mapping/Proof 2|Necessary Condition]] ===
+{{:Inverse of Injective and Surjective Mapping is Mapping/Proof 2}}{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f^{-1}: T \to S$ be a [[Definition:Mapping|mapping]].
+By [[Inverse Mapping is Bijection]], both $f$ and $f^{-1}$ are [[Definition:Bijection|bijections]].
+Hence, in particular, $f$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping is Extension iff Composite with Inclusion}
+Tags: Inclusion Mappings, Composite Mappings
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $A \subseteq S$.
+Let $f: S \to T$ and $g: A \to T$ be [[Definition:Mapping|mappings]].
+Then $f$ is an [[Definition:Extension of Mapping|extension]] of $g$ {{iff}}:
+:$f = g \circ i_A$
+where $i_A$ is the [[Definition:Inclusion Mapping|inclusion mapping]] on $A$.
+This can be illustrated using a [[Definition:Commutative Diagram|commutative diagram]] as follows:
+::$\begin {xy} \xymatrix@L + 2mu@ + 1em {
+ A \ar[r]^*{i_A}
+     \ar@{-->}[rd]_*{f = g \circ i_A}
+&
+ S \ar[d]^*{g}
+\\
+&
+ T
+} \end {xy}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f: S \to T$ be an [[Definition:Extension of Mapping|extension]] of $g: A \to T$.
+Then by definition:
+{{begin-eqn}}
+{{eqn | lo= \forall x \in A:
+      | l = \map f x
+      | r = \map g x
+      | c = 
+}}
+{{eqn | r = \map g {\map {i_A} x}
+      | c = {{Defof|Inclusion Mapping}}
+}}
+{{eqn | r = \map {\paren {g \circ i_A} } x
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{end-eqn}}
+That is:
+:$f = g \circ i_A$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f = g \circ i_A$ where $i_A$ is the [[Definition:Inclusion Mapping|inclusion mapping]] on $A$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall x \in A:
+      | l = \map f x
+      | r = \map {\paren {g \circ i_A} } x
+      | c = Definition of $f$
+}}
+{{eqn | r = \map g {\map {i_A} x}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map g x
+      | c = {{Defof|Inclusion Mapping}}
+}}
+{{end-eqn}}
+So, by definition, $f$ is an [[Definition:Extension of Mapping|extension]] of $g$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of Restrictions of Mapping}
+Tags: Restrictions
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let the [[Definition:Cardinality|cardinality]] of $S$ be $n$.
+Let $F$ be the [[Definition:Set|set]] of [[Definition:Restriction of Mapping|restrictions]] of $f$ to a [[Definition:Subset|subset]] of $S$.
+Then there are $2^n$ [[Definition:Element|elements]] of $F$.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Let $g: A \to T$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to $A$.
+By definition of [[Definition:Restriction of Mapping|restriction of mapping]]:
+:$\forall x \in A: \map g x = \map f x$
+and hence there is one [[Definition:Mapping|mapping]] $g: A \to T$ such that $g$ is a [[Definition:Restriction of Mapping|restriction]] of $f$.
+Therefore for each [[Definition:Subset|subset]] of $S$ there exists a [[Definition:Unique|unique]] [[Definition:Restriction of Mapping|restriction]] of $f$ to that [[Definition:Subset|subset]].
+Therefore there are as many [[Definition:Restriction of Mapping|restriction]] of $f$ as there are [[Definition:Subset|subsets]] of $S$.
+That is, by definition of [[Definition:Power Set|power set]], the same as the number of [[Definition:Element|elements]] of the [[Definition:Power Set|power set]] of $S$.
+From [[Cardinality of Power Set of Finite Set]], that is $2^n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projection is Surjection/Family of Sets}
+Tags: Surjections, Indexed Families, Projections
+
+\begin{theorem}
+Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets]].
+Let $\displaystyle \prod_{\alpha \mathop \in I} S_\alpha$ be the [[Definition:Cartesian Product of Family|Cartesian product]] of $\family {S_\alpha}_{\alpha \mathop \in I}$.
+Let each of $S_\alpha$ be [[Definition:Non-Empty Set|non-empty]].
+For each $\beta \in I$, let $\displaystyle \pr_\beta: \prod_{\alpha \mathop \in I} S_\alpha \to S_\beta$ be the [[Definition:Projection on Family of Sets|$\beta$th projection]] on $\displaystyle S = \prod_{\alpha \mathop \in I} S_\alpha$.
+Then $\pr_\beta$ is a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+Consider the $\beta$th projection.
+Let $x_\beta \in S_\beta$.
+Let $\map x \beta = x_\beta$
+Suppose $\gamma \in I: \gamma \ne \beta$.
+As $S_\gamma \ne \O$ it is possible to use the [[Axiom:Axiom of Choice|axiom of choice]] to choose $\map x \gamma \in S_\gamma$.
+Then:
+:$\displaystyle x \in \prod_{\alpha \mathop \in I} S_\alpha$
+and:
+:$\map {\pr_\beta} x = \map x \beta$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Taxicab Metric on Real Vector Space is Metric}
+Tags: Taxicab Metric, Taxicab Metric on Real Vector Space is Metric
+
+\begin{theorem}
+The [[Definition:Taxicab Metric/Real Vector Space|taxicab metric]] on the [[Definition:Real Vector Space|real vector space]] $\R^n$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+This is an instance of the [[Definition:Taxicab Metric/General Definition|taxicab metric]] on the [[Definition:Finite Cartesian Product|cartesian product]] of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
+This is proved in [[Taxicab Metric is Metric]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The [[Definition:Taxicab Metric/Real Vector Space|taxicab metric]] on $\R^n$ is:
+: $\displaystyle d_1 \left({x, y}\right) = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert$
+for $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \R^n$.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_1 \left({x, x}\right)
+      | r = \sum_{i \mathop = 1}^n \left\vert{x_i - x_i}\right\vert
+      | c = Definition of $d_1$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n 0
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_1$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+{{begin-eqn}}
+{{eqn | l = d_1 \left({x, y}\right) + d_1 \left({y, z}\right)
+      | r = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert + \sum_{i \mathop = 1}^n \left\vert{y_i - z_i}\right\vert
+      | c = Definition of $d_1$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \sum_{i \mathop = 1}^n \left\vert{x_i - z_i}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = d_1 \left({x, z}\right)
+      | c = Definition of $d_1$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_1$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_1 \left({x, y}\right)
+      | r = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert
+      | c = Definition of $d_1$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \left\vert{y_i - x_i}\right\vert
+      | c = 
+}}
+{{eqn | r = d \left({y, x}\right)
+      | c = Definition of $d_1$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_1$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | lo= \exists i \in \left\{ {1, 2, \ldots, n}\right\}:
+      | l = x_i
+      | o = \ne
+      | r = y_i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{x_i - y_i}\right\vert
+      | o = >
+      | r = 0
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | ll= \implies
+      | l = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_1 \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of $d_1$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Chebyshev Distance on Real Vector Space is Metric}
+Tags: Chebyshev Distance
+
+\begin{theorem}
+The [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$:
+: $\displaystyle \forall x, y \in \R^n: d_\infty \left({x, y}\right):= \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$
+is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+This is an instance of the [[Definition:Chebyshev Distance|Chebyshev distance]] on the [[Definition:Finite Cartesian Product|cartesian product]] of [[Definition:Metric Space|metric spaces]] $A_1, A_2, \ldots, A_3$.
+This is proved in [[Chebyshev Distance is Metric]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_\infty} {x, x}
+      | r = \max_{i \mathop = 1}^n \size {x_i - x_i}
+      | c = Definition of $d_\infty$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_\infty$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $k \in \closedint 1 n$ such that:
+{{begin-eqn}}
+{{eqn | l = \size {x_k - z_k}
+      | r = \max_{i \mathop = 1}^n \size {x_i - z_i}
+      | c = 
+}}
+{{eqn | r = \map {d_\infty} {x, z}
+      | c = 
+}}
+{{end-eqn}}
+Then by the [[Triangle Inequality for Real Numbers]]:
+:$\size {x_k - z_k} \le \size {x_k - y_k} + \size {y_k - z_k}$
+But by the nature of the $\max$ operation:
+:$\displaystyle \size {x_k - y_k} \le \max_{i \mathop = 1}^n \size {x_i - y_i}$
+and:
+:$\displaystyle \size {y_k - z_k} \le \max_{i \mathop = 1}^n \size {y_i - z_i}$
+Thus:
+:$\displaystyle \size {x_k - y_k} + \size {y_k - z_k} \le \max_{i \mathop = 1}^n \size {x_i - y_i} + \max_{i \mathop = 1}^n \size {y_i - z_i}$
+Hence:
+:$\map {d_\infty} {x, z} \le \map {d_\infty} {x, y} + \map {d_\infty} {y, z}$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_\infty$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_\infty} {x, y}
+      | r = \max_{i \mathop = 1}^n \size {x_i - y_i}
+      | c = Definition of $d_\infty$
+}}
+{{eqn | r = \max_{i \mathop = 1}^n \size {y_i - x_i}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \map {d_\infty} {y, x}
+      | c = Definition of $d_\infty$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_\infty$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+Let $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists k \in \closedint 1 n
+      | l = x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x_k - y_k}
+      | o = >
+      | r = 0
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \leadsto
+      | l = \max_{i \mathop = 1}^n \size {x_i - y_i}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_\infty} {x, y}
+      | o = >
+      | r = 0
+      | c = Definition of $d_\infty$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_\infty$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Chebyshev Distance is Metric}
+Tags: Chebyshev Distance
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+:$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
+where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \mathcal A$.
+Then $d_\infty$ is a [[Definition:Metric|metric]],
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_\infty} {x, x}
+      | r = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, x_i} }
+      | c = Definition of $d_\infty$
+}}
+{{eqn | r = 0
+      | c = as $d_i$ fulfils [[Definition:Metric Space Axioms|axiom $M1$]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_\infty$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $k \in \closedint 1 n$ such that:
+{{begin-eqn}}
+{{eqn | l = \map {d_k} {x_k, z_k}
+      | r = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, z_i} }
+      | c = 
+}}
+{{eqn | r = \map {d_\infty} {x, z}
+      | c = 
+}}
+{{end-eqn}}
+Then by application of [[Definition:Metric Space Axioms|axiom $M2$]] for [[Definition:Metric|metric]] $d_k$:
+:$\map {d_k} {x_k, z_k} \le \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k}$
+But by the nature of the $\max$ operation:
+:$\displaystyle \map {d_k} {x_k, y_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
+and:
+:$\displaystyle \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$
+Thus:
+:$\displaystyle \map {d_k} {x_k, y_k} + \map {d_k} {y_k, z_k} \le \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} } + \max_{i \mathop = 1}^n \set {\map {d_i} {y_i, z_i} }$
+Hence:
+:$\map {d_\infty} {x, z} \le \map {d_\infty} {x, y} + \map {d_\infty} {y, z}$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_\infty$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_\infty} {x, y}
+      | r = \max_{i \mathop = 1}^n \map {d_i} {x_i, y_i}
+      | c = Definition of $d_\infty$
+}}
+{{eqn | r = \max_{i \mathop = 1}^n \map {d_i} {y_i, x_i}
+      | c = as $d_i$ fulfils [[Definition:Metric Space Axioms|axiom $M3$]]
+}}
+{{eqn | r = \map {d_\infty} {y, x}
+      | c = Definition of $d_\infty$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_\infty$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+Let $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists k \in \closedint 1 n
+      | l = x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_k} {x_k, y_k}
+      | o = >
+      | r = 0
+      | c = as $d_k$ fulfils [[Definition:Metric Space Axioms|axiom $M4$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \max_{i \mathop = 1}^n \map {d_i} {x_i, y_i}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_\infty} {x, y}
+      | o = >
+      | r = 0
+      | c = Definition of $d_\infty$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_\infty$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Positive Multiple of Metric is Metric}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $k \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Let $d_k: A \times A \to \R$ be the [[Definition:Real-Valued Function|function]] defined as:
+:$\forall \left({x, y}\right) \in A: d_k \left({x, y}\right) = k \cdot d \left({x, y}\right)$
+Then $M_k = \left({A, d_k}\right)$ is a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_k \left({x, x}\right)
+      | r = k \cdot d \left({x, x}\right)
+      | c = Definition of $d_k$
+}}
+{{eqn | r = 0
+      | c = as $d$ fulfils [[Definition:Metric Space Axioms|axiom $M1$]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_k$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+{{begin-eqn}}
+{{eqn | l = d_k \left({x, y}\right) + d_k \left({y, z}\right)
+      | r = k \cdot d \left({x, y}\right) + k \cdot d \left({y, z}\right)
+      | c = Definition of $d_k$
+}}
+{{eqn | r = k \left({d \left({x, y}\right) + d \left({y, z}\right)}\right)
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = k \cdot d \left({x, z}\right)
+      | c = as $d$ fulfils [[Definition:Metric Space Axioms|axiom $M2$]]
+}}
+{{eqn | r = d_k \left({x, z}\right)
+      | c = Definition of $d_k$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_k$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_k \left({x, y}\right)
+      | r = k \cdot d \left({x, y}\right)
+      | c = Definition of $d_k$
+}}
+{{eqn | r = k \cdot d \left({y, x}\right)
+      | c = as $d$ fulfils [[Definition:Metric Space Axioms|axiom $M3$]]
+}}
+{{eqn | r = d_k \left({y, x}\right)
+      | c = Definition of $d_k$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_k$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = as $d$ fulfils [[Definition:Metric Space Axioms|axiom $M4$]]
+}}
+{{eqn | ll= \implies
+      | l = k \cdot d \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = as $k > 0$
+}}
+{{eqn | ll= \implies
+      | l = d_k \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of $d_k$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_k$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Successive Numbers contains Unique Multiple}
+Tags: Number Theory
+
+\begin{theorem}
+Let $m \in \Z_{\ge 1}$.
+Then $\set {m, m + 1, \ldots, m + n - 1}$ contains a [[Definition:Unique|unique]] [[Definition:Integer|integer]] that is a [[Definition:Multiple of Integer|multiple]] of $n$.
+That is, in any [[Definition:Set|set]] containing $n$ successive [[Definition:Integer|integers]], $n$ [[Definition:Divisor of Integer|divides]] [[Definition:Exactly One|exactly one]] of those [[Definition:Integer|integers]].
+\end{theorem}
+
+\begin{proof}
+Let $S_m = \set {m, m + 1, \ldots, m + n - 1}$ be a [[Definition:Set|set]] containing $n$ successive [[Definition:Integer|integers]].
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$, the smallest number in $S_m$.
+=== Basis for the Induction ===
+The case $m = 1$ is verified as follows:
+:$S_1 = \set {1, 2, \ldots, n}$
+This contains a [[Definition:Multiple of Integer|multiple]] of $n$, namely, $n$ itself.
+If $S_1$ is a [[Definition:Singleton|singleton]] then this [[Definition:Element|element]] is trivially unique.
+If there are any other [[Definition:Integer|integers]] in $S_1$, they are less than $n$ and more than $0$, and so cannot be a [[Definition:Multiple of Integer|multiple]] of $n$.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $m \in \N$ with $m \ge 1$.
+Assume that:
+:$S_m = \set {m, m + 1, \ldots, m + n - 1}$
+contains a unique [[Definition:Multiple of Integer|multiple]] of $n$.
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Consider the set:
+:$S_{m + 1} = \set {m + 1, m + 2, \ldots, m + n - 1, m + n}$
+As $n \equiv 0 \pmod n$, $m + n$ is a [[Definition:Multiple of Integer|multiple]] of $n$ {{iff}} $m$ is a [[Definition:Multiple of Integer|multiple]] of $n$.
+That means that we can replace $m + n$ with $m$ and instead analyze:
+:${S_{m + 1} }' = \set {m + 1, m + 2, \ldots, m + n - 1, m}$
+{{handwaving}}
+But ${S_{m + 1} }' = S_m$, which contains a unique [[Definition:Multiple of Integer|multiple]] of $n$ by the [[Set of Successive Numbers contains Unique Multiple#Induction Hypothesis|induction hypothesis]].
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+[[Category:Number Theory]]
+g2ax9b3le2fkujs6qx0wil67129zywp
+\end{proof}<|endoftext|>
+\section{Taxicab Metric is Topologically Equivalent to Chebyshev Distance on Real Vector Space}
+Tags: Chebyshev Distance, Taxicab Metric
+
+\begin{theorem}
+For $n \in \N$, let $\R^n$ be a [[Definition:Real Vector Space|real vector space]].
+Let $d_1$ be the [[Definition:Taxicab Metric/Real Vector Space|taxicab metric]] on $\R^n$.
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$.
+Then
+:$\forall x, y \in \R^n: d_\infty \left({x, y}\right) \le d_1 \left({x, y}\right) \le n \cdot d_\infty \left({x, y}\right)$
+It follows that $d_1$ and $d_\infty$ are [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$, we have:
+:$\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right)$ and $y = \left({y_1, y_2, \ldots, y_n}\right)$.
+Let $j$ be chosen so that:
+:$\displaystyle \left\vert{x_j - y_j}\right\vert = \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$
+Then:
+{{begin-eqn}}
+{{eqn | l = d_\infty \left({x, y}\right)
+      | r = \left\vert{x_j - y_j}\right\vert
+      | c = Definition of [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]]
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 1}^n \left\vert{x_i - y_i}\right\vert
+}}
+{{eqn | r = d_1 \left({x, y}\right)
+}}
+{{eqn | o = \le
+      | r = n \left\vert{x_j - y_j}\right\vert
+}}
+{{eqn | r = n \cdot d_\infty \left({x, y}\right)
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{L1 Metric on Closed Real Interval is Metric}
+Tags: Definite Integrals, L1 Metric
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Real Function|real functions]] which are [[Definition:Continuous Real Function|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$.
+Let $d: S \times S \to \R$ be the [[Definition:L1 Metric/Closed Real Interval|$L^1$ metric]] on $\left[{a \,.\,.\, b}\right]$:
+:$\displaystyle \forall f, g \in S: d \left({f, g}\right) := \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t$
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, f}\right)
+      | r = \int_a^b \left\vert{f \left({t}\right) - f \left({t}\right)}\right\vert \ \mathrm d t
+      | c = Definition of $d$
+}}
+{{eqn | r = \int_a^b 0 \ \mathrm d t
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | r = 0
+      | c = [[Definite Integral of Constant]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+{{begin-eqn}}
+{{eqn | l = \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert + \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert
+      | o = \ge
+      | r = \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | ll= \implies 
+      | l = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t + \int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert \ \mathrm d t
+      | o = \ge
+      | r = \int_a^b \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert \ \mathrm d t
+      | c = [[Relative Sizes of Definite Integrals]]
+}}
+{{eqn | ll= \implies 
+      | l = d \left({f, g}\right) + d \left({g, h}\right)
+      | o = \ge
+      | r = d \left({f, h}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t
+      | c = Definition of $d$
+}}
+{{eqn | r = \int_a^b \left\vert{g \left({t}\right) - f \left({t}\right)}\right\vert \ \mathrm d t
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | r = d \left({g, f}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | lo= \forall t \in \left[{a \,.\,.\, b}\right]:
+      | l = \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert
+      | o = \ge
+      | r = 0
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | ll= \implies
+      | l = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t
+      | o = \ge
+      | r = 0
+      | c = [[Sign of Function Matches Sign of Definite Integral]]
+}}
+{{end-eqn}}
+From [[Zero Definite Integral of Nowhere Negative Function implies Zero Function]] we have that:
+:$d \left({f, g}\right) = 0 \implies f = g$
+on $\left[{a \,.\,.\, b}\right]$.
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Metric is Metric}
+Tags: Euclidean Metric, Euclidean Metric is Metric
+
+\begin{theorem}
+Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_{i'}$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
+The [[Definition:Euclidean Metric|Euclidean metric]] on $\mathcal A$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Euclidean Metric|Euclidean metric]] on $\mathcal A$ is a special case of the [[Definition:P-Product Metric|$p$-product metric]].
+The result follows from [[P-Product Metric is Metric|$p$-Product Metric is Metric]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that the [[Definition:Euclidean Metric|Euclidean metric]] on $\mathcal A$ is defined as:
+:$\displaystyle \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}$
+where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \mathcal A$.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, x}
+      | r = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, x_i} }^2}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\sum_{i \mathop = 1}^n 0^2}^{\frac 1 2}
+      | c = as $d_{i'}$ fulfils [[Definition:Metric Space Axioms|axiom $M1$]]
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let:
+:$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
+:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
+:$(3): \quad \map {d_{i'} } {x_i, y_i} = r_i$
+:$(4): \quad \map {d_{i'} } {y_i, z_i} = s_i$.
+Thus we need to show that:
+:$\displaystyle \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, y} + \map {d_2} {y, z}
+      | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\sum r_i^2}^{\frac 1 2} + \paren {\sum s_i^2}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \paren {\sum \paren {r_i + s_i}^2}^{\frac 1 2}
+      | c = [[Minkowski's Inequality for Sums/Index 2|Minkowski's Inequality for Sums: index $2$]]
+}}
+{{eqn | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} + \map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2}
+      | c = Definition of $r_i$ and $s_i$
+}}
+{{eqn | o = \ge
+      | r = \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}
+      | c = as $d_{i'}$ fulfils [[Definition:Metric Space Axioms|axiom $M2$]]
+}}
+{{eqn | r = \map {d_2} {x, z}
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, y}
+      | r = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\sum \paren {\map {d_{i'} } {y_i, x_i} }^2}^{\frac 1 2}
+      | c = as $d_i$ fulfils [[Definition:Metric Space Axioms|axiom $M3$]]
+}}
+{{eqn | r = \map {d_2} {y, x}
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists k \in \closedint 1 n: x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_k} {x_k, y_k}
+      | o = >
+      | r = 0
+      | c = as $d_k$ fulfils [[Definition:Metric Space Axioms|axiom $M4$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_2} {x, y}
+      | o = >
+      | r = 0
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Space is Normed Space}
+Tags: Examples of Norms, Euclidean Space
+
+\begin{theorem}
+The [[Definition:Euclidean Norm|Euclidean norm]] is a [[Definition:Norm on Vector Space|norm]] on the [[Definition:Euclidean Space|Euclidean space]] $\R^n$.
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Norm on Vector Space|norm axioms]].
+=== Positive Definiteness ===
+As $\mathbf 0 = \tuple {0, \ldots, 0}$, it follows that:
+:$\norm {\mathbf 0} = \sqrt {0^2 + \ldots + 0^2} = 0$
+Suppose instead that $\norm {\mathbf v} = 0$ for some $\mathbf v = \tuple {v_1, \ldots, v_n}$ with $v_1, \ldots, v_n \in \R$.
+As $\displaystyle \sqrt {\sum_{i \mathop = 1}^n v_n^2 } = 0$, it follows from squaring both sides that:
+:$\displaystyle \sum_{i \mathop = 1}^n v_n^2 = 0$
+From [[Even Power is Non-Negative]], it follows that for all $i$: $v_i = 0$.
+Hence $\mathbf v = \mathbf 0$.
+{{qed|lemma}}
+=== Positive homogeneity ===
+{{ProofWanted|I refuse to believe this is not covered somewhere, but cannot find it}}
+=== Triangle Inequality ===
+Follows from [[Triangle Inequality for Vectors in Euclidean Space]].
+{{qed}}
+[[Category:Examples of Norms]]
+[[Category:Euclidean Space]]
+h0t1y2y2hps3keus5ajty6mep8p8nd5
+\end{proof}<|endoftext|>
+\section{Number of Multiples less than Given Number}
+Tags: Number Theory
+
+\begin{theorem}
+Let $m, n \in \N_{\ge 1}$.
+The number of [[Definition:Multiple of Integer|multiples]] of $m$ not greater than $n$ is given by:
+:$q = \floor {\dfrac n m}$
+where $\floor {\cdot}$ denotes the [[Definition:Floor Function|floor function]]
+\end{theorem}
+
+\begin{proof}
+By the [[Division Theorem]]:
+:$(1): \quad n = q m + r$
+where $0 \le r < q$.
+As $r < q$, it follows that the greatest [[Definition:Multiple of Integer|multiple]] of $m$ up to $n$ is $q m$.
+So all the [[Definition:Multiple of Integer|multiples]] of $m$ up to $n$ are:
+:$m, 2 m, 3 m, \ldots, q m$
+Dividing both sides of $(1)$ by $q$:
+:$(2): \quad \dfrac n m = q + \dfrac r m$
+Taking the [[Definition:Floor Function|floor]] of $(2)$:
+:$\floor {\dfrac n m} = \floor {q + \dfrac r m}$
+But as $0 \le \dfrac r m < 1$:
+:$\floor {q + \dfrac r m} = q$
+Recall that all the [[Definition:Multiple of Integer|multiples]] of $m$ up to $n$ are $m, 2 m, 3 m, \ldots, q m$.
+It follows that the number of [[Definition:Multiple of Integer|multiples]] of $m$ up to $n$ is:
+:$q = \floor {\dfrac n m}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Restriction of Non-Continuous Mapping on Metric Space to Subspace may be Continuous}
+Tags: Continuity, Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be a [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]].
+Let $Y \subseteq A_1$.
+Let $f {\restriction_Y}: Y \to A_2$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to $Y$.
+Let $f {\restriction_Y}$ be [[Definition:Continuous on Metric Subspace|$\left({d_Y, d_2}\right)$-continuous]].
+Then it is not necessarily the case that $f$ is also [[Definition:Continuous on Metric Subspace|$\left({d_1, d_2}\right)$-continuous]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $f: \R \to \R$ be given by:
+:$f \left({x}\right) = \begin{cases}
+0 & : x \in \Q \\
+1 & : x \in \R
+\end{cases}$
+where $\Q$ is the set of [[Definition:Rational Number|rational numbers]].
+Then $f {\restriction_\Q}: \Q \to \R$ is the [[Definition:Constant Mapping|constant function]] $f_0$ with value $0$.
+By [[Constant Mapping is Continuous]], $f {\restriction_\Q}$ is [[Definition:Continuous on Metric Subspace|continuous at every point]].
+However, $f$ is not [[Definition:Continuous Real Function at Point|continuous at any point]] on $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Bounded Real Functions on Closed Interval is Metric}
+Tags: Supremum Metric, Supremum Metric on Bounded Real Functions on Closed Interval is Metric
+
+\begin{theorem}
+Let $\closedint a b \subseteq \R$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Real Function|real functions]] $f: \closedint a b \to \R$.
+Let $d: A \times A \to \R$ be the [[Definition:Supremum Metric on Bounded Real Functions on Closed Interval|supremum metric]] on $A$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Supremum Metric on Bounded Real Functions on Closed Interval|supremum metric]] on $A \times A$ is defined as:
+:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$
+where $f$ and $g$ are [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Real Function|real functions]].
+So:
+:$\exists K, L \in \R: \left\vert{f \left({x}\right)}\right\vert \le K, \left\vert{g \left({x}\right)}\right\vert \le L$
+for all $x \in \left[{a \,.\,.\, b}\right]$.
+First note that we have:
+{{begin-eqn}}
+{{eqn | l = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert
+      | r = \left\vert{f \left({x}\right) + \left({- g \left({x}\right)}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{\left({- g \left({x}\right)}\right)}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{g \left({x}\right)}\right\vert
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | o = \le
+      | r = K + L
+      | c = 
+}}
+{{end-eqn}}
+and so the {{RHS}} exists.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, f}\right)
+      | r = \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - f \left({x}\right)}\right\vert
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{0}\right\vert
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $f, g, h \in A$.
+Let $c \in \left[{a \,.\,.\, b}\right]$.
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \in
+      | r = \left[{a \,.\,.\, b}\right]
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{f \left({c}\right) - h \left({c}\right)}\right\vert
+      | o = \le
+      | r = \left\vert{f \left({c}\right) - g \left({c}\right)}\right\vert + \left\vert{g \left({c}\right) - h \left({c}\right)}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({c}\right) - g \left({c}\right)}\right\vert + \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{g \left({c}\right) - h \left({c}\right)}\right\vert
+      | c = {{Defof|Supremum of Real-Valued Function}}
+}}
+{{eqn | r = d \left({f, g}\right) + d \left({g, h}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] for:
+:$S := \left\{ {\left\vert{f \left({c}\right) - h \left({c}\right)}\right\vert: c \in \left[{a \,.\,.\, b}\right]}\right\}$
+So:
+:$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup S = d \left({f, h}\right)$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{g \left({x}\right) - f \left({x}\right)}\right\vert
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = d \left({g, f}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+As $d$ is the [[Definition:Supremum of Real-Valued Function|supremum]] of the [[Definition:Absolute Value|absolute value]] of the [[Definition:Image of Mapping|image]] of the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise sum]] of $f$ and $g$:
+:$\forall f, g \in A: d \left({f, g}\right) \ge 0$
+Suppose $f, g \in A: d \left({f, g}\right) = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sup_{x \mathop \in \left[{a \,.\,.\, b}\right]} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert
+      | r = 0
+      | c = Definition of $d$
+}}
+{{eqn | ll= \implies
+      | lo= \forall x \in \left[{a \,.\,.\, b}\right]:
+      | l = f \left({x}\right) 
+      | r = g \left({x}\right)
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | ll= \implies
+      | l = f
+      | r = g
+      | c = [[Equality of Mappings]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The [[Definition:Closed Real Interval|interval]] is an instance of a [[Definition:Set|set]].
+Hence [[Supremum Metric on Bounded Real-Valued Functions is Metric]] can be directly applied.
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Continuous Real Functions is Subspace of Bounded}
+Tags: Supremum Metric
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\mathscr C \left[{a \,.\,.\, b}\right]$ be the [[Definition:Set|set]] of all [[Definition:Continuous Real Function on Closed Interval|continuous functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$.
+Let $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ be the [[Definition:Set|set]] of all [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Real Function|real functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$.
+Let $d$ be the [[Definition:Supremum Metric on Bounded Real Functions on Closed Interval|supremum metric]] on $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$.
+Then $\left({\mathscr C \left[{a \,.\,.\, b}\right], d_{\mathscr C} }\right)$ is a [[Definition:Metric Subspace|subspace]] of $\left({\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right), d}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $f \in \mathscr C \left[{a \,.\,.\, b}\right]$.
+Then by [[Image of Closed Real Interval is Bounded]], $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\left[{a \,.\,.\, b}\right]$.
+Thus $f \in \mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Definite Integral of Nowhere Negative Function implies Zero Function}
+Tags: Definite Integrals
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $h: \left[{a \,.\,.\, b}\right] \to \R$ be a [[Definition:Continuous Real Function on Closed Interval|continuous]] [[Definition:Real Function|real function]] such that:
+:$\forall x \in \left[{a \,.\,.\, b}\right]: h \left({x}\right) \ge 0$
+Let:
+:$\displaystyle \int_a^b h \left({x}\right) \rd x = 0$
+Then:
+:$\forall x \in \left[{a \,.\,.\, b}\right]: h \left({x}\right) = 0$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that:
+:$\exists c \in \left[{a \,.\,.\, b}\right]: h \left({c}\right) > 0$
+As $h$ is [[Definition:Continuous Real Function on Closed Interval|continuous]], there exists some [[Definition:Closed Real Interval|closed real interval]] $\left[{r \,.\,.\, s}\right] \subseteq \left[{a \,.\,.\, b}\right]$ where $r < s$ such that:
+:$\exists \epsilon \in \R_{>0}: \forall x \in \left[{r \,.\,.\, s}\right]: h \left({x}\right) > \dfrac {h \left({c}\right)} 2$
+From [[Sign of Function Matches Sign of Definite Integral]]:
+:$\displaystyle \int_r^s h \left({x}\right) \rd x > 0$
+This makes a [[Definition:Strictly Positive Real Number|strictly positive]] contribution to the [[Definition:Definite Integral|integral]].
+$h$ is still [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\left[{a \,.\,.\, r}\right]$ and $\left[{s \,.\,.\, b}\right]$.
+So, again from [[Sign of Function Matches Sign of Definite Integral]]:
+:$\forall x \in \left [{a \,.\,.\, r} \right]: f \left({x}\right) \ge 0 \implies \displaystyle \int_a^r f \left({x}\right) \rd x \ge 0$
+:$\forall x \in \left [{s \,.\,.\, b} \right]: f \left({x}\right) \ge 0 \implies \displaystyle \int_s^b f \left({x}\right) \rd x \ge 0$
+Thus as:
+:$\displaystyle \int_a^b h \left({x}\right) \rd x = \int_a^r f \left({x}\right) \rd x + \int_r^s f \left({x}\right) \rd x + \int_s^b f \left({x}\right) \rd x$
+it follows that:
+:$\displaystyle \int_a^b h \left({x}\right) \rd x > 0$
+But [[Definition:By Hypothesis|by hypothesis]]:
+:$\displaystyle \int_a^b h \left({x}\right) \rd x = 0$
+Thus by [[Proof by Contradiction|contradiction]], there can be no $c \in \left[{a \,.\,.\, b}\right]$ such that $h \left({c}\right) > 0$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{L2 Metric on Closed Real Interval is Metric}
+Tags: L2 Metric
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Real Function|real functions]] which are [[Definition:Continuous Real Function|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $d: S \times S \to \R$ be the [[Definition:L2 Metric/Closed Real Interval|$L^2$ metric]] on $\closedint a b$:
+:$\displaystyle \forall f, g \in S: \map {d_2} {f, g} := \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}$
+Then $d_2$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {f, f}
+      | r = \paren {\int_a^b \paren {\map f t - \map f t}^2 \rd t}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\int_a^b 0^2 \rd t}^{\frac 1 2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = 0
+      | c = [[Definite Integral of Constant]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+It is required to be shown:
+:$\map {d_2} {f, g} + \map {d_2} {g, h} \ge d_2 \map {d_2} {f, h}$
+for all $f, g, h \in S$.
+Let:
+:$(1): \quad t \in \closedint a b$
+:$(2): \quad \map f t - \map g t = \map r t$
+:$(3): \quad \map g t - \map h t = \map s t$
+Thus we need to show that:
+:$\displaystyle \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2} \ge \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {f, g} + \map {d_2} {g, h}
+      | r = \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map g t - \map h t}^2 \rd t}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\int_a^b \paren {\map r t}^2 \rd t}^{\frac 1 2} + \paren {\int_a^b \paren {\map s t}^2 \rd t}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \paren {\int_a^b \paren {\map r t + \map s t}^2 \rd t}^{\frac 1 2}
+      | c = [[Minkowski's Inequality for Integrals]]
+}}
+{{eqn | r = \paren {\int_a^b \paren {\map f t - \map g t + \map g t - \map h t}^2 \rd t}^{\frac 1 2}
+      | c = Definition of $\map r t$ and $\map s t$
+}}
+{{eqn | r = \paren {\int_a^b \paren {\map f t - \map h t}^2 \rd t}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | r = \map {d_2} {f, h}
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {f, g}
+      | r = \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \paren {\int_a^b \paren {\map g t - \map f t}^2 \rd t}^{\frac 1 2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \map {d_2} {g, f}
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | lo= \forall t \in \closedint a b:
+      | l = \map f t - \map g t
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\map f t - \map g t}^2
+      | o = \ge
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}
+      | o = \ge
+      | r = 0
+      | c = [[Sign of Function Matches Sign of Definite Integral]]
+}}
+{{end-eqn}}
+From [[Zero Definite Integral of Nowhere Negative Function implies Zero Function]] we have that:
+:$\map {d_2} {f, g} = 0 \implies f = g$
+on $\closedint a b$.
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element in Bounded Metric Space has Bound}
+Tags: Bounded Metric Spaces
+
+\begin{theorem}
+Let $M = \left({X, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $M' = \left({Y, d_Y}\right)$ be a [[Definition:Metric Subspace|subspace]] of $M$.
+Let $M'$ be [[Definition:Bounded Metric Space|bounded in $M$]].
+Then:
+:$\forall a' \in X: \exists K' \in \R: \forall x \in Y: d \left({x, a}\right) \le K'$
+That is, if there is one [[Definition:Element|element]] of $X$ which satisfies the condition for $Y$ to be [[Definition:Bounded Metric Space|bounded in $M$]], they ''all'' do.
+\end{theorem}
+
+\begin{proof}
+Let $a \in X$ such that $\exists K \in \R: \forall x \in Y: d \left({x, a}\right) \le K$.
+Let $a' \in X$.
+{{begin-eqn}}
+{{eqn | l =  \left({x, a'}\right)
+      | o = \le
+      | r = d \left({x, a}\right) + d \left({a, a'}\right)
+      | c = [[Definition:Metric Space Axioms|Metric Space Axiom $M2$]]
+}}
+{{eqn | o = \le
+      | r = K + d \left({a, a'}\right)
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = K'
+      | c = where $K' = K + d \left({a, a'}\right)$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergence of Square of Linear Combination of Sequences whose Squares Converge}
+Tags: Real Analysis, Series
+
+\begin{theorem}
+Let $\sequence {x_i}$ and $\sequence {y_i}$ be [[Definition:Real Sequence|real sequences]] such that the [[Definition:Series of Numbers|series]] $\ds \sum_{i \mathop \ge 0} x_i^2$ and $\ds \sum_{i \mathop \ge 0} y_i^2$ are [[Definition:Convergent Series of Numbers|convergent]].
+Let $\lambda, \mu \in \R$ be [[Definition:Real Number|real numbers]].
+Then $\ds \sum_{i \mathop \ge 0} \paren {\lambda x_i + \mu y_i}^2$ is [[Definition:Convergent Series of Numbers|convergent]].
+\end{theorem}
+
+\begin{proof}
+Let $n \in \N$.
+Then:
+:$\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2 = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \lambda \mu \sum_{i \mathop = 1}^n x_i y_i$
+By [[Cauchy's Inequality]]:
+:$\ds \sum_{i \mathop = 1}^n x_i y_i \le \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2
+      | o = \le
+      | r = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \size {\lambda \mu} \sum_{i \mathop = 1}^n x_i y_i
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \lambda^2 \sum_{i \mathop = 1}^n x_i^2 + \mu^2 \sum_{i \mathop = 1}^n y_i^2 + 2 \size {\lambda \mu} \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2}
+}}
+{{eqn | r = \paren {\size \lambda \paren {\sum_{i \mathop = 1}^n x_i^2}^{\frac 1 2} + \size \mu \paren {\sum_{i \mathop = 1}^n y_i^2}^{\frac 1 2} }^2
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \paren {\size \lambda \paren {\sum_{i \mathop \ge 0} x_i^2}^{\frac 1 2} + \size \mu \paren {\sum_{i \mathop \ge 0} y_i^2}^{\frac 1 2} }^2
+      | c = 
+}}
+{{end-eqn}}
+Thus the [[Definition:Sequence of Partial Sums|sequence of partial sums]] $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is [[Definition:Bounded Above Real Sequence|bounded above]].
+We also have that $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is also [[Definition:Increasing Real Sequence|increasing]].
+So by the [[Monotone Convergence Theorem (Real Analysis)|Monotone Convergence Theorem]], $\ds \sum_{i \mathop = 1}^n \paren {\lambda x_i + \mu y_i}^2$ is [[Definition:Convergent Series of Numbers|convergent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hilbert Sequence Space is Metric Space}
+Tags: Hilbert Sequence Space, Hilbert Sequence Space is Metric Space
+
+\begin{theorem}
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Real Sequence|real sequences]] $\left\langle{x_i}\right\rangle$ such that the [[Definition:Series of Numbers|series]] $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is [[Definition:Convergent Series of Numbers|convergent]].
+Let $\ell^2 = \left({A, d_2}\right)$ be the [[Definition:Hilbert Sequence Space|Hilbert sequence space on $\R$]].
+Then $\ell^2$ is a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+$\ell^2$ is a particular instance of the general [[Definition:P-Sequence Metric|$p$-sequence space]] $\ell^p$.
+Hence [[P-Sequence Space of Real Sequences is Metric Space|$p$-Sequence Space of Real Sequences is Metric Space]] can be applied directly.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of the [[Definition:Hilbert Sequence Space|Hilbert sequence space on $\R$]]:
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Real Sequence|real sequences]] $\left\langle{x_i}\right\rangle$ such that the [[Definition:Series of Numbers|series]] $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is [[Definition:Convergent Series of Numbers|convergent]].
+Then $\ell^2 := \left({A, d_2}\right)$ where $d_2: A \times A: \to \R$ is the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in A: d_2 \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$
+From [[Convergence of Square of Linear Combination of Sequences whose Squares Converge]] we have that $\displaystyle \sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2$ does actually converge.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_2 \left({x, x}\right)
+      | r = \left({\sum_{k \mathop \ge 0} \left({x_k - x_k}\right)^2}\right)^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \left({\sum_{k \mathop \ge 0} 0^2}\right)^{\frac 1 2}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $z = \left\langle{z_i}\right\rangle \in A$.
+{{MissingLinks|Minkowski's inequality currently only applies to finite sums}}
+{{begin-eqn}}
+{{eqn | l = d_2 \left({x, y}\right) + d_2 \left({y, z}\right)
+      | r = \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2} + \left({\sum_{k \mathop \ge 0} \left({y_k - z_k}\right)^2}\right)^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | o = \ge
+      | r = \left({\sum_{k \mathop \ge 0} \left({x_k - z_k}\right)^2}\right)^{\frac 1 2}
+      | c = [[Minkowski's Inequality for Sums/Index 2|Minkowski's Inequality for Sums: index $2$]]
+}}
+{{eqn | r = d_2 \left({x, z}\right)
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_2 \left({x, y}\right)
+      | r = \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \left({\sum_{k \mathop \ge 0} \left({y_k - x_k}\right)^2}\right)^{\frac 1 2}
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | r = d_2 \left({y, x}\right)
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | lo= \exists k \in \N:
+      | l = x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({x_k - y_k}\right)^2
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_2 \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Bounded Real-Valued Functions is Metric}
+Tags: Supremum Metric, Supremum Metric on Bounded Real-Valued Functions is Metric
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Bounded Real-Valued Function|bounded real-valued functions]] $f: X \to \R$.
+Let $d: A \times A \to \R$ be the [[Definition:Supremum Metric on Bounded Real-Valued Functions|supremum metric]] on $A$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Supremum Metric on Bounded Real-Valued Functions|supremum metric]] on $A \times A$ is defined as:
+:$\displaystyle \forall f, g \in A: \map d {f, g} := \sup_{x \mathop \in X} \size {\map f x - \map g x}$
+where $f$ and $g$ are [[Definition:Bounded Real-Valued Function|bounded real-valued functions]].
+So:
+:$\exists K, L \in \R: \size {\map f x} \le K, \size {\map g x} \le L$
+for all $x \in X$.
+First note that we have:
+{{begin-eqn}}
+{{eqn | l = \size {\map f x - \map g x}
+      | r = \size {\map f x + \paren {-\map g x} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \size {\map f x} + \size {\paren {-\map g x} }
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \size {\map f x} + \size {\map g x}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | o = \le
+      | r = K + L
+      | c = 
+}}
+{{end-eqn}}
+and so the {{RHS}} exists.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {f, f}
+      | r = \sup_{x \mathop \in X} \size {\map f x - \map f x}
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{x \mathop \in X} \size 0
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $f, g, h \in A$.
+Let $c \in X$.
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \in
+      | r = X
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map f c - \map h c}
+      | o = \le
+      | r = \size {\map f c - \map g c} + \size {\map g c - \map h c}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \sup_{x \mathop \in X} \size {\map f c - \map g c} + \sup_{x \mathop \in X} \size {\map g c - \map h c}
+      | c = {{Defof|Supremum of Real-Valued Function}}
+}}
+{{eqn | r = \map d {f, g} + \map d {g, h}
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+Thus $\map d {f, g} + \map d {g, h}$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] for:
+:$S := \set {\size {\map f c - \map h c}: c \in X}$
+So:
+:$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup S = d \left({f, h}\right)$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {f, g}
+      | r = \sup_{x \mathop \in X} \size {\map f x - \map g x}
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{x \mathop \in X} \size {\map g x - \map f x}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \map d {g, f}
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+As $d$ is the [[Definition:Supremum of Real-Valued Function|supremum]] of the [[Definition:Absolute Value|absolute value]] of the [[Definition:Image of Mapping|image]] of the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise sum]] of $f$ and $g$:
+:$\forall f, g \in A: \map d {f, g} \ge 0$
+Suppose $f, g \in A: \map d {f, g} = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {f, g}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sup_{x \mathop \in X} \size {\map f x - \map g x}
+      | r = 0
+      | c = Definition of $d$
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall x \in X:
+      | l = \map f x 
+      | r = \map g x
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \leadsto
+      | l = f
+      | r = g
+      | c = [[Equality of Mappings]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that the [[Definition:Supremum Metric on Bounded Real-Valued Functions|supremum metric]] on $A \times A$ is defined as:
+:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in X} \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert$
+where $f$ and $g$ are [[Definition:Bounded Real-Valued Function|bounded real-valued functions]].
+From [[Real Number Line is Metric Space]], the [[Definition:Real Number|real numbers]] $\R$ together with the [[Definition:Absolute Value|absolute value function]] form a [[Definition:Metric Space|metric space]].
+The result follows by [[Supremum Metric is Metric]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Bounded Real Sequences is Metric}
+Tags: Supremum Metric, Supremum Metric on Bounded Real Sequences is Metric
+
+\begin{theorem}
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Bounded Real Sequence|bounded real sequences]].
+Let $d: A \times A \to \R$ be the [[Definition:Supremum Metric on Bounded Real Sequences|supremum metric]] on $A$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Supremum Metric on Bounded Real Sequences|supremum metric]] on $A \times A$ is defined as:
+:$\displaystyle \forall x, y \in A: d \left({x, y}\right) := \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert$
+where $x = \left\langle{x_i}\right\rangle$ and $y = \left\langle{y_i}\right\rangle$ are [[Definition:Bounded Real Sequence|bounded real sequences]].
+So:
+:$\exists K, L \in \R: \left\vert{x_n}\right\vert \le K, \left\vert{y_n}\right\vert \le L$
+for all $n \in \N$.
+First note that we have:
+{{begin-eqn}}
+{{eqn | l = \left\vert{x_n - y_n}\right\vert
+      | r = \left\vert{x_n + \left({- y_n}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{x_n}\right\vert + \left\vert{\left({- y_n}\right)}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \left\vert{x_n}\right\vert + \left\vert{y_n}\right\vert
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | o = \le
+      | r = K + L
+      | c = 
+}}
+{{end-eqn}}
+and so the {{RHS}} exists.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({x, x}\right)
+      | r = \sup_{n \mathop \in \N} \left\vert{x_n - x_n}\right\vert
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{n \mathop \in \N} \left\vert{0}\right\vert
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $x, y, z \in A$.
+Let $n \in \N$.
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \in
+      | r = \N
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{x_n - z_n}\right\vert
+      | o = \le
+      | r = \left\vert{x_n - y_n}\right\vert + \left\vert{y_n - z_n}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert + \sup_{n \mathop \in \N} \left\vert{y_n - z_n}\right\vert
+      | c = {{Defof|Supremum of Real Sequence}}
+}}
+{{eqn | r = d \left({x, y}\right) + d \left({y, z}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+Thus $d \left({x, y}\right) + d \left({y, z}\right)$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] for:
+:$S := \left\{ {\left\vert{x_n - z_n}\right\vert: n \in \N}\right\}$
+So:
+:$d \left({x, y}\right) + d \left({y, z}\right) \ge \sup S = d \left({x, z}\right)$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({x, y}\right)
+      | r = \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{n \mathop \in \N} \left\vert{y_n - x_n}\right\vert
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = d \left({y, x}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+As $d$ is the [[Definition:Supremum of Real Sequence|supremum]] of the [[Definition:Absolute Value|absolute value]] of the difference of [[Definition:Term of Sequence|terms]]elements of the sequences $\left\langle{x_i}\right\rangle$ and $\left\langle{y_i}\right\rangle$:
+:$\forall x, y \in A: d \left({x, y}\right) \ge 0$
+Suppose $x, y \in A: d \left({x, y}\right) = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d \left({x, y}\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sup_{n \mathop \in \N} \left\vert{x_n - y_n}\right\vert
+      | r = 0
+      | c = Definition of $d$
+}}
+{{eqn | ll= \implies
+      | lo= \forall n \mathop \in \N:
+      | l = x_n 
+      | r = y_n
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = y
+      | c = [[Equality of Mappings]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, a [[Definition:Real Sequence|real sequence]] is a [[Definition:Mapping|mapping]] from the [[Definition:Natural Numbers|natural numbers]] $\N$ to the [[Definition:Real Number|real numbers]] $\R$.
+Thus a [[Definition:Bounded Real Sequence|bounded real sequence]] is a [[Definition:Bounded Real-Valued Function|bounded real-valued function]].
+The result follows from [[Supremum Metric on Bounded Real-Valued Functions is Metric]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric is Metric}
+Tags: Supremum Metric
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $M = \left({A', d'}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Bounded Mapping to Metric Space|bounded mappings]] $f: S \to M$.
+Let $d: A \times A \to \R$ be the [[Definition:Supremum Metric|supremum metric]] on $A$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Supremum Metric|supremum metric]] on $A \times A$ is defined as:
+:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)$
+where $f$ and $g$ are [[Definition:Bounded Mapping to Metric Space|bounded mappings]].
+First note that we have:
+{{begin-eqn}}
+{{eqn | l = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert
+      | r = \left\vert{f \left({x}\right) + \left({- g \left({x}\right)}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{\left({- g \left({x}\right)}\right)}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \left\vert{f \left({x}\right)}\right\vert + \left\vert{g \left({x}\right)}\right\vert
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | o = \le
+      | r = K + L
+      | c = 
+}}
+{{end-eqn}}
+and so the [[Definition:RHS|RHS]] exists.
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, f}\right)
+      | r = \sup_{x \mathop \in S} d' \left({f \left({x}\right), f \left({x}\right)}\right)
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{x \mathop \in S} \left\vert{0}\right\vert
+      | c = as $d'$ fulfils [[Definition:Metric Space Axioms|axiom $M1$]]
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $f, g, h \in A$.
+Let $x, y \in S$.
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \in
+      | r = S
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d' \left({f \left({c}\right), h \left({c}\right)}\right)
+      | o = \le
+      | r = d' \left({f \left({c}\right), g \left({c}\right)}\right) + d' \left({g \left({c}\right), h \left({c}\right)}\right)
+      | c = as $d'$ fulfils [[Definition:Metric Space Axioms|axiom $M2$]]
+}}
+{{eqn | o = \le
+      | r = \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right) + \sup_{x \mathop \in S} d' \left({g \left({x}\right), h \left({x}\right)}\right)
+      | c = Definition of [[Definition:Supremum of Real-Valued Function|Supremum]]
+}}
+{{eqn | r = d \left({f, g}\right) + d \left({g, h}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] for:
+:$X := \left\{ {d' \left({f \left({c}\right), g \left({c}\right)}\right): c \in X}\right\}$
+So:
+:$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup X = d \left({f, h}\right)$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{x \mathop \in S} d' \left({g \left({x}\right), f \left({x}\right)}\right)
+      | c = as $d'$ fulfils [[Definition:Metric Space Axioms|axiom $M3$]]
+}}
+{{eqn | r = d \left({g, f}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+As $d$ is the [[Definition:Supremum of Real-Valued Function|supremum]] of the [[Definition:Absolute Value|absolute value]] of the [[Definition:Image of Mapping|image]] of the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise sum]] of $f$ and $g$:
+:$\forall f, g \in A: d \left({f, g}\right) \ge 0$
+Suppose $f, g \in A: d \left({f, g}\right) = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sup_{x \mathop \in S} d' \left({f \left({x}\right), g \left({x}\right)}\right)
+      | r = 0
+      | c = Definition of $d$
+}}
+{{eqn | ll= \implies
+      | lo= \forall x \in S:
+      | l = d' \left({f \left({x}\right), g \left({x}\right)}\right)
+      | r = 0
+      | c = as $d'$ fulfils [[Definition:Metric Space Axioms|axioms $M1$ and $M4$]]
+}}
+{{eqn | ll= \implies
+      | l = f
+      | r = g
+      | c = as $d'$ fulfils [[Definition:Metric Space Axioms|axiom $M4$]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Bounded Continuous Mappings is Metric}
+Tags: Supremum Metric
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Continuous on Metric Space|continuous mappings]] $f: M_1 \to M_2$ which are also [[Definition:Bounded Mapping to Metric Space|bounded]].
+Let $d: A \times A \to \R$ be the [[Definition:Supremum Metric|supremum metric]] on $A$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+The set $A$ is a [[Definition:Subset|subset]] of the [[Definition:Set|set]] $A'$ of all [[Definition:Bounded Mapping to Metric Space|bounded mappings]] $f: M_1 \to M_2$.
+Let $d': A' \times A' \to \R$ be the [[Definition:Supremum Metric|supremum metric]] on $A'$.
+From [[Supremum Metric is Metric]], $\left({A', d'}\right)$ is a [[Definition:Metric Space|metric space]]. 
+By definition, $A$ is a [[Definition:Metric Subspace|metric subspace]] of $A'$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Harmonic Number is not Integer}
+Tags: Harmonic Numbers, Harmonic Number is not Integer
+
+\begin{theorem}
+Let $H_n$ be the $n$th [[Definition:Harmonic Number|harmonic number]].
+Then $H_n$ is not an [[Definition:Integer|integer]] for $n \ge 2$.
+That is, the only harmonic numbers that are integers are $H_0$ and $H_1$.
+\end{theorem}
+
+\begin{proof}
+As $H_0 = 0$ and $H_1 = 1$, they are [[Definition:Integer|integers]].
+The claim is that $H_n$ is not an [[Definition:Integer|integer]] for all $n \ge 2$.
+{{AimForCont}} otherwise:
+:$(\text P): \quad \exists m \in \N: H_m \in \Z$
+By the definition of the [[Definition:Harmonic Numbers|harmonic numbers]]:
+:$H_m = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 m$
+Let $2^t$ denote the highest [[Definition:Power|power]] of two in the [[Definition:Denominator|denominators]] of the [[Definition:Summand|summands]].
+Then:
+{{begin-eqn}}
+{{eqn | l = H_m
+      | r = 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 m
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = H_m - \frac 1 {2^t}
+      | r = 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^t - 1} + \frac 1 {2^t + 1} + \cdots + \frac 1 m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2^{t - 1} H_m - \frac 1 2
+      | r = 2^{t - 1} + \frac {2^{t - 1} } 2 + \frac {2^{t - 1} } 3 + \frac {2^{t - 1} } 4 + \frac {2^{t - 1} } 5 + \frac {2^{t - 1} } 6 + \cdots + \frac {2^{t - 1} } m
+      | c = multiplying by $2^{t-1}$
+}}
+{{eqn | n = 2
+      | r = 2^{t - 1} + 2^{t - 2} + \frac {2^{t - 1} } 3 + 2^{t - 3} + \frac {2^{t - 1} } 5 + \frac {2^{t - 2} } 3 + \cdots + \frac {2^{t - 1} } m
+      | c = cancelling powers of $2$
+}}
+{{end-eqn}}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Denominator|denominators]] on the {{RHS}} of $(2)$.
+Then no [[Definition:Element|element]] of $S$ can have $2$ as a [[Definition:Divisor of Integer|factor]], as follows.
+Consider an arbitrary [[Definition:Summand|summand]]:
+:$\dfrac {2^{t - 1} } {2^j \times k}$
+for some $k \in \Z$, where $j \ge 0$ is the highest [[Definition:Integer Power|power]] of $2$ that [[Definition:Divisor of Integer|divides]] the [[Definition:Denominator|denominator]].
+For any $2$ to remain after simplification, we would need $j > t - 1$.
+Were this to be so, then $2^j \times k$ would have $2^t$ as a [[Definition:Divisor of Integer|factor]], and some [[Definition:Denominator|denominator]] would be a [[Definition:Multiple of Integer|multiple]] of $2^t$.
+By [[Greatest Power of Two not Divisor]], the set of [[Definition:Denominator|denominators]] of the {{RHS}} of $(1)$:
+:$\set {1, 2, 3, \ldots, 2^t - 1, 2^t + 1, \ldots, m}$
+contains no [[Definition:Multiple of Integer|multiple]] of $2^t$.
+Therefore there can be no [[Definition:Multiple of Integer|multiple]] of $2$ in the [[Definition:Denominator|denominators]] of the {{RHS}} of $(2)$.
+Let:
+:$\ell = \map \lcm S$
+be the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of the [[Definition:Element|elements]] of $S$.
+Because $2$ is not a [[Definition:Divisor of Integer|divisor]] of any of the [[Definition:Element|elements]] of $S$, it will not be a [[Definition:Divisor of Integer|divisor]] of $\ell$.
+Hence $\ell$ is likewise [[Definition:Odd Integer|odd]].
+We have:
+{{begin-eqn}}
+{{eqn | l = 2^{t - 1} H_m - \frac 1 2
+      | r = 2^{t - 1} + 2^{t - 2} + \frac {2^{t - 1} } 3 + 2^{t - 3} + \frac {2^{t - 1} } 5 + \frac {2^{t - 2} } 3 + \ldots + \frac {2^{t - 1} } m
+      | c = from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2^t H_m - 1} 2
+      | r = \frac {2^{t - 1} \ell + 2^{t - 2} \ell + 2^{t - 1} \paren {\ell / 3} + 2^{t - 3} \ell + 2^{t - 1} \paren {\ell / 5} + \ldots + 2^{t - 1} \paren {\ell / m} } \ell
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $\ell$
+}}
+{{eqn | ll= \leadsto
+      | l = \ell \paren {2^t H_m - 1}
+      | r = 2 \paren {2^{t - 1} \ell + 2^{t - 2} \ell + 2^{t - 1} \paren {\ell / 3} + 2^{t - 3} \ell + 2^{t - 1} \paren {\ell / 5} + \ldots + 2^{t - 1} \paren {\ell / m} }
+      | c = multiplying both sides by $2 \ell$
+}}
+{{end-eqn}}
+But the {{LHS}} of that last equation is [[Definition:Odd Integer|odd]], while its {{RHS}} is [[Definition:Even Integer|even]].
+As this is a [[Proof by Contradiction|contradiction]], it follows that our assumption $(\text P)$ that such an $m$ exists is false.
+That is, there are no [[Definition:Harmonic Numbers|harmonic numbers]] apart from $0$ and $1$ which are [[Definition:Integer|integers]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{AimForCont}}:
+:$(\text P): \quad \exists m \in \N: H_m \in \Z$
+By the definition of the [[Definition:Harmonic Numbers|harmonic numbers]]:
+:$1 + \dfrac 1 2 + \dfrac 1 3 + \cdots +\dfrac 1 m = H_m$
+$m$ is either [[Definition:Prime Number|prime]] or [[Definition:Composite Number|composite]].
+If $m$ is [[Definition:Prime Number|prime]], we have that:
+{{begin-eqn}}
+{{eqn | l = 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 m
+      | r = H_m
+}}
+{{eqn | l = m! + \frac {m!} 2 + \frac {m!} 3 + \cdots + \paren {m - 1}!
+      | r = m!\cdot H_m
+      | c = multiplying by $m!$
+}}
+{{end-eqn}}
+If $k \le n$, then $k$ [[Definition:Divisor of Integer|divides]] $n!$, so all terms are [[Definition:Integer|integers]].
+Every term on the {{LHS}} is [[Definition:Divisor of Integer|divisible]] by $m$ except for one, namely $\paren {m - 1}!$.
+Hence, the {{LHS}} is not [[Definition:Divisor of Integer|divisible]] by $m$.
+But the {{RHS}} is [[Definition:Divisor of Integer|divisible]] by $m$, so this is a [[Proof by Contradiction|contradiction]].
+If $m$ is [[Definition:Composite Number|composite]], let $p$ be the largest [[Definition:Prime Number|prime]] which is less than $m$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 p + \cdots + \frac 1 m
+      | r = H_m
+}}
+{{eqn | l = m! + \frac {m!} 2 + \frac {m!} 3 + \cdots \frac {m!} p + \cdots + \paren {m - 1}!
+      | r = m! \cdot H_m
+      | c = multiplying by $m!$
+}}
+{{end-eqn}}
+For $k < p$, we have that $m!$ is [[Definition:Divisor of Integer|divisible]] by $p$ and $k$ is not.
+So from [[Euclid's Lemma]], every term $\dfrac {m!} k$ is an [[Definition:Integer|integer]] [[Definition:Divisor of Integer|divisible]] by $p$.
+{{AimForCont}} there exists $k \in \Z$, with $p< k< m$ such that $\dfrac {m!} k$ is not [[Definition:Divisor of Integer|divisible]] by $p$.
+Because $m!$ is [[Definition:Divisor of Integer|divisible]] by $p$ and $\dfrac{m!} k$ is not, it follows from [[Euclid's Lemma]] that $k$ is a [[Definition:Multiple of Integer|multiple]] of $p$.
+Since $k$ is a [[Definition:Multiple of Integer|multiple]] of $p$ which is greater than $p$:
+:$2 p \le k$
+Thus:
+:$p < 2 p \le k < m$
+From [[Bertrand's Postulate]], there exists a [[Definition:Prime Number|prime]] $q$ such that $p < q < 2 p$.
+Hence:
+:$p < q <m $
+This [[Proof by Contradiction|contradicts]] the fact that $p$ is the largest [[Definition:Prime Number|prime]] less than $m$.
+Hence, the assumption that there exists $k \in \Z$ with $p < k < m$ such that $\dfrac {m!} k$ is not [[Definition:Divisor of Integer|divisible]] by $p$ is false.
+Therefore, every term of the {{LHS}}, except perhaps $\dfrac{m!} p$, is a [[Definition:Multiple of Integer|multiple]] of $p$.
+From above:
+:$p < m < 2 p$
+so $p$ is the only [[Definition:Positive Integer|positive integer]] less than $m$ which is [[Definition:Divisor of Integer|divisible]] by $p$.
+Since $m!$ is the [[Definition:Integer Multiplication|product]] of [[Definition:Positive Integer|positive integer]] less than or equal to $m$, it follows that $m!$ is [[Definition:Divisor of Integer|divisible]] by $p$ [[Definition:Multiplicity of Prime Factor|only once]].
+Therefore, $\dfrac {m!} p$ is not [[Definition:Divisor of Integer|divisible]] by $p$.
+Because every term of the {{LHS}} is [[Definition:Divisor of Integer|divisible]] by $p$ except for one, it follows that the {{LHS}} is not [[Definition:Divisor of Integer|divisible]] by $p$.
+Because the {{RHS}} is a [[Definition:Multiple of Integer|multiple]] of $m!$, and hence a [[Definition:Multiple of Integer|multiple]] of $p$, this is a [[Proof by Contradiction|contradiction]].
+So the assumption that there exists $m \in \N: H_m \in \Z$ is false.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Differentiability Class is Metric}
+Tags: Supremum Metric
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $r \in \N$ be a [[Definition:Natural Number|natural number]].
+Let $A := \mathscr D^r \left[{a \,.\,.\, b}\right]$ be the [[Definition:Set|set]] of all [[Definition:Continuous Real Function on Closed Interval|continuous functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$ which are of [[Definition:Differentiability Class|differentiability class $r$]].
+Let $d: A \times A \to \R$ be the [[Definition:Supremum Metric/Differentiability Class|supremum metric]] on $A$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Supremum Metric/Differentiability Class|supremum metric]] on $A \times A$ is defined as:
+:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert$
+where $f$ and $g$ are [[Definition:Continuous Real Function on Closed Interval|continuous functions]] on $\left[{a \,.\,.\, b}\right]$ which are of [[Definition:Differentiability Class|differentiability class $r$]].
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, f}\right)
+      | r = \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - f^{\left({i}\right)} \left({x}\right)}\right\vert
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{0}\right\vert
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $f, g, h \in A$.
+Let $c \in \left[{a \,.\,.\, b}\right]$.
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \in
+      | r = \left[{a \,.\,.\, b}\right]
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{f^{\left({i}\right)} \left({c}\right) - h^{\left({i}\right)} \left({c}\right)}\right\vert
+      | o = \le
+      | r = \sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{f^{\left({i}\right)} \left({c}\right) - g^{\left({i}\right)} \left({c}\right)}\right\vert + \sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{g^{\left({i}\right)} \left({c}\right) - h^{\left({i}\right)} \left({c}\right)}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert + \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{g^{\left({i}\right)} \left({x}\right) - h^{\left({i}\right)} \left({x}\right)}\right\vert
+      | c = Definition of [[Definition:Supremum of Real-Valued Function|Supremum]]
+}}
+{{eqn | r = d \left({f, g}\right) + d \left({g, h}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+Thus $d \left({f, g}\right) + d \left({g, h}\right)$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] for:
+:$S := \left\{ {\sup_{i \in \left\{ {0, 1, 2, \ldots, r}\right\} } \left\vert{f^{\left({i}\right)} \left({c}\right) - g^{\left({i}\right)} \left({c}\right)}\right\vert: c \in \left[{a \,.\,.\, b}\right]}\right\}$
+So:
+:$d \left({f, g}\right) + d \left({g, h}\right) \ge \sup S = d \left({f, h}\right)$
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert
+      | c = Definition of $d$
+}}
+{{eqn | r = \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{g^{\left({i}\right)} \left({x}\right) - f^{\left({i}\right)} \left({x}\right)}\right\vert
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | r = d \left({g, f}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+As $d$ is the [[Definition:Supremum of Real-Valued Function|supremum]] of the [[Definition:Absolute Value|absolute value]] of the [[Definition:Image of Mapping|image]] of the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise sum]] of $f$ and $g$:
+:$\forall f, g \in A: d \left({f, g}\right) \ge 0$
+Suppose $f, g \in A: d \left({f, g}\right) = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d \left({f, g}\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - f^{\left({i}\right)} \left({x}\right)}\right\vert
+      | r = 0
+      | c = Definition of $d$
+}}
+{{eqn | ll= \implies
+      | lo= \forall x \in \left[{a \,.\,.\, b}\right]:
+      | l = f \left({x}\right) 
+      | r = g \left({x}\right)
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \implies
+      | l = f
+      | r = g
+      | c = [[Equality of Mappings]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Metric on Continuous Real Functions is Metric}
+Tags: Supremum Metric, Supremum Metric on Continuous Real Functions is Metric
+
+\begin{theorem}
+Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\mathscr C \left[{a \,.\,.\, b}\right]$ be the [[Definition:Set|set]] of all [[Definition:Continuous Real Function on Closed Interval|continuous functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$.
+Let $d$ be the [[Definition:Supremum Metric on Continuous Real Functions|supremum metric]] on  $\mathscr C \left[{a \,.\,.\, b}\right]$.
+Then $d$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ be the [[Definition:Set|set]] of all [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Real Function|real functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$.
+From [[Supremum Metric on Continuous Real Functions is Subspace of Bounded]], $\left({\mathscr C \left[{a \,.\,.\, b}\right], d_{\mathscr C} }\right)$ is a [[Definition:Metric Subspace|(metric) subspace]] of $\left({\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right), d}\right)$.
+The result follows from [[Subspace of Metric Space is Metric Space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $A := \mathscr D^r \left[{a \,.\,.\, b}\right]$ be the [[Definition:Set|set]] of all [[Definition:Continuous Real Function on Closed Interval|continuous functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$ which are of [[Definition:Differentiability Class|differentiability class $r$]].
+Let $d_r: A \times A \to \R$ be the [[Definition:Supremum Metric/Differentiability Class|supremum metric]] on $A$, defined as:
+:$\displaystyle \forall f, g \in A: d \left({f, g}\right) := \sup_{\substack {x \mathop \in \left[{a \,.\,.\, b}\right] \\ i \mathop \in \left\{ {0, 1, 2, \ldots, r}\right\} } } \left\vert{f^{\left({i}\right)} \left({x}\right) - g^{\left({i}\right)} \left({x}\right)}\right\vert$
+where:
+: $f$ and $g$ are [[Definition:Continuous Real Function on Closed Interval|continuous functions]] on $\left[{a \,.\,.\, b}\right]$ which are [[Definition:Differentiability Class|$r$ times differentiable]]
+: $r \in \N$ is a [[Definition:Natural Number|natural number]].
+From [[Supremum Metric on Differentiability Class is Metric]], $d_r$ is a [[Definition:Metric|metric]] on $A$.
+By definition, [[Definition:Real Function|real functions]] which are [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\left[{a \,.\,.\, b}\right]$ are of [[Definition:Differentiability Class|differentiability class]] $0$.
+Thus setting $r = 0$ it follows that $\mathscr C \left[{a \,.\,.\, b}\right]$ is $\mathscr D^0 \left[{a \,.\,.\, b}\right]$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Metric Space is Metric Space}
+Tags: Metric Subspaces
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$.
+Let $d_H: H \times H \to \R$ be the [[Definition:Restriction of Mapping|restriction]] $d \restriction_{H \times H}$ of $d$ to $H$.
+Let $\left({H, d_H}\right)$ be a [[Definition:Metric Subspace|metric subspace]] of $\left({A, d}\right)$.
+Then $d_H$ is a [[Definition:Metric|metric]] on $H$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Restriction of Mapping|restriction]]:
+:$\forall x, y \in H: d_H \left({x, y}\right) = d \left({x, y}\right)$
+As $d$ is a [[Definition:Metric|metric]], the [[Definition:Metric Space Axioms|metric space axioms]] are all fulfilled by all $x, y \in A$ under $d$.
+As $H \subseteq A$, by definition of [[Definition:Subset|subset]], all $x, y \in H$ are also [[Definition:Element|elements]] of $A$.
+Therefore the [[Definition:Metric Space Axioms|metric space axioms]] are all fulfilled by all $x, y \in H$ under $d_H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lp Metric on Closed Real Interval is Metric}
+Tags: Lp Metrics
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Real Function|real functions]] which are [[Definition:Continuous Real Function|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$.
+Let $p \in \R_{\ge 1}$.
+Let $d_p: S \times S \to \R$ be the [[Definition:Lp Metric|$L^p$ metric]] on $\left[{a \,.\,.\, b}\right]$:
+:$\displaystyle \forall f, g \in S: d \left({f, g}\right) := \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}$
+Then $d_p$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({f, f}\right)
+      | r = \left({\int_a^b \left\vert{f \left({t}\right) - f \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\int_a^b 0^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = 0
+      | c = [[Definite Integral of Constant]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+It is required to be shown:
+:$d_p \left({f, g}\right) + d_p \left({g, h}\right) \ge d_p \left({f, h}\right)$
+for all $f, g, h \in S$.
+Let:
+:$(1): \quad t \in \left[{a \,.\,.\, b}\right]$
+:$(2): \quad \left\vert{f \left({t}\right) - g \left({t}\right) = r \left({t}\right)}\right\vert$
+:$(3): \quad \left\vert{g \left({t}\right) - h \left({t}\right) = s \left({t}\right)}\right\vert$
+Thus we need to show that:
+:$\displaystyle \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p} + \left({\int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p} \ge \left({\int_a^b \left\vert{f \left({t}\right) - h \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}$
+We have:
+{{begin-eqn}}
+{{eqn | l = d_p \left({f, g}\right) + d_p \left({g, h}\right)
+      | r = \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p} + \left({\int_a^b \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\int_a^b \left({r \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p} + \left({\int_a^b \left({s \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \left({\int_a^b \left({r \left({t}\right) + s \left({t}\right)}\right)^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = [[Minkowski's Inequality for Integrals]]
+}}
+{{eqn | r = \left({\int_a^b \left({\left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert + \left\vert{g \left({t}\right) - h \left({t}\right)}\right\vert}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}
+      | c = Definition of $r \left({t}\right)$ and $s \left({t}\right)$
+}}
+{{eqn | r = \left({\int_a^b \left\vert{f \left({t}\right) - h \left({t}\right\vert}\right)^2 \ \mathrm d t}\right)^{\frac 1 2}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = d_p \left({f, h}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({f, g}\right)
+      | r =  \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r =  \left({\int_a^b \left\vert{g \left({t}\right) - f \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = d_p \left({g, f}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | lo= \forall t \in \left[{a \,.\,.\, b}\right]:
+      | l = f \left({t}\right)
+      | o = \ne
+      | r = g \left({t}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{f \left({t}\right) - g \left({t}\right\vert}\right)^p
+      | o = \ge
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}
+      | o = \ge
+      | r = 0
+      | c = [[Sign of Function Matches Sign of Definite Integral]]
+}}
+{{end-eqn}}
+From [[Zero Definite Integral of Nowhere Negative Function implies Zero Function]] we have that:
+:$d_p \left({f, g}\right) = 0 \implies f = g$
+on $\left[{a \,.\,.\, b}\right]$.
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Sequence Space of Real Sequences is Metric Space}
+Tags: P-Sequence Metrics
+
+\begin{theorem}
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Real Sequence|real sequences]] $\left\langle{x_i}\right\rangle$ such that the [[Definition:Series of Numbers|series]] $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is [[Definition:Convergent Series of Numbers|convergent]].
+Let $d_p$ be the [[Definition:P-Sequence Metric on Real Sequences|$p$-sequence metric on $\R$]].
+Then $\ell^p := \left({A, d_p}\right)$ is a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:P-Sequence Metric on Real Sequences|$p$-sequence metric on $\R$]]:
+Let $A$ be the [[Definition:Set|set]] of all [[Definition:Real Sequence|real sequences]] $\left\langle{x_i}\right\rangle$ such that the [[Definition:Series of Numbers|series]] $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is [[Definition:Convergent Series of Numbers|convergent]].
+Then $\ell^p := \left({A, d_2}\right)$ where $d_p: A \times A: \to \R$ is the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in A: d_p \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}$
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, x}\right)
+      | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - x_k}\right\vert^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum_{k \mathop \ge 0} 0^p}\right)^{\frac 1 p}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $z = \left\langle{z_i}\right\rangle \in A$.
+{{MissingLinks|Minkowski's inequality currently only applies to finite sums}}
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, y}\right) + d_p \left({y, z}\right)
+      | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p} + \left({\sum_{k \mathop \ge 0} \left\vert{y_k - z_k}\right\vert^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | o = \ge
+      | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - z_k}\right\vert^p}\right)^{\frac 1 p}
+      | c = [[Minkowski's Inequality for Sums]]
+}}
+{{eqn | r = d_p \left({x, z}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, y}\right)
+      | r = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \left({\sum_{k \mathop \ge 0} \left\vert{y_k - x_k}\right\vert^p}\right)^{\frac 1 p}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = d_p \left({y, x}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | lo= \exists k \in \N:
+      | l = x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{x_k - y_k}\right\vert^p
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({\sum_{k \mathop \ge 0} \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_p \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Peano Structure is Unique}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $\left({P, s, 0}\right)$ and $\left({P', s', 0'}\right)$ be [[Definition:Peano Structure|Peano structures]].
+Then there is a [[Definition:Unique|unique]] [[Definition:Bijection|bijection]] $f: P \to P'$ such that:
+{{begin-eqn}}
+{{eqn|l = f \left({0}\right)
+     |r = 0'
+}}
+{{eqn|l = \forall n \in P: f \left({s \left({n}\right)}\right)
+     |r = s' \left({f \left({n}\right)}\right)
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+First to establish [[Definition:Unique|uniqueness]] of $f$.
+Suppose that $f, g: P \to P'$ both satisfy the conditions.
+Define $A \subseteq P$ as:
+:$A := \left\{{n \in P: f \left({n}\right) = g \left({n}\right)}\right\}$
+Then the first condition implies that $0 \in A$.
+Now suppose that $n \in A$. Then:
+{{begin-eqn}}
+{{eqn|l = f \left({s \left({n}\right)}\right)
+     |r = s' \left({f \left({n}\right)}\right)
+     |c = Assumption on $f$
+}}
+{{eqn|r = s' \left({g \left({n}\right)}\right)
+     |c = By hypothesis
+}}
+{{eqn|r = g \left({s \left({n}\right)}\right)
+     |c = Assumption on $g$
+}}
+{{end-eqn}}
+Hence $s \left({n}\right) \in A$.
+Since $P$ is a [[Definition:Peano Structure|Peano structure]], it follows that $A = P$.
+Hence by [[Equality of Mappings]], $f = g$.
+For the existence of such an $f$, we apply the [[Principle of Recursive Definition/Proof 1|Principle of Recursive Definition]] with $0' \in P'$ and $s': P' \to P'$.
+Lastly, it is to be established that the $f$ so obtained is a [[Definition:Bijection|bijection]].
+Let $A' \subseteq P$ be defined as:
+:$A' := \left\{{n \in P: \forall m \in P: f \left({n}\right) = f \left({m}\right) \implies n = m}\right\}$
+It follows from [[Axiom:Peano's Axioms|Axiom $(P4)$]] that $f \left({n}\right) = 0'$ implies $n = 0$.
+Hence $0 \in A'$.
+Suppose now that $n \in A'$, but $s \left({n}\right) \notin A'$.
+Then, since $f \left({s \left({n}\right)}\right) \ne 0'$, there exists some $m \in P, m \ne n$ such that:
+:$f \left({s \left({n}\right)}\right) = f \left({s \left({m}\right)}\right)$
+That is, such that:
+:$s' \left({f \left({n}\right)}\right) = s' \left({f \left({m}\right)}\right)$
+But by [[Axiom:Peano's Axioms|Axiom $(P3)$]], $s'$ is [[Definition:Injection|injective]].
+Hence $f \left({n}\right) = f \left({m}\right)$, contradicting the assumption that $n \in A'$.
+Therefore $s \left({n}\right) \in A'$, so that $A' = P$ by [[Axiom:Peano's Axioms|Axiom $(P5)$]].
+That is, $f$ is an [[Definition:Injection|injection]].
+Next, define $A'' \subseteq P'$ as:
+:$A'' := \left\{{n' \in P': \exists n \in P: f \left({n}\right) = n'}\right\}$
+Since $f \left({0}\right) = 0'$, we have $0' \in A''$.
+Suppose now that $n' \in A''$, and let $f \left({n}\right) = n'$.
+Then:
+{{begin-eqn}}
+{{eqn|l = f \left({s \left({n}\right)}\right)
+     |r = s' \left({f \left({n}\right)}\right)
+     |c = Definition of $f$
+}}
+{{eqn|r = s' \left({n'}\right)
+     |c = Definition of $n$
+}}
+{{end-eqn}}
+meaning $s' \left({n'}\right) \in A''$.
+Hence by [[Axiom:Peano's Axioms|Axiom $(P5)$]], $A'' = P'$.
+That is, $f$ is [[Definition:Surjection|surjective]].
+It follows that $f$ is the sought [[Definition:Bijection|bijection]].
+{{qed}}
+[[Category:Abstract Algebra]]
+19j09orgblsnntox1mcxhjb7duzqpc5
+\end{proof}<|endoftext|>
+\section{P-adic Metric is Metric}
+Tags: P-adic Metrics
+
+\begin{theorem}
+Let $p \in \N$ be a [[Definition:Prime Number|prime]].
+Let $\norm{\,\cdot\,}_p: \Q \to \R_{\ge 0}$ be the [[Definition:P-adic Norm|$p$-adic norm]] on $\Q$.
+Let $d_p$ be the [[Definition:P-adic Metric|$p$-adic metric]] on $\Q$:
+:$\forall x, y \in \Q: \map {d_p} {x, y} = \norm{x - y}_p$
+Then $d_p$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:P-adic Metric|$p$-adic metric]] on $\Q$ is defined as the [[Definition:Metric Induced by Norm|metric induced]] by the [[Definition:P-adic Norm|$p$-adic norm]] on $\Q$.
+It follows from [[Metric Induced by Norm is Metric]] that $d_p$ is a [[Definition:Metric|metric]].
+{{qed}}
+[[Category:P-adic Metrics]]
+fkm4lqmfi69cxlte36doik06uiubwpw
+\end{proof}<|endoftext|>
+\section{Constant Function is Continuous/Metric Space/Proof 2}
+Tags: Constant Mappings, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] from $A_1$ to $A_2$:
+:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
+That is, every [[Definition:Element|point]] in $A_1$ maps to the same [[Definition:Element|point]] $c$ in $A_2$.
+Then $f_c$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+\end{theorem}
+
+\begin{proof}
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] between two [[Definition:Metric Space|metric spaces]] $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
+Let $\epsilon \in \R_{>0}$.
+Let $x \in A_1$.
+Pick any $\delta \in \R_{>0}$.
+Let $y \in A_1$ such that $d_1 \left({x, y}\right) < \delta$.
+Now we have:
+:$f_c \left({x}\right) = c = f_c \left({y}\right)$
+Hence:
+:$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$
+By definition of a [[Definition:Metric|metric]]:
+:$d_2 \left({c, c}\right) = 0 < \epsilon$
+Thus we have that:
+:$\forall \epsilon > 0: \forall x \in A_1: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$
+Hence by definition of [[Definition:Continuous Mapping (Metric Spaces)|continuity on a metric space]], $f_c$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constant Function is Continuous/Metric Space/Proof 1}
+Tags: Constant Mappings, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] from $A_1$ to $A_2$:
+:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
+That is, every [[Definition:Element|point]] in $A_1$ maps to the same [[Definition:Element|point]] $c$ in $A_2$.
+Then $f_c$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+\end{theorem}
+
+\begin{proof}
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] between two [[Definition:Metric Space|metric spaces]] $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
+From [[Constant Function is Uniformly Continuous/Metric Space|Constant Function is Uniformly Continuous]], $f_c$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+The result follows from [[Uniformly Continuous Function is Continuous/Metric Space|Uniformly Continuous Function is Continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constant Function is Continuous/Metric Space}
+Tags: Constant Mappings, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f_c: A_1 \to A_2$ be the [[Definition:Constant Mapping|constant mapping]] from $A_1$ to $A_2$:
+:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
+That is, every [[Definition:Element|point]] in $A_1$ maps to the same [[Definition:Element|point]] $c$ in $A_2$.
+Then $f_c$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous throughout $A_1$ with respect to $d_1$ and $d_2$]].
+\end{theorem}<|endoftext|>
+\section{Constant Function is Continuous/Real Function}
+Tags: Constant Mappings, Real Analysis, Continuous Functions
+
+\begin{theorem}
+Let $f_c: \R \to \R$ be the [[Definition:Constant Mapping|constant mapping]]:
+:$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$
+Then $f_c$ is [[Definition:Continuous Real Function|continuous on $\R$]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Constant Function is Uniformly Continuous/Real Function|Constant Function is Uniformly Continuous: Real Function]]
+:[[Uniformly Continuous Function is Continuous/Real Function|Uniformly Continuous Function is Continuous: Real Function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping is Continuous/Metric Space}
+Tags: Continuous Mappings on Metric Spaces, Identity Mappings
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+The [[Definition:Identity Mapping|identity mapping]] $I_A: A \to A$ defined as:
+:$\forall x \in A: I_A \left({x}\right) = x$
+is a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in A$.
+Let $\epsilon \in \R_{>0}$.
+Let $\delta  = \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d \left({x, a}\right)
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d \left({I_A \left({x}\right), I_A \left({a}\right)}\right)
+      | r = d \left({x, a}\right)
+      | c = 
+}}
+{{eqn | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping on Real Vector Space from Chebyshev to Euclidean Metric is Continuous}
+Tags: Continuous Mappings on Metric Spaces, Identity Mappings, Euclidean Metric, Chebyshev Distance
+
+\begin{theorem}
+Let $\R^n$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Real Vector Space|real vector space]].
+Let $d_2$ be the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] on $\R^n$.
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$.
+Let $I: \R^n \to \R^n$ be the [[Definition:Identity Mapping|identity mapping]] from $\R^n$ to itself.
+Then the [[Definition:Mapping|mapping]]:
+:$I: \struct {\R^n, d_\infty} \to \struct {\R^n, d_2}$
+is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_\infty, d_2}$-continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$.
+Let $\epsilon \in \R_{>0}$.
+Let $\delta = \dfrac {\epsilon} {\sqrt n}$.
+Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_\infty} {x, a} < \delta$.
+That is:
+:$\ds \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } < \delta$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, a}
+      | r = \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - x_i} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \sqrt {n \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } }
+      | c = 
+}}
+{{eqn | o = <
+      | r = \sqrt {n \delta^2}
+      | c = 
+}}
+{{eqn | r = \sqrt {\epsilon^2}
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_\infty, d_2}$-continuity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping on Real Vector Space from Euclidean to Chebyshev Distance is Continuous}
+Tags: Continuous Mappings on Metric Spaces, Identity Mappings, Euclidean Metric, Chebyshev Distance
+
+\begin{theorem}
+Let $\R^n$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Real Vector Space|real vector space]].
+Let $d_2$ be the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] on $\R^n$.
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$.
+Let $I: \R^n \to \R^n$ be the [[Definition:Identity Mapping|identity mapping]] from $\R^n$ to itself.
+Then the [[Definition:Mapping|mapping]]:
+:$I: \left({\R^n, d_2}\right) \to \left({\R^n, d_\infty}\right)$
+is [[Definition:Continuous Mapping (Metric Spaces)|$\left({d_2, d_\infty}\right)$-continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $a = \left({a_1, a_2, \ldots, a_n}\right) \in \R^n$.
+Let $\epsilon \in \R_{>0}$.
+Let $\delta = \epsilon$.
+Let $x = \left({x_1, x_2, \ldots, x_n}\right)$ be such that $d_2 \left({x, a}\right) < \delta$.
+That is:
+:$\displaystyle \sqrt{\sum_{i \mathop = i}^n \left({a_i - x_i}\right)} < \delta$
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = i}^n \left({a_i - x_i}\right)
+      | o = <
+      | r = \delta^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | lo= \forall i \in \left\{ {1, 2, \ldots, n}\right\}:
+      | l = \left({a_i - x_i}\right)^2
+      | o = <
+      | r = \delta^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | lo= \forall i \in \left\{ {1, 2, \ldots, n}\right\}:
+      | l = \left\vert{a_i - x_i}\right\vert
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \max_{i \mathop \le i \mathop \le n} \left\{ {\left\vert{a_i - x_i}\right\vert}\right\}
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_\infty \left({x, a}\right)
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Continuous Mapping (Metric Spaces)|$\left({d_2, d_\infty}\right)$-continuity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point}
+Tags: Metric Spaces, Continuous Mappings on Metric Spaces, Composite Mappings
+
+\begin{theorem}
+Let $M_1 = \left({X_1, d_1}\right), M_2 = \left({X_2, d_2}\right), M_3 = \left({X_3, d_3}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: M_1 \to M_2$ be [[Definition:Continuous at Point of Metric Space|continuous at $a \in X_1$]].
+Let $g: M_2 \to M_3$ be [[Definition:Continuous at Point of Metric Space|continuous at $f \left({a}\right) \in X_2$]].
+Then their [[Definition:Composition of Mappings|composite]] $g \circ f: M_1 \to M_3$ is [[Definition:Continuous at Point of Metric Space|continuous at $a \in X_1$]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+The strategy is to find a $\delta \in \R_{>0}$ such that:
+: $d_1 \left({x, a}\right) < \delta \implies d_3 \left({g \left({f \left({x}\right)}\right), g \left({f \left({a}\right)}\right)}\right) < \epsilon$
+As $g$ is [[Definition:Continuous at Point of Metric Space|continuous at $f \left({a}\right)$]]:
+:$\exists \eta \in \R_{>0}: \forall y \in X_2: d_2 \left({y, f \left({a}\right)}\right) < \eta \implies d_3 \left({g \left({y}\right), g \left({f \left({a}\right)}\right)}\right) < \epsilon$
+As $f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]]:
+:$\forall \eta \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in X_1: d_1 \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \eta$
+Hence:
+:$d_3 \left({g \left({f \left({x}\right)}\right), g \left({f \left({a}\right)}\right)}\right) < \epsilon$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composite of Continuous Mappings between Metric Spaces is Continuous}
+Tags: Metric Spaces, Continuous Mappings on Metric Spaces, Composite Mappings, Composite of Continuous Mappings between Metric Spaces is Continuous
+
+\begin{theorem}
+Let $M_1 = \left({X_1, d_1}\right), M_2 = \left({X_2, d_2}\right), M_3 = \left({X_3, d_3}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be [[Definition:Continuous Mapping (Metric Spaces)|continuous mappings]].
+Then their [[Definition:Composition of Mappings|composite]] $g \circ f: M_1 \to M_3$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+From [[Metric Induces Topology]], the [[Definition:Metric Space|metric spaces]] described are [[Definition:Topological Space|topological spaces]].
+The result follows from [[Composite of Continuous Mappings is Continuous]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $f$ and $g$ be [[Definition:Continuous Mapping (Metric Spaces)|continuous mappings]].
+By definition:
+:$f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a \in X_1$]] for all $a \in X_1$
+:$g$ is [[Definition:Continuous at Point of Metric Space|continuous at $\map f a \in X_2$]] for all $\map f a \in X_2$.
+The result follows from [[Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space Completeness is not Preserved by Homeomorphism}
+Tags: Complete Metric Spaces, Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi: M_1 \to M_2$ be a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+If $M_1$ is [[Definition:Complete Metric Space|complete]] then it is not necessarily the case that so is $M_2$.
+\end{theorem}
+
+\begin{proof}
+Let $\Z_{>0}$ be the [[Definition:Set|set]] of [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]] on $\Z_{>0}$.
+Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the [[Definition:Metric|metric]] on $\Z_{>0}$ defined as:
+:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
+Let $\tau_d$ denote the [[Definition:Topology Induced by Metric|metric topology]] for $d$.
+Let $\tau_\delta$ denote the [[Definition:Topology Induced by Metric|metric topology]] for $\delta$.
+From [[Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic]]:
+:$\tau_d$ and $\tau_\delta$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+From [[Positive Integers under Usual Metric is Complete Metric Space]]:
+:$\struct {\Z_{>0}, d}$ is a [[Definition:Complete Metric Space|complete metric space]].
+However, from [[Positive Integers under Scaled Euclidean Metric is not Complete Metric Space]]:
+:$\struct {\Z_{>0}, \delta}$ is not a [[Definition:Complete Metric Space|complete metric space]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping from L1 Space to Real Number Space is Continuous}
+Tags: Continuous Mappings, L1 Metric
+
+\begin{theorem}
+Let $\left({\R, d}\right)$ be the [[Definition:Real Number Line|real number line]] under the [[Definition:Usual Metric|usual metric]] $d$.
+Let $X$ be the [[Definition:Set|set]] of [[Definition:Continuous Real Function|continuous real functions]] $f: \left[{a \,.\,.\, b}\right] \to \R$.
+Let $d_1$ be the [[Definition:L1 Metric/Closed Real Interval|$L^1$ metric]] on $X$.
+Let $I: X \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\displaystyle \forall f \in X: I \left({f}\right) := \int_a^b f \left({t}\right) \ \mathop d t$
+Then the [[Definition:Mapping|mapping]]:
+:$I : \left({X, d_1}\right) \to \left({\R, d}\right)$
+is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:L1 Metric/Closed Real Interval|$L^1$ metric]] on $X$ is defined as:
+:$\displaystyle \forall f, g \in S: d_1 \left({f, g}\right) := \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t$
+Let $\epsilon \in \R_{>0}$.
+Let $f \in X$.
+Let $\delta = \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall g \in X:
+      | l = d_1 \left({f, g}\right)
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert \ \mathrm d t
+      | o = <
+      | r = \delta
+      | c = Definition of [[Definition:L1 Metric/Closed Real Interval|$L^1$ Metric]]
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{\int_a^b \left({f \left({t}\right) - g \left({t}\right)}\right) \ \mathrm d t}\right\vert
+      | o = <
+      | r = \delta
+      | c = [[Absolute Value of Definite Integral]]
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{\int_a^b f \left({t}\right) \ \mathrm d t - \int_a^b g \left({t}\right) \ \mathrm d t}\right\vert
+      | o = <
+      | r = \delta
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{I \left({f}\right) - I \left({g}\right)}\right\vert
+      | o = <
+      | r = \delta
+      | c = Definition of $I$
+}}
+{{eqn | ll= \implies
+      | l = d \left({I \left({f}\right), I \left({g}\right)}\right)
+      | o = <
+      | r = \delta
+      | c = Definition of $d$
+}}
+{{eqn | ll= \implies
+      | l = d \left({I \left({f}\right), I \left({g}\right)}\right)
+      | o = <
+      | r = \epsilon
+      | c = Definition of $\delta$
+}}
+{{end-eqn}}
+Thus it has been demonstrated that:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall g \in X: d_1 \left({f, g}\right) < \delta \implies d \left({I \left({f}\right), I \left({g}\right)}\right) < \epsilon$
+Hence by definition of [[Definition:Continuous at Point of Metric Space|continuity at a point]], $I$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $f$.
+As $f$ is chosen arbitrarily, it follows that $I$ is [[Definition:Continuous at Point of Metric Space|continuous]] for all $f \in X$.
+The result follows by definition of [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cartesian Product under Chebyshev Distance of Continuous Mappings between Metric Spaces is Continuous}
+Tags: Continuous Mappings on Metric Spaces, Chebyshev Distance
+
+\begin{theorem}
+Let $n \in \N_{>0}$.
+Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_n = \left({A_n, d_n}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $N_1 = \left({B_1, d'_1}\right), N_2 = \left({B_2, d'_2}\right), \ldots, N_n = \left({B_n, d'_n}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f_i: M_i \to N_i$ be [[Definition:Continuous Mapping (Metric Spaces)|continuous mappings]] for all $i \in \left\{{1, 2, \ldots, n}\right\}$.
+Let $\displaystyle \mathcal M = \prod_{i \mathop = 1}^n M_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $\displaystyle \mathcal N = \prod_{i \mathop = 1}^n N_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $B_1, B_2, \ldots, B_n$.
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/General Definition|Chebyshev distance]] on $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$, and $\displaystyle \mathcal B = \prod_{i \mathop = 1}^n B_i$, defined as:
+: $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$
+: $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d'_i \left({x_i, y_i}\right)}\right\}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$ or $\mathcal B$.
+Let $F: M \to N$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \mathcal A: F \left({x_1, x_2, \ldots, x_n}\right) = \left({f \left({x_1}\right), \left({x_2}\right), \ldots,  \left({x_n}\right)}\right)$
+Then $F$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+Let $x \in \mathcal A$.
+Let $k \in \left\{{1, 2, \ldots, n}\right\}$.
+Then as $f_k$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]]:
+:$(1): \quad \exists \delta_k \in \R_{>0}: \forall y_k \in A_k: d_k \left({x_k, y_k}\right) < \delta_k \implies d' \left({f_k \left({x_k}\right), f_k \left({y_k}\right)}\right) < \epsilon$
+Let $\delta = \max \left\{ {\delta_k: k \in  \left\{{1, 2, \ldots, n}\right\} }\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall y \in \mathcal A:
+      | l = d_\infty \left({x, y}\right)
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}
+      | o = <
+      | r = \delta
+      | c = Definition of [[Definition:Chebyshev Distance/General Definition|Chebyshev Distance]]
+}}
+{{eqn | ll= \implies
+      | lo= \forall k \in \left\{ {1, 2, \ldots, n}\right\}:
+      | l = d_k \left({x_k, y_k}\right)
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d'_k \left({f_k \left({x_k}\right), f_k \left({y_k}\right)}\right)
+      | o = <
+      | r = \epsilon
+      | c = from $(1)$
+}}
+{{eqn | ll= \implies
+      | l = \max_{i \mathop = 1}^n \left\{ {d'_i \left({f_i \left({x_i}\right), f_i \left({y_i}\right)}\right)}\right\}
+      | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_\infty \left({F \left({x}\right), F \left({y}\right)}\right)
+      | o = <
+      | r = \epsilon
+      | c = Definition of [[Definition:Chebyshev Distance/General Definition|Chebyshev Distance]]
+}}
+{{end-eqn}}
+Thus it has been demonstrated that:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in X: d_\infty \left({x, y}\right) < \delta \implies d_\infty \left({F \left({x}\right), F \left({y}\right)}\right) < \epsilon$
+Hence by definition of [[Definition:Continuous at Point of Metric Space|continuity at a point]], $F$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $x$.
+As $x$ is chosen arbitrarily, it follows that $F$ is [[Definition:Continuous at Point of Metric Space|continuous]] for all $x \in X$.
+The result follows by definition of [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Addition of Coordinates on Cartesian Plane under Chebyshev Distance is Continuous Function}
+Tags: Continuous Mappings on Metric Spaces, Chebyshev Distance
+
+\begin{theorem}
+Let $\R^2$ be the [[Definition:Real Number Plane|real number plane]].
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/Real Number Plane|Chebyshev distance]] on $\R^2$.
+Let $f: \R^2 \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\forall \left({x_1, x_2}\right) \in \R^2: f \left({x_1, x_2}\right) = x_1 + x_2$
+Then $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+First we note that:
+{{begin-eqn}}
+{{eqn | l = \left\vert{\left({x_1 + x_2}\right) - \left({y_1 + y_2}\right)}\right\vert
+      | r = \left\vert{\left({x_1 - y_1}\right) + \left({x_2 - y_2}\right)}\right\vert
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left\vert{x_1 - y_1}\right\vert + \left\vert{x_2 - y_2}\right\vert
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | n = 1
+      | o = \le
+      | r = 2 \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}
+      | c = 
+}}
+{{end-eqn}}
+Let $\epsilon \in \R_{>0}$.
+Let $x = \left({x_1, x_2}\right) \in \R^2$.
+Let $\delta = \dfrac \epsilon 2$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall y =  \left({y_1, y_2}\right) \in \R^2:
+      | l = d_\infty \left({x, y}\right)
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}
+      | o = <
+      | r = \delta
+      | c = Definition of [[Definition:Chebyshev Distance/Real Number Plane|Chebyshev Distance]]
+}}
+{{eqn | ll= \implies
+      | l = 2 \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}
+      | o = <
+      | r = \epsilon
+      | c = Definition of $\epsilon$
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{\left({x_1 + x_2}\right) - \left({y_1 + y_2}\right)}\right\vert
+      | o = <
+      | r = \epsilon
+      | c = from $(1)$
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{f \left({x}\right) -  f \left({y}\right)}\right\vert
+      | o = <
+      | r = \epsilon
+      | c = Definition of $f$
+}}
+{{eqn | ll= \implies
+      | l = d \left({f \left({x}\right), f \left({y}\right)}\right)
+      | o = <
+      | r = \epsilon
+      | c = Definition of [[Definition:Usual Metric|Usual Metric]] on $\R$
+}}
+{{end-eqn}}
+Thus it has been demonstrated that:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: d_\infty \left({x, y}\right) < \delta \implies d \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$
+Hence by definition of [[Definition:Continuous at Point of Metric Space|continuity at a point]], $f$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $x$.
+As $x$ was chosen arbitrarily, it follows that $f$ is [[Definition:Continuous at Point of Metric Space|continuous]] for all $x \in \R^2$.
+The result follows by definition of [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Addition of Coordinates on Euclidean Plane is Continuous Function}
+Tags: Continuous Mappings, Real Number Plane with Euclidean Metric
+
+\begin{theorem}
+Let $\struct {\R^2, d_2}$ be the [[Definition:Real Number Plane with Euclidean Metric|real number plane with the usual (Euclidean) metric]].
+Let $f: \R^2 \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = x_1 + x_2$
+Then $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+First we note that:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }
+      | c = 
+}}
+{{eqn | r = \size {\paren {x_1 - y_1} + \paren {x_2 - y_2} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \size {x_1 - y_1} + \size {x_2 - y_2}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | n = 1
+      | o = \le
+      | r = \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
+      | c = [[P-Product Metrics on Real Vector Space are Topologically Equivalent|$p$-Product Metrics on Real Vector Space are Topologically Equivalent]]
+}}
+{{end-eqn}}
+Let $\epsilon \in \R_{>0}$.
+Let $x = \tuple {x_1, x_2} \in \R^2$.
+Let $\delta = \dfrac \epsilon {\sqrt 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall y = \tuple {y_1, y_2} \in \R^2:
+      | l = \map {d_2} {x, y}
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
+      | o = <
+      | r = \delta
+      | c = {{Defof|Euclidean Metric on Real Number Plane}}
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt 2 \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
+      | o = <
+      | r = \delta \sqrt 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\paren {x_1 + x_2} - \paren {y_1 + y_2} }t
+      | o = <
+      | r = \epsilon
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map f x - \map f y}
+      | o = <
+      | r = \epsilon
+      | c = Definition of $f$
+}}
+{{eqn | ll= \leadsto
+      | l = \map d {\map f x, \map f y}
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Euclidean Metric on Real Number Line}}
+}}
+{{end-eqn}}
+Thus it has been demonstrated that:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R^2: \map {d_2} {x, y} < \delta \implies \map d {\map f x, \map f y} < \epsilon$
+Hence by definition of [[Definition:Continuous at Point of Metric Space|continuity at a point]], $f$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $x$.
+As $x$ is chosen arbitrarily, it follows that $f$ is [[Definition:Continuous at Point of Metric Space|continuous]] for all $x \in \R^2$.
+The result follows by definition of [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space Continuity by Inverse of Mapping between Open Balls}
+Tags: Open Balls, Continuity
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$.
+Let $a \in A_1$ be a point in $A_1$.
+$f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]] with respect to the [[Definition:Metric|metrics]] $d_1$ and $d_2$ [[Definition:Iff|iff]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: B_\delta \left({a; d_1}\right) \subseteq f^{-1} \left[{B_\epsilon \left({f \left({a}\right); d_2}\right)}\right]$
+where $B_\epsilon \left({f \left({a}\right); d_2}\right)$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $f \left({a}\right)$ with respect to the [[Definition:Metric|metric]] $d_2$, and similarly for $B_\delta \left({a; d_1}\right)$.
+\end{theorem}
+
+\begin{proof}
+By definition, $f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]] with respect to the [[Definition:Metric|metrics]] $d_1$ and $d_2$ [[Definition:Iff|iff]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left[{B_\delta \left({a; d_1}\right)}\right] \subseteq B_\epsilon \left({f \left({a}\right); d_2}\right)$
+For a [[Definition:Mapping|mapping]] $f: X \to Y$ we have:
+:$f \left[{U}\right] \subseteq V \iff U \subseteq f^{-1} \left[{V}\right]$
+where $U \subseteq X$ and $V \subseteq Y$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball is Neighborhood of all Points Inside}
+Tags: Open Balls, Neighborhoods
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $a \in A$.
+Let $B_\epsilon \left({a}\right)$ be an [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $M$.
+Let $x \in B_\epsilon \left({a}\right)$.
+Then $B_\epsilon \left({a}\right)$ is a [[Definition:Neighborhood (Metric Space)|neighborhoods]] of $x$ in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Open Ball of Point Inside Open Ball]]:
+: $\exists \delta \in \R: B_\delta \left({x}\right) \subseteq B_\epsilon \left({a}\right)$
+Thus by definition $B_\delta \left({x}\right)$ is a [[Definition:Neighborhood (Metric Space)|neighborhoods]] of $x$ in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space Continuity by Neighborhood}
+Tags: Neighborhoods, Continuity
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$.
+Let $a \in A_1$ be a point in $A_1$.
+Then the following definitions of [[Definition:Continuous at Point of Metric Space|continuity of $f$ at $a$ with respect to $d_1$ and $d_2$]] are [[Definition:Logical Equivalence|equivalent]]:
+\end{theorem}
+
+\begin{proof}
+=== $\epsilon$-Ball Definition implies Definition by Neighborhood ===
+Suppose that [[Definition:Continuous Mapping (Metric Space)/Point/Definition 3|$f$ is $\tuple {d_1, d_2}$-continuous at $a$]] in the sense that:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \sqbrk {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
+where $\map {B_\epsilon} {\map f a; d_2}$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $\map f a$ with respect to the [[Definition:Metric|metric]] $d_2$, and similarly for $\map {B_\delta} {a; d_1}$.
+Let $N'$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $\map f a$ in $M_2$.
+Then :
+{{begin-eqn}}
+{{eqn | lo= \exists \epsilon \in \R_{>0}:
+      | l = \map {B_\epsilon} {\map f a; d_2}
+      | o = \subseteq
+      | r = N'
+      | c = {{Defof|Neighborhood (Metric Space)}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists \delta \in \R_{>0}:
+      | l = f \sqbrk {\map {B_\delta} {a; d_1} }
+      | o = \subseteq
+      | r = \map {B_\epsilon} {\map f a; d_2}
+      | c = as $f$ is [[Definition:Continuous Mapping (Metric Space)/Point/Definition 3|$\tuple {d_1, d_2}$-continuous at $a$]]
+}}
+{{end-eqn}}
+But by [[Open Ball is Neighborhood of all Points Inside]], $\map {B_\delta} {a; d_1}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M_1$.
+Thus letting $N := \map {B_\delta} {a; d_1}$:
+:$f \sqbrk N \subseteq \map {B_\epsilon} {\map f a; d_2} \subseteq N'$
+and so the condition for $f$ to be [[Definition:Continuous Mapping (Metric Space)/Point/Definition 4|$\tuple {d_1, d_2}$-continuous at $a$ by Neighborhood]] is fulfilled.
+{{qed|lemma}}
+=== Definition by Neighborhood implies $\epsilon$-Ball Definition ===
+Suppose that [[Definition:Continuous Mapping (Metric Space)/Point/Definition 4|$f$ is $\tuple {d_1, d_2}$-continuous at $a$]] in the sense that:
+:for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $N'$ of $\map f a$ in $M_2$ there exists a corresponding [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ of $a$ in $M_1$ such that $f \sqbrk N \subseteq N'$.
+Let $\epsilon \in \R_{>0}$.
+Consider the [[Definition:Open Ball|open $\epsilon$-ball]] $\map {B_\epsilon} {\map f a; d_2}$ of $\map f a$.
+By [[Open Ball is Neighborhood of all Points Inside]], $\map {B_\epsilon} {\map f a; d_2}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $\map f a$ in $M_2$.
+Let $N' := \map {B_\epsilon} {\map f a; d_2}$.
+By definition of [[Definition:Continuous Mapping (Metric Space)/Point/Definition 4|$\tuple {d_1, d_2}$-continuity at $a$]], there exists a [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ of $a$ in $M_1$ such that:
+:$f \sqbrk N \subseteq N'$
+By definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+:$\exists \delta \in \R_{>0}: \map {B_\delta} {a; d_1} \subseteq N$.
+Therefore:
+:$\map f {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$
+and so the condition for $f$ to be [[Definition:Continuous Mapping (Metric Space)/Point/Definition 3|$\tuple {d_1, d_2}$-continuous at $a$ by open $\epsilon$-ball]] is fulfilled.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence Characterization of Open Sets}
+Tags: Open Sets
+
+\begin{theorem}
+Let $\left({X, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $G \subseteq X$.
+Then the following are equivalent:
+:$(1): \quad G \subseteq X$ is an [[Definition:Open Set (Metric Space)|open set]] of $\left({X, d}\right)$
+:$(2): \quad \forall x \in G: \forall \left\langle{x_n}\right\rangle \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \left\langle{x_n}\right\rangle \in G$
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Let $G \subseteq X$ be [[Definition:Open Set (Metric Space)|open]] in $\left({X, d}\right)$.
+Let $x \in G$.
+Let $\left\langle{x_n}\right\rangle$ be a [[Definition:Sequence|sequence]] in $X$ such that $x_n \to x$.
+By definition of [[Definition:Open Set (Metric Space)|open set]], there exists $\epsilon > 0$ such that:
+: $B_\epsilon \left({x}\right) \subseteq G$
+where $B_\epsilon \left({x}\right)$ is the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$.
+Since $x_n \to x$, there exists $n_0 \in \N$ such that:
+: $\forall n \ge n_0: d \left({x_n, x}\right) < \epsilon$
+Thus:
+: $\forall n \ge n_0: x_n \in B_\epsilon \left({x}\right) \subseteq G$
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $G \subseteq X$.
+Let:
+:$\forall x \in G: \forall \left\langle{x_n}\right\rangle \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \left\langle{x_n}\right\rangle \in G$
+Aiming for a [[Definition:Contradiction|contradiction]], suppose $G$ is ''not'' [[Definition:Open Set (Metric Space)|open]] in $\left({X, d}\right)$.
+Then:
+: $\exists x \in G: \forall \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \nsubseteq G$
+Therefore, for $n = 1, 2, \ldots$ we can [[Principle of Mathematical Induction|inductively]] find a [[Definition:Sequence|sequence]]:
+: $x_n \in B_{1/n} \left({x}\right) \cap \left({X \setminus G}\right) \implies x_n \notin G$
+and:
+: $\forall n \in \N: d \left({x_n, x}\right) < \dfrac 1 n \implies x_n \to x \in G$ and $\forall n \in \N: x_n \notin G$
+This contradicts hypothesis $(2)$.
+Thus, by [[Proof by Contradiction]], $G$ is [[Definition:Open Set (Metric Space)|open]] in $\left({X, d}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space Continuity by Inverse of Mapping between Neighborhoods}
+Tags: Neighborhoods, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$.
+Let $a \in A_1$ be a point in $A_1$.
+$f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]] with respect to the [[Definition:Metric|metrics]] $d_1$ and $d_2$ {{iff}}:
+:for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ of $\map f a$ in $M_2$, $f^{-1} \sqbrk N$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+\end{theorem}
+
+\begin{proof}
+By definition, $f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]] with respect to the [[Definition:Metric|metrics]] $d_1$ and $d_2$ {{iff}}:
+:for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ of $\map f a$ in $M_2$ there exists a corresponding [[Definition:Neighborhood (Metric Space)|neighborhood]] $N'$ of $a$ in $M_1$ such that $f \sqbrk {N'} \subseteq N$.
+For a [[Definition:Mapping|mapping]] $f: X \to Y$ we have:
+:$f \sqbrk U \subseteq V \iff U \subseteq f^{-1} \sqbrk V$
+where $U \subseteq X$ and $V \subseteq Y$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Metric Space has Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $a \in A$ be a point in $M$.
+Then there exists some [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+\end{theorem}
+
+\begin{proof}
+Let $a \in A$.
+Then $A$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Metric Space is Element of its Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then $a \in N$.
+\end{theorem}
+
+\begin{proof}
+Trivially follows by definition of [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Superset of Neighborhood in Metric Space is Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Let $N \subseteq N' \subseteq A$.
+Then $N'$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+: $\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq N$
+where $B_\epsilon \left({a}\right)$ is the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $M$.
+By [[Subset Relation is Transitive]]:
+:$B_\epsilon \left({a}\right) \subseteq N'$
+The result follows by definition of [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Neighborhoods in Metric Space is Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $N, N'$ be [[Definition:Neighborhood (Metric Space)|neighborhoods]] of $a$ in $M$.
+Then $N \cap N'$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+:$\exists \epsilon_1 \in \R_{>0}: \map {B_{\epsilon_1} } a \subseteq N$
+where $\map {B_{\epsilon_1} } a$ is the [[Definition:Open Ball|open $\epsilon_1$-ball]] of $a$ in $M$.
+:$\exists \epsilon_2 \in \R_{>0}: \map {B_{\epsilon_2} } a \subseteq N'$
+where $\map {B_{\epsilon_2} } a$ is the [[Definition:Open Ball|open $\epsilon_2$-ball]] of $a$ in $M$.
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$\map {B_\epsilon} a \subseteq N \cap N'$
+where $\epsilon = \min \set {\epsilon_1, \epsilon_2}$
+The result follows by definition of [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood in Metric Space has Subset Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then there exists a [[Definition:Neighborhood (Metric Space)|neighborhood]] $N'$ of $a$ such that:
+:$(1): \quad N' \subseteq N$
+:$(2): \quad N'$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of each of its points.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+: $\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq N$
+By [[Open Ball is Neighborhood of all Points Inside]], $N' = B_\epsilon \left({a}\right)$ fulfils the conditions of the statement.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basic Properties of Neighborhood in Metric Space}
+Tags: Neighborhoods
+
+\begin{theorem}
+This page gathers together the basic properties of a [[Definition:Neighborhood (Metric Space)|neighborhood]] in a [[Definition:Metric Space|metric space]].
+\end{theorem}<|endoftext|>
+\section{Basis for Element of Real Number Line}
+Tags: Real Intervals, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $M = \struct {\R, d}$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $a \in \R$ be a point in $M$.
+Then the [[Definition:Set|set]] of all [[Definition:Open Real Interval|open intervals]] containing $a$ is a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then by definition:
+:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$
+where $\map {B_\epsilon} a$ is the [[Definition:Open Ball|open $\epsilon$-ball at $a$]].
+The result follows from [[Open Ball in Real Number Line is Open Interval]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball in Real Number Line is Open Interval}
+Tags: Real Intervals, Open Balls, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\struct {\R, d}$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line $\R$ with the usual (Euclidean) metric $d$]].
+Let $x \in \R$ be a point in $\R$.
+Let $\map {B_\epsilon} x$ be the [[Definition:Open Ball|open $\epsilon$-ball at $x$]].
+Then $\map {B_\epsilon} x$ is the [[Definition:Open Real Interval|open interval]] $\openint {x - \epsilon} {x + \epsilon}$.
+\end{theorem}
+
+\begin{proof}
+Let $S = \map {B_\epsilon} x$ be an [[Definition:Open Ball of Metric Space|open $\epsilon$-ball at $x$]].
+Let $y \in \map {B_\epsilon} x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | m = \map {B_\epsilon} x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map d {y, x}
+      | o = <
+      | m = \epsilon
+      | c = {{Defof|Open Ball of Metric Space|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \size {y - x}
+      | o = <
+      | m = \epsilon
+      | c = {{Defof|Euclidean Metric on Real Number Line}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -\epsilon
+      | o = <
+      | m = y - x
+      | mo= <
+      | r = \epsilon
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x - \epsilon
+      | o = <
+      | m = y
+      | mo= <
+      | r = x + \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y
+      | o = \in
+      | m = \openint {x - \epsilon} {x + \epsilon}
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+As the implications go both ways:
+:$\map {B_\epsilon} x \subseteq \openint {x - \epsilon} {x + \epsilon}$
+and
+:$\map {B_\epsilon} x \supseteq \openint {x - \epsilon} {x + \epsilon}$
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\map {B_\epsilon} x = \openint {x - \epsilon} {x + \epsilon}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Real Interval is Open Ball}
+Tags: Real Intervals, Open Balls, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $I := \openint a b \subseteq \R$ be an [[Definition:Open Real Interval|open real interval]].
+Then $I$ is the [[Definition:Open Ball|open $\epsilon$-ball]] $\map {B_\epsilon} \alpha$ of some $\alpha \in \R$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \alpha
+      | r = \frac {a + b} 2
+}}
+{{eqn | l = \epsilon
+      | r = \frac {b - a} 2
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = \alpha - \epsilon
+}}
+{{eqn | l = b
+      | r = \alpha + \epsilon
+}}
+{{end-eqn}}
+Thus:
+:$\openint a b = \openint {\alpha - \epsilon} {\alpha + \epsilon}$
+From [[Open Ball in Real Number Line is Open Interval]]:
+:$\openint {\alpha - \epsilon} {\alpha + \epsilon} = \map {B_\epsilon} \alpha$
+where $\map {B_\epsilon} \alpha$ is the [[Definition:Open Ball|open $\epsilon$-ball]] of $\alpha$ in $\R$.
+Hence the result
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhoods in Standard Discrete Metric Space}
+Tags: Discrete Metrics, Neighborhoods
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]] where $d$ is the [[Definition:Standard Discrete Metric|standard discrete metric]].
+Let $a \in A$.
+Then $\left\{ {a}\right\}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ which forms a [[Definition:Basis for Neighborhood System|basis]] for the [[Definition:System of Neighborhoods|system of neighborhoods]] of $a$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Standard Discrete Metric|standard discrete metric]]:
+:$d \left({x, y}\right) = \begin{cases}
+0 & : x = y  \\ 
+1 & : x \ne y
+\end{cases}$
+Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = B_\epsilon \left({a}\right)
+      | r = \left\{ {x \in A: d \left({x, a}\right) < \epsilon}\right\}
+      | c = Definition of [[Definition:Open Ball|Open $\epsilon$-Ball]] of $a$
+}}
+{{eqn | ll= \implies
+      | l = B_\epsilon \left({a}\right)
+      | r = \left\{ {a}\right\}
+      | c = 
+}}
+{{end-eqn}}
+So by definition $\left\{ {a}\right\}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+Let $\mathcal N_a$ be the [[Definition:System of Neighborhoods|system of neighborhoods]] of $a$.
+Let $N \in \mathcal N_a$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \in
+      | r = N
+      | c = [[Point in Metric Space is Element of its Neighborhood]]
+}}
+{{eqn | ll= \implies
+      | l = \left\{ {a}\right\} 
+      | o = \subseteq
+      | r = N
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = B_\epsilon \left({a}\right)
+      | o = \subseteq
+      | r = N
+      | c = for all $\epsilon \in \R_{>0}$ such that $\epsilon < 1$
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Basis for Neighborhood System|basis of a system of neighborhoods]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Standard Discrete Metric Space is Neighborhood of Each Point}
+Tags: Discrete Metrics, Neighborhoods
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]] where $d$ is the [[Definition:Standard Discrete Metric|standard discrete metric]].
+Let $S \subseteq A$.
+Let $a \in S$.
+Then $S$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+That is, every [[Definition:Subset|subset]] of $A$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of each of its points.
+\end{theorem}
+
+\begin{proof}
+Let $S \subseteq A$.
+Let $a \in S$.
+From [[Neighborhoods in Standard Discrete Metric Space]], $\left\{{a}\right\}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+As $a \in S$ it follows from [[Singleton of Element is Subset]] that $\left\{{a}\right\} \subseteq S$.
+The result follows from [[Superset of Neighborhood in Metric Space is Neighborhood]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Heaviside Step Function}
+Tags: Continuity, Heaviside Step Function
+
+\begin{theorem}
+Let $\mu_c: \R \to \R$ be the [[Definition:Heaviside Step Function|Heaviside step function]]:
+:$\mu_c \left({x}\right) = \begin{cases}
+0 & : x < c \\
+1 & : x > c \\
+\text{arbitrary} & : x = c
+\end{cases}$
+Then $\mu_c$ is [[Definition:Continuous Real Function at Point|continuous]] at every point of $\R$ except at $c$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R: x \ne c$.
+Let $\epsilon \in \R_{>0}$.
+Let $\delta < \left\vert{x - c}\right\vert$.
+Then by definition of the [[Definition:Heaviside Step Function|Heaviside step function]]:
+:$\forall y \in \left[{x - \delta \,.\,.\, x + \delta}\right]: \mu_c \left({x}\right) = \begin{cases}
+0 & : x < c \\
+1 & : x > c
+\end{cases}$
+Thus:
+:$\forall y \in \left[{x - \delta \,.\,.\, x + \delta}\right]: \mu_c \left({y}\right) = \mu_c \left({x}\right)$
+Thus:
+:$\left\vert{y - x}\right\vert \implies : \mu_c \left({y}\right) - \mu_c \left({x}\right) < \epsilon$
+and so $\mu_c$ is [[Definition:Continuous Real Function at Point|continuous]] at $\R$ except at $x$.
+{{qed|lemma}}
+Now suppose $x = c$.
+Let $\epsilon = \dfrac 1 3$.
+Let $\delta \in \R_{>0}$.
+Let $\alpha \in \R_{>0}: \alpha < \delta$.
+Let $y = x + \alpha$.
+Let $y \in \R: \left\vert{y - x}\right\vert < \delta$ and $y \ne x$.
+Suppose $\mu_c \left({y}\right) - \mu_c \left({x}\right) < \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = x + \alpha
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \mu_c \left({y}\right)
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+while:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = x - \alpha
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \mu_c \left({y}\right)
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So whatever the value of $f \left({x}\right)$, either:
+:$f \left({x + \alpha}\right) - f \left({x}\right) > \epsilon$
+or:
+:$f \left({x - \alpha}\right) - f \left({x}\right) > \epsilon$
+It follows that $\mu_c$ is not [[Definition:Continuous Real Function at Point|continuous]] at $c$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space Continuity by Neighborhood Basis}
+Tags: Neighborhood Bases, Continuity
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$.
+Let $a \in A_1$ be a point in $A_1$.
+Let $\BB_{\map f a}$ be a [[Definition:Basis for Neighborhood System|basis for the neighborhood system at $\map f a$]].
+$f$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]] with respect to the [[Definition:Metric|metrics]] $d_1$ and $d_2$ {{iff}}:
+:for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\BB_{\map f a}$ be a [[Definition:Basis for Neighborhood System|basis for the neighborhood system at $\map f a$]] {{iff}}:
+:$\forall N_a \subseteq M_2: \exists N \in \BB_{\map f a}: N \subseteq N_a$
+where $N_a$ denotes a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $\map f a$ in $M_2$.
+=== Necessary Condition ===
+Let $f$ be [[Definition:Continuous at Point of Metric Space|continuous at $a$]] with respect to the [[Definition:Metric|metrics]] $d_1$ and $d_2$.
+Then by [[Metric Space Continuity by Inverse of Mapping between Neighborhoods]]:
+:for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ of $\map f a$ in $M_2$, $f^{-1} \\sqbrk N$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+In particular, let $N \in \BB_{\map f a}$.
+Then $f^{-1} \sqbrk N$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f$ be such that:
+:for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+Let $N'$ be any [[Definition:Neighborhood (Metric Space)|neighborhood]] of $\map f a$.
+Then by definition:
+:$\exists B \in \BB_{\map f a}: B \subseteq N'$
+From [[Image of Subset under Relation is Subset of Image/Corollary 3|Image of Subset under Relation is Subset of Image: Corollary 3]]:
+:$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {N'}$
+[[Definition:By Hypothesis|By hypothesis]] $f^{-1} \sqbrk B$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+From [[Superset of Neighborhood in Metric Space is Neighborhood]] it follows that $f^{-1} \sqbrk {N'}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+As $N'$ is arbitrary, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Intervals form Neighborhood Basis in Real Number Line}
+Tags: Real Intervals, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $a \in \R$ be a point in $\R$.
+Let $\BB_a$ be defined as:
+:$\BB_a := \set {\closedint {a - \epsilon} {a + \epsilon}: \epsilon \in \R_{>0} }$
+that is, the [[Definition:Set|set]] of all [[Definition:Closed Real Interval|closed intervals]] of $\R$ with $a$ as a [[Definition:Midpoint of Interval|midpoint]].
+Then $\BB_a$ is a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then by definition:
+:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
+where $\map {B_{\epsilon'} } a$ is the [[Definition:Open Ball|open $\epsilon'$-ball at $a$]].
+From [[Open Ball in Real Number Line is Open Interval]]:
+:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
+Let $\epsilon \in \R_{>0}: \epsilon < \epsilon'$.
+Then by definition of [[Definition:Closed Real Interval|closed interval]] and [[Definition:Open Real Interval|open interval]]:
+:$\closedint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Subset Relation is Transitive]]:
+:$\closedint {a - \epsilon} {a + \epsilon} \subseteq N$
+Also by definition of [[Definition:Closed Real Interval|closed interval]] and [[Definition:Open Real Interval|open interval]]:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq \closedint {a - \epsilon} {a + \epsilon}$
+From [[Open Real Interval is Open Ball]], $\openint {a - \epsilon} {a + \epsilon}$ is the [[Definition:Open Ball|open $\epsilon$-ball at $a$]].
+Thus by definition, $\closedint {a - \epsilon} {a + \epsilon}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Hence there exists a [[Definition:Neighborhood (Metric Space)|neighborhood]] $\closedint {a - \epsilon} {a + \epsilon}$ of $a$ which is a [[Definition:Subset|subset]] of $N$.
+Hence the result by definition of [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Rational-Number Balls form Neighborhood Basis in Real Number Line}
+Tags: Real Intervals, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $a \in \R$ be a point in $\R$.
+Let $\BB_a$ be defined as:
+:$\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \Q_{>0} }$
+that is, the [[Definition:Set|set]] of all [[Definition:Open Ball|open $\epsilon$-balls]] of $a$ for [[Definition:Rational Number|rational]] $\epsilon$.
+Then the $\BB_a$ is a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then by definition:
+:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
+where $\map {B_{\epsilon'} } a$ is the [[Definition:Open Ball|open $\epsilon'$-ball at $a$]].
+From [[Open Ball in Real Number Line is Open Interval]]:
+:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Between two Real Numbers exists Rational Number]]:
+:$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$
+and so:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Open Real Interval is Open Ball]]
+:$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
+is the [[Definition:Open Ball|open $\epsilon$-ball at $a$]].
+From [[Subset Relation is Transitive]]:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq N$
+From [[Open Ball is Neighborhood of all Points Inside]], $\openint {a - \epsilon} {a + \epsilon}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Hence there exists a [[Definition:Neighborhood (Metric Space)|neighborhood]] $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a [[Definition:Subset|subset]] of $N$.
+Hence the result by definition of [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Reciprocal-N Balls form Neighborhood Basis in Real Number Line}
+Tags: Real Intervals, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $a \in \R$ be a point in $\R$.
+Let $\BB_a$ be defined as:
+:$\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N} }$
+that is, the [[Definition:Set|set]] of all [[Definition:Open Ball|open $\epsilon$-balls]] of $a$ for $\epsilon$ which are [[Definition:Reciprocal|reciprocals]] of [[Definition:Integer|integers]].
+Then $\BB_a$ is a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then by definition:
+:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
+where $\map {B_{\epsilon'} } a$ is the [[Definition:Open Ball|open $\epsilon'$-ball at $a$]].
+From [[Open Ball in Real Number Line is Open Interval]]:
+:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Between two Real Numbers exists Rational Number]]:
+:$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$
+Let $\epsilon''$ be expressed in [[Definition:Canonical Form of Rational Number|canonical form]]:
+:$\epsilon'' = \dfrac p q$
+Let $\epsilon = \dfrac 1 q$
+Then $\epsilon \le \epsilon'' < \epsilon'$
+and so:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Open Real Interval is Open Ball]]
+:$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
+is the [[Definition:Open Ball|open $\epsilon$-ball at $a$]].
+From [[Subset Relation is Transitive]]:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq N$
+From [[Open Ball is Neighborhood of all Points Inside]], $\openint {a - \epsilon} {a + \epsilon}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Hence there exists a [[Definition:Neighborhood (Metric Space)|neighborhood]] $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a [[Definition:Subset|subset]] of $N$.
+Hence the result by definition of [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Open Reciprocal-N Balls forms Neighborhood Basis in Real Number Line}
+Tags: Real Intervals, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $a \in \R$ be a point in $\R$.
+Let $k \in \Z$ be some fixed [[Definition:Integer|integer]].
+Let $\BB_a$ be defined as:
+:$\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$
+that is, the [[Definition:Set|set]] of all [[Definition:Open Ball|open $\epsilon$-balls]] of $a$ for $\epsilon$ which are [[Definition:Reciprocal|reciprocals]] of [[Definition:Integer|integers]] greater than $k$.
+Then the $\BB_a$ is a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then by definition:
+:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
+where $\map {B_{\epsilon'} } a$ is the [[Definition:Open Ball|open $\epsilon'$-ball at $a$]].
+From [[Open Ball in Real Number Line is Open Interval]]:
+:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Between two Real Numbers exists Rational Number]]:
+:$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$
+Let $\epsilon''$ be expressed in [[Definition:Canonical Form of Rational Number|canonical form]]:
+:$\epsilon'' = \dfrac p q$
+Let $\epsilon''' = \dfrac 1 q$
+Then $\epsilon''' \le \epsilon'' < \epsilon'$
+If $q \le k$, let $\epsilon = \dfrac 1 {k + 1}$
+Otherwise, let $\epsilon = \epsilon'''$.
+Then:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
+From [[Open Real Interval is Open Ball]]
+:$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
+is the [[Definition:Open Ball|open $\epsilon$-ball at $a$]].
+By its method of construction:
+:$\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$
+From [[Subset Relation is Transitive]]:
+:$\openint {a - \epsilon} {a + \epsilon} \subseteq N$
+From [[Open Ball is Neighborhood of all Points Inside]], $\openint {a - \epsilon} {a + \epsilon}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Hence the result by definition of [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood Basis in Real Number Line is Infinite}
+Tags: Real Intervals, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Let $a \in \R$ be a point in $\R$.
+Let $\BB_a$ be a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+Then $\BB_a$ is an [[Definition:Infinite Set|infinite set]].
+\end{theorem}
+
+\begin{proof}
+Suppose $\BB_a$ be [[Definition:Finite|finite]].
+Let the [[Definition:Element|elements]] of $\BB_a$ be [[Definition:Finite Enumeration|enumerated]] as $N_1, N_2, \ldots, N_n$.
+For each $N_k \in \BB_a$, let $B_{\epsilon_k} \left({a}\right)$ be the [[Definition:Open Ball|open $\epsilon_k$-ball]] of $a$ for some $\epsilon_k \in \R_{>0}$.
+Let $\alpha = \min \set {\epsilon_k: k \in \set {1, 2, \ldots, n} }$.
+Consider the [[Definition:Open Real Interval|open interval]]:
+:$I = \openint {a - \beta} {a + \beta}$
+where $\beta = \dfrac \alpha 2$
+By the method of construction of $I$, it follows that $\map {B_{\epsilon_k} } a \nsubseteq I$ for any $k$.
+Thus $I$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ which does not have an element of $\BB_a$ as a [[Definition:Subset|subset]].
+So $\BB_a$ cannot be a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distinct Points in Metric Space have Disjoint Neighborhoods}
+Tags: Neighborhoods, Metric Spaces
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $x, y \in M: x \ne y$.
+Then there exist [[Definition:Neighborhood (Metric Space)|neighborhoods]] $N_x$ and $N_y$ of $x$ and $y$ respectively such that $N_x \cap N_y = \varnothing$, that is, that are [[Definition:Disjoint Sets|disjoint]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in A: x \ne y$.
+From [[Distinct Points in Metric Space have Disjoint Open Balls]], there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Ball|open $\epsilon$-balls]] $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ containing $x$ and $y$ respectively.
+From [[Open Ball is Neighborhood of all Points Inside]] it follows that $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ are [[Definition:Neighborhood (Metric Space)|neighborhoods]] of $x$ and $y$ respectively.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball in Cartesian Product under Chebyshev Distance}
+Tags: Chebyshev Distance, Open Balls
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \AA = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \AA \times \AA \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\AA$:
+:$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
+where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
+Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
+Let $\epsilon \in \R_{>0}$.
+Let $\map {B_\epsilon} {a; d_\infty}$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $M = \struct {\AA, d_\infty}$.
+Then:
+:$\displaystyle \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+Let $x = \tuple {x_1, x_2, \ldots, x_n} \in \AA$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \map {B_\epsilon} {a; d_\infty}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_\infty} {x, a}
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadsto
+      | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, a_i} }
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Chebyshev Distance}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall i \in \set {1, 2, \ldots, n}
+      | l = \map {d_i} {x_i, a_i}
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Maximum Element}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall i \in \set {1, 2, \ldots, n}
+      | l = x_i
+      | o = \in
+      | r = \map {B_\epsilon} {a_i; d_i}
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
+      | c = {{Defof|Finite Cartesian Product}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map {B_\epsilon} {a; d_\infty}
+      | o = \subseteq
+      | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
+      | c = {{Defof|Subset}}
+}}
+{{end-eqn}}
+And then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall i \in \set {1, 2, \ldots, n}
+      | l = x_i
+      | o = \in
+      | r = \map {B_\epsilon} {a_i; d_i}
+      | c = {{Defof|Finite Cartesian Product}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall i \in \set {1, 2, \ldots, n}
+      | l = \map {d_i} {x_i, a_i}
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadsto
+      | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, a_i} }
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Maximum Element}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_\infty} {x, a}
+      | o = <
+      | r = \epsilon
+      | c = {{Defof|Chebyshev Distance}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \map {B_\epsilon} {a; d_\infty}
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadsto
+      | l = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}
+      | o = \subseteq
+      | r = \map {B_\epsilon} {a; d_\infty}
+      | c = {{Defof|Subset}}
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Metric Space has Countable Neighborhood Basis}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $a \in A$.
+Then there exists a [[Definition:Basis for Neighborhood System|basis for the neighborhood system]] of $a$ which is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Countable Set|countable set]]:
+:$\mathcal B_a := \left\{ {B_\epsilon \left({a}\right): \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}\right\}$
+That is, let $\mathcal B_a$ be the [[Definition:Set|set]] of all [[Definition:Open Ball|open $\epsilon$-balls]] of $a$ such that $\epsilon$ is of the form $\dfrac 1 n$ for [[Definition:Strictly Positive Integer|(strictly) positive integral]] $n$.
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+Then by definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+:$\exists \epsilon' \in \R_{>0}: B_{\epsilon'} \left({a}\right) \subseteq N$
+From [[Between two Real Numbers exists Rational Number]]:
+:$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$
+Let $\epsilon''$ be expressed in [[Definition:Canonical Form of Rational Number|canonical form]] as:
+:$\epsilon'' = \dfrac p q$
+where $p$ and $q$ are [[Definition:Coprime Integers|coprime integers]] and $q > 0$.
+Then:
+:$\epsilon := \dfrac 1 q \le \dfrac p q$
+Thus $0 < \epsilon < \epsilon'$.
+So:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = B_\epsilon \left({a}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d \left({x, a}\right)
+      | o = < 
+      | r = \epsilon
+      | c = Definition of [[Definition:Open Ball|Open $\epsilon$-Ball]]
+}}
+{{eqn | ll= \implies
+      | l = d \left({x, a}\right)
+      | o = < 
+      | r = \epsilon'
+      | c = as $\epsilon < \epsilon'$
+}}
+{{eqn | ll= \implies
+      | l = x
+      | o = \in 
+      | r = B_{\epsilon'} \left({a}\right)
+      | c = Definition of [[Definition:Open Ball|Open $\epsilon$-Ball]]
+}}
+{{eqn | ll= \implies
+      | l = B_\epsilon \left({a}\right)
+      | o = \subseteq 
+      | r = B_{\epsilon'} \left({a}\right)
+      | c = Definition of [[Definition:Subset|Subset]]
+}}
+{{eqn | ll= \implies
+      | l = B_\epsilon \left({a}\right)
+      | o = \subseteq 
+      | r = N
+      | c = [[Subset Relation is Transitive]]
+}}
+{{end-eqn}}
+From [[Open Ball is Neighborhood of all Points Inside]], $B_\epsilon \left({a}\right)$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+But as $\epsilon = \dfrac 1 q$ where $q \in \Z_{>0}$ it follows that:
+: $B_\epsilon \left({a}\right) \in \left\{ {B_\epsilon \left({a}\right): \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}\right\}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood Basis in Cartesian Product under Chebyshev Distance}
+Tags: Chebyshev Distance, Neighborhoods
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_n = \left({A_n, d_n}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+: $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.
+Let $a = \left({a_1, a_2, \ldots, a_n}\right) \in \mathcal A$.
+For all $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $\mathcal B_{a_i}$ be a [[Definition:Basis for Neighborhood System|basis for the system of neighborhoods]] for $a_i$ in $M_i$.
+Let $\mathcal B_a$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $\mathcal B_{a_1}, \mathcal B_{a_2}, \ldots, \mathcal B_{a_n}$:
+:$\displaystyle \mathcal B_a = \prod_{i \mathop = 1}^n \mathcal B_{a_i}$
+Then $\mathcal B_a$ is a [[Definition:Basis for Neighborhood System|basis for the system of neighborhoods]] for $a$ in $M$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ in $M$.
+Then by definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+:$\exists \epsilon \in \R_{>0}: B_\epsilon \left({a, d_\infty}\right) \subseteq N$
+where $B_\epsilon \left({a}\right)$ is the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ under $d_\infty$.
+From [[Open Ball in Cartesian Product under Chebyshev Distance]]:
+:$\displaystyle B_\epsilon \left({a; d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right)$
+From [[Open Ball is Neighborhood of all Points Inside]], each $B_\epsilon \left({a_i; d_i}\right)$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a_i$.
+By definition of [[Definition:Basis for Neighborhood System|basis for system of neighborhoods]]:
+:$\exists N_i \in \mathcal B_{a_i}: N_i \subseteq B_\epsilon \left({a_i; d_i}\right)$
+From [[Cartesian Product of Subsets]]:
+:$\displaystyle \prod_{i \mathop = 1}^n N_i \subseteq \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right) \subseteq N$
+But $\displaystyle \prod_{i \mathop = 1}^n N_i \in \mathcal B_a = \prod_{i \mathop = 1}^n \mathcal B_{a_i}$
+Hence the result by definition of [[Definition:Basis for Neighborhood System|basis for system of neighborhoods]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projection from Cartesian Product under Chebyshev Distance is Continuous}
+Tags: Chebyshev Distance, Projections, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \AA = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \AA \times \AA \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\AA$:
+:$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
+where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
+For all $i \in \set {1, 2, \ldots, n}$, let $\pr_i: \AA \to A_i$ be the [[Definition:Projection (Mapping Theory)|$i$th projection]] on $\AA$:
+:$\forall a \in \AA: \map {\pr_i} a = a_i$
+where $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
+Then for all $i \in \set {1, 2, \ldots, n}$, $p_i$ is [[Definition:Continuous on Metric Space|continuous on $\AA$]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
+Let $\map {B_\epsilon} {a_i; d_i}$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $a_i$ in $M_i$.
+From [[Open Ball in Cartesian Product under Chebyshev Distance]]:
+:$\displaystyle \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$
+By definition of [[Definition:Projection (Mapping Theory)|$i$th projection]]:
+:$\forall x \in \map {B_\epsilon} {a; d_\infty}: \map {\pr_i} x \in \map {B_\epsilon} {a_i; d_i}$
+Thus by definition of [[Definition:Image of Subset under Mapping|image of subset of $\AA$ under $\pr_i$]]:
+:$\pr_i \sqbrk {\map {B_\epsilon} {a; d_\infty} } \subseteq  \map {B_\epsilon} {a_i; d_i}$
+Thus by definition of [[Definition:Continuous at Point of Metric Space|continuity at a point]], $\pr_i$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $a$.
+As $a$ is arbitrary, it follows that $\pr_i$ is [[Definition:Continuous at Point of Metric Space|continuous]] at all points of $\AA$.
+Hence, by definition, $\pr_i$ is [[Definition:Continuous on Metric Space|continuous from $\AA$ to $A_i$]].
+As $i$ is arbitrary, it follows that the result holds for all $i \in \set {1, 2, \ldots, n}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Mapping to Cartesian Product under Chebyshev Distance}
+Tags: Chebyshev Distance, Projections, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_n = \left({A_n, d_n}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+: $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.
+For all $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $\operatorname{pr}_i: \mathcal A \to A_i$ be the [[Definition:Projection (Mapping Theory)|$i$th projection]] on $\mathcal A$:
+:$\forall a \in \mathcal A: \operatorname{pr}_i \left({a}\right) = a_i$
+where $a = \left({a_1, a_2, \ldots, a_n}\right) \in \mathcal A$.
+Let $M' = \left({X, d'}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $f: X \to \mathcal A$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Continuous on Metric Space|continuous on $X$]] {{iff}} each of $\operatorname{pr}_i \circ f: X \to A_i$ is [[Definition:Continuous on Metric Space|continuous on $X$]].
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, let $i \in \left\{ {1, 2, \ldots, n}\right\}$ be arbitrary.
+=== Necessary Condition ===
+Let $f$ be [[Definition:Continuous Mapping (Topology)|continuous]].
+From [[Projection from Cartesian Product under Chebyshev Distance is Continuous]], $\operatorname{pr}_i: \mathcal A \to A_i$ is [[Definition:Continuous on Metric Space|continuous on $\mathcal A$]].
+From [[Composite of Continuous Mappings between Metric Spaces is Continuous]] it follows that $\operatorname{pr}_1 \circ f$ is [[Definition:Continuous on Metric Space|continuous on $\mathcal A$]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\operatorname{pr}_i \circ f$ be [[Definition:Continuous on Metric Space|continuous on $X$]].
+Let $a \in X$.
+By definition of [[Definition:Continuous at Point of Metric Space|continuity at a point]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \operatorname{pr}_i \circ f \left[{B_\delta \left({a; d'}\right)}\right] \subseteq B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$
+where $B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $\operatorname{pr}_i \circ f \left({a}\right)$ with respect to the [[Definition:Metric|metric]] $d_i$, and similarly for $B_\delta \left({a; d'}\right)$.
+From [[Projection from Cartesian Product under Chebyshev Distance is Continuous]], we have that:
+:$\forall \epsilon \in \R_{>0}: \operatorname{pr}_i \left[{B_\epsilon \left({f \left({a}\right); d_\infty}\right)}\right] \subseteq B_\epsilon \left({\operatorname{pr}_i \left({f \left({a}\right)}\right); d_i}\right)$
+From [[Open Ball in Cartesian Product under Chebyshev Distance]]:
+:$\displaystyle B_\epsilon \left({f \left({a}\right); d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$
+From [[Metric Space Continuity by Inverse of Mapping between Open Balls]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: B_\delta \left({a; d'}\right) \subseteq \left({\operatorname{pr}_i \circ f}\right)^{-1} \left[{B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)}\right]$
+where $B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $\operatorname{pr}_i \circ f \left({a}\right)$ with respect to the [[Definition:Metric|metric]] $d_i$, and similarly for $B_\delta \left({a; d'}\right)$.
+{{finish|not going anywhere, need to rethink}}
+\end{proof}<|endoftext|>
+\section{Positive Image of Point of Continuous Real Function implies Positive Closed Interval of Domain}
+Tags: Continuous Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Continuous Real Function|continuous real function]].
+Let $a \in \R$ such that $f \left({a}\right) > 0$.
+Then:
+:$\exists k \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \left[{a - \delta \,.\,.\, a + \delta}\right]: f \left({x}\right) \ge k$
+\end{theorem}
+
+\begin{proof}
+Let $f \left({a}\right) = l$ where $l > 0$.
+As $f$ is [[Definition:Continuous Real Function|continuous]]:
+:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left\vert{y - x}\right\vert < \delta \implies \left\vert{f \left({y}\right) - f \left({x}\right)}\right\vert < \epsilon$
+Let $\epsilon = \dfrac l 2 = k$.
+Then:
+:$\exists \delta' \in \R_{>0}: \forall y \in \R: \left\vert{y - x}\right\vert < \delta' \implies \left\vert{f \left({y}\right) - f \left({a}\right)}\right\vert < \dfrac l 2$
+Thus:
+:$\forall x \in \left({a - \delta' \,.\,.\, a + \delta'}\right): f \left({x}\right) > \dfrac l 2$
+Let $\delta = \dfrac {\delta'} 2$
+Then $a - \delta \in \left({a - \delta' \,.\,.\, a + \delta'}\right)$ and $a + \delta \in \left({a - \delta' \,.\,.\, a + \delta'}\right)$.
+Thus:
+:$\left[{a - \delta \,.\,.\, a + \delta}\right] \subseteq \left({a - \delta' \,.\,.\, a + \delta'}\right)$
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Sequence in Metric Space in Neighborhood}
+Tags: Metric Spaces, Limits of Sequences, Neighborhoods
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\left\langle{a_n}\right\rangle$ be a [[Definition:Sequence|sequence in $A$]].
+Then $\displaystyle \lim_{n \mathop \to \infty} a_n = a$ [[Definition:Iff|iff]] for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $V$ of $a$:
+:$\exists N \in \N: n > N \implies a_n \in V$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\displaystyle \lim_{n \mathop \to \infty} a_n = a$.
+Let $V$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+By definition of [[Definition:Neighborhood (Metric Space)|neighborhood]]:
+:$\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq V$
+where $B_\epsilon \left({a}\right)$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $M$.
+By definition of [[Definition:Limit of Sequence (Metric Space)|limit]]:
+:$\exists N \in \N: n > N \implies d \left({a, a_n}\right) < \epsilon$
+Hence $a_n \in V$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\left\langle{a_n}\right\rangle$ be such that for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $V$ of $a$:
+:$\exists N \in \N: n > N \implies a_n \in V$
+Let $\epsilon \in \R_{>0}$.
+Then by [[Open Ball is Neighborhood of all Points Inside]], $B_\epsilon \left({a}\right)$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+Let $N \in \N$ be such that:
+:$\forall n > N: a_n \in B_\epsilon \left({a}\right)$
+Then:
+: $d \left({a, a_n}\right) < \epsilon$
+and so by definition of [[Definition:Limit of Sequence (Metric Space)|limit]]:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = a$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square of Real Number is Non-Negative}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $x \in \R$.
+Then:
+: $0 \le x^2$
+\end{theorem}
+
+\begin{proof}
+There are two cases to consider:
+: $(1): \quad x = 0$
+: $(2): \quad x \ne 0$
+Let $x = 0$.
+Then:
+:$x^2 = 0$
+and thus
+:$0 \le x^2$
+{{qed|lemma}}
+Let $x \ne 0$.
+From [[Square of Non-Zero Real Number is Strictly Positive]] it follows that:
+:$0 < x^2$
+and so by definition:
+:$0 \le x^2$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Real Number between Zero and One is Greater than Square}
+Tags: Real Analysis, Inequalities, Real Number between Zero and One is Greater than Square
+
+\begin{theorem}
+Let $x \in \R$.
+Let $0 < x < 1$.
+Then:
+: $0 < x^2 < x$
+\end{theorem}
+
+\begin{proof}
+We are given that $0 < x < 1$.
+By direct application of [[Real Number Ordering is Compatible with Multiplication]], it follows that:
+: $0 \times x < x \times x < 1 \times x$
+and the result follows.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+We have that [[Real Numbers form Ordered Integral Domain]].
+Thus [[Square of Element Less than Unity in Ordered Integral Domain]] applies directly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Greater than One is Less than Square}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $x \in \R$.
+Let $x > 1$.
+Then:
+: $x^2 > x$
+\end{theorem}
+
+\begin{proof}
+As $x > 1$ it follows that $x > 0$.
+Thus by [[Real Number Ordering is Compatible with Multiplication]]:
+: $x \times x > 1 \times x$
+and the result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Power of Real Number greater than One is Unbounded Above}
+Tags: Powers
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 1$.
+Let [[Definition:Set|set]] $S = \set {x^n: n \in \N}$.
+Then $S$ is [[Definition:Unbounded Above Set|unbounded above]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $S$ were [[Definition:Bounded Above Set|bounded above]].
+Then $S$ has a [[Definition:Supremum of Set|supremum]] $B$.
+As $x > 1$, it follows that $\dfrac B x < B$ and so therefore $\dfrac B x$ can not be an [[Definition:Upper Bound of Set|upper bound]].
+Therefore:
+:$\exists n \in \N: x^n > \dfrac B x \implies x^{n + 1} > B$
+So $B$ can not be an [[Definition:Upper Bound of Set|upper bound]].
+From that [[Definition:Contradiction|contradiction]] it can be concluded that $S$ can not have an [[Definition:Upper Bound of Set|upper bound]].
+So by [[Proof by Contradiction]] it follows that $S$ is [[Definition:Bounded Above Set|unbounded above]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Real Number between Zero and One is Bounded}
+Tags: Powers
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $0 < x < 1$.
+Let [[Definition:Set|set]] $S = \set {x^n: n \in \N}$.
+Then:
+:$\inf S = 0$
+and:
+:$\sup S = 1$
+where $\inf S$ and $\sup S$ are the [[Definition:Infimum of Set|infimum]] and [[Definition:Supremum of Set|supremum]] of $S$ respectively.
+\end{theorem}
+
+\begin{proof}
+When $n = 0$ it follows that $x^n = 1$ and so $\sup S \ge 1$.
+Let $x^k \in S$.
+As $x < 1$, it follows from [[Real Number Ordering is Compatible with Multiplication]] that:
+:$x^{k + 1} < x^k$
+So:
+:$\forall x \in S: x \le 1$
+Hence it follows that $\sup S = 1$.
+As $x > 0$, it follows by [[Real Number Ordering is Compatible with Multiplication]] that:
+:$\forall x \in S: x \ge 0$
+Therefore $x$ is a [[Definition:Lower Bound of Set|lower bound]] of $S$.
+Now suppose $h > 0$ is also a [[Definition:Lower Bound of Set|lower bound]] of $S$.
+Then:
+:$\forall n \in \N: x^n \ge h$
+and:
+:$\forall n \in \N: \paren {\dfrac 1 x}^n \le \dfrac 1 h$
+But as $0 < x < 1$ it follows that:
+:$\dfrac 1 x > 1$
+Thus $\dfrac 1 h$ is an [[Definition:Upper Bound of Set|upper bound]] for $\set {\paren {\dfrac 1 x}^n: n \in \N}$.
+From [[Power of Real Number greater than One is Unbounded Above]], this has been shown to be [[Definition:Unbounded Above Set|unbounded above]].
+Therefore there can be no such [[Definition:Lower Bound of Set|lower bound]] $h > 0$ of $S$.
+Hence $\inf S = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Function by Convergent Sequences/Real Number Line}
+Tags: Real Analysis, Limits of Sequences
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Open Real Interval|open interval]] $\openint a b$, except possibly at the point $c \in \openint a b$.
+Then $\displaystyle \lim_{x \mathop \to c} \map f x = l$ {{iff}}:
+:for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \mathop \infty} x_n = c$
+it is true that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\displaystyle \lim_{x \mathop \to c} \map f x = l$.
+Let $\epsilon > 0$.
+Then by the definition of the [[Definition:Limit of Real Function|limit of a function]]:
+:$\exists \delta > 0: \size {\map f x - l} < \epsilon$
+provided $0 < \size {x - c} < \delta$.
+Now suppose that $\sequence {x_n}$ is a [[Definition:Real Sequence|sequence]] of points of $\openint a b$ such that:
+:$\forall n \in \N_{>0}: x_n \ne c$
+and:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = c$
+Since $\delta > 0$, from the definition of the [[Definition:Limit of Real Function|limit of a function]]:
+:$\exists N: \forall n > N: \size {x_n - c} < \delta$
+But:
+:$\forall n \in \N_{>0}: x_n \ne c$
+That means:
+:$0 < \size {x_n - c} < \delta$
+But that implies:
+:$\size {\map f {x_n} - l} < \epsilon$
+That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
+:$\forall n > N: \size {\map f {x_n} - l} < \epsilon$
+Thus:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = c$, it is true that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
+{{AimForCont}} it is ''not'' true that:
+:$\displaystyle \lim_{x \mathop \to c} \map f x = l$
+Thus:
+:$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < \size {x - c} < \delta: \size {\map f {x_n} - l} \ge \epsilon$
+In particular, if $\delta = \dfrac 1 n$, we can find an $x_n$ where $0 < \size {x - c} < \dfrac 1 n$ such that:
+:$\size {\map f {x_n} - l} \ge \epsilon$
+But then $\sequence {x_n}$ is a [[Definition:Real Sequence|sequence]] of points of $\openint a b$ such that:
+:$\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = c$
+but for which it is ''not'' true that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Mapping between Metric Spaces by Convergent Sequence}
+Tags: Continuous Mappings on Metric Spaces, Limits of Sequences
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $a \in X$ [[Definition:Iff|iff]]:
+: whenever $\displaystyle \lim_{n \mathop \to \infty} x_n = a$ for a [[Definition:Sequence|sequence]] $\left \langle {x_n} \right \rangle$ of points of $A_1$  
+it is true that:
+: $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Continuous at Point of Metric Space|continuous]] at $a \in A_1$.
+Let $\displaystyle \lim_{n \mathop \to \infty} x_n = a$.
+Let $V$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $f \left({a}\right)$.
+Then by [[Metric Space Continuity by Inverse of Mapping between Neighborhoods]] $f^{-1} \left[{V}\right]$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+By [[Limit of Sequence in Metric Space in Neighborhood]]:
+:$\exists N \in \N: n > N \implies x_n \in f^{-1} \left[{V}\right]$
+Thus for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $V$ of $f \left({a}\right)$
+:$\exists N \in \N: n > N \implies f \left({x_n}\right) \in V$
+By [[Limit of Sequence in Metric Space in Neighborhood]]:
+: $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Aiming for a [[Proof by Contradiction|contradiction]], suppose $f$ is not [[Definition:Continuous at Point of Metric Space|continuous]] at $a \in A_1$.
+Then there is a [[Definition:Neighborhood (Metric Space)|neighborhood]] $V$ of $f \left({a}\right)$ such that for each [[Definition:Neighborhood (Metric Space)|neighborhood]] $U$ of $a$:
+:$f \left[{U}\right] \nsubseteq V$
+In particular, for each [[Definition:Open Ball|open $\epsilon$-ball]] $B_\epsilon \left({a}\right)$ such that $\forall n \in \N_{>0}: \epsilon = \dfrac 1 n$:
+:$f \left({B_\epsilon \left({a}\right)}\right) \nsubseteq V$
+Thus for each $n \in \N_{>0}$ there exists a point $x_n$ such that:
+:$x_n \subseteq B_\epsilon \left({a}\right)$
+and:
+:$f \left({x_n}\right) \notin V$
+Now:
+:$d_1 \left({a, x_n}\right) < \dfrac 1 n$
+and therefore:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = a$
+whereas $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$ is impossible, as:
+:$\forall n \in \N_{>0}: f \left({x_n}\right) \notin V$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Image of Sequence/Real Number Line}
+Tags: Real Analysis, Limits of Sequences
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Real Interval|interval]] $\Bbb I$.
+Let $\left \langle {x_n} \right \rangle$ be a [[Definition:Real Sequence|sequence]] of points in $\Bbb I$ such that:
+: $\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$
+where:
+: $(1): \quad \xi \in \Bbb I$
+: $(2): \quad \displaystyle \lim_{n \mathop \to \infty} x_n$ denotes the [[Definition:Limit of Real Sequence|limit of $x_n$]].
+Then:
+: $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({\xi}\right)$
+That is:
+: $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({\lim_{n \mathop \to \infty} x_n}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Limit of Function by Convergent Sequences/Real Number Line|Limit of Function by Convergent Sequences]], we have:
+:$\displaystyle \lim_{x \mathop \to \xi} f \left({x}\right) = l$
+{{iff}}:
+: for each [[Definition:Real Sequence|sequence]] $\left \langle {x_n} \right \rangle$ of points of $\left({a \,.\,.\, b}\right)$ such that:
+:: $\forall n \in \N_{>0}: x_n \ne \xi$
+:and:
+:: $\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$
+:it is true that:
+:: $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = l$
+The result follows directly from this and the definition of [[Definition:Continuous on Interval|continuity on an interval]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Subset of Real Numbers is Arbitrarily Close}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $b$ be an [[Definition:Infimum of Subset of Real Numbers|infimum]] of $A$.
+Let $\epsilon \in \R_{>0}$.
+Then:
+:$\exists x \in A: x - b < \epsilon$
+\end{theorem}
+
+\begin{proof}
+Note that $A$ is [[Definition:Non-Empty Set|non-empty]] as the [[Definition:Empty Set|empty set]] does not admit an [[Definition:Infimum of Subset of Real Numbers|infimum]] (in $\R$).
+Suppose $\epsilon \in \R_{>0}$ such that:
+:$\forall x \in A: x - b \ge \epsilon$
+Then:
+:$\forall x \in A: b + \epsilon \le x$
+and so $b + \epsilon$ would be a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] of $A$ which is greater than $b$.
+But since $b$ is an [[Definition:Infimum of Subset of Real Numbers|infimum]] of $A$ there can be no such $b + \epsilon$.
+From that [[Proof by Contradiction|contradiction]] it follows that:
+:$\exists x \in A: x - b < \epsilon$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Sequence in Set of Real Numbers whose Limit is Infimum}
+Tags: Real Analysis, Limits of Sequences
+
+\begin{theorem}
+Let $A \subseteq \R$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $b$ be an [[Definition:Infimum of Subset of Real Numbers|infimum]] of $A$.
+Then there exists a [[Definition:Real Sequence|sequence]] $\sequence {a_n}$ in $\R$ such that:
+:$(1): \quad \forall n \in \N: a_n \in A$
+:$(2): \quad \displaystyle \lim_{n \mathop \to \infty} a_n = b$
+\end{theorem}
+
+\begin{proof}
+From [[Infimum of Subset of Real Numbers is Arbitrarily Close]]:
+For $\epsilon = \dfrac 1 n$ there exists an $a_n \in A$ such that:
+:$a_n - b < \dfrac 1 n$
+Since $b$ is an [[Definition:Infimum of Subset of Real Numbers|infimum]] of $A$:
+:$0 \le a_n - b$
+Therefore:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Sequence in Subset of Metric Space whose Limit is Infimum}
+Tags: Metric Spaces, Limits of Sequences
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $a \in A$.
+Let $S \subseteq A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $A$.
+Then there exists a [[Definition:Sequence|sequence]] $\sequence {a_n}$ of points of $S$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map d {a, a_n} = \map d {a, S}$
+\end{theorem}
+
+\begin{proof}
+From [[Existence of Sequence in Set of Real Numbers whose Limit is Infimum]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \map d {a, a_n} = b$
+where $b$ is an [[Definition:Infimum of Subset of Real Numbers|infimum]] of $\map d {a, a_n}$.
+Hence the result by definition of [[Definition:Distance between Element and Subset of Metric Space|distance to a subset of a metric space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Sequence in Product of Metric Spaces under Chebyshev Distance}
+Tags: Chebyshev Distance, Limits of Sequences
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_k = \struct {A_n, d_k}$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \AA = \prod_{i \mathop = 1}^k A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_k$.
+Let $d_\infty: \AA \times \AA \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\AA$:
+:$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^k \set {\map {d_i} {x_i, y_i} }$
+where $x = \tuple {x_1, x_2, \ldots, x_k}, y = \tuple {y_1, y_2, \ldots, y_k} \in \AA$.
+Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] of points of $\AA$:
+:$a_n = \tuple {a_{n 1}, a_{n 2}, \ldots, a_{n k} } \in \AA$
+Let $c = \tuple {c_1, c_2, \ldots, c_k} \in \AA$.
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = c \iff \forall i \in \set {1, 2, \ldots, k}: \lim_{n \mathop \to \infty} a_{n i} = c_i$
+where $\displaystyle \lim_{n \mathop \to \infty}$ denotes a [[Definition:Limit of Sequence (Metric Space)|limit]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} a_n
+      | r = c
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{n \mathop \to \infty} \map {d_\infty} {a_n, c}
+      | r = 0
+      | c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \lim_{n \mathop \to \infty} \max_{i \mathop = 1}^k \set {\map {d_i} {a_{n i}, c_i} }
+      | r = 0
+      | c = {{Defof|Chebyshev Distance}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \max_{i \mathop = 1}^k \set {\lim_{n \mathop \to \infty} \map {d_i} {a_{n i}, c_i} }
+      | r = 0
+      | c = [[Limit of Image of Sequence]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall i \in \set {1, 2, \ldots, k}:
+      | l = \lim_{n \mathop \to \infty} \map {d_i} {a_{n i}, c_i}
+      | r = 0
+      | c = {{Defof|Max Function}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Recurrence Formula for Sum of Sequence of Fibonacci Numbers}
+Tags: Fibonacci Numbers
+
+\begin{theorem}
+Let $g_n$ be the sum of the first $n$ [[Definition:Fibonacci Numbers|Fibonacci numbers]].
+Then:
+:$\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$
+\end{theorem}
+
+\begin{proof}
+Let $F_n$ be the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]].
+By definition:
+:$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$
+Hence, $g_n$, the sum of the first $n$ [[Definition:Fibonacci Numbers|Fibonacci numbers]] is
+:$g_0 = 0, g_1 = 1, g_2 = 2, g_3 = 4, g_4 = 7, \ldots$
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_{j \mathop = 0}^n F_j = g_n = g_{n - 1} + g_{n - 2} + 1$
+=== Basis for the Induction ===
+$P(2)$ is the case $g_2 = g_1 + g_0 + 1$, which holds:
+:$2 = 1 + 0 + 1$
+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]]:
+:$\displaystyle \sum_{j \mathop = 0}^k F_j = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1$
+Then we need to show:
+:$\displaystyle \sum_{j \mathop = 0}^{k + 1} F_j = \sum_{j \mathop = 0}^k F_j + \sum_{j \mathop = 0}^{k - 1} F_j + 1$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = g_{k + 1}
+      | r = \sum_{j \mathop = 0}^{k + 1} F_j
+      | c = Definition of $g_k$
+}}
+{{eqn | r = \sum_{j \mathop = 0}^k F_j + F_{k + 1}
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1 + F_{k + 1}
+      | c = [[Recurrence Formula for Sum of Sequence of Fibonacci Numbers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1 + (F_k + F_{k - 1})
+      | c = {{Defof|Fibonacci Numbers}}: $F_{k + 1} = F_k + F_{k - 1}$
+}}
+{{eqn | r = \sum_{j \mathop = 0}^k F_j + \sum_{j \mathop = 0}^{k - 1} F_j + 1
+      | c = pulling terms into summations
+}}
+{{eqn | r = g_k + g_{k - 1} + 1
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$
+where $g_n$ is the sum of the first $n$ [[Definition:Fibonacci Numbers|Fibonacci numbers]].
+{{qed}}
+[[Category:Fibonacci Numbers]]
+dku20szhkbm8map6jy77dy7ntvtzh76
+\end{proof}<|endoftext|>
+\section{Limit of Subsequence equals Limit of Sequence/Metric Space}
+Tags: Metric Spaces, Convergence, Limits of Sequences
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\left \langle {x_n} \right \rangle$ be a [[Definition:Sequence|sequence in $A$]].
+Let $l \in A$ such that:
+: $\displaystyle \lim_{n \mathop \to \infty} x_n = l$
+Let $\left \langle {x_{n_r}} \right \rangle$ be a [[Definition:Subsequence|subsequence]] of $\left \langle {x_n} \right \rangle$.
+Then:
+: $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+Since $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of [[Definition:Limit of Sequence (Metric Space)|limit]] that:
+:$\exists N: \forall n > N: d \left({x_n, l}\right) < \epsilon$
+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:
+:$d \left({x_n, l}\right) < \epsilon$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Real Sequence is Bounded}
+Tags: Real Analysis, Limits of Sequences, Convergent Real Sequence is Bounded
+
+\begin{theorem}
+Let $\left \langle {x_n} \right \rangle$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $l \in A$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.
+Then $\left \langle {x_n} \right \rangle$ is [[Definition:Bounded Real Sequence|bounded]].
+\end{theorem}
+
+\begin{proof}
+From [[Real Number Line is Metric Space]], the set $\R$ under the [[Definition:Usual Metric|usual metric]] is a [[Definition:Metric Space|metric space]].
+By [[Convergent Sequence in Metric Space is Bounded]] it follows that:
+:$\exists M > 0: \forall n, m \in \N: \size {x_n - x_m} \le M$
+Then for $n \in \N$, by the [[Triangle Inequality for Real Numbers]]:
+{{begin-eqn}}
+{{eqn | l = \size {x_n}
+      | r = \size {x_n - x_1 + x_1}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \size {x_n - x_1} + \size {x_1}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = M + \size {x_1}
+      | c = 
+}}
+{{end-eqn}}
+Hence $\sequence {x_n}$ is [[Definition:Bounded Real Sequence|bounded]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $x_n \to l$ as $n \to \infty$.
+To show that $\sequence {x_n}$ is [[Definition:Bounded Sequence|bounded sequence]], we need to find $K$ such that:
+:$\forall n \in \N: \size {x_n} \le K$
+Because $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]]:
+:$\forall \epsilon > 0: \exists N: n > N \implies \size {x_n - l} < \epsilon$
+In particular, this is true when $\epsilon = 1$.
+That is:
+:$\exists N_1: \forall n > N_1: \size {x_n - l} < 1$
+By [[Backwards Form of Triangle Inequality]]:
+:$\forall n > N_1: \size {x_n} - \size l \le \size {x_n - l} < 1$
+That is:
+:$\size {x_n} < \size l + 1$
+So we set:
+:$K = \max \set {\size {x_1}, \size {x_2}, \ldots, \size {x_{N_1} }, \size l + 1}$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monotone Convergence Theorem (Real Analysis)/Decreasing Sequence}
+Tags: Monotone Convergence Theorem (Real Analysis)
+
+\begin{theorem}
+Let $\sequence {x_n}$ be [[Definition:Decreasing Real Sequence|decreasing]] and [[Definition:Bounded Below Real Sequence|bounded below]].
+Then $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to its [[Definition:Infimum of Sequence|infimum]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ be [[Definition:Decreasing Real Sequence|decreasing]] and [[Definition:Bounded Below Real Sequence|bounded below]].
+Then $\sequence {-x_n}$ is [[Definition:Increasing Real Sequence|increasing]] and [[Definition:Bounded Above Real Sequence|bounded above]].
+Thus the [[Monotone Convergence Theorem (Real Analysis)/Increasing Sequence|Monotone Convergence Theorem for Increasing Sequence]] applies and the proof follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Monotone Convergence Theorem (Real Analysis)/Increasing Sequence}
+Tags: Monotone Convergence Theorem (Real Analysis)
+
+\begin{theorem}
+Let $\sequence {x_n}$ be [[Definition:Increasing Real Sequence|increasing]] and [[Definition:Bounded Above Real Sequence|bounded above]].
+Then $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to its [[Definition:Supremum of Sequence|supremum]].
+\end{theorem}
+
+\begin{proof}
+Suppose $\sequence {x_n}$ is [[Definition:Increasing Real Sequence|increasing]] and [[Definition:Bounded Above Real Sequence|bounded above]].
+By the [[Continuum Property]], it has a [[Definition:Supremum of Sequence|supremum]], $B$.
+We need to show that $x_n \to B$ as $n \to \infty$.
+Let $\epsilon > 0$.
+By the definition of [[Definition:Supremum of Sequence|supremum]], $B - \epsilon$ is not an [[Definition:Upper Bound of Sequence|upper bound]].
+Thus:
+:$\exists x_N: x_N > B - \epsilon$
+But $\sequence {x_n}$ is [[Definition:Increasing Sequence|increasing]].
+Hence:
+:$\forall n > N: x_n \ge x_N > B - \epsilon$
+But $B$ ''is'' still an [[Definition:Upper Bound of Sequence|upper bound]] for $\sequence {x_n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \forall n > N: B - \epsilon
+      | o = <
+      | r = x_n \le B
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \forall n > N: B - \epsilon
+      | o = <
+      | r = x_n < B + \epsilon
+      | c = [[Real Plus Epsilon]]
+}}
+{{eqn | ll= \leadsto
+      | l = \forall n > N: \size {x_n - B}
+      | o = <
+      | r = \epsilon
+      | c = [[Negative of Absolute Value/Corollary 1|Negative of Absolute Value: Corollary 1]]
+}}
+{{end-eqn}}
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Triangle Inequality for Distance from Element to Set}
+Tags: Triangle Inequality, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $S \subseteq A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $A$.
+Let $x, y \in A$.
+Then:
+:$\map d {x, S} \le \map d {x, y} + \map d {y, S}$
+where $\map d {x, S}$ is the [[Definition:Distance between Element and Subset of Metric Space|distance from $x$ to $S$]].
+\end{theorem}<|endoftext|>
+\section{Open Ball in Euclidean Plane is Interior of Circle}
+Tags: Open Balls, Real Number Plane with Euclidean Metric
+
+\begin{theorem}
+Let $\R^2$ be the [[Definition:Real Number Plane with Euclidean Metric|real number plane with the usual (Euclidean) metric]].
+Let $x = \tuple {x_1, x_2} \in \R^2$ be a point in $\R^2$.
+Let $\map {B_\epsilon} x$ be the [[Definition:Open Ball|open $\epsilon$-ball at $x$]].
+Then $\map {B_\epsilon} x$ is the [[Definition:Interior of Region|interior]] of the [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $x$ and whose [[Definition:Radius of Circle|radius]] is $\epsilon$.
+\end{theorem}
+
+\begin{proof}
+Let $S = \map {B_\epsilon} x$ be an [[Definition:Open Ball|open $\epsilon$-ball at $x$]].
+Let $y = \tuple {y_1, y_2} \in \map {B_\epsilon} x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | m = \map {B_\epsilon} x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map d {y, x}
+      | o = <
+      | m = \epsilon
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2}
+      | o = <
+      | m = \epsilon
+      | c = {{Defof|Real Number Plane with Euclidean Metric}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2
+      | o = <
+      | m = \epsilon^2
+}}
+{{end-eqn}}
+But from [[Equation of Circle]]:
+:$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 = \epsilon^2$
+is the equation of a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $\tuple {x_1, x_2}$ and whose [[Definition:Radius of Circle|radius]] is $\epsilon$.
+The result follows by definition of [[Definition:Interior of Region|interior]] and [[Open Ball of Point Inside Open Ball]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball in Euclidean 3-Space is Interior of Sphere}
+Tags: Open Balls, Euclidean Space
+
+\begin{theorem}
+Let $\R^3$ be the [[Definition:Real Euclidean Space|real Euclidean $3$-space]] considered as a [[Definition:Metric Space|metric space]] under the [[Definition:Usual Metric|usual metric]].
+Let $x = \tuple {x_1, x_2, x_3} \in \R^3$ be a point in $\R^3$.
+Let $\map {B_\epsilon} x$ be the [[Definition:Open Ball|open $\epsilon$-ball at $x$]].
+Then $\map {B_\epsilon} x$ is the [[Definition:Interior of Region|interior]] of the [[Definition:Sphere (Geometry)|sphere]] whose [[Definition:Center of Sphere|center]] is $x$ and whose [[Definition:Radius of Sphere|radius]] is $\epsilon$.
+\end{theorem}
+
+\begin{proof}
+Let $S = \map {B_\epsilon} x$ be an [[Definition:Open Ball|open $\epsilon$-ball at $x$]].
+Let $y = \tuple {y_1, y_2, y_3} \in \map {B_\epsilon} x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | m = \map {B_\epsilon} x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map d {y, x}
+      | o = <
+      | m = \epsilon
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2}
+      | o = <
+      | m = \epsilon
+      | c = {{Defof|Euclidean Metric|subdef = Ordinary Space}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2
+      | o = <
+      | m = \epsilon^2
+}}
+{{end-eqn}}
+But from [[Equation of Sphere in Rectangular Coordinates]]:
+:$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2 = \epsilon^2$
+is the equation of a [[Definition:Sphere (Geometry)|sphere]] whose [[Definition:Center of Sphere|center]] is $x$ and whose [[Definition:Radius of Sphere|radius]] is $\epsilon$.
+The result follows by definition of [[Definition:Interior of Region|interior]] and [[Open Ball of Point Inside Open Ball]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball in Real Number Plane under Chebyshev Distance}
+Tags: Open Balls, Chebyshev Distance
+
+\begin{theorem}
+Let $\R^2$ be the [[Definition:Real Number Plane|real number plane]].
+Let $d_\infty: \R^2 \times \R^2 \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev Distance]] on $\R^2$:
+: $\displaystyle d_\infty \left({x, y}\right):= \max \left\{ {\left\vert{x_1 - y_1}\right\vert, \left\vert{x_2 - y_2}\right\vert}\right\}$
+where $x = \left({x_1, x_2}\right), y = \left({y_1, y_2}\right) \in \R^2$.
+For $a \in \R^2$, let $B_\epsilon \left({a}\right)$ be the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball at $a$]].
+Then $B_\epsilon \left({a}\right)$ is the [[Definition:Interior of Region|interior]] of the [[Definition:Square (Geometry)|square]] centered at $a$ and whose [[Definition:Side of Polygon|sides]] are of [[Definition:Length (Linear Measure)|length]] $2 \epsilon$ parallel to the [[Definition:Coordinate Axis|coordinate axes]].
+\end{theorem}
+
+\begin{proof}
+Let $a = \left({a_1, a_2}\right)$.
+From [[Open Ball in Cartesian Product under Chebyshev Distance]]:
+:$B_\epsilon \left({a; d_\infty}\right) = B_\epsilon \left({a_1; d}\right) \times B_\epsilon \left({a_2; d}\right)$
+where $d$ is the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+From [[Open Ball in Real Number Line is Open Interval]]:
+:$B_\epsilon \left({a_1; d}\right) \times B_\epsilon \left({a_2; d}\right) = \left({a_1 - \epsilon \,.\,.\, a_1 + \epsilon}\right) \times \left({a_2 - \epsilon \,.\,.\, a_2 + \epsilon}\right)$
+That is:
+:$x \in B_\epsilon \left({a; d_\infty}\right) \iff \left({a_2 - \epsilon < x_2 < a_2 + \epsilon}\right) \land \left({a_2 - \epsilon < x_2 < a_2 + \epsilon}\right)$
+from which the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball in Standard Discrete Metric Space}
+Tags: Open Balls, Discrete Metrics
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $d$ be the [[Definition:Standard Discrete Metric|standard discrete metric]] on $M$.
+Let $a \in A$.
+Let $\map {B_\epsilon} {a; d}$ be an [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $M$.
+Then:
+:$\map {B_\epsilon} {a; d} = \begin{cases}
+\set a & : \epsilon \le 1 \\
+A & : \epsilon > 1
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}: \epsilon \le 1$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall x \in A:
+      | l = x
+      | o = \ne
+      | r = a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map d {x, a}
+      | o = \ge
+      | r = \epsilon
+      | c = {{Defof|Standard Discrete Metric}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \notin
+      | r = \map {B_\epsilon} {a; d}
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map {B_\epsilon} {a; d}
+      | r = \set a
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $\epsilon \in \R_{>0}: \epsilon > 1$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall x \in A:
+      | l = \map d {x, a}
+      | o = >
+      | r = \epsilon
+      | c = {{Defof|Standard Discrete Metric}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \map {B_\epsilon} {a; d}
+      | c = {{Defof|Open Ball|Open $\epsilon$-Ball}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map {B_\epsilon} {a; d}
+      | r = A
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance from Point to Subset is Continuous Function}
+Tags: Continuous Mappings, Distance Function
+
+\begin{theorem}
+Let $M = \struct {X, d}$ be a [[Definition:Metric Space|metric space]].
+Let $A \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
+Let $f: X \to \R$ be the [[Definition:Real-Valued Function|function]] defined as:
+:$\forall x \in X: \map f x = \map d {x, A}$
+where $\map d {x, A}$ denotes the [[Definition:Distance between Element and Subset of Metric Space|distance from $x$ to $A$]].
+Then $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+$\forall x, y \in X, \forall z \in A$, by the definition of the [[Definition:Distance between Element and Subset of Metric Space|distance from $x$ to $A$]], we have:
+:$\map d {x, A} \le \map d {x, z} \le \map d {x, y} + \map d {y, z}$
+Taking infimum with respect to $z \in A$ to above inequality, we get:
+:$\map d {x, A} \le \map d {x, y} + \map d {y, A}$
+Similarly, we have:
+:$\map d {y, A} \le \map d {x, y} + \map d {x, A}$
+Therefore:
+:$\norm {\map d {x, A} - \map d {y, A} } \le \map d {x, y}$
+We can conclude that $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighbourhood of Point Contains Point of Subset iff Distance is Zero}
+Tags: Neighborhoods, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {X, d}$ be a [[Definition:Metric Space|metric space]].
+Let $A \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
+Let $x \in X$.
+Then every [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$ contains a point of $A$ {{iff}}:
+:$\map d {x, A} = 0$
+where $\map d {x, A}$ denotes the [[Definition:Distance between Element and Subset of Metric Space|distance from $x$ to $A$]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \in X$.
+Let every [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$ contain a point in $A$.
+Every [[Definition:Open Ball|open ball]] $\map {B_\epsilon} x$ [[Definition:Center of Open Ball|centered]] at $x$ is seen to be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$ in $M$.
+But then this implies that for every $\epsilon \in \R_{\gt 0}$ there must exists a $y \in A$ such that:
+:$\map d {x, y} < \epsilon$
+From this it is seen that $\map d {x, A} = 0$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\map d {x, A} = 0$ for $x \in X$.
+Let $S$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$ in $M$.
+From the definition of a [[Definition:Neighborhood (Metric Space)|neighborhood]], there exists an [[Definition:Open Ball|open ball]] $\map {B_\epsilon} x$ [[Definition:Center of Open Ball|centered]] at $x$ such that:
+:$\map {B_\epsilon} x \subseteq S$
+We have that:
+:$\epsilon < 0$
+and
+$\map d {x, A} = 0$
+Hence there must exists a $y \in A$ such that:
+:$\map d {x, y} < \epsilon \implies \map {B_\epsilon} x \cap A \ne \O$
+If otherwise then $\epsilon$ would be a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] for $\set {\map d {x, y} : y \in A}$ that is greater than $0$, which is a [[Definition:Contradiction|contradiction]].
+Then from [[Set Intersection Preserves Subsets]] we have:
+:$\map {B_\epsilon} x \cap A \subseteq S \cap A$
+and so:
+:$\map {B_\epsilon} x \cap A \ne \O \implies S \cap A \ne \O$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping from Cartesian Product under Chebyshev Distance to Real Number Line is Continuous}
+Tags: Continuous Mappings, Chebyshev Distance
+
+\begin{theorem}
+Let $M = \struct {A, d'}$ be a [[Definition:Metric Space|metric space]].
+Let $\displaystyle \AA = A \times A$ be the [[Definition:Cartesian Product|cartesian product]] of $A$ with itself.
+Let $d_\infty: \AA \times \AA \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\AA$:
+: $\displaystyle \map {d_\infty} {x, y} = \max \set {\map {d'} {x_1, y_1}, \map {d'} {x_2, y_2} }$
+where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
+Then the [[Definition:Mapping|mapping]]:
+:$d': \struct {A \times A, d_\infty} \to \struct {\R, d}$
+where $d$ is the [[Definition:Usual Metric|usual metric]], is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+From definition of [[Definition:Continuous Mapping (Metric Space)|continuous mapping]]:
+We just need to show that:
+:$\forall \tuple {a, b} \in A \times A: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall \tuple {x, y} \in A \times A: \map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \delta \implies \map d {\map {d'} {x, y}, \map {d'} {a, b}} < \epsilon$
+Let $\tuple {a, b} \in A \times A$ and $\epsilon \in \R_{>0}$.
+Let $\tuple {x, y} \in A \times A$.
+Suppose $\map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \dfrac \epsilon 2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {\map {d'} {x, y}, \map {d'} {a, b} }
+      | r = \size {\map {d'} {x, y} - \map {d'} {a, b} }
+      | c = {{Defof|Euclidean Metric on Real Number Line}}
+}}
+{{eqn | o = \le
+      | r = \size {\map {d'} {x, y} - \map {d'} {a, y} } + \size {\map {d'} {a, y} - \map {d'} {a, b} }
+      | c = [[Triangle Inequality]]
+}}
+{{eqn | o = \le
+      | r = \map {d'} {x, a} + \map {d'} {y, b}
+      | c = [[Reverse Triangle Inequality]] on $d'$
+}}
+{{eqn | o = \le
+      | r = 2 \max \set {\map {d'} {x, a}, \map {d'} {y, b} }
+      | c =
+}}
+{{eqn | o = \le
+      | r = 2 \ \map {d_\infty} {\tuple {x, y}, \tuple {a, b} }
+      | c = {{Defof|Chebyshev Distance}}
+}}
+{{eqn | o =  <
+      | r = 2 \cdot \frac \epsilon 2
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Therefore $d'$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set is Open iff Union of Open Balls}
+Tags: Metric Spaces, Open Sets, Open Balls
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $U \subseteq A$.
+Then $U$ is [[Definition:Open Set (Metric Space)|open]] in $M$ {{iff}} it is a [[Definition:Set Union|union]] of [[Definition:Open Ball|open balls]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $U$ be [[Definition:Open Set (Metric Space)|open]] in $M$.
+Let $a \in U$.
+Then by definition of [[Definition:Open Set (Metric Space)|open set]]:
+:$\exists \delta_a \in \R_{>0}: \map {B_{\delta_a} } {a, d} \subseteq U$
+where $\map {B_{\delta_a} } {a, d}$ is the [[Definition:Open Ball|open $\delta_a$-ball]] of $a$ in $M$.
+Therefore:
+:$\displaystyle U = \bigcup_{a \mathop \in U} \map {B_{\delta_a} } {a, d}$
+and so $U$ is a [[Definition:Set Union|union]] of [[Definition:Open Ball|open balls]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $U$ be a [[Definition:Set Union|union]] of [[Definition:Open Ball|open balls]] of $M$.
+Let the [[Definition:Center of Open Ball|centers]] of these [[Definition:Open Ball|open balls]] be the [[Definition:Element|elements]] of an [[Definition:Indexing Set|indexing set]] $I$.
+Then $U$ can be written as:
+:$\displaystyle U = \bigcup_{a \mathop \in I} \map {B_{\delta_a} } {a, d}$
+where $\delta_a \in \R_{>0}$ is the [[Definition:Radius of Open Ball|radius]] of the [[Definition:Open Ball|open ball]] of $a$.
+Let $x \in U$.
+Then by definition of [[Definition:Set Union|union]]:
+:$\exists a \in I: x \in \map {B_{\delta_a} } {a, d}$
+From [[Open Ball is Neighborhood of all Points Inside]], $\map {B_{\delta_a} } {a, d}$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$.
+By [[Set is Subset of Union/Family of Sets|Set is Subset of Union]]:
+:$\map {B_{\delta_a} } {a, d} \subseteq U$
+From [[Superset of Neighborhood in Metric Space is Neighborhood]], it follows that $U$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$.
+Since $x$ is arbitrary, it follows that $U$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of each of its points.
+Hence by definition, $U$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Open and Closed in Metric Space}
+Tags: Metric Spaces, Open Sets, Closed Sets, Empty Set
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Then the [[Definition:Empty Set|empty set]] $\varnothing$ is both [[Definition:Open Set (Metric Space)|open]] and [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Empty Set is Open in Metric Space]], $\varnothing$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+From [[Empty Set is Closed in Metric Space]], $\varnothing$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Closed in Metric Space}
+Tags: Metric Spaces, Closed Sets, Empty Set
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then the [[Definition:Empty Set|empty set]] $\varnothing$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Metric Space is Open in Itself]], $A$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+But:
+:$\varnothing = \relcomp A A$
+where $\complement_A$ denotes the [[Definition:Relative Complement|set complement relative to $A$]].
+The result follows by definition of [[Definition:Closed Set/Metric Space/Definition 1|closed set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Closed in Itself}
+Tags: Metric Spaces, Closed Sets
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $A$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Empty Set is Open in Metric Space]], $\O$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+But:
+:$A = \relcomp A \O$
+where $\complement_A$ denotes the [[Definition:Relative Complement|set complement relative to $A$]].
+The result follows by definition of [[Definition:Closed Set/Metric Space/Definition 1|closed set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Open and Closed in Itself}
+Tags: Metric Spaces, Open Sets, Closed Sets
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Then $A$ is both [[Definition:Open Set (Metric Space)|open]] and [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Metric Space is Open in Itself]], $A$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+From [[Metric Space is Closed in Itself]], $A$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Metric Space contains Limits of Sequences iff Closed}
+Tags: Metric Spaces, Closed Sets
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$.
+Then $H$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$ [[Definition:Iff|iff]]:
+:for each [[Definition:Sequence|sequence]] $\left\langle{a_n}\right\rangle$ of points of $H$ that [[Definition:Convergent Sequence (Metric Space)|converges]] to a point $a \in A$, it follows that $a \in H$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $H$ be [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+Suppose that:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = a$
+and:
+:$\forall n \in \N_{>0}: a_n \in H$
+If the set $\left\{ {a_1, a_2, \ldots}\right\}$ is [[Definition:Infinite Set|infinite]] then every [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$ contains infinitely many points of $H$.
+Thus $a$ is a [[Definition:Limit Point (Metric Space)|limit point]] of $H$.
+So by definition of [[Definition:Closed Set/Metric Space/Definition 2|closed set]], $a \in H$.
+On the other hand, if $\left\{ {a_1, a_2, \ldots}\right\}$ is [[Definition:Finite Set|finite]], then for some $N \in \N$:
+:$n, m > N \implies a_n = a_m$
+Since:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = a$
+Then:
+:$\forall n > N: d \left({a_n, a}\right) = 0$
+Thus:
+:$a_n = a$
+and so:
+:$a \in H$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $H$ be a [[Definition:Set|set]] such that:
+:for each [[Definition:Sequence|sequence]] $\left\langle{a_n}\right\rangle$ such that $\displaystyle \lim_{n \mathop \to \infty} a_n = a$, it follows that $a \in H$.
+Let $b$ be a [[Definition:Limit Point (Metric Space)|limit point]] of $H$.
+Then $b$ is the [[Definition:Limit of Sequence (Metric Space)|limit]] of a [[Definition:Convergent Sequence (Metric Space)|convergent sequence]] of points of $H$.
+[[Definition:By Hypothesis|By hypothesis]], $b \in H$.
+Thus $H$ is a [[Definition:Closed Set/Metric Space/Definition 2|closed set]] by definition.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Metric Space is Closed iff contains all Zero Distance Points}
+Tags: Metric Spaces, Closed Sets
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$.
+Then $H$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$ {{iff}}:
+:$\forall x \in A: \map d {x, H} = 0 \implies x \in H$
+where $\map d {x, H}$ denotes the [[Definition:Distance between Element and Subset of Metric Space|distance between $x$ and $H$]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $H$ be [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+Let $x \in A: \map d {x, H} = 0 \implies x \in H$.
+By [[Existence of Sequence in Subset of Metric Space whose Limit is Infimum]] there exists a [[Definition:Sequence|sequence]] $\sequence {a_n}$ of points of $H$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map d {x, a_n} = 0$
+So every [[Definition:Neighborhood (Metric Space)|neighborhood]] of $x$ contains points of $H$.
+If some $a_n \in H$ then $x \in H$.
+Otherwise each $a_n$ is different from $x$
+Then $x$ is the [[Definition:Limit of Sequence (Metric Space)|limit]] of $\sequence {a_n}$ and hence of $H$.
+Thus $x \in H$ by definition of [[Definition:Closed Set/Metric Space/Definition 2|closed set]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $H$ is such that:
+:$\forall x \in A: \map d {x, H} = 0 \implies x \in H$
+Let $x$ be a [[Definition:Limit Point (Metric Space)|limit point]] of $H$.
+Then $\map d {x, H} = 0$.
+{{explain|There should be a result proving the statement: $x$ is a limit point of $H$ implies $\map d {x, H} {{=}} 0$. I thought it was already there but I can't find it.}}
+Thus $H$ contains all its [[Definition:Limit Point (Metric Space)|limit points]] and so is by definition a [[Definition:Closed Set/Metric Space/Definition 2|closed set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Mapping between Metric Spaces by Closed Sets}
+Tags: Metric Spaces, Closed Sets, Continuous Mappings
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]] [[Definition:Iff|iff]]:
+:for every $V \subseteq A_2$ which is [[Definition:Closed Set (Metric Space)|closed in $M_2$]], $f^{-1} \left({V}\right)$ is [[Definition:Closed Set (Metric Space)|closed in $M_1$]].
+\end{theorem}
+
+\begin{proof}
+First the following is established.
+Let $W \in A_2$.
+We note that:
+: $f^{-1} \left[{A_2}\right] = A_1$
+Hence, from [[Preimage of Set Difference under Mapping]]:
+:$f^{-1} \left[{A_2 \setminus W}\right] = A_1 \setminus f^{-1} \left[{W}\right]$
+=== Necessary Condition ===
+Suppose that:
+:for every $V \subseteq A_2$ which is [[Definition:Closed Set (Metric Space)|closed in $M_2$]], $f^{-1} \left({V}\right)$ is [[Definition:Closed Set (Metric Space)|closed in $M_1$]].
+Let $U$ be [[Definition:Open Set (Metric Space)|open]] in $M_2$.
+Then by definition $A_2 \setminus U$ is [[Definition:Closed Set (Metric Space)|closed]] in $M_2$.
+By hypothesis, $f^{-1} \left[{A_2 \setminus U}\right] = A_1 \setminus f^{-1} \left[{U}\right]$ is [[Definition:Closed Set (Metric Space)|closed]] in $M_1$.
+So $f^{-1} \left[{U}\right]$ is [[Definition:Open Set (Metric Space)|open]] in $M_1$.
+This is true for any $U \in A_2$ as $U$ is arbitrary.
+It follows by definition that $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now let $f$ be [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+Let $V$ be [[Definition:Closed Set (Metric Space)|closed]] in $M_2$.
+Then $A_2 \setminus V$ is [[Definition:Open Set (Metric Space)|open]] in $M_2$.
+As $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]], $f^{-1} \left[{A_2 \setminus V}\right] = A_1 \setminus f^{-1} \left[{V}\right]$ is [[Definition:Open Set (Metric Space)|open]] in $M_1$.
+So $f^{-1} \left[{V}\right]$ is [[Definition:Closed Set (Metric Space)|closed]] in $M_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space defined by Closed Sets}
+Tags: Metric Spaces, Closed Sets
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then:
+{{begin-axiom}}
+{{axiom | n = C1
+        | lc= $A$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$
+}}
+{{axiom | n = C2
+        | lc= $\O$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$
+}}
+{{axiom | n = C3
+        | lc= The [[Definition:Set Union|union]] of a [[Definition:Finite Set|finite number]] of [[Definition:Closed Set (Metric Space)|closed sets]] of $M$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $M$
+}}
+{{axiom | n = C4
+        | lc= The [[Definition:Set Intersection|intersection]] of arbitrarily many [[Definition:Closed Set (Metric Space)|closed sets]] of $M$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $M$
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+From [[Metric Space is Closed in Itself]], $C1$ holds.
+{{qed|lemma}}
+From [[Empty Set is Closed in Metric Space]], $C2$ holds.
+{{qed|lemma}}
+Let $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ be the [[Definition:Set Union|union]] of a [[Definition:Finite|finite]] number of [[Definition:Closed Set (Metric Space)|closed sets]] of $M$.
+Then from [[De Morgan's Laws (Set Theory)|De Morgan's laws]]:
+:$\displaystyle A \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$
+By definition of [[Definition:Closed Set (Metric Space)|closed set]], each of the $A \setminus V_i$ is by definition [[Definition:Open Set (Metric Space)|open]] in $M$.
+We have that $\displaystyle \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$ is the [[Definition:Set Intersection|intersection]] of a [[Definition:Finite|finite]] number of [[Definition:Open Set (Metric Space)|open sets]] of $M$.
+Therefore, by [[Finite Intersection of Open Sets of Metric Space is Open]], $\displaystyle \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i} = A \setminus \bigcup_{i \mathop = 1}^n V_i$ is likewise [[Definition:Open Set (Metric Space)|open]] in $M$.
+Then by definition of [[Definition:Closed Set (Metric Space)|closed set]], $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ is [[Definition:Closed Set (Topology)|closed]] in $M$.
+Thus $C3$ holds.
+{{qed|lemma}}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] (either [[Definition:Finite|finite]] or [[Definition:Infinite|infinite]]).
+Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the [[Definition:Set Intersection|intersection]] of a [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Closed Set (Metric Space)|closed sets]] of $M$ indexed by $I$.
+Then from [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection|De Morgan's laws]]:
+:$\displaystyle A \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {A \setminus V_i}$
+By definition of [[Definition:Closed Set (Metric Space)|closed set]], each of $A \setminus V_i$ are by definition [[Definition:Open Set (Metric Space)|open]] in $M$.
+We have that $\displaystyle \bigcup_{i \mathop \in I} \paren {A \setminus V_i}$ is the [[Definition:Set Union|union]] of a [[Definition:Indexed Family of Subsets|family]] of [[Definition:Open Set (Metric Space)|open sets]] of $M$ indexed by $I$.
+Therefore, by definition of a [[Definition:Topology|topology]], $\displaystyle \bigcup_{i \mathop \in I} \paren {A \setminus V_i} = A \setminus \bigcap_{i \mathop \in I} V_i$ is likewise [[Definition:Open Set (Metric Space)|open]] in $M$.
+Then by definition of [[Definition:Closed Set (Metric Space)|closed set]], $\displaystyle \bigcap_{i \mathop \in I} V_i$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+Thus $C4$ holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Union of Closed Sets of Metric Space may not be Closed}
+Tags: Closed Sets, Set Union
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $V_1, V_2, V_3, \ldots$ be an [[Definition:Infinite Set|infinite set]] of [[Definition:Closed Set (Metric Space)|closed sets]] of $M$.
+Then it is not necessarily the case that $\displaystyle \bigcup_{n \mathop \in \N} V_n$ is itself a [[Definition:Closed Set (Metric Space)|closed set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Closed Real Interval|closed real interval]]:
+:$\closedint {\dfrac 1 n} 1 \subseteq \R$
+From [[Closed Real Interval is Closed Set]], $\closedint {\dfrac 1 n} 1$  is [[Definition:Closed Set (Metric Space)|closed]] in $\R$ for all $n \in \N$.
+But:
+:$\displaystyle \bigcup_{n \mathop \in \N} \closedint {\dfrac 1 n} 1 = \hointl 0 1$
+The result follows from [[Half-Open Real Interval is neither Open nor Closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Sets of Cartesian Product of Metric Spaces under Chebyshev Distance}
+Tags: Chebyshev Distance, Open Sets
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_n = \left({A_n, d_n}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+: $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.
+For $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $U_i$ be [[Definition:Open Set (Metric Space)|open]] in $M_i$.
+Then $\displaystyle \prod_{i \mathop = 1}^n U_i$ is [[Definition:Open Set (Metric Space)|open]] in $M = \left({\mathcal A, d_\infty}\right)$.
+\end{theorem}<|endoftext|>
+\section{Basis for Product of Metric Spaces under Chebyshev Distance}
+Tags: Chebyshev Distance, Open Sets
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right), \ldots, M_n = \left({A_n, d_n}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_1, A_2, \ldots, A_n$.
+Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+: $\displaystyle d_\infty \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {d_i \left({x_i, y_i}\right)}\right\}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.
+For $i \in \left\{ {1, 2, \ldots, n}\right\}$, let $U_i$ be [[Definition:Open Set (Metric Space)|open]] in $M_i$.
+Then $\left\{ {\displaystyle \prod_{i \mathop = 1}^n U_i}\right\}$ is a [[Definition:Basis for Open Sets (Metric Space)|basis for the open sets]] of $M$.
+\end{theorem}<|endoftext|>
+\section{Open Balls form Basis for Open Sets of Metric Space}
+Tags: Open Sets, Open Balls
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\mathcal B$ be the [[Definition:Set|set]] of all [[Definition:Open Ball|open balls]] of $M$.
+Then $\mathcal B$ is a [[Definition:Basis for Open Sets (Metric Space)|basis for the open sets]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Let $U$ be an [[Definition:Open Set (Metric Space)|open set]] of $M$.
+Then by definition:
+:$\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$
+Thus:
+:$\displaystyle U = \bigcup_{y \mathop \in U} B_\epsilon \left({y}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Graph of Continuous Mapping between Metric Spaces is Closed in Chebyshev Product}
+Tags: Chebyshev Distance, Continuous Mappings
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = A_1 \times A_2$ be the [[Definition:Cartesian Product|cartesian product]] of $A_1$ and $A_2$.
+Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+: $\displaystyle d_\infty \left({x, y}\right) = \max \left\{ {d_1 \left({x_1, y_1}\right), d_2 \left({x_2, y_2}\right)}\right\}$
+where $x = \left({x_1, x_2}\right), y = \left({y_1, y_2}\right) \in \mathcal A$.
+Let $\Gamma_f$ be the [[Definition:Graph of Mapping|graph of $f$]].
+Then $\Gamma_f$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $\left({\mathcal A, d_\infty}\right)$.
+\end{theorem}<|endoftext|>
+\section{Closed Subset of Real Numbers with Lower Bound contains Infimum}
+Tags: Boundedness, Real Analysis
+
+\begin{theorem}
+Consider the [[Definition:Real Number Line|real number line]] as a [[Definition:Metric Space|metric space]] under the [[Definition:Usual Metric|usual metric]].
+Let $A \subseteq \R$ such that $A$ is [[Definition:Closed Set (Metric Space)|closed in $\R$]] and $A \neq \varnothing$.
+Let $A$ be [[Definition:Bounded Below Subset of Real Numbers|bounded below]].
+Then $A$ contains its [[Definition:Infimum of Subset of Real Numbers|infimum]].
+\end{theorem}
+
+\begin{proof}
+From [[Infimum of Bounded Below Set of Reals is in Closure]]:
+:$\map \inf A \in \map \cl A$
+From [[Set is Closed iff Equals Topological Closure]]:
+:$A = \map \cl A$
+Therefore $\map \inf A \in A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point of Metric Space is Isolated iff not Limit Point}
+Tags: Limit Points, Isolated Points
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Let $H'$ be the [[Definition:Set|set]] of [[Definition:Limit Point (Metric Space)|limit points]] of $H$.
+Let $H^i$ be the [[Definition:Set|set]] of [[Definition:Isolated Point (Metric Space)/Subset|isolated points]] of $H$.
+Then:
+:$A' \cap A^i = \O$
+\end{theorem}<|endoftext|>
+\section{Set of Isolated Points of Metric Space is Disjoint from Limit Points}
+Tags: Limit Points, Isolated Points
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Let $H'$ be the [[Definition:Set|set]] of [[Definition:Limit Point (Metric Space)|limit points]] of $H$.
+Let $H^i$ be the [[Definition:Set|set]] of [[Definition:Isolated Point (Metric Space)/Subset|isolated points]] of $H$.
+Then:
+:$H' \cap H^i = \varnothing$
+\end{theorem}
+
+\begin{proof}
+Let $a \in H_i$.
+Then by definition of [[Definition:Isolated Point (Metric Space)/Subset|isolated point]]:
+:$\exists \epsilon \in \R_{>0}: \left\{{x \in H: d \left({x, a}\right) < \epsilon}\right\} = \left\{{a}\right\}$
+But by [[Definition:Metric Space Axioms|metric space axiom $M1$]]:
+:$d \left({a, a}\right) = 0$
+and so:
+:$\left\{{x \in H: 0 < d \left({x, a}\right) < \epsilon}\right\} = \varnothing$
+So by definition $a$ is not a [[Definition:Limit Point (Metric Space)|limit point]] of $H$.
+That is:
+:$a \notin H'$
+or:
+:$a \in \complement_A \left({H'}\right)$
+It follows from [[Intersection with Complement is Empty iff Subset]] that:
+:$H' \cap H^i = \varnothing$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Metric Space is Subset of its Closure}
+Tags: Set Closures
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Then:
+:$H \subseteq H^-$
+where $H^-$ denotes the [[Definition:Closure (Metric Space)|closure]] of $H$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Closure (Metric Space)|closure]]:
+:$H^- = H' \cup H^i$
+where:
+:$H'$ denotes the [[Definition:Set|set]] of [[Definition:Limit Point (Metric Space)|limit points]] of $H$
+:$H^i$ denotes the [[Definition:Set|set]] of [[Definition:Isolated Point (Metric Space)/Subset|isolated points]] of $H$.
+Let $a \in H$.
+If $a$ is a [[Definition:Limit Point (Metric Space)|limit point]] of $H$ then $a \in H'$.
+Suppose $a \notin H'$.
+Then by definition of [[Definition:Limit Point (Metric Space)|limit point]]:
+:$\neg \forall \epsilon \in \R_{>0}: \left\{{x \in A: 0 < d \left({x, a}\right) < \epsilon}\right\} \ne \varnothing$
+That is:
+:$\exists \epsilon \in \R_{>0}: \left\{{x \in A: 0 < d \left({x, a}\right) < \epsilon}\right\} = \varnothing$
+and so as $d \left({a, a}\right) = 0$:
+:$\exists \epsilon \in \R_{>0}: \left\{{x \in A: d \left({x, a}\right) < \epsilon}\right\} = \left\{{a}\right\}$
+Thus, by definition, $a$ is an [[Definition:Isolated Point (Metric Space)/Subset|isolated point]] of $H$.
+So $a \in H'$ or $a \in H^i$.
+Hence by definition of [[Definition:Set Union|set union]]:
+:$a \in H \subseteq H^-$
+By definition of [[Definition:Subset|subset]]:
+:$H \subseteq H \subseteq H^-$
+and hence the result by definition of [[Definition:Closure (Metric Space)|closure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Closure of Subset of Metric Space iff Limit of Sequence}
+Tags: Set Closures
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Let $H^-$ denote the [[Definition:Closure (Metric Space)|closure]] of $H$.
+Let $a \in A$.
+Then $a \in H^-$ {{iff}} there exists a [[Definition:Sequence|sequence]] $\sequence {x_n}$ of points of $H$ which [[Definition:Convergent Sequence in Metric Space|converges]] to the [[Definition:Limit of Sequence (Metric Space)|limit]] $a$.
+\end{theorem}
+
+\begin{proof}
+From definition of [[Definition:Closure (Metric Space)|closure]], $H^- = H' \cup H^i$.
+Suppose that $a \in H^-$.
+If $a \in H^i$, then $a \in H$ and so $\sequence {a, a, \ldots}$ is a [[Definition:Sequence|sequence]] in $H$ that converges to $a$.
+If $a \in H'$, then by [[Limit Point is Limit of Convergent Sequence]] there exist a [[Definition:Sequence|sequence]] in $H$ that [[Definition:Convergent Sequence in Metric Space|converges]] to $a$.
+{{qed|lemma}}
+For the converse, let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $H$ that [[Definition:Convergent Sequence in Metric Space|converges]] to a point $a \in A$.
+So let $U$ be an [[Definition:Open Set of Metric Space|open set]] that contains $a$.
+By the definition of an [[Definition:Open Set of Metric Space|open set]], it follows that there exists an [[Definition:Open Ball|open ball]] [[Definition:Center of Open Ball|centered]] at $a$ such that $\map {B_\epsilon} a \subseteq U$.
+But then because $\sequence {x_n}$ [[Definition:Convergent Sequence in Metric Space|converges]] to $a$ there exists $m \in \N_{\gt 0}$ such that:
+{{begin-eqn}}
+{{eqn | l = \map d {x_m, a}
+      | o = <
+      | r = \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = H \cap \map {B_\epsilon} a
+      | o = \ne
+      | r = \O
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = H \cap U
+      | o = \ne
+      | r = \O
+      | c = 
+}}
+{{end-eqn}}
+Because from [[Set Intersection Preserves Subsets]] it is seen that $H \cap \map {B_\epsilon} a \subseteq H \cap U$.
+It has been shown that every [[Definition:Open Set of Metric Space|open set]] that contains $a$ also contains a point in $H$.
+So by [[Condition for Point being in Closure]], it is seen that $a \in H^-$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Subset of Closed Set of Metric Space is Subset}
+Tags: Set Closures, Closure of Subset of Closed Set of Metric Space is Subset
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $F$ be a [[Definition:Closed Set (Metric Space)|closed set]] of $M$.
+Let $H \subseteq F$ be a [[Definition:Subset|subset]] of $F$.
+Let $H^-$ denote the [[Definition:Closure (Metric Space)|closure]] of $H$.
+Then $H^- \subseteq F$.
+\end{theorem}
+
+\begin{proof}
+From [[Metric Induces Topology]], the [[Definition:Topology Induced by Metric|topology $\tau$ induced]] by the [[Definition:Metric|metric]] $d$ is a [[Definition:Topology|topology]] on $M$.
+From [[Metric Closure and Topological Closure of Subset are Equivalent]], it is sufficient to show that the [[Definition:Topological Closure|topological closure]] of $H$ is contained in $F$.
+From [[Set is Closed in Metric Space iff Closed in Induced Topological Space]]:
+:$F$ is [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Topological Space|topological space]] $\struct{A, \tau}$ [[Definition:Topology Induced by Metric|induced]] by the [[Definition:Metric|metric]] $d$.
+We have:
+{{begin-eqn}}
+{{eqn | l = H
+      | o = \subseteq
+      | r = F
+}}
+{{eqn | ll= \leadsto
+      | l = H^-
+      | o = \subseteq
+      | r = F^-
+      | c = [[Topological Closure of Subset is Subset of Topological Closure]]
+}}
+{{eqn | r = F
+      | c = [[Set is Closed iff Equals Topological Closure]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x \in H^-$.
+From [[Point in Closure of Subset of Metric Space iff Limit of Sequence]]
+:there exists a [[Definition:Sequence|sequence]] $\sequence {a_n}$ of points of $H$ which [[Definition:Convergent Sequence in Metric Space|converges]] to the [[Definition:Limit of Sequence (Metric Space)|limit]] $x$.
+By assumption:
+:$\sequence {a_n}$ is also a [[Definition:Sequence|sequence]] of points of $F$
+From [[Subset of Metric Space contains Limits of Sequences iff Closed]]:
+:$x \in F$
+Thus it has been shown:
+:$H^- \subseteq F$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Subset of Metric Space is Intersection of Closed Supersets}
+Tags: Set Closures
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Let $H^-$ denote the [[Definition:Closure (Metric Space)|closure]] of $H$.
+Then $H^-$ is the [[Definition:Set Intersection|intersection]] of all [[Definition:Closed Set (Metric Space)|closed sets]] of $M$ of which $H$ is a [[Definition:Subset|subset]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbb K$ be the [[Definition:Set|set]] of all [[Definition:Closed Set (Metric Space)|closed sets]] $K$ of $M$ such that $H \subseteq K$.
+From [[Closure of Subset of Closed Set of Metric Space is Subset]]:
+:$\forall K \in \mathbb K: H^- \subseteq K$
+From [[Intersection is Largest Subset]]:
+:$\displaystyle H^- \subseteq \bigcap \mathbb K$
+{{qed|lemma}}
+From [[Closure of Subset of Metric Space is Closed]], $H^-$ is [[Definition:Closed Set (Metric Space)|closed]].
+From [[Subset of Metric Space is Subset of its Closure]], $H \subseteq H^-$.
+So $H^-$ is, by definition, a [[Definition:Closed Set (Metric Space)|closed set]] of $M$ which contains $H$.
+But $\displaystyle \bigcap \mathbb K$ is the [[Definition:Set Intersection|intersection]] of '''all''' [[Definition:Closed Set (Metric Space)|closed sets]] of $M$ which contain $H$.
+So from [[Intersection is Subset]] it follows that:
+: $\displaystyle \bigcap \mathbb K \subseteq H^-$
+{{qed|lemma}}
+Finally, we have that:
+: $\displaystyle H^- \subseteq \bigcap \mathbb K$
+: $\displaystyle \bigcap \mathbb K \subseteq H^-$
+So by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+: $\displaystyle H^- = \bigcap \mathbb K$
+which is what we needed to prove.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Metric Subspace of Real Numbers under Euclidean Metric}
+Tags: Rational Number Space, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\struct {\Q, d_\Q}$ be the [[Definition:Rational Number|set of rational numbers]] under the [[Definition:Real-Valued Function|function]] $d_\Q: \Q \times \Q \to \R$ defined as:
+:$\forall x, y \in \Q: \map d {x, y} = \size {x - y}$
+Let $\struct {\R, d}$ denote the [[Definition:Real Number Line with Euclidean Metric|real number line with the usual (Euclidean) metric]].
+Then $\struct {\Q, d_\Q}$ is a [[Definition:Metric Subspace|metric subspace]] of $\struct {\R, d}$, where:
+\end{theorem}
+
+\begin{proof}
+From [[Rational Numbers form Subfield of Real Numbers]]:
+:$\Q \subseteq \R$
+By the definition of the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]] on $\R$:
+:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
+and so $d_\Q$ is a [[Definition:Restriction of Mapping|restriction]] of $d$:
+:$d_\Q = d {\restriction}_{\Q \times \Q}$
+From [[Euclidean Metric on Real Number Line is Metric]], $d$ is a [[Definition:Metric|metric]] on $\R$.
+The result follows by definition of [[Definition:Metric Subspace|metric subspace]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Metric on Real Number Line is Metric}
+Tags: Euclidean Metric, Real Number Line with Euclidean Metric, Euclidean Metric on Real Number Line is Metric
+
+\begin{theorem}
+The [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]] on the [[Definition:Real Number Line|real number line]] $\R$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]] on the [[Definition:Real Number Line|real number line]] is a special case of the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] on the [[Definition:Real Vector Space|real vector space]] $\R^n$.
+The result follows from [[Euclidean Metric on Real Vector Space is Metric]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]]:
+:$M = \struct {\R, d}$
+where $d$ is the [[Definition:Distance Function|distance function]] given by:
+:$\map d {x, y} = \size {x - y}$
+=== Proof of $\text M 1$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {x, x}
+      | r = \size {x - x}
+      | c = Definition of $d$
+}}
+{{eqn | r = \size 0
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $\text M 2$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {x, y} + \map d {y, z}
+      | r = \size {x - y} + \size {y - z}
+      | c = Definition of $d$
+}}
+{{eqn | o = \ge
+      | r = \size {\paren {x - y} + \paren {y - z} }
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \size {x - z}
+      | c = 
+}}
+{{eqn | r = \map d {x, z}
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $\text M 3$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {x, y}
+      | r = \size {x - y}
+      | c = Definition of $d$
+}}
+{{eqn | r = \size {y - x}
+      | c = 
+}}
+{{eqn | r = \map d {y, x}
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $\text M 4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x - y
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x - y}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map d {x, y}
+      | o = >
+      | r = 0
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 4$]] holds for $d$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Unit n-Cube}
+Tags: Definitions: Real Analysis
+
+\begin{theorem}
+Let $n \in \N$.
+The '''unit $n$-cube''' $I^n$ is the [[Definition:Finite Cartesian Product|Cartesian product]] of $n$ instances of the [[Definition:Closed Real Interval|closed real interval]] $\left\{ {x \in \R: 0 \le x \le 1}\right\}$:
+:$I^n = \left[{0 \,.\,.\, 1}\right]^n$
+\end{theorem}<|endoftext|>
+\section{Unit n-Cube under Chebyshev Distance is Subspace of Real Vector Space}
+Tags: Metric Subspaces
+
+\begin{theorem}
+Let $n \in \N$.
+Let $I^n$ denote the [[Definition:Unit n-Cube|unit $n$-cube]]:
+:$I^n = \left[{0 \,.\,.\, 1}\right]^n$
+that is, the [[Definition:Finite Cartesian Product|Cartesian product]] of $n$ instances of the [[Definition:Closed Real Interval|closed real interval]] $\left\{ {x \in \R: 0 \le x \le 1}\right\}$.
+Let $d_c: I^n \times I^n \to \R$ be defined as:
+:$\displaystyle d_c \left({x, y}\right) = \max_{i \mathop = 1}^n \left\{ {\left\lvert{x_i - y_i}\right\rvert}\right\}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({x_1, x_2, \ldots, x_n}\right) \in I^n$.
+Then $\left({I_n, d_c}\right)$ is a [[Definition:Metric Subspace|metric subspace]] of $\left({\R^n, d_\infty}\right)$, where $d_\infty$ is the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on the [[Definition:Real Vector Space|real vector space]] $\R_n$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Straightforward but a bit tedious.}}
+\end{proof}<|endoftext|>
+\section{Unit n-Sphere under Euclidean Metric is Metric Subspace of Euclidean Real Vector Space}
+Tags: Metric Subspaces
+
+\begin{theorem}
+Let $\Bbb S^n$ be the [[Definition:Unit Sphere|unit $n$-sphere]].
+Let $d_S: \Bbb S^n \times \Bbb S^n \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\displaystyle \forall x, y \in \Bbb S^n: d_S \left({x, y}\right) = \sqrt {\sum_{i \mathop = 1}^{n + 1} \left({x_i - y_i}\right)^2}$
+where $x = \left({x_1, x_2, \ldots, x_{n + 1} }\right), y = \left({y_1, y_2, \ldots, y_{n + 1} }\right)$.
+Then $\left({\Bbb S^n, d_S}\right)$ is a [[Definition:Metric Subspace|metric subspace]] of $\left({\R^{n + 1}, d}\right)$, where $d$ is the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] on the [[Definition:Real Vector Space|real vector space]] $\R^{n + 1}$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Straightforward but tedious.}}
+\end{proof}<|endoftext|>
+\section{Vector Subspace of Real Vector Space under Chebyshev Metric is Metric Subspace}
+Tags: Metric Subspaces, Chebyshev Distance
+
+\begin{theorem}
+Let $n \in \N$.
+Let $A$ be the set of all [[Definition:Ordered Tuple|ordered $n+1$-tuples]] $\tuple {x_1, x_2, \ldots, x_{n + 1} }$ of [[Definition:Real Number|real numbers]] such that $x_{n + 1} = 0$.
+Let $d: A \times A \to \R$ be the [[Definition:Real-Valued Function|function]] defined as:
+:$\displaystyle \forall x, y \in A: \map d {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$
+where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.
+Then $\struct {A, d}$ is a [[Definition:Metric Subspace|metric subspace]] of $\struct {\R^{n + 1}, d_\infty}$ where $d_\infty$ is the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on the [[Definition:Real Vector Space|real vector space]] $\R^{n + 1}$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Straightforward but tedious.}}
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping on Metric Space is Continuous}
+Tags: Metric Subspaces, Continuous Mappings
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\left({H, d_H}\right)$ be a [[Definition:Metric Subspace|metric subspace]] of $M$.
+Then the [[Definition:Inclusion Mapping|inclusion mapping]] $i_H: H \to A$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in H$.
+Let $\epsilon \in \R_{>0}$.
+Let $\delta = \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d_H \left({a, y}\right)
+      | o = <
+      | r = \delta
+      | c = for some $y \in H$
+}}
+{{eqn | ll= \implies
+      | l = d \left({i_H \left({a}\right), i_H \left({y}\right)}\right)
+      | r = d \left({a, y}\right)
+      | c = 
+}}
+{{eqn | r = d_H \left({a, y}\right)
+      | c = 
+}}
+{{eqn | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+So by definition $d_H$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]].
+As $a \in H$ is arbitrary, it follows that $d_H$ is [[Definition:Continuous on Metric Space|continuous on $H$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bertrand-Chebyshev Theorem}
+Tags: Number Theory, Bertrand-Chebyshev Theorem
+
+\begin{theorem}
+For all $n \in \N_{>0}$, there exists a [[Definition:Prime Number|prime number]] $p$ with $n < p \le 2 n$.
+\end{theorem}
+
+\begin{proof}
+We will first prove the theorem for the case $n \le 2047$. 
+Consider the following sequence of [[Definition:Prime Number|prime numbers]]:
+:$2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503$
+Each of these [[Definition:Prime Number|prime number]] is smaller than twice the previous one.
+Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 2047$, contains one of these [[Definition:Prime Number|prime numbers]].
+=== [[Bertrand-Chebyshev Theorem/Lemma 1|Lemma 1]] ===
+{{:Bertrand-Chebyshev Theorem/Lemma 1}}{{qed|lemma}}
+=== [[Bertrand-Chebyshev Theorem/Lemma 2|Lemma 2]] ===
+{{:Bertrand-Chebyshev Theorem/Lemma 2}}{{qed|lemma}}
+=== [[Bertrand-Chebyshev Theorem/Lemma 3|Lemma 3]] ===
+{{:Bertrand-Chebyshev Theorem/Lemma 3}}{{qed|lemma}}
+In particular, if $p > \sqrt {2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$.
+For $n \ge 3$, there is no [[Definition:Prime Factor|prime factor]] $p$ with $\dfrac 2 3 n < p \le n$, for such a [[Definition:Prime Number|prime number]] divides $n!$ exactly once and $\paren {2 n}!$ exactly twice.
+Therefore, by [[Bertrand-Chebyshev Theorem/Lemma 1|Lemma 1]]:
+:$\displaystyle \dfrac {2^{2 n} } {2 n + 1} \le \dbinom {2 n} n \le \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt{2 n} \mathop < p \mathop \le \frac 2 3 n} p \prod_{n \mathop < p \mathop \le 2 n} p$
+for $n \ge 3$.
+Now assume there is no prime number $p$ with $n < p \le 2 n$. 
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \frac {2^{2 n} } {2 n + 1} 
+      | r = \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac 2 3 n} p 
+      | o = \le
+      | c = 
+}}
+{{eqn | r = \paren {2 n}^{\sqrt {2 n} } \prod_{p \mathop \le \frac 2 3 n} p
+      | o = \le
+      | c = 
+}}
+{{eqn | r = \paren {2 n}^{\sqrt {2 n} } 2^{\frac 4 3 n}
+      | o = \le
+      | c = by [[Bertrand-Chebyshev Theorem/Lemma 2|Lemma 2]]
+}}
+{{end-eqn}}
+This is a contradiction if $n$ is large enough.
+Indeed, we have:
+:$2^{\frac 2 3 n} \le \paren {2 n + 1} \paren {2 n}^{\sqrt {2 n} }$
+Now:
+:$2 n + 1 \le \paren {2 n}^2 \le \paren {2 n}^{\frac 1 3 \sqrt {2 n} }$
+for $n \ge 18$.
+So:
+:$2^{2 n} \le \paren {2 n}^{4 \sqrt {2 n} }$
+Put $r = \sqrt {2 n}$.
+Then:
+:$2^{r^2} \le r^{8 r}$
+or equivalently:
+:$2^r \le r^8$
+This fails when $r = 2^6 = 64$.
+It fails thereafter, since $2^r$ increases faster than $r^8$.
+So our proof works if:
+:$n \ge 2^{11} = 2048$
+and the examples show it is true for smaller $n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $\dbinom n k$ be the [[Definition:Binomial Coefficient|binomial coefficient]] of $n$ over $k$ for a [[Definition:Positive Integer|positive integer]] $k \in \Z_{\ge 0}$.
+Let $S_n = \left\langle{x_k}\right\rangle$ be the [[Definition:Sequence|sequence]] defined as:
+:$x_k = \dbinom n k$
+Then $S_n$ is [[Definition:Strictly Increasing Sequence|strictly increasing]] exactly where $0 \le k < \dfrac n 2$.
+\end{theorem}
+
+\begin{proof}
+When $k \ge 0$, we have:
+{{begin-eqn}}
+{{eqn | l = \binom n {k + 1}
+      | r = \frac {n!} {\left({k + 1}\right)! \left({n - k - 1}\right)!}
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{eqn | r = \frac {n - k} {n - k} \frac {n!} {\left({k + 1}\right)! \left({n - k - 1}\right)!}
+      | c = 
+}}
+{{eqn | r = \frac {n - k} {\left({k + 1}\right) \left({n - k}\right)} \frac {n!} {k! \left({n - k - 1}\right)!}
+      | c = extracting $k + 1$ from its [[Definition:Factorial|factorial]]
+}}
+{{eqn | r = \frac {n - k} {k + 1} \frac {n!} {k! \left({n - k}\right)!}
+      | c = inserting $n - k$ into its [[Definition:Factorial|factorial]]
+}}
+{{eqn | r = \frac {n - k} {k + 1} \binom n k
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{end-eqn}}
+In order for $S_n$ to be [[Definition:Strictly Increasing Sequence|strictly increasing]], it is necessary for $\dfrac {n - k} {k + 1} > 1$.
+So:
+{{begin-eqn}}
+{{eqn | l = \dfrac {n - k} {k + 1}
+      | o = >
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = n - k
+      | o = >
+      | r = k + 1
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = n
+      | o = >
+      | r = 2 k + 1
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = n
+      | o = >
+      | r = 2 \left({k + 1}\right) - 1
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dbinom n {k + 1} > \dbinom n k$ {{iff}} $k + 1$ is less than half of $n$.
+Hence the result.
+{{Qed}}
+[[Category:Binomial Coefficients]]
+h1tjmwqkc19n3s0v7w6d6nvdds38hk9
+\end{proof}<|endoftext|>
+\section{Isometry between Metric Spaces is Continuous}
+Tags: Isometries
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi: M_1 \to M_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
+Then $\phi: M_1 \to M_2$ is a [[Definition:Continuous on Metric Space|continuous mapping]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in A_1$.
+Let $\epsilon \in \R_{>0}$.
+Let $\delta = \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d_1 \left({a, y}\right)
+      | o = <
+      | r = \delta
+      | c = for some $y \in A_1$
+}}
+{{eqn | ll= \implies
+      | l = d_2 \left({\phi \left({a}\right), \phi \left({y}\right)}\right)
+      | o = <
+      | r = \delta
+      | c = as $d_2 \left({\phi \left({a}\right), \phi \left({y}\right)}\right) = d_1 \left({a, y}\right)$
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+So by definition $\phi$ is [[Definition:Continuous at Point of Metric Space|continuous at $a$]].
+As $a \in H$ is arbitrary, it follows that $d_H$ is [[Definition:Continuous on Metric Space|continuous on $H$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Topologically Equivalent Metrics}
+Tags: Topologically Equivalent Metrics
+
+\begin{theorem}
+{{TFAE|def = Topologically Equivalent Metrics}}
+Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $d_1$ and $d_2$ be [[Definition:Topologically Equivalent Metrics/Definition 1|topologically equivalent by definition 1]].
+Let $\struct {B, d}$ and $\struct {C, d'}$ be [[Definition:Metric Space|metric spaces]]:
+Let $f: B \to A$ and $g: A \to C$ be [[Definition:Mapping|mappings]].
+Then by definition:
+:$f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d, d_1}$-continuous]] {{iff}} $f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d, d_2}$-continuous]]
+:$g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d'}$-continuous]] {{iff}} $g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_2, d'}$-continuous]].
+Suppose $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+Let $h: A \to A$ be the [[Definition:Identity Mapping|identity mapping]].
+That is:
+:$\forall a \in A: \map h a = a$
+From [[Identity Mapping is Continuous]], $h$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_2, d_2}$-continuous]].
+Hence because $h$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d'}$-continuous]] {{iff}} $h$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_2, d'}$-continuous]], $h$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d_2}$-continuous]].
+From [[Definition:Metric Space Continuity by Open Set|Metric Space Continuity by Open Set]], it follows that $h^{-1} \sqbrk U$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+But $h^{-1} \sqbrk U = U$, so $U$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+A similar argument, still starting with the same proposition that $g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d'}$-continuous]] {{iff}} $g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_2, d'}$-continuous]], shows that if $U$ is [[Definition:Open Set (Metric Space)|$d_1$-open]] then it is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+Thus $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics/Definition 2|topologically equivalent by definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $d_1$ and $d_2$ be [[Definition:Topologically Equivalent Metrics/Definition 2|topologically equivalent by definition 2]].
+Then by definition:
+:$U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_1$-open]] {{iff}} $U \subseteq A$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+From [[Definition:Metric Space Continuity by Open Set|Metric Space Continuity by Open Set]], we have that:
+:$f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d, d_1}$-continuous]] {{iff}} for every set $U \subseteq A$ which is [[Definition:Open Set (Metric Space)|open]] in $\struct {A, d_1}$, $f^{-1} \sqbrk U$ is [[Definition:Open Set (Metric Space)|open]] in $\struct {B, d}$
+:$f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d, d_2}$-continuous]] {{iff}} for every set $U \subseteq A$ which is [[Definition:Open Set (Metric Space)|open]] in $\struct {A, d_2}$, $f^{-1} \sqbrk U$ is [[Definition:Open Set (Metric Space)|open]] in $\struct {B, d}$.
+Hence $f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d, d_1}$-continuous]] {{iff}} $f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d, d_2}$-continuous]].
+Similarly:
+:$g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d'}$-continuous]] {{iff}} for every set $U \subseteq C$ which is [[Definition:Open Set (Metric Space)|open]] in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is [[Definition:Open Set (Metric Space)|open]] in $\struct {A, d_1}$
+:$g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_2, d'}$-continuous]] {{iff}} for every set $U \subseteq C$ which is [[Definition:Open Set (Metric Space)|open]] in $\struct {C, d'}$, $g^{-1} \sqbrk U$ is [[Definition:Open Set (Metric Space)|open]] in $\struct {A, d_2}$.
+Hence $g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_1, d'}$-continuous]] {{iff}} $g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\tuple {d_2, d'}$-continuous]].
+Thus $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics/Definition 1|topologically equivalent by definition 1]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isometric Metric Spaces are Homeomorphic}
+Tags: Isometries, Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $M_1$ and $M_2$ be [[Definition:Isometric Metric Spaces|isometric]].
+Then $M_1$ and $M_2$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+By the definition of an [[Definition:Isometry (Metric Spaces)|isometry]], there exists a [[Definition:Bijection|bijection]] $f: A_1 \to A_2$ such that:
+: $\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$
+From [[Isometry of Metric Spaces is Homeomorphism]], $f$ is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]] from $M_1$ to $M_2$.
+The result follows by definition of [[Definition:Homeomorphic Metric Spaces|homeomorphic metric spaces]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homeomorphic Metric Spaces are not necessarily Isometric}
+Tags: Isometries, Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $M_1$ and $M_2$ be [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+Then it is not necessarily the case that $M_1$ and $M_2$ are [[Definition:Isometric Metric Spaces|isometric]].
+\end{theorem}
+
+\begin{proof}
+Consider the spaces $\struct {\openint 0 4, d}$ and $\struct {\openint 0 1, d}$, where $d$ is the [[Definition:Euclidean Metric on Real Number Line|Euclidean Metric]].
+By [[Open Real Intervals are Homeomorphic]], they are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+Let $\phi: \openint 0 4 \to \openint 0 1$ be a mapping.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {\map \phi 1, \map \phi 3}
+      | o = \le
+      | r = \map d {0, 1}
+}}
+{{eqn | r = 1
+}}
+{{eqn | o = < 
+      | r = 2
+}}
+{{eqn | r = \map d {1, 3}
+}}
+{{end-eqn}}
+showing that $\phi$ cannot be an [[Definition:Isometry (Metric Spaces)|isometry]].
+Therefore the two spaces above are not [[Definition:Isometric Metric Spaces|isometric]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Mapping between Metrics separated by Scale Factor is Continuous}
+Tags: Identity Mappings, Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
+Let $d_1$ and $d_2$ be such that:
+:$\forall x, y \in A: \map {d_2} {x, y} \le K \map {d_2} {x, y}$
+Let $I_A: A \to A$ be the [[Definition:Identity Mapping|identity mapping]] on $A$.
+Then $I_A$ is [[Definition:Continuous on Metric Space|continuous]] from $M_1$ to $M_2$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$.
+Let $a \in A$.
+Set $\delta = \dfrac \epsilon K$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {d_1} {x, a}
+      | o = <
+      | r = \delta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_2} {\map {I_A} x, \map {I_A} a}
+      | o = \le
+      | r = K \map {d_1} {x, a}
+      | c = 
+}}
+{{eqn | o = <
+      | r = K \delta
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+Hence by definition $I_A$ is [[Definition:Continuous at Point of Metric Space|continuous]] at $a$.
+As $a$ is arbitrary, it follows that this is true for all $a \in A$.
+Thus $I_A$ is [[Definition:Continuous on Metric Space|continuous]] on the whole of $M_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic}
+Tags: Topologically Equivalent Metrics, Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A, d_1}\right)$ and $M_2 = \left({A, d_2}\right)$ be [[Definition:Metric Space|metric spaces]] on the same [[Definition:Underlying Set of Metric Space|underlying set]] $A$.
+Let $d_1$ and $d_2$ be [[Definition:Topologically Equivalent Metrics|topologically equivalent]].
+Then $M_1$ and $M_2$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+By definition, $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent]] on $A$ {{iff}}:
+:For all [[Definition:Metric Space|metric spaces]] $\left({B, d}\right)$ and $\left({C, d'}\right)$:
+:For all [[Definition:Mapping|mappings]] $f: B \to A$ and $g: A \to C$:
+:: $(1): \quad f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\left({d, d_1}\right)$-continuous]] {{iff}} $f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\left({d, d_2}\right)$-continuous]]
+:: $(2): \quad g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\left({d_1, d'}\right)$-continuous]] {{iff}} $g$ is [[Definition:Continuous Mapping (Metric Spaces)|$\left({d_2, d'}\right)$-continuous]].
+Set $B = C = A$.
+The result follows by definition of [[Definition:Homeomorphic Metric Spaces|homeomorphic metric spaces]].
+{{qed}}
+[[Category:Topologically Equivalent Metrics]]
+[[Category:Homeomorphic Metric Spaces]]
+k46uy5cad2030usp0vcgec9fka4wwox
+\end{proof}<|endoftext|>
+\section{Real Vector Space under Chebyshev Distance is Homeomorphic to that under Euclidean Metric}
+Tags: Homeomorphic Metric Spaces, Chebyshev Distance, Euclidean Metric
+
+\begin{theorem}
+Let $\R^n$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Real Vector Space|real vector space]].
+Let $d_\infty: \R^n \times \R^n \to \R$ be the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance on $\R^n$]].
+Let $d_2: \R^n \times \R^n \to \R$ be the [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric on $\R^n$]].
+Let $M_1 = \left({\R^n, d_\infty}\right)$ and $M_2 = \left({\R^n, d_2}\right)$ be the corresponding [[Definition:Metric Space|metric spaces]].
+Then $M_1$ and $M_2$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+From [[Relation between P-Product Metric and Chebyshev Distance on Real Vector Space|Relation between $p$-Product Metric and Chebyshev Distance on Real Vector Space]]:
+:$\forall x, y \in \R^n: d_\infty \left({x, y}\right) \le d_p \left({x, y}\right) \le n^{1/p} d_\infty \left({x, y}\right)$
+The [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] $d_2$ is a special case of the [[Definition:P-Product Metric/Real Vector Space|$p$-product metric]] $d_p$ for $p = 2$.
+It follows by definition that $d_\infty$ and $d_2$ are [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]].
+It follows from [[Lipschitz Equivalent Metrics are Topologically Equivalent]] that $d_\infty$ and $d_2$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent]].
+The result follows from [[Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Homeomorphic Metric Spaces}
+Tags: Homeomorphic Metric Spaces
+
+\begin{theorem}
+{{TFAE|def = Homeomorphic Metric Spaces}}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+In order to prove the assertion it is sufficient to prove that the conditions for [[Definition:Homeomorphic Metric Spaces|homeomorphism]] in definitions $2$ to $4$ are [[Definition:Necessary and Sufficient|necessary and sufficient conditions]] for $f$ and $f^{-1}$ to be [[Definition:Continuous on Metric Space|continuous]] on $M_1$ and $M_2$ respectively.
+=== Definition 1 implies Definition 2 ===
+Let $f$ and $f^{-1}$ both be [[Definition:Continuous on Metric Space|continuous by definition $2$]].
+Let $U \subseteq A_1$ be [[Definition:Open Set of Metric Space|open in $M_1$]].
+As $f^{-1}$ is [[Definition:Continuous on Metric Space|continuous]], $\left({f^{-1}}\right)^{-1} \left[{U}\right] = f \left[{U}\right]$ is [[Definition:Open Set of Metric Space|open in $M_2$]].
+That is, $f \left[{U}\right] = f \left[{U}\right]$ is [[Definition:Open Set of Metric Space|open in $M_2$]].
+Let $f \left[{U}\right] \subseteq A_2$ be [[Definition:Open Set of Metric Space|open in $M_2$]].
+Then as $f$ is [[Definition:Continuous on Metric Space|continuous]], $f^{-1} \left[{f \left[{U}\right]}\right] = U$ is [[Definition:Open Set (Metric Space)|open in $M_1$]].
+Thus:
+: for all $U \subseteq A_1$, $U$ is an [[Definition:Open Set of Metric Space|open set]] of $M_1$ {{iff}} $f \left[{U}\right]$ is an [[Definition:Open Set of Metric Space|open set]] of $M_2$.
+{{qed|lemma}}
+=== Definition 2 implies Definition 4 ===
+Let [[Definition:Homeomorphism/Metric Spaces/Definition 2|definition $2$]] hold.
+Let $a \in A_1$.
+Let $N \subseteq A_1$.
+Then:
+:$N$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$
+{{iff}}:
+:$N$ contains an [[Definition:Open Set of Metric Space|open set]] $U$ containing $a$
+{{iff}}:
+:$f \left[{N}\right]$ contains an [[Definition:Open Set of Metric Space|open set]] $U' = f \left[{U}\right]$ containing $f \left({a}\right)$
+{{iff}}:
+:$f \left[{N}\right]$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] $f \left({a}\right)$.
+{{qed|lemma}}
+=== Definition 4 implies Definition 1 ===
+Let [[Definition:Homeomorphism/Metric Spaces/Definition 4|definition $4$]] hold.
+Let $a \in A_1$.
+Let $U \subseteq A_2$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $f \left({a}\right)$.
+Then $f \left[{f^{-1} \left[{U}\right]}\right]$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] $f \left({a}\right)$.
+Hence $f^{-1} \left[{U}\right]$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $a$.
+Thus $f$ is [[Definition:Continuous on Metric Space|continuous]].
+Similarly, let $b \in A_2$.
+Let $V \subseteq A_1$ be a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $f^{-1} \left({b}\right)$.
+Then $f^{-1} \left[{f \left[{V}\right]}\right]$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] $f^{-1} \left({b}\right)$.
+Hence $f \left[{V}\right]$ is a [[Definition:Neighborhood (Metric Space)|neighborhood]] of $f \left({f^{-1} \left({b}\right)}\right) = b$.
+Thus $f^{-1}$ is [[Definition:Continuous on Metric Space|continuous]].
+{{qed|lemma}}
+=== Definition 2 iff Definition 3 ===
+This is demonstrated in [[Continuity of Mapping between Metric Spaces by Closed Sets]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Liouville's Theorem (Number Theory)}
+Tags: Number Theory, Transcendental Number Theory, Liouville's Theorem (Number Theory)
+
+\begin{theorem}
+Let $x$ be an [[Definition:Irrational Number|irrational number]] that is [[Definition:Algebraic Number|algebraic]] of [[Definition:Degree of Algebraic Number|degree]] $n$.
+Then there exists a constant $c > 0$ (which can depend on $x$) such that:
+:$\size {x - \dfrac p q} \ge \dfrac c {q^n}$
+for every pair $p, q \in \Z$ with $q \ne 0$.
+\end{theorem}
+
+\begin{proof}
+Let $r_1, r_2, \ldots, r_k$ be the [[Definition:Rational Number|rational]] [[Definition:Root of Polynomial|roots]] of a [[Definition:Polynomial (Analysis)|polynomial]] $P$ of [[Definition:Degree of Polynomial|degree]] $n$ that has $x$ as a [[Definition:Root of Polynomial|root]].
+Since $x$ is [[Definition:Irrational Number|irrationa]]l, it does not equal any $r_i$. 
+Let $c_1 > 0$ be the minimum of $\size {x - r_i}$.
+If there are no $r_i$, let $c_1 = 1$.
+Now let $\alpha = \dfrac p q$ where $\alpha \notin \set {r_1, \ldots, r_k}$. 
+Then:
+{{begin-eqn}}
+{{eqn | l = \map P \alpha
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map P \alpha}
+      | o = \ge
+      | r = \frac 1 {q^n}
+      | c = as $\map P \alpha$ is a multiple of $\dfrac 1 {q^n}$
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map P x - \map P \alpha}
+      | o = \ge
+      | r = \frac 1 {q^n}
+      | c = because $\map P x = 0$
+}}
+{{end-eqn}}
+Suppose:
+:$\displaystyle \map P x = \sum_{k \mathop = 0}^n a_k x^k$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map P x - \map P \alpha
+      | r = \sum_{k \mathop = 0}^n a_k x^k - \sum_{k \mathop = 0}^n a_k {\alpha }^k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | r = \sum_{k \mathop = 0}^n a_k \paren {x^k - {\alpha }^k }
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | r = \sum_{k \mathop = 1}^n a_k \paren {x^k - {\alpha }^k }
+      | c = $x^0 - {\alpha }^0 = 0$
+}}
+{{eqn | ll= \leadsto
+      | r = \sum_{k \mathop = 1}^n a_k \paren {x - \alpha} \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i
+      | c = [[Difference of Two Powers]]
+}}
+{{eqn | ll= \leadsto
+      | r = \paren {x - \alpha} \sum_{k \mathop = 1}^n a_k \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i
+      | c = 
+}}
+{{end-eqn}}
+'''Case 1''': If $\size {x - \alpha} \le 1, \alpha \notin \set {r_1, \ldots, r_k}$, then
+{{begin-eqn}}
+{{eqn | l = \size {\alpha } - \size{x }
+      | o = \le
+      | r = \size {x - \alpha}
+      | c = [[Reverse Triangle Inequality]]
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\alpha } - \size{x }
+      | o = \le
+      | r = 1
+      | c = Transitive
+}}
+{{eqn | ll= \leadsto
+      | l = \size \alpha
+      | o = \le
+      | r = \size x + 1
+      | c = rearranging
+}}
+{{end-eqn}}
+Therefore
+{{begin-eqn}}
+{{eqn | l = \size {\displaystyle \map P x - \map P \alpha }
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \alpha^i }
+      | c = [[Triangle Inequality]]
+}}
+{{eqn | ll= \leadsto
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \paren {\size x + 1 }^i }
+      | c = substituting $\size x + 1$ for $\alpha$
+}}
+{{eqn | ll= \leadsto
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 } \paren {\frac {\size x + 1 } {x } }^i }
+      | c = moving $ x^{-i} $ over
+}}
+{{eqn | ll= \leadsto
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k x^{k - 1 } } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {\paren {1 + \frac 1 x }^i }
+      | c = moving $x^{k - 1 }$ out of the sum
+}}
+{{eqn | ll= \leadsto
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k x^{k - 1 } } \frac {\paren {1 + \frac 1 x }^k - 1 } {\paren {1 + \frac 1 x } - 1 }
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k x^{k } } \paren {\paren {1 + \frac 1 x }^k - 1 }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o = \le
+      | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \paren {\paren {\size {x } + 1 }^k - \size {x }^{k } }
+      | c = Distribute
+}}
+{{end-eqn}}
+To summarize:
+:$\size {\map P x - \map P \alpha} \le \size {x - \alpha} c_x$
+where:
+:$\displaystyle c_x = \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \paren {\paren {\size {x } + 1 }^k - \size {x }^{k } }$
+So for such $\alpha$:
+:$\size {x - \alpha} \ge \dfrac {\size {\map P x - \map P \alpha} } {c_x} \ge \dfrac 1 {c_x q^n}$
+'''Case 2''': If $\size {x - \alpha} \gt 1, \alpha \notin \set {r_1, \ldots, r_k}$ then:
+:$\size {x - \alpha} \gt 1 \ge \dfrac 1 {q^n}$
+'''Case 3''': If $\alpha \in \set {r_1, \ldots, r_k}$, then:
+:$\size {x - \alpha} \ge c_1 \ge \dfrac {c_1} {q^n}$
+$c = \begin{cases} \dfrac 1 {c_x} & : \size {x - \alpha} \le 1, \alpha \notin \set {r_1, \ldots, r_k} \\ 1 & : \size {x - \alpha} \gt 1, \alpha \notin \set {r_1, \ldots, r_k} \\ c_1 & : \alpha \in \set {r_1, \ldots, r_k} \end{cases}$
+Then:
+:$\size {x - \dfrac p q} \ge \dfrac c {q^n}$
+for all $\dfrac p q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Liouville Numbers are Irrational}
+Tags: Number Theory
+
+\begin{theorem}
+[[Definition:Liouville Number|Liouville numbers]] are [[Definition:Irrational Number|irrational]].
+\end{theorem}
+
+\begin{proof}
+Let $x$ be a [[Definition:Liouville Number|Liouville number]]. 
+Aiming for a [[Proof by Contradiction|contradiction]], suppose $x$ were [[Definition:Rational Number|rational]], that is:
+: $x = \dfrac a b$
+with $a, b \in \Z$ and $b > 0$.
+By definition of a [[Definition:Liouville Number|Liouville number]], for all $n \in \N$, there exist $p,q \in \Z$ (which may depend on $n$) with $q > 1$ such that:
+:$0 < \left \lvert x - \dfrac p q \right \rvert < \dfrac 1 {q^n}$
+Let $n$ be a [[Definition:Positive Integer|positive integer]] such that $2^{n - 1} > b$.
+Let $p$ and $q$ be any [[Definition:Integer|integers]] with $q > 1$.
+We have:
+:$\left \lvert x - \dfrac p q \right \rvert = \dfrac {\left \lvert a q - b p \right \rvert} {b q}$
+If $\left \lvert a q - b p \right \rvert = 0$, this would violate the first inequality.
+If $\left \lvert a q - b p \right \rvert \ne 0$, then:
+{{begin-eqn}}
+{{eqn |l = \left \lvert x - \frac p q \right \rvert
+      |r = \frac 1 {b q}
+      |c = as $\left \lvert a q - b p \right \rvert$ is a [[Definition:Positive Integer|positive integer]]
+      |o = \ge
+}}
+{{eqn |r = \frac 1 {2^{n - 1} q}
+      |c = by our choice of $n$
+      |o = >
+}}
+{{eqn |r = \frac 1 {q^n}
+      |c = as $q > 1$ by definition
+      |o = \ge
+}}
+{{end-eqn}}
+which would violate the second inequality.
+Therefore, in any case, if $n$ is sufficiently large, there cannot exist integers $p$ and $q$ with $q > 1$ satisfying the two inequalities.
+This is a contradiction.
+Thus $x$ is [[Definition:Irrational Number|irrational]].
+{{qed}}
+[[Category:Number Theory]]
+6szpys8wrdmc3pufjkt99dsohygqay5
+\end{proof}<|endoftext|>
+\section{Cardinality of Power Set of Natural Numbers Equals Cardinality of Real Numbers}
+Tags: Set Theory
+
+\begin{theorem}
+The [[Definition:Cardinality|cardinality]] of the [[Definition:Power Set|power set]] of the [[Definition:Natural Numbers|natural numbers]] is equal to the [[Definition:Cardinality of Continuum|cardinality of the real numbers]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Set Theory]]
+1fjmusthm13xq6mbsj08iw4zyuc2rn0
+\end{proof}<|endoftext|>
+\section{Liouville's Constant is Transcendental}
+Tags: Transcendental Number Theory, Transcendental Numbers
+
+\begin{theorem}
+[[Definition:Liouville's Constant|Liouville's constant]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop \ge 1} \dfrac 1 {10^{n!} }
+      | r = \frac 1 {10^1} + \frac 1 {10^2} + \frac 1 {10^6} +  \frac 1 {10^{24} } + \cdots
+      | c = 
+}}
+{{eqn | r = 0.11000 \, 10000 \, 00000 \, 00000 \, 00010 \, 00 \ldots
+      | c = 
+}}
+{{end-eqn}}
+is [[Definition:Transcendental|transcendental]].
+\end{theorem}
+
+\begin{proof}
+Let $q = 10^{n!}$ and write:
+:$\displaystyle L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }$
+for some suitable $p \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {L - \frac p q}
+      | r = \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }
+      | c =  
+}}
+{{eqn | r = \frac 2 {10^{\paren {n + 1}!} }
+      | o = \le
+      | c = 
+}}
+{{eqn | r = \frac 2 {q^{n + 1} }
+      | c = 
+}}
+{{eqn | r = \frac 1 {q^n}
+      | o = <
+      | c = as $q \ge 10$ for all $n \in \N$
+}}
+{{end-eqn}}
+Thus, by definition, $L$ is a [[Definition:Liouville Number|Liouville number]].
+Therefore, by [[Liouville's Theorem (Number Theory)|Liouville's Theorem]], $L$ is [[Definition:Transcendental Number|transcendental]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isometry of Metric Spaces is Equivalence Relation}
+Tags: Isometries, Equivalence Relations
+
+\begin{theorem}
+Let $M_1$ and $M_2$ be [[Definition:Metric Space|metric spaces]].
+Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are [[Definition:Isometric Metric Spaces|isometric]].
+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 $M$ be a [[Definition:Topological Space|topological space]].
+From [[Identity Mapping on Metric Space is Isometry]], the [[Definition:Identity Mapping|identity mapping]] $I_M: M \to M$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+So $M \sim M$, and $\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $M_1$ and $M_2$ be [[Definition:Topological Space|topological spaces]] such that $M_1 \sim M_2$.
+By definition, there exists an [[Definition:Isometry (Metric Spaces)|isometry]] $f: M_1 \to M_2$.
+From [[Inverse of Isometry of Metric Spaces is Isometry]] it follows that $f^{-1}: M_2 \to M_1$ is also an [[Definition:Isometry (Metric Spaces)|isometry]].
+So $M_2 \sim M_1$, and $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $M_1, M_2, M_3$ be [[Definition:Metric Space|metric spaces]] such that $M_1 \sim M_2$ and $M_2 \sim M_3$.
+By definition, there exist [[Definition:Isometry (Metric Spaces)|isometries]] $f: M_1 \to M_2$ and $g: M_2 \to M_3$.
+From [[Composite of Isometries is Isometry]] it follows that $g \circ f: M_1 \to M_3$ is also an [[Definition:Isometry (Metric Spaces)|isometry]].
+So $M_1 \sim M_3$, 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{Homeomorphism of Metric Spaces is Equivalence Relation}
+Tags: Homeomorphic Metric Spaces, Equivalence Relations
+
+\begin{theorem}
+Let $M_1$ and $M_2$ be [[Definition:Metric Space|metric spaces]].
+Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are [[Definition:Homeomorphic Metric 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 $M$ be a [[Definition:Metric Space|metric space]].
+From [[Identity Mapping on Metric Space is Homeomorphism]], the [[Definition:Identity Mapping|identity mapping]] $I_M: M \to M$ is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+So $M \sim M$, and $\sim$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $M_1$ and $M_2$ be [[Definition:Topological Space|topological spaces]] such that $M_1 \sim M_2$.
+By definition, there exists a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]] $f: M_1 \to M_2$.
+From [[Inverse of Homeomorphism between Metric Spaces is Homeomorphism]] it follows that $f^{-1}: M_2 \to M_1$ is also a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+So $M_2 \sim M_1$, and $\sim$ has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $M_1, M_2, M_3$ be [[Definition:Metric Space|metric spaces]] such that $M_1 \sim M_2$ and $M_2 \sim M_3$.
+By definition, there exist [[Definition:Homeomorphism (Metric Spaces)|homeomorphisms]] $f: M_1 \to M_2$ and $g: M_2 \to M_3$.
+From [[Composite of Homeomorphisms between Metric Spaces is Homeomorphism]] it follows that $g \circ f: M_1 \to M_3$ is also a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+So $M_1 \sim M_3$, 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{Isometry of Metric Spaces is Homeomorphism}
+Tags: Isometries, Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: M_1 \to M_2$ be an [[Definition:Isometry (Metric Spaces)|isometry]].
+Then $f$ is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]] from $M_1$ to $M_2$.
+\end{theorem}
+
+\begin{proof}
+By the definition of an [[Definition:Isometry (Metric Spaces)|isometry]], $f$ is a [[Definition:Bijection|bijection]] $f: A_1 \to A_2$ such that:
+: $\forall a, b \in A_1: d_1 \left({a, b}\right) = d_2 \left({\phi \left({a}\right), \phi \left({b}\right)}\right)$
+By [[Isometry between Metric Spaces is Continuous]], $f$ is a [[Definition:Continuous on Metric Space|continuous mapping]] from $M_1$ to $M_2$.
+By [[Isometry between Metric Spaces is Continuous/Corollary|the corollary to Isometry between Metric Spaces is Continuous]], $f^{-1}$ is a [[Definition:Continuous on Metric Space|continuous mapping]] from $M_2$ to $M_1$.
+Thus $f$ is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]] from $M_1$ to $M_2$.
+Hence the result.
+{{qed}}
+[[Category:Isometries]]
+[[Category:Homeomorphic Metric Spaces]]
+mygfttyy9exz01uppqpaf2h4p3jq0ky
+\end{proof}<|endoftext|>
+\section{Equivalence Class of Isometries is Subset of Equivalence Class of Homeomorphisms}
+Tags: Homeomorphic Metric Spaces, Isometries, Equivalence Classes
+
+\begin{theorem}
+Let $\mathcal C_I$ be an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Isometry (Metric Spaces)|isometries]] on the [[Definition:Set|set]] of [[Definition:Metric Space|metric spaces]].
+Then $\mathcal C_I$ is a [[Definition:Subset|subset]] of an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Homeomorphic Metric Spaces|homeomorphisms]] on the [[Definition:Set|set]] of [[Definition:Metric Space|metric spaces]].
+\end{theorem}
+
+\begin{proof}
+From [[Isometry of Metric Spaces is Equivalence Relation]], all [[Definition:Isometry (Metric Spaces)|isometries]] can be partitioned into [[Definition:Equivalence Class|equivalence classes]].
+From [[Homeomorphism of Metric Spaces is Equivalence Relation]], all [[Definition:Homeomorphic Metric Spaces|homeomorphisms]] can be partitioned into [[Definition:Equivalence Class|equivalence classes]].
+Let $\mathcal C_I$ be an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Isometry (Metric Spaces)|isometries]].
+Let $f: M_1 \to M_2$ be an [[Definition:Element|element]] of $\mathcal C_I$ where $M_1$ and $M_2$ are some [[Definition:Metric Space|metric spaces]].
+From [[Isometry of Metric Spaces is Homeomorphism]], $\mathcal C_I$ is a [[Definition:Homeomorphic Metric Spaces|homeomorphism]].
+Thus $\mathcal C_I$ is an [[Definition:Element|element]] of some [[Definition:Equivalence Class|equivalence class]] of [[Definition:Homeomorphic Metric Spaces|homeomorphisms]] on the [[Definition:Set|set]] of [[Definition:Metric Space|metric spaces]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Local Uniform Convergence and Compact Convergence}
+Tags: Complex Analysis, Uniform Convergence
+
+\begin{theorem}
+Let $U \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open set]] of the [[Definition:Complex Plane|complex plane]].
+Let $f_n: U \to \C$ be a [[Definition:Sequence|sequence]] of functions which [[Definition:Pointwise Convergence|converges pointwise]] to $f: U \to \C$.
+Then $f_n$ [[Definition:Uniform Convergence|converges to $f$ uniformly]] on all [[Definition:Compact Subset of Complex Plane|compact subsets]] of $U$ {{iff}} $f_n$ [[Definition:Locally Uniform Convergence|converges locally uniformly]] on $U$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Suppose that $f_n$ [[Definition:Uniform Convergence|converges to $f$ uniformly]] on all [[Definition:Compact Subset of Complex Plane|compact subsets]] of $U$.
+Let $z \in U$.
+Since $U$ is [[Definition:Open Set (Complex Analysis)|open]], there is an [[Definition:Open Ball|open ball]] $B_\varepsilon \left({z}\right)$, around $z$, of [[Definition:Radius of Open Ball|radius]] $\varepsilon$, with the [[Definition:Closed Ball|closure of the ball]], $\overline{B_\varepsilon \left({z}\right)}$, contained in $U$.
+{{proofread}}
+Since $\overline {B_\varepsilon \left({z}\right)}$ is [[Definition:Closed Set (Complex Analysis)|closed]] and [[Definition:Bounded Subset of Complex Plane|bounded]], it is [[Definition:Compact Subset of Complex Plane|compact]], by the [[Heine-Borel Theorem/Euclidean Space|Heine-Borel Theorem]]. 
+We then have by assumption that $f_n$ [[Definition:Uniform Convergence|converges uniformly]] on $\overline{B_\varepsilon \left({z}\right)}$.
+{{proofread}}
+Thus $f_n$ [[Definition:Locally Uniform Convergence|converges locally uniformly]] on $U$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Now suppose that $f_n$ [[Definition:Locally Uniform Convergence|converges to $f$ locally uniformly]].
+That is, for every $z \in U$, there is an $r$ so that $B_r \left({z}\right) \subset U$ and $f_n$ [[Definition:Uniform Convergence|converges uniformly]] on $B_r \left({z}\right)$.
+Let $K \subset U$ be an arbitrary [[Definition:Compact Subset of Complex Plane|compact subset]].
+Let $\left\{ {B_{r_z} \left({z}\right)}\right\}_{z \mathop \in K}$ be an [[Definition:Open Cover|open cover]] of $K$ by [[Definition:Open Ball|open balls]], one centered at each $z \in K$, such that $f_n$ [[Definition:Uniform Convergence|converges uniformly]] on each [[Definition:Open Ball|ball]].
+Since $K$ is [[Definition:Compact Subset of Complex Plane|compact]], there is a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $\left\{ {B_{r_i} \left({z_i}\right) }\right\}_{i \mathop = 1}^n$
+which [[Definition:Open Cover|covers]] $K$.
+Let $\varepsilon > 0$, and $1 \le i \le n$. 
+Then there is an $N_i$ large enough so that:
+:$\displaystyle \sup_{z \mathop \in B_{r_i} \left({z_i}\right)} \left\vert{f_n \left({z}\right) - f \left({z}\right)}\right\vert < \varepsilon$
+for all $n > N_i$, since the [[Definition:Uniform Convergence|convergence is uniform]].
+Let $N = \displaystyle \max_{1 \mathop \le i \mathop \le n} \left\{N_i\right\}$.
+Let $B = \displaystyle \bigcup_{i \mathop = 1}^n B_{r_i} \left( z_i \right)$.
+Then:
+: $\displaystyle \sup_{z \in B} \left\vert f_n \left({z}\right) - f \left({z}\right) \right\vert < \varepsilon$
+for all $n > N$.
+Thus $f_n$ [[Definition:Uniform Convergence|converges uniformly]] on $B$, and so also on $K$.
+{{qed}}
+[[Category:Complex Analysis]]
+[[Category:Uniform Convergence]]
+rkvkyq8spt2njdh00fbbgiyc18kj8lv
+\end{proof}<|endoftext|>
+\section{Frattini's Argument}
+Tags: Group Theory, Normalizers
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $K$ be a [[Definition:Finite Group|finite]] [[Definition:Normal Subgroup|normal subgroup]] of $G$, and $p$ a [[Definition:Prime Number|prime]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Structure|order]] of $K$.
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $K$, and $\map {N_G} P$ the [[Definition:Normalizer|normalizer]] of $P$ in $G$.
+Then:
+:$G = \map {N_G} P \circ K = K \circ \map {N_G} P$
+\end{theorem}
+
+\begin{proof}
+Let $g \in G$.
+Since $K$ is [[Definition:Normal Subgroup|normal]] in $G$, and $P \subset K$, the [[Definition:Conjugate of Group Subset|conjugate]] $g \circ P \circ g^{-1}$ of $P$ is also a [[Definition:Subset|subset]] of $K$.
+From [[Inner Automorphism is Automorphism]], $g \circ P \circ g^{-1}$ is a [[Definition:Subgroup|subgroup]] of $K$ of the same [[Definition:Order of Structure|order]] as $P$.
+Thus $g \circ P \circ g^{-1}$ is also a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $K$.
+By [[Second Sylow Theorem]]:
+:$\exists k \in K: g \circ P \circ g^{-1} = k \circ P \circ k^{-1}$
+Then:
+:$g^{-1} \circ k \circ P \circ k^{-1} \circ g = P$
+and so by definition of [[Definition:Normalizer|normalizer]]:
+:$k^{-1} \circ g \in \map {N_G} P$
+We can then write $g$ as
+:$g = k \circ \paren {k^{-1} \circ g} \in K \circ \map {N_G} P$
+Since $K$ is [[Definition:Normal Subgroup|normal]]:
+:$\map {N_G} P \circ K = K \circ \map {N_G} P$
+by [[Subset Product with Normal Subgroup as Generator]].
+{{qed}}
+{{Namedfor|Giovanni Frattini|cat = Frattini}}
+\end{proof}<|endoftext|>
+\section{Sum of Integrals on Complementary Sets}
+Tags: Measure Theory
+
+\begin{theorem}
+Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
+Let $A, E \in \Sigma$ with $A \subseteq E$.
+Let $f$ be a [[Definition:Integrable Function on Measure Space|$\mu$-integrable function]] on $X$.
+Then 
+:$\displaystyle \int_E f \rd \mu = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu$
+\end{theorem}
+
+\begin{proof}
+Let $\chi_E$ be the [[Definition:Characteristic Function of Set|characteristic function]] of $E$.
+Because $A$ and $E \setminus A$ are [[Definition:Disjoint Sets|disjoint]]:
+:$A \cap \paren {E \setminus A} = \O$
+By [[Characteristic Function of Union/Variant 2|Characteristic Function of Union]]:
+:$\chi_E = \chi_A + \chi_{E \mathop \setminus A}$ 
+[[Definition:Integral of Integrable Function|Integrating]] $f$ over $E$ gives:
+{{begin-eqn}}
+{{eqn | l = \int_E f \rd \mu
+      | r = \int \chi_E \cdot f \rd \mu
+      | c = {{Defof|Integral of Integrable Function over Measurable Set}}
+}}
+{{eqn | r = \int \paren {\chi_A + \chi_{E \mathop \setminus A} } \cdot f \rd \mu
+      | c = by the above argument
+}}
+{{eqn | r = \int \paren {\chi_A \cdot f + \chi_{E \mathop \setminus A} \cdot f} \rd \mu
+      | c = [[Pointwise Operation on Distributive Structure is Distributive]]
+}}
+{{eqn | r = \int \chi_A \cdot f \rd \mu + \int \chi_{E \mathop \setminus A} \cdot f \rd \mu
+      | c = [[Integral of Integrable Function is Additive]]
+}}
+{{eqn | r = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu
+      | c = {{Defof|Integral of Integrable Function over Measurable Set}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Pointwise Operation on Algebraic Structure}
+Tags: Pointwise Operations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] such that $S \ne \varnothing$.
+Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $T^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$.
+Let $f, g \in T^S$, that is, let $f: S \to T$ and $g: S \to T$ be [[Definition:Mapping|mappings]].
+Let $\oplus: T^S \to T^S$ be the [[Definition:Pointwise Operation|pointwise operation on $T^S$ induced by $\circ$]].
+Then $\oplus$ is [[Definition:Closed Operation|closed]] on $T^S$ {{iff}} $\struct {T, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\struct {T, \circ}$ be [[Definition:Closed Algebraic Structure|closed]].
+Let $x \in S$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall f, g \in T^S:
+      | l = \map {f \oplus g} x
+      | r = \map f x \circ \map g x
+      | c = {{Defof|Pointwise Operation}}
+}}
+{{eqn | o = \in
+      | r = T
+      | c = as $\struct {T, \circ}$ is [[Definition:Closed Algebraic Structure|closed]]
+}}
+{{end-eqn}}
+So $\oplus$ is [[Definition:Closed Operation|closed]] on $T^S$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\oplus$ is [[Definition:Closed Operation|closed]] on $T^S$.
+{{AimForCont}} $\struct {T, \circ}$ is not [[Definition:Closed Algebraic Structure|closed]].
+Then:
+:$(1): \quad \exists s, t \in T: s \circ t \notin T$
+By definition, $T^S$ is the [[Definition:Set of All Mappings|set of all mappings]] from $S$ \to $T$.
+As $S \ne \O$ it follows that $\exists x \in S$.
+Thus, let $x \in S$ be arbitrary.
+Let $f, g \in T^S$ such that:
+:$(2): \quad \map f x = s, \map g x = t$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f \oplus g} x
+      | r = \map f x \circ \map g x
+      | c = {{Defof|Pointwise Operation}}
+}}
+{{eqn | r = s \circ t
+      | c = from $(2)$
+}}
+{{eqn | o = \notin
+      | r = T
+      | c = from $(1)$
+}}
+{{end-eqn}}
+That is, $\oplus$ is not [[Definition:Closed Operation|closed]] on $T^S$.
+From that [[Proof by Contradiction|contradiction]] it follows that $\struct {T, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+[[Category:Pointwise Operations]]
+aawr0dnrannkhib34n0pcxs1d4xq9t9
+\end{proof}<|endoftext|>
+\section{Pointwise Operation on Distributive Structure is Distributive}
+Tags: Pointwise Operations, Distributive Operations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\left({T, +, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] with two [[Definition:Binary Operation|operations]] $+$ and $\circ$.
+Let $T^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$.
+Let $\left({T^S, +', \circ'}\right)$ be the [[Definition:Induced Structure|structure induced]] on $T^S$ by $+$ and $\circ$.
+Let $\circ$ be [[Definition:Distributive Operation|distributive]] over $+$ in $T$.
+Then $\circ'$ is [[Definition:Distributive Operation|distributive]] over $+'$ in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $f, g, h: S \to T$ be elements of $T^S$.
+Let $x \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left ({f \circ' \left({g +' h}\right)}\right) \left({x}\right)
+      | r = f \left({x}\right) \circ \left({g \left({x}\right) + h \left({x}\right)}\right)
+      | c = {{Defof|Induced Structure}}
+}}
+{{eqn | r = \left({f \left({x}\right) \circ g \left({x}\right)}\right) + \left({f \left({x}\right) \circ h \left({x}\right)}\right)
+      | c = $\circ$ is [[Definition:Distributive Operation|distributive]] over $+$ on $T$
+}}
+{{eqn | r = \left ({\left ({f \circ' g}\right) +' \left ({f \circ' h}\right)}\right) \left({x}\right)
+      | c = {{Defof|Induced Structure}}
+}}
+{{end-eqn}}
+Similarly:
+:$\left ({\left({g +' h}\right) \circ' f}\right) \left({x}\right) = \left ({\left ({g \circ' f}\right) +' \left ({h \circ' f}\right)}\right) \left({x}\right)$
+Hence the result.
+{{qed}}
+[[Category:Pointwise Operations]]
+[[Category:Distributive Operations]]
+i7v04c2cehmt02hfx4jsqykh9wx9boy
+\end{proof}<|endoftext|>
+\section{Identity Mapping on Metric Space is Homeomorphism}
+Tags: Homeomorphic Metric Spaces, Topologically Equivalent Metrics, Identity Mappings
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+The [[Definition:Identity Mapping|identity mapping]] $I_A: M \to M$ defined as:
+:$\forall x \in A: \map {I_A} x = x$
+is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+We have [[Identity Mapping is Bijection]].
+We also have [[Identity Mapping is Continuous/Metric Space|Identity Mapping is Continuous]].
+Hence, by definition, $I_T$ is a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+{{qed}}
+[[Category:Homeomorphic Metric Spaces]]
+[[Category:Topologically Equivalent Metrics]]
+[[Category:Identity Mappings]]
+sxh0koi2n9pjcxoqy02l6jki7ki9tly
+\end{proof}<|endoftext|>
+\section{Inverse of Homeomorphism between Metric Spaces is Homeomorphism}
+Tags: Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M = \left({A_1, d_1}\right), M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: M_1 \to M_2$ be a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+Then $f^{-1}: M_2 \to M_1$ is also a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Homeomorphism (Metric Spaces)|homeomorphism]] is a [[Definition:Bijection|bijection]] such that both $f$ and $f^{-1}$ are [[Definition:Continuous on Metric Space|continuous]].
+As $f$ is a [[Definition:Bijection|bijection]] then by [[Bijection iff Inverse is Bijection]], so is $f^{-1}$.
+So by definition $f^{-1}$ is a [[Definition:Bijection|bijection]] such that both $f^{-1}$ and $\left({f^{-1}}\right)^{-1}$ are [[Definition:Continuous on Metric Space|continuous]].
+The result follows from [[Inverse of Inverse of Bijection]].
+{{qed}}
+[[Category:Homeomorphic Metric Spaces]]
+2f2r3swyhcm2fzhrr2czygk18dmhpsd
+\end{proof}<|endoftext|>
+\section{Composite of Homeomorphisms between Metric Spaces is Homeomorphism}
+Tags: Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1, M_2, M_3$ be [[Definition:Metric Space|metric spaces]]. 
+Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be [[Definition:Homeomorphic Metric Spaces|homeomorphisms]].
+Then $g \circ f: M_1 \to M_3$ is also a [[Definition:Homeomorphic Metric Spaces|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Homeomorphic Metric 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:Homeomorphic Metric Spaces|homeomorphism]], $f$ and $g$ are both [[Definition:Continuous on Metric Space|continuous mappings]].
+From [[Composite of Continuous Mappings on Metric Spaces is Continuous]] it follows that $g \circ f$ is also a [[Definition:Continuous on Metric Space|continuous mapping]].
+Hence the result, from definition of [[Definition:Homeomorphic Metric Spaces|homeomorphism]].
+{{qed}}
+[[Category:Homeomorphic Metric Spaces]]
+pzwqca5p5a19hs7lun1ktkbdqz5vqg8
+\end{proof}<|endoftext|>
+\section{Metrizable Space is Hausdorff}
+Tags: Hausdorff Spaces, Metrizable Topologies
+
+\begin{theorem}
+Let $T$ be a [[Definition:Metrizable Space|metrizable]] [[Definition:Topological Space|topological space]].
+Then $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Homeomorphic Topological Spaces|homeomorphic]] to a [[Definition:Topological Space|topological space]] $\struct {S, \tau_d}$ such that $\tau_d$ is the [[Definition:Topology Induced by Metric|topology induced]] by a [[Definition:Metric|metric]] $d$ on $S$.
+From [[Metric Space is Hausdorff]], $\struct {S, d}$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+As the [[Definition:Open Set (Metric Space)|open sets]] of $\struct {S, d}$ are the same as the [[Definition:Open Set (Topology)|open sets]] of $\struct {S, \tau_d}$, it follows that $\struct {S, \tau_d}$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+From [[T2 Space is Preserved under Homeomorphism|$T_2$ Space is Preserved under Homeomorphism]] it follows that $T$ is also a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood of Point in Metrizable Space contains Closed Neighborhood}
+Tags: Metrizable Topologies, Neighborhoods
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Metrizable Space|metrizable]] [[Definition:Topological Space|topological space]].
+Let $x \in S$ be an arbitrary point of $T$.
+Let $N$ be a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+Then $N$ has as a [[Definition:Subset|subset]] a [[Definition:Neighborhood of Point|neighborhood]] $V$ of $x$ such that $V$ is [[Definition:Closed Set (Topology)|closed]].
+\end{theorem}
+
+\begin{proof}
+Since $N$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$, there exists an [[Definition:Open Set (Topology)|open]] set $U \subseteq N$ containing $x$ by definition.
+As $\struct {S, \tau}$ is [[Definition:Metrizable Space|metrizable]], the set $U$ is [[Definition:Open Set (Metric Space)|open]] with respect to some [[Definition:Metric Space|metric space]] $\struct {S, d}$.
+Hence:
+:$\exists \epsilon > 0: \map {\mathcal B_\epsilon} x \subseteq U$
+where $\map {\mathcal B_\epsilon} x$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$.
+Taking $\zeta = \dfrac \epsilon 2$, we have that the [[Definition:Closed Ball|closed $\zeta$-ball]] of $x$ is contained in $\map {\mathcal B_\epsilon} x$, and hence is a [[Definition:Subset|subset]] of $U \subseteq N$. 
+This [[Definition:Closed Ball|closed ball]] is a [[Definition:Closed Neighborhood|closed neighborhood]] of $x$ in $\struct {S, \tau}$, and is a [[Definition:Subset|subset]] of $N$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Underlying Set of Topological Space is Closed}
+Tags: Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then the [[Definition:Underlying Set of Topological Space|underlying set]] $S$ of $T$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Closed Set (Topology)|closed set]], $U$ is [[Definition:Open Set (Topology)|open]] in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Empty Set is Element of Topology]], $\varnothing$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Set Difference with Empty Set is Self]], we have $S \setminus \varnothing = S$, so $S$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basic Properties of Neighborhood in Topological Space}
+Tags: Neighborhoods
+
+\begin{theorem}
+This page gathers together the basic properties of a [[Definition:Neighborhood of Point|neighborhood of a point]] in a [[Definition:Topological Space|topological space]].
+\end{theorem}<|endoftext|>
+\section{Point in Topological Space has Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Then there exists in $T$ at least one [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Then $S$ itself is a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Topological Space is Element of its Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Let $N$ be a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $T$.
+Then $a \in N$.
+\end{theorem}
+
+\begin{proof}
+Trivially follows by definition of [[Definition:Neighborhood of Point|neighborhood]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Superset of Neighborhood in Topological Space is Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Let $N$ be a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $T$.
+Let $N \subseteq N' \subseteq S$.
+Then $N'$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $T$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Neighborhood of Point|neighborhood]]:
+:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
+where $U$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+By [[Subset Relation is Transitive]]:
+:$U \subseteq N'$
+The result follows by definition of [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Neighborhoods in Topological Space is Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Let $M, N$ be a [[Definition:Neighborhood of Point|neighborhoods]] of $x$ in $T$.
+Then $M \cap N$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $T$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Neighborhood of Point|neighborhood]]:
+: $\exists U_1 \in \tau: x \in U_1 \subseteq M$
+where $U_1$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+: $\exists U_2 \in \tau: x \in U_2 \subseteq N$
+where $U_2$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+Thus by [[Set Intersection Preserves Subsets]]:
+: $U \subseteq M \cap N$
+where $U = U_1 \cap U_2$
+The result follows by definition of [[Definition:Neighborhood of Point|neighborhood]] of $a$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood in Topological Space has Subset Neighborhood}
+Tags: Neighborhoods
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Let $N$ be a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $T$.
+Then there exists a [[Definition:Neighborhood of Point|neighborhood]] $N'$ of $x$ such that:
+:$(1): \quad N' \subseteq N$
+:$(2): \quad N'$ is a [[Definition:Neighborhood of Point|neighborhood]] of each of its points.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Neighborhood of Point|neighborhood]]:
+:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
+where $U$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+By [[Set is Open iff Neighborhood of all its Points]], $N' = U$ fulfils the conditions of the statement.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Open in Neighborhood Space}
+Tags: Neighborhood Spaces, Empty Set, Open Sets
+
+\begin{theorem}
+Let $\struct {S, \mathcal N}$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Then the [[Definition:Empty Set|empty set]] $\varnothing$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \mathcal N}$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\varnothing$ were not an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \mathcal N}$.
+Then $\exists x \in \varnothing$ such that $\varnothing$ is not a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $\varnothing$.
+By definition of [[Definition:Empty Set|empty set]], such an $x$ does not exist.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Whole Space is Open in Neighborhood Space}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $\struct {S, \NN}$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Then $S$ itself is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Then by [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $\text N 1$]] there exists a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] $N$ of $x$.
+As $N \subseteq S$ it follows from [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $\text N 3$]] that $S$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+As this holds for all $x \in S$ it follows that $S$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of two Open Sets of Neighborhood Space is Open}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $\left({S, \mathcal N}\right)$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $U$ and $V$ be [[Definition:Open Set (Neighborhood Space)|open sets]] of $\left({S, \mathcal N}\right)$.
+Then $U \cap V$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\left({S, \mathcal N}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $U$ and $V$ be [[Definition:Open Set (Neighborhood Space)|open sets]] of $\left({S, \mathcal N}\right)$.
+Let $x \in S$ such that $x \in U \cap V$.
+Then $x \in U$ and $x \in V$, both of which are [[Definition:Neighborhood (Neighborhood Space)|neighborhoods]] of $x$ by definition of [[Definition:Open Set (Neighborhood Space)|open set]].
+By [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $N 4$]] it follows that $U \cap V$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+As $x$ is arbitrary, it follows that the above is true for all $x \in U \cap V$.
+It follows by definition that $U \cap V$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\left({S, \mathcal N}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Open Sets of Neighborhood Space is Open}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $S$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $I$ be an [[Definition:Indexing Set|indexing set]].
+Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family]] of [[Definition:Open Set (Neighborhood Space)|open sets]] of $\struct {S, \NN}$ indexed by $I$.
+Then their [[Definition:Union of Family|union]] $\ds \bigcup_{\alpha \mathop \in I} U_i$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$.
+\end{theorem}
+
+\begin{proof}
+Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
+Then $x \in U_\beta$ for some $\beta \in I$.
+By definition of [[Definition:Open Set (Neighborhood Space)|open set]], $U_\beta$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+But from [[Set is Subset of Union]]:
+:$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$
+By [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $N 3$]] it follows that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+As $x$ is arbitrary, it follows that the above is true for all $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
+It follows by definition that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\struct {S, \NN}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood Space is Topological Space}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $\left({S, \mathcal N}\right)$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $\tau = \left\{{N: N \in \mathcal N}\right\}$ be the [[Definition:Set|set]] of all [[Definition:Open Set (Neighborhood Space)|open sets]] of $\left({S, \mathcal N}\right)$.
+Then $\left({S, \tau}\right)$ forms a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+Each of the [[Definition:Open Set Axioms|open set axioms]] is examined in turn:
+=== $O1$: Union of Open Sets ===
+From [[Union of Open Sets of Neighborhood Space is Open]], it follows that [[Definition:Open Set Axioms|open set axiom $O1$]] is fulfilled.
+{{qed|lemma}}
+=== $O2$: Intersection of Open Sets ===
+From [[Intersection of two Open Sets of Neighborhood Space is Open]], it follows that [[Definition:Open Set Axioms|open set axiom $O2$]] is fulfilled.
+{{qed|lemma}}
+=== $O3$: Underlying Set ===
+From [[Whole Space is Open in Neighborhood Space]], it follows that [[Definition:Open Set Axioms|open set axiom $O3$]] is fulfilled.
+{{qed|lemma}}
+All the [[Definition:Open Set Axioms|open set axioms]] are fulfilled, and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset in Neighborhood Space is Neighborhood iff it contains Open Set}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $\left({S, \mathcal N}\right)$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $x \in S$ be a [[Definition:Element|point]] of $S$.
+Let $N \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+Then $N$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$ {{iff}} there exists an [[Definition:Open Set (Neighborhood Space)|open set]] $U$ of $\left({S, \mathcal N}\right)$ such that $x \in U \subseteq N$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $N$ be a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+Then by [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $N 5$]], $N$ contains a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] $U$ of $x$ such that $U$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of each of its [[Definition:Element|points]].
+By [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $N 2$]], $x \in U$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $N$ be such that:
+:$\exists U \in \mathcal N: x \in U \subseteq N$
+By definition of [[Definition:Open Set (Neighborhood Space)|open set]], $U$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+By [[Axiom:Neighborhood Space Axioms|neighborhood space axiom $N 3$]], $N$ is then a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Neighborhood Space Induced by Topological Space}
+Tags: Definitions: Neighborhoods
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\tau$ be a [[Definition:Topology|topology]] on $S$, thus forming the [[Definition:Topological Space|topological space]] $\left({S, \tau}\right)$.
+For all $x \in S$, let $\mathcal N_x$ be the [[Definition:Neighborhood Filter of Point|neighborhood filter]] of $x$ in $S$.
+Let $\mathcal N$ be the [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$ such that:
+:$\forall N \in \mathcal N: N$ is a [[Definition:Neighborhood of Point|neighborhood]] of each of its [[Definition:Element|points]]
+Then $\left({S, \mathcal N}\right)$ is the '''[[Definition:Neighborhood Space|neighborhood space]] induced by $\left({S, \tau}\right)$'''.
+\end{theorem}<|endoftext|>
+\section{Induced Neighborhood Space is Neighborhood Space}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\tau$ be a [[Definition:Topology|topology]] on $S$, thus forming the [[Definition:Topological Space|topological space]] $\left({S, \tau}\right)$.
+Let $\left({S, \mathcal N}\right)$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\left({S, \tau}\right)$]].
+Then $\left({S, \mathcal N}\right)$ is a [[Definition:Neighborhood Space|neighborhood space]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Let $\mathcal N_x$ be the [[Definition:Neighborhood Filter of Point|neighborhood filter of $x$]].
+From [[Basic Properties of Neighborhood in Topological Space]], $\mathcal N_x$ fulfils the [[Axiom:Neighborhood Space Axioms|neighborhood space axioms]].
+Hence the result.
+{{qed}}
+[[Category:Neighborhood Spaces]]
+hpgov4auvakl03167go55u84y7pcjmg
+\end{proof}<|endoftext|>
+\section{Topological Space Induced by Neighborhood Space is Topological Space}
+Tags: Neighborhood Spaces
+
+\begin{theorem}
+Let $\left({S, \mathcal N}\right)$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $\left({S, \tau}\right)$ be the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\left({S, \mathcal N}\right)$]].
+Then $\left({S, \tau}\right)$ is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+From [[Neighborhood Space is Topological Space]], $\left({S, \mathcal N}\right)$ is a [[Definition:Topological Space|topological space]].
+Consequently, the [[Definition:Open Set (Topology)|open sets]] of $\left({S, \tau}\right)$ are exactly the [[Definition:Open Set (Neighborhood Space)|open sets]] of $\left({S, \mathcal N}\right)$.
+{{qed}}
+[[Category:Neighborhood Spaces]]
+n5jwkuh9ttiaatf3sqat7cdai2am1q8
+\end{proof}<|endoftext|>
+\section{Topological Space induced by Neighborhood Space induced by Topological Space}
+Tags: Neighborhood Spaces, Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\tau$ be a [[Definition:Topology|topology]] on $S$, thus forming the [[Definition:Topological Space|topological space]] $\left({S, \tau}\right)$.
+Let $\left({S, \mathcal N}\right)$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]].
+Let $\left({S, \tau'}\right)$ be the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\mathcal N$ on $S$]].
+Then $\tau = \tau'$.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $\left({S, \tau}\right)$.
+By [[Set is Open iff Neighborhood of all its Points]], $U$ is a [[Definition:Neighborhood of Point|neighborhood]] of each of its [[Definition:Element|points]].
+By definition, $U$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\left({S, \mathcal N}\right)$.
+Thus by definition of [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]]:
+:$U \in \mathcal N$
+Then, by definition of the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\mathcal N$ on $S$]]:
+: $U \in \tau'$.
+Thus $\tau \subseteq \tau'$.
+Now let $U \in \tau'$ be an [[Definition:Open Set (Topology)|open set]] of $\left({S, \tau'}\right)$.
+Then, by definition, in the [[Definition:Neighborhood Space|neighborhood space]] $\left({S, \mathcal N}\right)$, $U$ is an [[Definition:Open Set (Neighborhood Space)|open set]] of $\left({S, \mathcal N}\right)$.
+That is, $U$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] in $\left({S, \mathcal N}\right)$ of each of its [[Definition:Element|points]].
+But the [[Definition:Neighborhood (Neighborhood Space)|neighborhoods]] of $\left({S, \mathcal N}\right)$ are the [[Definition:Neighborhood of Point|neighborhoods]] of $\left({S, \tau}\right)$.
+Thus by [[Set is Open iff Neighborhood of all its Points]], $U$ is an [[Definition:Open Set (Topology)|open set]] of $\left({S, \tau}\right)$.
+That is:
+: $U \in \tau$.
+It follows by definition of [[Definition:Set Equality|set equality]] that $\tau = \tau'$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Correspondence between Neighborhood Space and Topological Space}
+Tags: Neighborhood Spaces, Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $\left({S, \mathcal N}\right)$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]].
+Let $\phi: \left({S, \tau}\right) \to \left({S, \mathcal N}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in S: \phi \left({x}\right) = x$
+:$\forall U \in \tau: \phi \left({U}\right) = U \in \mathcal N$
+Let $\left({T, \mathcal N'}\right)$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $\left({T, \tau'}\right)$ be the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\mathcal N$ on $S$]].
+Let $\psi: \left({T, \mathcal N'}\right) \to \left({T, \tau'}\right)$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall y \in T: \psi \left({y}\right) = y$
+:$\forall V \in \mathcal N': \psi \left({V}\right) = V \in \tau'$
+Then:
+:$(1): \quad \phi^{-1} \left({\phi \left({S, \tau}\right)}\right) = \left({S, \tau}\right)$
+:$(2): \quad \psi^{-1} \left({\psi \left({T, \mathcal N'}\right)}\right) = \left({T, \mathcal N'}\right)$
+\end{theorem}
+
+\begin{proof}
+From the construction of:
+:the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]]
+:the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\mathcal N$ on $S$]]
+the mappings $\phi$ and $\psi$ are [[Definition:Well-Defined Mapping|well-defined mappings]].
+From [[Topological Space induced by Neighborhood Space induced by Topological Space]], $\phi$ is a [[Definition:Bijection|bijection]].
+From [[Neighborhood Space induced by Topological Space induced by Neighborhood Space]], $\psi$ is a [[Definition:Bijection|bijection]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood Space induced by Topological Space induced by Neighborhood Space}
+Tags: Neighborhood Spaces, Topology
+
+\begin{theorem}
+Let $\left({S, \mathcal N}\right)$ be a [[Definition:Neighborhood Space|neighborhood space]].
+Let $\left({S, \tau}\right)$ be the [[Definition:Topological Space Induced by Neighborhood Space|topological space induced by $\mathcal N$ on $S$]].
+Let $\left({S, \mathcal N'}\right)$ be the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]].
+Then $\mathcal N = \mathcal N'$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Let $\mathcal N_x$ be the [[Definition:Set|set]] of all [[Definition:Neighborhood (Neighborhood Space)|neighborhoods]] of $x$.
+Let $N \in \mathcal N_x$ be a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$.
+From [[Subset in Neighborhood Space is Neighborhood iff it contains Open Set]], $N$ is the [[Definition:Superset|superset]] of some [[Definition:Open Set (Neighborhood Space)|open set]] $U$ in $\left({S, \mathcal N}\right)$.
+By [[Neighborhood Space is Topological Space]] we have that $U$ is an [[Definition:Open Set (Topology)|open set]] of $\left({S, \tau}\right)$.
+Thus, by definition, $N$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in the context of the [[Definition:Topological Space|topological space]] $\left({S, \tau}\right)$.
+Thus by definition of the [[Definition:Neighborhood Space Induced by Topological Space|neighborhood space induced by $\tau$ on $S$]], it follows that:
+: $N \in \mathcal N'_x$
+where $\mathcal N'_x$ is the [[Definition:Set|set]] of all [[Definition:Neighborhood (Neighborhood Space)|neighborhoods]] of $x$ in $\left({S, \mathcal N'}\right)$.
+Thus $\mathcal N_x \subseteq \mathcal N'_x$.
+{{qed|lemma}}
+Now suppose that $N \in \mathcal N'_x$.
+Then, by definition of [[Definition:Neighborhood (Neighborhood Space)|neighborhood]], $N$ is the [[Definition:Superset|superset]] of some [[Definition:Open Set (Topology)|open set]] $U$ in $\left({S, \tau}\right)$ such that $x \in U$.
+Since $U \in \tau$ it follows that $U$ is an [[Definition:Open Set (Neighborhood Space)|open set]]in $\left({S, \mathcal N}\right)$.
+So by [[Subset in Neighborhood Space is Neighborhood iff it contains Open Set]], $N$ is a [[Definition:Neighborhood (Neighborhood Space)|neighborhood]] of $x$ in $\left({S, \mathcal N}\right)$.
+That is:
+: $N \in \mathcal N_x$
+Thus it follows that $\mathcal N_x \subseteq \mathcal N'_x$
+{{qed|lemma}}
+It follows by definition of [[Definition:Set Equality|set equality]] that $\mathcal N_x = \mathcal N'_x$.
+As $x$ is arbitrary, it applies to all $x \in S$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient with Zero/Integer Coefficients}
+Tags: Examples of Binomial Coefficients
+
+\begin{theorem}
+:$\forall n \in \N: \dbinom n 0 = 1$
+where $\dbinom n 0$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+From the definition:
+{{begin-eqn}}
+{{eqn | l = \binom n 0
+      | r = \frac {n!} {0! \ n!}
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = \frac {n!} {1 \cdot n!}
+      | c = {{Defof|Factorial}} of $0$
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Form of Geometric Sequence of Integers in Lowest Terms}
+Tags: Geometric Sequences of Integers
+
+\begin{theorem}
+Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence|geometric sequence]] of [[Definition:Length of Sequence|length]] $n$ consisting of [[Definition:Positive Integer|positive integers]] only.
+Let $r$ be the [[Definition:Common Ratio|common ratio]] of $G_n$.
+Let the elements of $G_n$ be the smallest [[Definition:Positive Integer|positive integers]] such that $G_n$ has [[Definition:Common Ratio|common ratio]] $r$.
+Then the $j$th [[Definition:Term of Geometric Sequence|term]] of $G_n$ is given by:
+:$a_j = p^{n - j} q^j$
+where $r = \dfrac q p$.
+That is:
+:$G_n = \tuple {p^n, p^{n - 1} q, p^{n - 2} q^2, \ldots, p q^{n - 1}, q^n}$
+\end{theorem}
+
+\begin{proof}
+From [[Form of Geometric Sequence of Integers]] the $j$th [[Definition:Term of Geometric Sequence|term]] of $G_n$ is given by:
+:$a_j = k p^{n - j} q^j$
+where the [[Definition:Common Ratio|common ratio]] is $\dfrac q p$.
+Thus:
+:$a_0 = k p^n$
+:$a_n = k q^n$
+From [[Geometric Sequence in Lowest Terms has Coprime Extremes]] it follows that $k = 1$.
+Hence the result.
+{{qed}}
+[[Category:Geometric Sequences of Integers]]
+cwfob7tfrj3rzjzscijwag34jsinvz6
+\end{proof}<|endoftext|>
+\section{Form of Geometric Sequence of Integers with Coprime Extremes}
+Tags: Geometric Sequences of Integers
+
+\begin{theorem}
+Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence|geometric sequence]] of [[Definition:Length of Sequence|length]] $n$ consisting of [[Definition:Positive Integer|positive integers]] only.
+Let $a_1$ and $a_n$ be [[Definition:Coprime Integers|coprime]].
+Then the $j$th [[Definition:Term of Geometric Sequence|term]] of $Q_n$ is given by:
+:$a_j = q^j p^{n - j}$
+\end{theorem}
+
+\begin{proof}
+Let $r$ be the [[Definition:Common Ratio|common ratio]] of $Q_n$.
+Let the elements of $Q_n$ be the smallest [[Definition:Positive Integer|positive integers]] such that $Q_n$ has [[Definition:Common Ratio|common ratio]] $r$.
+From [[Geometric Sequence with Coprime Extremes is in Lowest Terms]], the elements of $Q_n$ are the smallest [[Definition:Positive Integer|positive integers]] such that $Q_n$ has [[Definition:Common Ratio|common ratio]] $r$.
+From [[Form of Geometric Sequence of Integers in Lowest Terms]] the $j$th [[Definition:Term of Geometric Sequence|term]] of $P$ is given by:
+:$a_j = q^j p^{n - j}$
+where $r = \dfrac p q$.
+Hence the result.
+{{qed}}
+[[Category:Geometric Sequences of Integers]]
+m62ikkujr35cgadosyuwq1eegs6vevx
+\end{proof}<|endoftext|>
+\section{Naturally Ordered Semigroup forms Peano Structure}
+Tags: Naturally Ordered Semigroup, Peano's Axioms
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $0 \in S$ be the [[Definition:Zero of Naturally Ordered Semigroup|zero]] of $S$.
+Let $1 \in S$ be the [[Definition:One of Naturally Ordered Semigroup|one]] of $S$.
+Let $s: S \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$s \left({n}\right) := n \circ 1$
+Then $\left({S, 0, s}\right)$ is a [[Definition:Peano Structure|Peano structure]].
+\end{theorem}
+
+\begin{proof}
+We verify [[Axiom:Peano's Axioms|Peano's axioms]] in turn.
+First, suppose that $s \left({m}\right) = s \left({n}\right)$ for some $m, n \in S$.
+That is:
+:$m \circ 1 = n \circ 1$
+By [[Definition:Naturally Ordered Semigroup|Axiom $(NO2)$]], it follows that $m = n$.
+Hence [[Axiom:Peano's Axioms|Axiom $(P3)$]] holds.
+Next, suppose that $s \left({n}\right) = 0$ for some $n \in S$.
+That is:
+:$n \circ 1 = 0$
+By [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]], this would imply:
+:$n \prec 0$
+However, this contradicts the [[Definition:Zero of Naturally Ordered Semigroup|definition of $0$]].
+Consequently, [[Axiom:Peano's Axioms|Axiom $(P4)$]] is satisfied.
+Finally, let $A \subseteq S$ be such that $0 \in A$ and, for all $n \in S$:
+:$n \in A \implies s \left({n}\right) \in A$
+Define $T := S \setminus A$.
+Suppose that $T \ne \varnothing$.
+Then by [[Definition:Naturally Ordered Semigroup|Axiom $(NO3)$]], $T$ has a [[Definition:Minimal Element|minimal element]] $n$.
+By assumption, $0 \in A$, so that $n \ne 0$.
+Therefore, by [[Definition:One of Naturally Ordered Semigroup|definition of $1$]]:
+:$1 \preceq n$
+By [[Ordering in terms of Addition]], there exists a $p \in S$ such that:
+:$p \circ 1 = n$
+since $\circ$ is [[Definition:Commutative Operation|commutative]].
+By [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]:
+:$p \prec n$
+Since $n$ is the [[Definition:Minimal Element|minimal element]] of $T$, it follows that $p \notin T$.
+Therefore, by definition, $p \in A$.
+But by assumption on $A$, this means that $n = p \circ 1 \in A$.
+This is a contradiction, hence $T = \varnothing$.
+That is to say, $A = S$, and [[Axiom:Peano's Axioms|Axiom $(P5)$]] holds as well.
+Having verified all the axioms, we conclude that $\left({S, 0, s}\right)$ is a [[Definition:Peano Structure|Peano structure]].
+{{qed}}
+[[Category:Naturally Ordered Semigroup]]
+[[Category:Peano's Axioms]]
+bzvkvibijvhiu90m46w9gnrd53xyz86
+\end{proof}<|endoftext|>
+\section{Divisibility of Elements in Geometric Sequence of Integers}
+Tags: Geometric Sequences of Integers
+
+\begin{theorem}
+Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence of Integers|geometric sequence of integers]].
+Let $j \ne k$.
+Then:
+:$\paren {\exists j \in \set {0, 1, \ldots, n - 1}: a_j \divides a_{j + 1} } \iff \paren {\forall j, k \in \set {0, 1, \ldots, n}, j < k: a_j \divides a_k}$
+where $\divides$ denotes [[Definition:Divisor of Integer|integer divisibility]].
+That is:
+:One [[Definition:Term of Geometric Sequence|term]] of a [[Definition:Geometric Sequence of Integers|geometric sequence of integers]] is the [[Definition:Divisor of Integer|divisor]] of the next [[Definition:Term of Geometric Sequence|term]]
+{{iff}}
+:All [[Definition:Term of Geometric Sequence|terms]] are [[Definition:Divisor of Integer|divisors]] of all later [[Definition:Term of Geometric Sequence|terms]].
+\end{theorem}
+
+\begin{proof}
+Let $a_j \divides a_{j + 1}$ for some $j \in \set {0, 1, \ldots, n - 1}$.
+Then by definition of [[Definition:Divisor of Integer|integer divisibility]]:
+:$\exists r \in \Z: r a_j = a_{j + 1}$
+Thus the [[Definition:Common Ratio|common ratio]] of $Q_n$ is $r$.
+So by definition of [[Definition:Geometric Sequence|geometric sequence]]:
+:$\forall j, k \in \set {0, 1, \ldots, n}, j < k: r^{k - j} a_j = a_k$
+and so $a_j \divides a_k$.
+The [[Definition:Converse Statement|converse]] is trivial.
+{{qed}}
+[[Category:Geometric Sequences of Integers]]
+9wvw4e48rd7y1uyoccxqq5hzi2ue3tc
+\end{proof}<|endoftext|>
+\section{Form of Geometric Sequence of Integers from One}
+Tags: Geometric Sequences of Integers
+
+\begin{theorem}
+Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a [[Definition:Geometric Sequence|geometric sequence]] consisting of [[Definition:Integer|integers]] only.
+Let $a_0 = 1$.
+Then the $j$th [[Definition:Term of Geometric Sequence|term]] of $Q_n$ is given by:
+:$a_j = a^j$
+where:
+:the [[Definition:Common Ratio|common ratio]] of $Q_n$ is $a$
+:$a = a_1$.
+Thus:
+:$Q_n = \tuple {1, a, a^2, \ldots, a^n}$
+\end{theorem}
+
+\begin{proof}
+From [[Form of Geometric Sequence of Integers]], the $j$th [[Definition:Term of Geometric Sequence|term]] of $Q_n$ is given by:
+:$a_j = k q^j p^{n - j}$
+where:
+:the [[Definition:Common Ratio|common ratio]] of $Q_n$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]] is $\dfrac q p$
+:$k$ is an [[Definition:Integer|integer]].
+As $a_0 = 1$ it follows that:
+:$1 = k p^{n - j}$
+from which it follows that:
+:$k = 1$
+:$p = 1$
+and the [[Definition:Common Ratio|common ratio]] of $Q_n$ is $q$.
+Thus:
+:$a_1 = q$
+Setting $a = a_1$ yields the result as stated.
+{{qed}}
+[[Category:Geometric Sequences of Integers]]
+oe6s3lsppor1abhbcntfeh882woz8za
+\end{proof}<|endoftext|>
+\section{Divisors of Power of Prime}
+Tags: Prime Numbers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then the only [[Definition:Divisor of Integer|divisors]] of $p^n$ are $1, p, p^2, \ldots, p^{n - 1}, p^n$.
+\end{theorem}
+
+\begin{proof}
+First it is necessary to establish that every [[Definition:Element|element]] of the [[Definition:Set|set]] $\set {1, p, p^2, \ldots, p^{n - 1}, p^n}$ is in fact a [[Definition:Divisor of Integer|divisor]] of $p^n$.
+For any $j \in \set {1, 2, \ldots, n}$:
+:$p^n = p^j p^{n - j}$
+and so each of $1, p, p^2, \ldots, p^{n - 1}, p^n$ is a [[Definition:Divisor of Integer|divisor]] of $p^n$.
+{{qed|lemma}}
+Let:
+:$a \in \Z_{>0}: a \notin \set {1, p, p^2, \ldots, p^{n - 1}, p^n}$
+Let $a = p^j$ where $j \in \Z: j > n$.
+Then:
+:$p^j = p^n p^{j - n}$
+and so $p^n$ is a [[Definition:Divisor of Integer|divisor]] of $p^j$.
+Hence $p_j \nmid p^n$.
+Now let:
+:$a \notin \set {p^k: k \in \Z_{>0} }$
+Then:
+:$\exists q \in \Bbb P: q \divides a$
+where:
+:$\Bbb P$ is the [[Definition:Set|set]] of all [[Definition:Prime Number|prime numbers]]
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+{{AimForCont}} $a \divides p^n$.
+From [[Divisor Relation is Transitive]] it follows that $q \divides p^n$.
+From [[Euclid's Lemma for Prime Divisors/General Result|Euclid's Lemma for Prime Divisors: General Result]] it follows that:
+:$q \divides p$
+As $p$ is a [[Definition:Prime Number|prime]], by definition its only [[Definition:Divisor of Integer|divisors]] are $1$ and $p$.
+This [[Proof by Contradiction|contradicts]] the supposition that $q$ is a [[Definition:Divisor of Integer|divisor]] of $p^n$.
+Hence $a \nmid p^n$.
+{{qed}}
+[[Category:Prime Numbers]]
+7rk579ji8oqt9bogdoui6eysst01x2q
+\end{proof}<|endoftext|>
+\section{Chebyshev Distance is Limit of P-Product Metric}
+Tags: Chebyshev Distance, P-Product Metrics
+
+\begin{theorem}
+Let $M_{1'} = \left({A_{1'}, d_{1'}}\right)$ and $M_{2'} = \left({A_{2'}, d_{2'}}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\mathcal A = A_{1'} \times A_{2'}$ be the [[Definition:Cartesian Product|cartesian product]] of $A_{1'}$ and $A_{2'}$.
+Let $p \in \R_{\ge 1}$.
+Let $d_p: \mathcal A \times \mathcal A \to \R$ be the [[Definition:P-Product Metric|$p$-product metric]] on $\mathcal A$
+: $d_p \left({x, y}\right) := \left({\left({d_{1'} \left({x_1, y_1}\right)}\right)^p + \left({d_{2'} \left({x_2, y_2}\right)}\right)^p}\right)^{1/p}$
+and let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the [[Definition:Chebyshev Distance|Chebyshev distance]] on $\mathcal A$:
+: $\displaystyle d_\infty \left({x, y}\right) := \max \left\{ {d_{1'} \left({x_1, y_1}\right), d_{2'} \left({x_2, y_2}\right)}\right\}$
+where $x = \left({x_1, x_2}\right), y = \left({y_1, y_2}\right) \in \mathcal A$.
+Then:
+:$\displaystyle d_\infty = \lim_{p \mathop \to \infty} d_p$
+in the sense that:
+:$\displaystyle \max \left\{ {d_{1'} \left({x_1, y_1}\right), d_{2'} \left({x_2, y_2}\right)}\right\} = \lim_{p \mathop \to \infty} \left({\left({d_{1'} \left({x_1, y_1}\right)}\right)^p + \left({d_{2'} \left({x_2, y_2}\right)}\right)^p}\right)^{1/p}$
+\end{theorem}
+
+\begin{proof}
+Let $x$ and $y$ be arbitrary.
+Let $a = d_{1'} \left({x_1, y_1}\right), b = d_{2'} \left({x_2, y_2}\right)$.
+{{WLOG}}, suppose that $\max \left\{ {a, b}\right\} = a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \lim_{p \mathop \to \infty} \left({a^p + b^p}\right)^{1/p}
+      | o = \ge
+      | r = \lim_{p \mathop \to \infty} \left({a^p}\right)^{1/p}
+      | c = 
+}}
+{{eqn | r = \lim_{p \mathop \to \infty} a
+      | c = 
+}}
+{{eqn | r = a
+      | c = 
+}}
+{{eqn | r = \max \left\{ {a, b}\right\}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \lim_{p \mathop \to \infty} \left({a^p + b^p}\right)^{1/p}
+      | o = \le
+      | r = \lim_{p \mathop \to \infty} \left({a^p + a^p}\right)^{1/p}
+      | c = 
+}}
+{{eqn | r = \lim_{p \mathop \to \infty} \left({2 a^p}\right)^{1/p}
+      | c = 
+}}
+{{eqn | r = \lim_{p \mathop \to \infty} a \cdot 2^{1/p}
+      | c = 
+}}
+{{eqn | r = a \lim_{p \mathop \to \infty} 2^{1/p}
+      | c = 
+}}
+{{eqn | r = a
+      | c = 
+}}
+{{eqn | r = \max \left\{ {a, b}\right\}
+      | c = 
+}}
+{{end-eqn}}
+The result follows by the [[Squeeze Theorem/Functions|Squeeze Theorem]].
+{{qed}}
+[[Category:Chebyshev Distance]]
+[[Category:P-Product Metrics]]
+fsb3qf1xl03y3jluddwc0hl5tnkivtl
+\end{proof}<|endoftext|>
+\section{Bertrand's Theorem}
+Tags: Classical Mechanics
+
+\begin{theorem}
+Let $U: \R_{>0} \to \R$ be [[Definition:Analytic Function|analytic]] for $r > 0$.
+Let $M > 0$ be a [[Definition:Non-Vanishing|nonvanishing]] [[Definition:Angular Momentum|angular momentum]] such that a [[Definition:Stable Orbit|stable]] [[Definition:Circular Orbit|circular]] [[Definition:Orbit (Phase Space)|orbit]] exists.
+Suppose that every [[Definition:Orbit (Phase Space)|orbit]] sufficiently close to the [[Definition:Circular Orbit|circular orbit]] is [[Definition:Closed Orbit|closed]].
+Then $U$ is either $k r^2$ or $-\dfrac k r$ (for $k > 0$) up to an [[Definition:Additive Constant|additive constant]].
+\end{theorem}
+
+\begin{proof}
+For simplicity we set $m = 1$, so that the [[Definition:Effective Potential|effective potential]] becomes:
+:$U_M = U + \dfrac {M^2} {2 r^2}$
+{{explain|What is $m$ in this context?}}
+Consider the [[Definition:Apsidial Angle|apsidial angle]]:
+:$\displaystyle \Phi = \sqrt 2 \int_{r_\min}^{r_\max} \frac {M \rd r} {r^2 \sqrt {E - U_M} }$
+where:
+:$E$ is the [[Definition:Energy|energy]]
+:$r_\min, r_\max$ are solutions to $\map {U_M} r = E$. 
+By definition, this is the [[Definition:Angle|angle]] between adjacent [[Definition:Apocenter|apocenters]] ([[Definition:Pericenter|pericenters]]).
+{{explain|Apocenters or pericenters? Or either? This could be clarified.}}
+Recall that if $\Phi$ is [[Definition:Commensurable|commensurable]] with $\pi$, then an [[Definition:Orbit (Phase Space)|orbit]] is [[Definition:Closed Orbit|closed]].
+{{explain|Link to a proof demonstrating this.}}
+=== Non-Perturbative Proof ===
+In general $U_M$ is not [[Definition:Monotonic Real Function|monotonic]] on $\openint {r_\min} {r_\max}$.
+{{explain|Link to a proof of that statement.}}
+Therefore a unique [[Definition:Inverse Function|inverse]] $\map r {U_M}$ does not exist.
+However, suppose it is possible to construct separate [[Definition:Inverse Function|inverse functions]] $r_{1, 2}$ for the intervals $\openint {r_0} {r_\min}$ and $\openint {r_0} {r_\max}$, where $r_0$ is the [[Definition:Minimum Value|minimum]] of $U_M$ on $\openint {r_\min} {r_\max}$.
+{{explain|What does $r_{1, 2}$ mean in this context?}}
+Note that the [[Definition:Orbit (Phase Space)|orbit]] corresponding to $r_0$ is the [[Definition:Stable Orbit|stable]] [[Definition:Circular Orbit|circular]] [[Definition:Orbit (Phase Space)|orbit]].
+{{explain|Link to a proof of the above.}}
+Therefore this will be possible for [[Definition:Orbit (Phase Space)|orbits]] sufficiently close to it. 
+Now write:
+:$\displaystyle \Phi = \int_{U_0}^E F \left({U_M}\right) \frac {\d U_{eff} } {\sqrt {E - U_M} }$
+where:
+:$F = \sqrt 2 M \, \map {\dfrac {\d} {\d U_M} } {\dfrac 1 {r_1} - \dfrac 1 {r_2} }$
+:$U_0 \equiv \map {U_M} {r_0}$
+{{explain|What does $U_{eff}$ mean?}}
+This is [[Abel's Integral Equation]] which can be solved for $F \left({U_M}\right)$, giving:
+:$\displaystyle F \left({U}\right) = \dfrac 1 \pi \int_{U_0}^{U_M} \frac {\Phi \rd E} {\sqrt{U_M - E} }$
+$\Phi$ may have [[Definition:Energy|energy]] dependence.
+However, we require:
+:$\Phi \left({E}\right) = 2 \pi q \left({E}\right)$
+such that $q: \R \to \Q$ is [[Definition:Continuous Real Function|continuous]], and a [[Definition:Rational Function|rational]] [[Definition:Continuous Real Function|continuous function]] must be [[Definition:Constant Mapping|constant]].
+{{Explain|Link to a result explaining why a [[Definition:Rational Function|rational]] [[Definition:Continuous Real Function|continuous function]] must be [[Definition:Constant Mapping|constant]].}}
+Finally, [[Definition:Integration|integrating]] gives:
+:$\dfrac 1 {r_1} - \dfrac 1 {r_2} = \gamma \sqrt{U_M - U_0}, \gamma = \dfrac {\sqrt 2 \Phi} {\pi M}$
+{{explain|Provide a link demonstrating the solution of that integral.}}
+Now consider defining the {{LHS}} to be a single [[Definition:Real Function|function]] $\dfrac 1 {\map r U}$, [[Definition:Meromorphic Function|meromorphic]] with a [[Definition:Simple Pole|simple pole]] at $r = 0$.
+{{explain|The definitions for [[Definition:Meromorphic Function|meromorphic]] and [[Definition:Simple Pole|simple pole]] as given on {{ProofWiki}} are specifically for [[Definition:Complex Function|complex functions]]. Either these need to be expanded to define them in the specific context of [[Definition:Real Function|real functions]], or the above definition is actually for a [[Definition:Complex Function|complex function]]. In any case, the domain of all the functions defined in this proof need to be clarified.}}
+Write:
+:$\dfrac 1 r = \gamma \sqrt {U_M - U_0} + \Omega \left({U_0, U}\right)$
+where $\Omega$ is an [[Definition:Analytic Function|analytic function]] in $U$ in an [[Definition:Open Neighborhood|open neighborhood]] of $U_0$ such that:
+:$\Omega \left({U_0, U_0}\right) = \dfrac 1 {r_0}$
+{{explain|What are $\left({U_0, U}\right)$ and $\left({U_0, U_0}\right)$ in this context?}}
+Explicitly:
+:$\sqrt {U_M - U_0} = \sqrt {U + \dfrac {M^2} {2 r^2} - U_0} = \dfrac 1 r \sqrt{\dfrac {M^2} 2 + \left({U - U_0}\right) r^2}$
+It can be seen that the {{RHS}} also has a [[Definition:Simple Pole|simple pole]] at $r=0$.
+However, it also has a [[Definition:Branch Point|branching point]] at $r = r_0$.
+To avoid this, the [[Definition:Radicand|radicand]] must be the [[Definition:Square (Algebra)|square]] of an [[Definition:Analytic Function|analytic function]] with a [[Definition:Zero of Function|zero]] at $r_0$.
+The only possibilities are:
+: $U \propto r^2, -\dfrac 1 {r^2}$
+{{qed}}
+=== Proof based on Asymptotic Behaviour ===
+The substitution $x = M / r$ gives:
+:$\displaystyle \Phi = \sqrt 2 \int_{x_\min}^{x_\max} \frac{\d x} {\sqrt{E - W \left({x}\right)}}$
+where $W \left({x}\right) \equiv U \left({\dfrac M x}\right) + \dfrac 1 2 x^2$.
+In general, the frequency of oscillations around a stable equilibrium at $x = x_0$ for a particle of mass $m$ in a potential $V$ is given by:
+:$\omega = \sqrt{\dfrac {V'' \left({x_0}\right)} m}$
+{{explain|Provide a link to a proof of the above, and explain what $m$ is.}}
+which means that:
+:$\Phi \approx 2 \pi \dfrac M {x_0^2} \sqrt {W'' \left({x_0}\right)} = 2 \pi \sqrt{\dfrac {U'} {3 U' + x_0 U''}}$
+Suppose $\Phi$ were [[Definition:Constant|constant]].
+Then the solutions are:
+: $U \left({r}\right) = k r^\alpha, \alpha \ge - 2, \alpha \ne 0$
+and:
+: $U \left({r}\right) = b \log r$
+Therefore:
+: $\Phi = \dfrac {2 \pi} {\sqrt{\alpha + 2}}$
+with $\alpha = 0$ corresponding to the logarithmic solution.
+This is discarded, since in that case $\Phi$ is not commensurable with $\pi$.
+Now consider $\alpha > 0$, so that $U \left({r}\right) \to \infty$ as $r \to \infty$.
+The substitution $x = y x_\max$ reduces $\Phi$ to:
+:$\displaystyle \Phi = \sqrt 2 \int_{y_\min}^1 \frac {\d y} {\sqrt {W \left({1}\right) - W \left({y}\right)} }$
+where:
+:$W \left({y}\right) \equiv \dfrac {y^2} 2 + \dfrac 1 { {x_\max}^2} U \left({\dfrac M {y x_\max}}\right)$
+Therefore, when $E \to \infty$:
+:$x_\max \to \infty$ and $y_\min \to 0$
+This means that:
+:$\displaystyle \lim_{E \mathop \to \infty} \Phi = \pi$
+Equating this with the previous result gives:
+:$\dfrac {2 \pi} {\sqrt {\alpha + 2}} = \pi$
+Therefore:
+: $\alpha=2$
+Now let $U \left({r}\right) = -k r^{-\beta}, 0 < \beta < 2$.
+A similar approach gives:
+:$\displaystyle \lim_{E \mathop \to -0} \Phi = \frac {2 \pi} {2 - \beta}$
+so that:
+:$\dfrac {2 \pi} {2 + \alpha} = \dfrac {2 \pi} {\sqrt {2 + \alpha} }$
+It follows that:
+:$\alpha = -1$
+and the result follows.
+{{qed}}
+{{Namedfor|Joseph Louis François Bertrand|cat = Bertrand}}
+\end{proof}<|endoftext|>
+\section{Cube Root of 2 is Irrational}
+Tags: Number Theory
+
+\begin{theorem}
+:$\sqrt [3] 2$ is [[Definition:Irrational Number|irrational]].
+\end{theorem}
+
+\begin{proof}
+Suppose to the contrary that $\sqrt [3] 2$ is [[Definition:Rational Number|rational]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \sqrt [3] 2
+      | r = \frac p q
+      | c = for some [[Definition:Integer|integer]] $p$ and $q$
+}}
+{{eqn | ll= \implies
+      | l = p^3
+      | r = q^3 + q^3
+      | c = 
+}}
+{{end-eqn}}
+which contradicts [[Fermat's Last Theorem]].
+{{qed}}
+[[Category:Number Theory]]
+8uo0v173rbvw5i3fyyqoqap1jy9fz41
+\end{proof}<|endoftext|>
+\section{Piecewise Continuously Differentiable Function/Definition 2 is Continuous}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $f$ satisfy [[Definition:Piecewise Continuously Differentiable Function/Definition 2|Piecewise Continuously Differentiable Function/Definition 2]].
+Then $f$ is [[Definition:Continuous Real Function|continuous]].
+\end{theorem}
+
+\begin{proof}
+$f$ satisfies the conditions of a [[Definition:Piecewise Continuously Differentiable Function/Definition 2|Piecewise Continuously Differentiable 
+Function by Definition 2]].
+Therefore, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_n = b$, such that $f$ is [[Definition:Continuously Differentiable Real Function|continuously differentiable]] on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+Here, the [[Definition:Derivative of Real Function|derivatives]] at $x_{i - 1}$ and $x_i$ are understood as [[Definition:One-Sided Derivative|one-sided derivatives]].
+$f$ is [[Definition:Differentiable Real Function|differentiable]] on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+Therefore, by [[Differentiable Function is Continuous]], $f$ is [[Definition:Continuous Real Function|continuous]] at every point of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+Here, the [[Definition:Continuous Real Function|continuities]] at $x_{i - 1}$ and $x_i$ are [[Definition:One-Sided Continuity|one-sided]].
+We use this result to go through every point of every $\closedint {x_{i - 1} } {x_i}$ in order to establish the [[Definition:Continuous Real Function|continuity]] of $f$ there:
+$f$ is [[Definition:Continuous Real Function|continuous]] on $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+$\openint {x_{i - 1} } {x_i}$ is a [[Definition:Subset|subset]] of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+Therefore, $f$ is [[Definition:Continuous Real Function at Point|continuous at every point]] of $\openint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+$x_0$ ($= a$) is the leftmost point in the [[Definition:Domain of Mapping|domain]] of $f$.
+$f$ is [[Definition:Right-Continuous at Point|right-continuous]] at $x_0$.
+Therefore, $f$ is [[Definition:Continuous Real Function|continuous]] at $x_0$.
+$x_n$ ($= b$) is the rightmost point in the [[Definition:Domain of Mapping|domain]] of $f$.
+$f$ is [[Definition:Left-Continuous at Point|left-continuous]] at $x_n$.
+Therefore, $f$ is [[Definition:Continuous Real Function|continuous]] at $x_n$.
+$f$ is [[Definition:Continuous Real Function at Point|continuous at every point]] $x_i$, $i \in \set {1, \ldots, n - 1}$, because $f$ is both [[Definition:Left-Continuous at Point|left-]] and [[Definition:Right-Continuous at Point|right-continuous]] at these points.
+All in all, we have found that $f$ is [[Definition:Continuous Real Function at Point|continuous at every point]] of $\closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.
+$f$ is therefore [[Definition:Continuous Real Function at Point|continuous at every point]] of the [[Definition:Set Union|union]] of those [[Definition:Real Interval|intervals]]: $\displaystyle \bigcup_{i \mathop = 1}^n \closedint {x_{i - 1}} {x_i}$.
+$\displaystyle \bigcup_{i \mathop = 1}^n \closedint {x_{i - 1}} {x_i}$ equals $\closedint a b$, the [[Definition:Domain of Mapping|domain]] of $f$.
+Therefore, $f$ is [[Definition:Continuous Real Function|continuous]] as $f$ is [[Definition:Continuous Real Function at Point|continuous at every point]] of its [[Definition:Domain of Mapping|domain]].
+{{qed}}
+[[Category:Real Analysis]]
+4dnzpl7tuw8rcwjsdhf6kem98mzi2bk
+\end{proof}<|endoftext|>
+\section{Homomorphism of Chain Complexes induces Homomorphism of Homology}
+Tags: Homological Algebra
+
+\begin{theorem}
+Let $A_\bullet$ and $B_\bullet$ be [[Definition:Complex (Homological Algebra)|chain complexes]] of abelian groups.
+Let $f: A_\bullet \to B_\bullet$ be a [[Definition:Homomorphism of Complexes|homomorphism]].
+Then for every $n$, $f$ induces a morphism $H_n \left({A_\bullet}\right) \to H_n \left({B_\bullet}\right)$ of [[Definition:Homology of Chain Complex|homology groups]].
+{{explain|Domain of $n$}}
+\end{theorem}
+
+\begin{proof}
+Let $\partial^A_\bullet$, $\partial^B_\bullet$ denote the differential on $A_\bullet$, respectively $B_\bullet$. 
+First it will be demonstrated that:
+:$\forall a \in \ker \left({\partial^A_n}\right) \subseteq A_n: f_n \left({a}\right) \in \ker \left({\partial^B_n}\right)$
+Thus:
+:$\partial^B_n f_n \left({a}\right) = f_{n - 1} \left({\partial^A_n a}\right) = f_{n - 1} \left({0}\right) = 0$
+Thus there exists a map:
+:$\bar f_n: \ker \left({\partial^A_n}\right) \to \ker \left({\partial^B_n}\right)$
+via the restriction of $f$.
+{{explain|Restriction of $f$ to what?}}
+Next will be shown that:
+:$\forall a \in \operatorname {Im} \left({\partial^A_{n + 1} }\right): f_n \left({a}\right) \in \operatorname {Im} \left({\partial^B_{n + 1} }\right)$
+Let $a = \partial^A_{n + 1}a'$
+Then:
+:$\partial^B_{n + 1} \left({f_{n + 1} \left({a'}\right)}\right) = f_n \left({\partial^A_n a'}\right) = f_n \left({a}\right)$.
+Let $\pi: \ker \left({\partial^B_n}\right) \to H_n \left({B_\bullet}\right)$ be the quotient map.
+Let $\rho = \pi \circ \bar f_n$.
+From above:
+:$\operatorname{Im} \left({\partial^A_{n + 1} }\right) \subseteq \ker \left({\rho}\right)$
+Thus $\rho$ factors through a map:
+:$\tilde f_n: H_n \left({A_\bullet}\right) \to H_n \left({B_\bullet}\right)$
+completing the proof.
+{{qed}}
+[[Category:Homological Algebra]]
+6jscnmx37cpml3xy9rzvr19jleqx6rr
+\end{proof}<|endoftext|>
+\section{Homotopic Chain Maps Induce Equal Maps on Homology}
+Tags: Homological Algebra
+
+\begin{theorem}
+Let $A_\bullet$, $B_\bullet$ be [[Definition:Complex (Homological Algebra)|chain complexes]] of abelian groups.
+Let $f, g: A_\bullet \to B_\bullet$ be [[Definition:Homomorphism of Complexes|chain maps]] which are [[Definition:Homotopic (Homological Algebra)|homotopic]].
+Then $f$ and $g$ [[Homomorphism of Chain Complexes induces Homomorphism of Homology|induce]] equal maps on homology.
+\end{theorem}
+
+\begin{proof}
+Let $\partial^A_\bullet, \partial^B_\bullet$ be the differentials on $A_\bullet$ and $B_{\bullet}$ respectively.
+Let $h$ be a homotopy between $f$ and $g$. 
+Let:
+:$a \in H_n \left({A}\right) \cong \ker \left({\partial^A_n}\right) / \operatorname{Im} \left({\partial^A_{n + 1} }\right)$
+There exists $\tilde a \in \ker \left({\partial^A_n}\right)$ representing $a$.
+It is enough to show that:
+:$f_n \left({\tilde a}\right) \sim g_n \left({\tilde a}\right)$
+Equivalently:
+:$f_n \left({\tilde a}\right) - g_n \left({\tilde a}\right) \in \operatorname{Im} \left({\partial^B_{n + 1} }\right)$
+We know that:
+:$\partial h - h \partial = f - g$
+This means:
+:$\partial^B_{n + 1} h_n - h_{n - 1} \partial^A_n = f_n - g_n$
+Plugging in $\tilde a$: 
+:$\partial^B_{n + 1} \left({h_n \left({\tilde a}\right)}\right) - h_{n - 1} \left({\partial^A_n \left({\tilde a}\right)}\right) = f_n \left({\tilde a}\right) - g_n \left({\tilde a}\right)$
+Since $\tilde a \in \ker \left({\partial^A_n}\right)$:
+:$\partial^B_{n + 1} \left({h_n \left({\tilde a}\right)}\right) = f_n \left({\tilde a}\right) - g_n \left({\tilde a}\right)$
+Therefore:
+:$f_n \left({\tilde a}\right) - g_n \left({\tilde a}\right) \in \operatorname{Im} \left({\partial^B_{n + 1} }\right)$
+{{qed}}
+[[Category:Homological Algebra]]
+lafbeuh2iho2kuiu0wumng4kjezcm0s
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Piecewise Continuously Differentiable Function}
+Tags: Piecewise Continuously Differentiable Functions
+
+\begin{theorem}
+{{TFAE|def = Piecewise Continuously Differentiable Function}}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+=== Definition 2 implies Definition 1 ===
+Assume that $f$ satisfies the requirement of [[Definition:Piecewise Continuously Differentiable Function/Definition 2|Definition 2]].
+We need to prove that $f$ satisfies the requirements of [[Definition:Piecewise Continuously Differentiable Function/Definition 1|Definition 1]].
+$f$ is [[Definition:Continuous Real Function|continuous]] by [[Piecewise Continuously Differentiable Function/Definition 2 is Continuous]].
+This proves $(1)$ in [[Definition:Piecewise Continuously Differentiable Function/Definition 1|Definition 1]].
+$f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\closedint {x_{i - 1} } {x_i}$, the [[Definition:Derivative of Real Function|derivatives]] at $x_{i - 1}$ and $x_i$ understood as [[Definition:One-Sided Derivative|one-sided derivatives]], for every $i \in \set {1, \ldots, n}$, by definition.
+Therefore, $f$ is continuously differentiable at every point of every $\openint {x_{i - 1} } {x_i}$ since $\openint {x_{i - 1} } {x_i}$ is a [[Definition:Subset|subset]] of $\closedint {x_{i - 1} } {x_i}$.
+This proves $(2.1)$ in [[Definition:Piecewise Continuously Differentiable Function/Definition 1|Definition 1]].
+$f$ has a [[Definition:Continuous Real Function|continuous]] [[Definition:Right-Hand Derivative|right-hand derivative]] at $x_{i - 1}$ for every $i \in \set {1, \ldots, n}$.
+This means that:
+:$\map {f'_+} {x_{i - 1} } = \displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$
+This implies that $\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ exists for every $i \in \set {1, \ldots, n}$.
+Also, $f$ has a continuous [[Definition:Left-Hand Derivative|left-hand derivative]] at $x_i$ for every $i \in \set {1, \ldots, n}$.
+This means that
+:$\map {f'_-} {x_i} = \displaystyle \lim_{x \mathop \to {x_i}^-} \map {f'} x$
+This implies that $\displaystyle \lim_{x \mathop \to {x_i}^-} \map {f'} x$ exists for every $i \in \set {1, \ldots, n}$.
+These two conclusions prove that requirement $(2.2)$ in [[Definition:Piecewise Continuously Differentiable Function/Definition 1|Definition 1]] is satisfied.
+This finishes the proof that $f$ satisfies the requirements of [[Definition:Piecewise Continuously Differentiable Function/Definition 1|Definition 1]].
+{{qed|lemma}}
+=== Definition 1 implies Definition 2 ===
+Assume that $f$ satisfies the requirements of [[Definition:Piecewise Continuously Differentiable Function/Definition 1|Definition 1]].
+We need to prove that $f$ satisfies the requirement of [[Definition:Piecewise Continuously Differentiable Function/Definition 2|Definition 2]].
+$f$ is [[Definition:Continuously Differentiable|continuously differentiable]] on $\openint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$ by definition.
+Also, the one-sided [[Definition:Limit of Real Function|limits]] $\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map {f'} x$ exist for every $i \in \set {1, \ldots, n}$.
+$f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$, the [[Definition:Derivative of Real Function|derivatives]] at $x_{i - 1}$ and $x_i$ understood as [[Definition:One-Sided Derivative|one-sided derivatives]], for every $i \in \set {1, \ldots, n}$, by [[Extendability Theorem for Derivatives Continuous on Open Intervals]].
+This is exactly the requirement of [[Definition:Piecewise Continuously Differentiable Function/Definition 2|Definition 2]].
+{{qed}}
+[[Category:Piecewise Continuously Differentiable Functions]]
+rbncfcw1bu62mjlmgotfd7i7zm49mkm
+\end{proof}<|endoftext|>
+\section{Bounded Function Continuous on Open Interval is Darboux Integrable}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $\closedint a b$ such that $a < b$.
+Let $f$ be [[Definition:Continuous Real Function on Interval|continuous]] on $\closedint a b$.
+Let $f$ be [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$.
+Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+By [[Condition for Darboux Integrability]], it suffices to show that, for a given [[Definition:Strictly Positive Real Number|strictly positive]] $\epsilon$, a [[Definition:Subdivision (Real Analysis)|subdivision]] $S$ of $\closedint a b$ exists such that:
+:$\map U S – \map L S < \epsilon$
+where $\map U S$ and $\map L S$ are respectively the [[Definition:Upper Sum|upper]] and [[Definition:Lower Sum|lower sums]] of $f$ on $\closedint a b$ with respect to the subdivision $S$.
+Since $f$ is [[Definition:Bounded Real-Valued Function|bounded]], a [[Definition:Strictly Positive Real Number|strictly positive]] [[Definition:Bound of Real-Valued Function|bound]] $K$ exists for $f$ on $\closedint a b$.
+Let a [[Definition:Strictly Positive Real Number|strictly positive]] $\epsilon$ be given, and choose a $\delta$ that satisfies:
+:$0 < \delta < \min \left({\dfrac \epsilon {6 K}, \dfrac {b - a} 2}\right)$
+We have that $f$ is [[Definition:Continuity|continuous]] on $\openint a b$.
+As $\delta > 0$, $\closedint {a + \delta} {b - \delta}$ is a [[Definition:Subset|subset]] of $\openint a b$.
+Thus $f$ is continuous on the [[Definition:Real Interval|interval]] $\closedint {a + \delta} {b - \delta}$.
+By [[Continuous Real Function is Darboux Integrable]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {a + \delta} {b - \delta}$.
+Since $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {a + \delta} {b - \delta}$, there exists a [[Definition:Subdivision (Real Analysis)|subdivision]] $S_\delta$ of $\closedint {a + \delta} {b - \delta}$ that satisfies:
+:$\map U {S_\delta} – \map L {S_\delta} < \dfrac \epsilon 3$
+where $\map U {S_\delta}$ and $\map L {S_\delta}$ are respectively the [[Definition:Upper Sum|upper]] and [[Definition:Lower Sum|lower sums]] of $f$ on $\closedint {a + \delta} {b - \delta}$ with respect to the [[Definition:Subdivision (Real Analysis)|subdivision]] $S_\delta$.
+Define the following [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$:
+:$S = S_\delta \cup \set {a, b}$
+The [[Definition:Upper Sum|upper sum]] of $f$ on $\closedint a b$ with respect to $S$ is per definition:
+:$\map U S = M_a \delta + \map U {S_\delta} + M_b \delta$
+where:
+:$M_a$ is the [[Definition:Supremum of Real-Valued Function|supremum]] of $f$ on $\closedint a {a + \delta}$
+:$M_b$ is the [[Definition:Supremum of Real-Valued Function|supremum]] of $f$ on $\closedint {b - \delta} b$.
+$M_a$ and $M_b$ exist by [[Continuum Property|the least upper bound property of the real numbers]] because $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a {a + \delta}$ and $\closedint {b - \delta} b$.
+The [[Definition:Lower Sum|lower sum]] of $f$ on $\closedint a b$ with respect to $S$ is per definition:
+:$\map L S = m_a \delta + \map L {S_\delta} + m_b \delta$
+where:
+:$m_a$ is the [[Definition:Infimum of Real-Valued Function|infimum]] of $f$ on $\closedint a {a + \delta}$
+:$m_b$ is the [[Definition:Infimum of Real-Valued Function|infimum]] of $f$ on $\closedint {b - \delta} b$
+$m_a$ and $m_b$ exist by the [[Continuum Property]], because $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a {a + \delta}$ and $\closedint {b - \delta} b$.
+Define the sum:
+{{begin-eqn}}
+{{eqn | l = U'
+      | r = K \delta + \map U {S_\delta} + K \delta
+}}
+{{eqn | o = \ge
+      | r = M_a \delta + \map U {S_\delta} + M_b \delta
+      | c = by $K \ge M_a$ and $K \ge M_b$
+}}
+{{eqn | r = \map U S
+}}
+{{end-eqn}}
+Define the sum:
+{{begin-eqn}}
+{{eqn | l = L'
+      | r = -K \delta + \map L {S_\delta} - K \delta
+}}
+{{eqn | o = \le
+      | r = m_a \delta + \map L {S_\delta} + m_b \delta
+      | c = by $-K \le m_a$ and $-K \le m_b$
+}}
+{{eqn | r = \map L S
+}}
+{{end-eqn}}
+Therefore, $U'$ and $L'$ satisfy:
+:$U' \ge \map U S$
+:$L' \le \map L S$
+From these two inequalities follows:
+{{begin-eqn}}
+{{eqn | l = \map U S – \map L S
+      | o = \le
+      | r = U' - L'
+}}
+{{eqn | r = K \delta + \map U {S_\delta} + K \delta - \paren {-K \delta + \map L {S_\delta} -K \delta}
+      | c = Definitions of $U'$ and $L'$
+}}
+{{eqn | r = 4 K \delta + \map U {S_\delta} - \map L {S_\delta}
+}}
+{{eqn | o = <
+      | r = 4 K \min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2} + \map U {S_\delta} - \map L {S_\delta}
+      | c = by $\delta < \min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2}$
+}}
+{{eqn | o = \le
+      | r = 4 K \dfrac \epsilon {6 K} + \map U {S_\delta} - \map L {S_\delta}
+      | c = by $\min \set {\dfrac \epsilon {6 K}, \dfrac {b - a} 2} \le \dfrac \epsilon {6 K}$
+}}
+{{eqn | o = <
+      | r = 4 K \dfrac \epsilon {6 K} + \dfrac \epsilon 3
+      | c = by $\map U {S_\delta} - \map L {S_\delta} < \dfrac \epsilon 3$
+}}
+{{eqn | r = \frac {2 \epsilon} 3 + \frac \epsilon 3
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Hence:
+:$\map U S – \map L S < \epsilon$
+{{qed}}
+[[Category:Integral Calculus]]
+ls000s394m0i1klb4scbrn5l04zhamt
+\end{proof}<|endoftext|>
+\section{Square Inscribed in Circle is greater than Half Area of Circle}
+Tags: Circles, Squares
+
+\begin{theorem}
+A [[Definition:Square (Geometry)|square]] [[Definition:Polygon Inscribed in Circle|inscribed]] in a [[Definition:Circle|circle]] has an [[Definition:Area|area]] greater than half that of the [[Definition:Circle|circle]].
+\end{theorem}
+
+\begin{proof}
+:[[File:InscribedCircumscribedSquare.png|300px]]
+Let $ABCD$ be a [[Definition:Square (Geometry)|square]] [[Definition:Polygon Inscribed in Circle|inscribed]] in a [[Definition:Circle|circle]].
+Let $EFGH$ be a [[Definition:Square (Geometry)|square]] [[Definition:Polygon Circumscribed around Circle|circumscribed]] around the same [[Definition:Circle|circle]].
+We have that:
+:$ABCD$ is twice the [[Definition:Area|area]] of the [[Definition:Triangle (Geometry)|triangle]] $ADB$.
+:$EFGH$ is twice the [[Definition:Area|area]] of the [[Definition:Rectangle|rectangle]] $EFBD$.
+From [[Area of Rectangle]], the [[Definition:Area|area]] of $EFBD$ is $ED \cdot DB$.
+From [[Area of Triangle in Terms of Side and Altitude]], the [[Definition:Area|area]] of $ADB$ is $\dfrac 1 2 \cdot ED \cdot DB$.
+Thus:
+:$EFGH = 2 \cdot ABCD$
+But the [[Definition:Area|area]] of $EFGH$ is greater than the [[Definition:Area|area]] of the [[Definition:Circle|circle]] around which it is [[Definition:Polygon Circumscribed around Circle|circumscribed]].
+Therefore half of the [[Definition:Area|area]] of $EFGH$ is greater than half of the [[Definition:Area|area]] of the [[Definition:Circle|circle]] around which it is [[Definition:Polygon Circumscribed around Circle|circumscribed]].
+Therefore the [[Definition:Area|area]] of $ABCD$ is greater than half of the [[Definition:Area|area]] of the [[Definition:Circle|circle]] within which it is [[Definition:Polygon Inscribed in Circle|inscribed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bounded Piecewise Continuous Function is Darboux Integrable}
+Tags: Piecewise Continuous Functions, Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $f$ be [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
+Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+We are given that $f$ is [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
+Therefore, there exists a [[Definition:Finite Subdivision|finite subdivision]] $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
+:$f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint {x_{i - 1} } {x_i}$.
+Note that $n$ is the number of [[Definition:Open Real Interval|intervals]] $\openint {x_{i - 1} } {x_i}$ defined from the [[Definition:Finite Subdivision|(finite) subdivision]] $\set {x_0, x_1, \ldots, x_n}$.
+We shall use [[Principle of Mathematical Induction|proof by induction]] on these $n$ [[Definition:Open Real Interval|intervals]].
+For all $k \in \set {1, 2, \ldots, n}$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$.
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_{i - 1} } {x_i}$
+for an arbitrary $i \in \set {1, 2, \ldots, k}$. 
+$f$ is [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] for the case $n = 1$ means that:
+:$f$ is [[Definition:Continuous Real Function on Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint {x_{i - 1} } {x_i}$.
+By [[Bounded Function Continuous on Open Interval is Darboux Integrable]], $f$ is [[Definition:Definite Integral|Darboux integrable]] on $\closedint {x_{i - 1} } {x_i}$.
+Thus $\map P 1$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$
+from which it is to be shown that:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_{k + 1} }$.
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+By definition of a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]], for every $i \in \set {1, 2, \ldots, k, k + 1}$:
+:$f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\openint {x_{i - 1} } {x_i}$
+:$f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint {x_{i - 1} } {x_i}$.
+By the [[Bounded Piecewise Continuous Function is Darboux Integrable#Induction Hypothesis|induction hypothesis]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$.
+From the [[Bounded Piecewise Continuous Function is Darboux Integrable#Basis for the Induction|basis for the induction]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_k} {x_{k + 1} }$.
+We have that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k}$ and $\closedint {x_k} {x_{k + 1} }$.
+Therefore, $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$ by [[Existence of Integral on Union of Adjacent Intervals]].
+We have that:
+:$\closedint {x_0} {x_{k + 1} } = \closedint {x_0} {x_k} \cup \closedint {x_k} {x_{k + 1} }$.
+Accordingly, $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint {x_0} {x_{k + 1} }$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bounded Piecewise Continuous Function may not have One-Sided Limits}
+Tags: Piecewise Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$, $a < b$. 
+Let $f$ be a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]].
+{{:Definition:Bounded Piecewise Continuous Function}}
+Then it is not necessarily the case that $f$ is a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]]:
+{{:Definition:Piecewise Continuous Function with One-Sided Limits}}
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Real Function|function]]:
+:$f \left({x}\right) = \begin{cases}
+0 & : x = a \\
+\sin \left({\dfrac 1 {x - a} }\right) & : x \in \left({a \,.\,.\, b}\right]
+\end{cases}$
+Consider the [[Definition:Finite Subdivision|(finite) subdivision]] $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.
+We observe that $\sin \left({\dfrac 1 {x - a} }\right)$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\left({a \,.\,.\, b}\right)$.
+Since $f \left({x}\right) = \sin \left({\dfrac 1 {x - a} }\right)$ on $\left({a \,.\,.\, b}\right)$, it follows that $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\left({a \,.\,.\, b}\right)$.
+Also, $f$ is [[Definition:Bounded Real-Valued Function|bounded]] by the [[Definition:Bound of Real-Valued Function|bound]] $1$ on $\left[{a \,.\,.\, b}\right]$.
+Therefore $f$ is a [[Definition:Bounded Piecewise Continuous Function|bounded piecewise continuous function]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$.
+We now investigate whether $f$ has [[Definition:One-Sided Limit of Real Function|one-sided limits]] at the [[Definition:Endpoint of Real Interval|endpoints]] of $\left[{a \,.\,.\, b}\right]$.
+The function $\sin \left({\dfrac 1 {x - a} }\right)$ varies between $-1$ and $+1$ as $x$ approaches $a$ from above.
+Thus it does not [[Definition:Convergent Real Function|converge]]. 
+Since $f \left({x}\right) = \sin \left({\dfrac 1 {x - a} }\right)$ when $x > a$ we conclude that $\displaystyle \lim_{x \mathop \to a^+} f \left({x}\right)$ does not exist either.
+$f$ has no [[Definition:One-Sided Limit of Real Function|one-sided limit]] at $a$.
+Hence $f$ is not a [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous function with one-sided limits]].
+{{qed}}
+[[Category:Piecewise Continuous Functions]]
+kkhpxp4jlyfs9es6kwm8a87h9dy5z1y
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Subset of Real Numbers}
+Tags: Rational Numbers, Real Numbers
+
+\begin{theorem}
+The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] forms a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] $\R$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \Q$, where $\Q$ denotes the [[Definition:Rational Number|set of rational numbers]].
+Consider the [[Definition:Rational Sequence|rational sequence]]:
+:$x, x, x, \ldots$
+This [[Definition:Rational Sequence|sequence]] is trivially [[Definition:Cauchy Sequence|Cauchy]].
+Thus there exists a [[Definition:Cauchy Sequence|Cauchy sequence]] $\eqclass {\sequence {x_n} } {}$ which is identified with a [[Definition:Rational Number|rational number]] $x \in \Q$ such that:
+So by the [[Definition:Real Number|definition of a real number]]:
+:$x \in \R$
+where $\R$ denotes the [[Definition:Real Number|set of real numbers]].
+Thus, by definition of [[Definition:Subset|subset]]:
+:$\Q \subseteq \R$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number is not necessarily Rational Number}
+Tags: Real Numbers, Rational Numbers
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]].
+Then it is not necessarily the case that $x$ is also a [[Definition:Rational Number|rational number]].
+\end{theorem}
+
+\begin{proof}
+By [[Proof by Counterexample]]:
+Let $x = \sqrt 2$.
+From [[Square Root of 2 is Irrational]]:
+:$\sqrt 2$ is an [[Definition:Irrational Number|irrational number]].
+By definition:
+:$x \in \R \setminus \Q$
+where:
+:$\R$ is the [[Definition:Real Number|set of real numbers]]
+:$\Q$ is the [[Definition:Rational Number|set of rational numbers]]
+:$\setminus$ denotes the [[Definition:Set Difference|set difference]].
+Thus $x$, while being a [[Definition:Real Number|real number]], is not also a [[Definition:Rational Number|rational number]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero is both Positive and Negative}
+Tags: Numbers
+
+\begin{theorem}
+The [[Definition:Number|number]] $0$ ([[Definition:Zero (Number)|zero]]) is the only [[Definition:Real Number|(real) number]] which is both:
+:a [[Definition:Positive Real Number|positive (real) number]]
+and
+:a [[Definition:Negative Real Number|negative (real) number]].
+\end{theorem}
+
+\begin{proof}
+Let $x$ be a [[Definition:Real Number|real number]] which is both [[Definition:Positive Real Number|positive]] and [[Definition:Negative Real Number|negative]].
+Thus:
+:$x \in \set {x \in \R: x \ge 0}$
+and:
+:$x \in \set {x \in \R: x \le 0}$
+and so:
+:$0 \le x \le 0$
+from which:
+:$x = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binary Operation on Subset is Binary Operation}
+Tags: Operations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\circ$ be a [[Definition:Binary Operation|binary operation]] on $S$.
+Let $T \subseteq S$.
+Let $\circ {\restriction}_T$ be the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $T$.
+Then $\circ {\restriction}_T$ is a [[Definition:Binary Operation|binary operation]] on $T$.
+\end{theorem}
+
+\begin{proof}
+Let $\Bbb U$ be a [[Definition:Universal Set|universal set]].
+Let $\circ: S \times S \to \Bbb U$ be a [[Definition:Binary Operation|binary operation]] on $S$.
+Let $T \subseteq S$.
+Let $\left({a, b}\right) \in T \times T$.
+By definition of [[Definition:Ordered Pair|ordered pair]] and [[Definition:Cartesian Product|cartesian product]]:
+:$a \in T$ and $b \in T$
+As $T \subseteq S$, it follows that:
+:$a \in S$ and $b \in S$
+Thus:
+:$\left({a, b}\right) \in S \times S$
+As $\circ$ is a [[Definition:Binary Operation|binary operation]] on $S$, it follows that:
+:$\circ \left({a, b}\right) \in \Bbb U$
+But by definition of [[Definition:Restriction of Operation|restriction]] of $\circ$:
+:$\circ \left({a, b}\right) = \circ {\restriction}_T \left({a, b}\right)$
+Thus:
+:$\circ {\restriction}_T \left({a, b}\right) \in \Bbb U$
+As $a$ and $b$ are arbitrary elements of $T$, it follows that this holds for all $a, b \in T$.
+Hence the result by definition of [[Definition:Binary Operation|binary operation]].
+{{qed}}
+[[Category:Operations]]
+owksju1jdr9447lowaf79mvn8kii969
+\end{proof}<|endoftext|>
+\section{Matrix Entrywise Addition over Ring is Associative}
+Tags: Matrix Entrywise Addition, Associativity, Matrix Entrywise Addition is Associative
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
+For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition over Ring|matrix entrywise sum]] of $\mathbf A$ and $\mathbf B$.
+The operation $+$ is [[Definition:Associative Operation|associative]] on $\map {\MM_R} {m, n}$.
+That is:
+:$\paren {\mathbf A + \mathbf B} + \mathbf C = \mathbf A + \paren {\mathbf B + \mathbf C}$
+for all $\mathbf A$, $\mathbf B$ and $\mathbf C$ in $\map {\MM_R} {m, n}$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be [[Definition:Element|elements]] of the [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {\mathbf A + \mathbf B} + \mathbf C
+      | r = \paren {\sqbrk a_{m n} + \sqbrk b_{m n} } + \sqbrk c_{m n}
+      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
+}}
+{{eqn | r = \sqbrk {a + b}_{m n} + \sqbrk c_{m n}
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \sqbrk {\paren {a + b} + c}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \sqbrk {a + \paren {b + c} }_{m n}
+      | c = {{Ring-axiom|A1}}
+}}
+{{eqn | r = \sqbrk a_{m n} + \sqbrk {b + c}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \sqbrk a_{m n} + \paren {\sqbrk b_{m n} + \sqbrk c_{m n} }
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \mathbf A + \paren {\mathbf B + \mathbf C}
+      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is the '''[[Definition:Hadamard Product|Hadamard product]]''' of $\mathbf A$ and $\mathbf B$ with respect to [[Definition:Ring Addition|ring addition]].
+We have from {{Ring-axiom|A1}} that [[Definition:Ring Addition|ring addition]] is [[Definition:Associative Operation|associative]].
+The result then follows directly from [[Associativity of Hadamard Product]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Final Topology with respect to Mapping}
+Tags: Topology
+
+\begin{theorem}
+Let $\left({X, \tau_X}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $Y$ be a [[Definition:Set|set]].
+Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
+Let $\tau_Y$ be the [[Definition:Final Topology|final topology]] on $Y$ with respect to $f$.
+Then:
+:$\tau_Y = \left\{ U \subseteq Y : f^{-1} \left({U}\right) \in \tau_X \right\}$
+Observe that the set on the right-hand side of the equality is sometimes denoted $f\tau$.
+Further, the following holds true:
+# $\forall \left(Z, \tau_Z \right)$ [[Definition:Topological Space|topological space]] $: \forall g: \left({Y, f\tau}\right) \to \left({Z, \tau_Z}\right)$ [[Definition:Mapping|mapping]] $: g$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] $\iff g \circ f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] (where $f$ is defined from $\left({X, \tau_X}\right)$ to $\left(Y, f\tau \right)$).
+# $\forall \tau_Y$ [[Definition:Topology|topology]] in $Y : \forall \left(Z, \tau_Z \right)$ [[Definition:Topological Space|topological space]], the following are equivalent:
+##$\tau_Y = f\tau$
+##$\forall g: \left({Y, \tau_Y}\right) \to \left({Z, \tau_Z}\right) : g$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] $\iff g \circ f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] (where $f$ is defined from $\left({X, \tau_X}\right)$ to $\left(Y, \tau_Y \right)$).
+That is, the [[Definition:Topology|topology]] $f\tau$ is ''characterised'' by the fact that it is the only one which makes all [[Definition:Mapping|mappings]] defined from it to any topological space [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Topology]]
+mwydy50vwd2g234im8t9ozvldbja70p
+\end{proof}<|endoftext|>
+\section{Composition of Symmetries is Symmetry}
+Tags: Symmetries
+
+\begin{theorem}
+Let $P$ be a [[Definition:Geometric Figure|geometric figure]].
+Let $S_P$ be the [[Definition:Set|set]] of all [[Definition:Symmetry Mapping|symmetries]] of $P$.
+Let $\circ$ denote [[Definition:Composition of Mappings|composition of mappings]].
+Let $\phi$ and $\psi$ be [[Definition:Symmetry Mapping|symmetries]] of $P$.
+Then $\phi \circ \psi$ is also a [[Definition:Symmetry Mapping|symmetry]] of $P$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$\phi \circ \psi$ is a [[Definition:Mapping|mapping]].
+We have by definition of [[Definition:Symmetry Mapping|symmetry]] that:
+:$\map \phi P$ is [[Definition:Congruence (Geometry)|congruent]] to $P$
+and:
+:$\map \psi {\map \phi P}$ is [[Definition:Congruence (Geometry)|congruent]] to $\map \phi P$
+Therefore:
+:$\phi \circ \psi$ is [[Definition:Congruence (Geometry)|congruent]] to $P$
+Thus $\phi \circ \psi$ is a [[Definition:Symmetry Mapping|symmetry]] of $P$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Symmetries is Associative}
+Tags: Symmetries
+
+\begin{theorem}
+Let $P$ be a [[Definition:Geometric Figure|geometric figure]].
+Let $S_P$ be the [[Definition:Set|set]] of all [[Definition:Symmetry Mapping|symmetries]] of $P$.
+Let $\circ$ denote [[Definition:Composition of Mappings|composition of mappings]].
+Let $\phi, \psi, \chi$ be [[Definition:Symmetry Mapping|symmetries]] of $P$.
+Then:
+: $\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$
+That is, [[Definition:Composition of Mappings|composition]] of [[Definition:Symmetry Mapping|symmetries]] is [[Definition:Associative|associative]].
+\end{theorem}
+
+\begin{proof}
+From [[Composition of Symmetries is Symmetry]]:
+:$\paren {\phi \circ \psi} \circ \chi$ is a [[Definition:Symmetry Mapping|symmetry]]
+and:
+:$\phi \circ \paren {\psi \circ \chi}$ is a [[Definition:Symmetry Mapping|symmetry]].
+It follows from [[Composition of Mappings is Associative]] that:
+: $\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square of Modulo less One equals One}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]].
+Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]]:
+:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
+Then:
+:$\eqclass {m - 1} m \times_m \eqclass {m - 1} m = \eqclass 1 m$
+where $\times_m$ denotes [[Definition:Modulo Multiplication|multiplication modulo $m$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \eqclass {m - 1} m \times_m \eqclass {m - 1} m
+      | r = \eqclass {\paren {m - 1}^2} m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = \eqclass {m^2 - 2 m + 1} m
+      | c = 
+}}
+{{eqn | r = \eqclass 1 m
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Non-Square Residue}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]] such that $m > 2$.
+Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]]:
+:$\Z_m = \left\{{\left[\!\left[{0}\right]\!\right]_m, \left[\!\left[{1}\right]\!\right]_m, \ldots, \left[\!\left[{m - 1}\right]\!\right]_m}\right\}$
+Then there exists at least one [[Definition:Residue (Modulo Arithmetic)|residue]] in $\Z_m$ which is not the [[Definition:Modulo Multiplication|product modulo $m$]] of a [[Definition:Residue (Modulo Arithmetic)|residue]] with itself:
+:$\exists p \in \Z_M: \forall x \in \Z_m: x \cdot_m x \ne p$
+\end{theorem}
+
+\begin{proof}
+We have that $1 \in \Z_m$ and that:
+:$1 \cdot_m 1 = 1$
+We have that $m - 1 \in \Z_m$ and that:
+:$\left({m - 1}\right) \cdot_m \left({m - 1}\right) = 1$
+Thus unless $m - 1 = 1$, that is, $m = 2$, there exist $2$ [[Definition:Residue (Modulo Arithmetic)|residues]] of $\Z_m$ whose [[Definition:Modulo Multiplication|product modulo $m$]] with itself equals $1$.
+There are $m - 2$ [[Definition:Residue (Modulo Arithmetic)|residues]] which, when [[Definition:Modulo Multiplication|multiplied modulo $m$]] with themselves have as a result a [[Definition:Residue (Modulo Arithmetic)|residue]].
+Thus there can be at maximum $m - 2$ [[Definition:Residue (Modulo Arithmetic)|residues]] (excluding $1$) which can be the [[Definition:Modulo Multiplication|product modulo $m$]] of a [[Definition:Residue (Modulo Arithmetic)|residue]] with itself.
+But there are $m - 1$ [[Definition:Residue (Modulo Arithmetic)|residues]] (excluding $1$).
+So at least $1$ [[Definition:Residue (Modulo Arithmetic)|residues]] is not the [[Definition:Modulo Multiplication|product modulo $m$]] of a [[Definition:Residue (Modulo Arithmetic)|residue]] with itself.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetry Group of Equilateral Triangle is Symmetric Group}
+Tags: Symmetry Group of Equilateral Triangle, Symmetric Group on 3 Letters
+
+\begin{theorem}
+Let $D_3$ denote the [[Definition:Symmetry Group of Equilateral Triangle|symmetry group of the equilateral triangle]].
+Let $S_3$ denote the [[Symmetric Group on 3 Letters|symmetric group on $3$ letters]].
+Then $D_3$ is [[Definition:Group Isomorphism|isomorphic]] to $S_3$.
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\ds \relcomp S {\bigcap \mathbb T} = \bigcup_{H \mathop \in \mathbb T} \relcomp S H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \relcomp S {\bigcap \mathbb T}
+      | r = S \setminus \paren {\bigcap \mathbb T}
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {S \setminus H}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \relcomp S H
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle \relcomp S {\bigcup \mathbb T} = \bigcap_{H \mathop \in \mathbb T} \relcomp S H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \relcomp S {\bigcup \mathbb T}
+      | r = S \setminus \paren {\bigcup \mathbb T}
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {S \setminus H}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \relcomp S H
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle \relcomp S {\bigcup_{i \mathop \in I} \mathbb S_i} = \bigcap_{i \mathop \in I} \relcomp S {S_i}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \relcomp S {\bigcup_{i \mathop \in I} S_i}
+      | r = S \setminus \paren {\bigcup_{i \mathop \in I} S_i}
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | r = \bigcap_{i \mathop \in I} \paren {S \setminus S_i}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union|De Morgan's Laws for Set Difference: Difference with Union]]
+}}
+{{eqn | r = \bigcap_{i \mathop \in I} \relcomp S {S_i}
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle \relcomp S {\bigcap_{i \mathop \in I} \mathbb S_i} = \bigcup_{i \mathop \in I} \relcomp S {S_i}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \relcomp S {\bigcap_{i \mathop \in I} S_i}
+      | r = S \setminus \paren {\bigcap_{i \mathop \in I} S_i}
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | r = \bigcup_{i \mathop \in I} \paren {S \setminus S_i}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection|De Morgan's Laws for Set Difference: Difference with Intersection]]
+}}
+{{eqn | r = \bigcup_{i \mathop \in I} \relcomp S {S_i}
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\relcomp S {T_1 \cup T_2} = \relcomp S {T_1} \cap \relcomp S {T_2}$
+\end{theorem}
+
+\begin{proof}
+Let $T_1, T_2 \subseteq S$.
+Then from [[Union is Smallest Superset]]:
+:$T_1 \cup T_2 \subseteq S$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \relcomp S {T_1 \cup T_2}
+      | r = S \setminus \paren {T_1 \cup T_2}
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | r = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{eqn | r = \relcomp S {T_1} \cap \relcomp S {T_2}
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\relcomp S {T_1 \cap T_2} = \relcomp S {T_1} \cup \relcomp S {T_2}$
+\end{theorem}
+
+\begin{proof}
+Let $T_1, T_2 \subseteq S$.
+Then from [[Intersection is Subset]] and [[Subset Relation is Transitive]]:
+:$T_1 \cap T_2 \subseteq S$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \relcomp S {T_1 \cap T_2}
+      | r = S \setminus \paren {T_1 \cap T_2}
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | r = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \relcomp S {T_1} \cup \relcomp S {T_2}
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection}
+Tags: De Morgan's Laws, Indexed Families
+
+\begin{theorem}
+:$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
+where:
+:$\displaystyle \bigcup_{i \mathop \in I} T_i := \set {x: \exists i \in I: x \in T_i}$
+that is, the [[Definition:Union of Family|union]] of $\family {T_i}_{i \mathop \in I}$.
+\end{theorem}
+
+\begin{proof}
+Suppose:
+:$\displaystyle x \in S \setminus \bigcap_{i \mathop \in I} T_i$
+Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = S \setminus \bigcap_{i \mathop \in I} T_i
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \notin
+      | r = \bigcap_{i \mathop \in I} T_i
+      | c = {{Defof|Set Difference}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \neg \leftparen {\forall i \in I: x}
+      | o = \in
+      | r = \rightparen {T_i}
+      | c = {{Defof|Intersection of Family}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists i \in I: \neg \leftparen x
+      | o = \in
+      | r = \rightparen {T_i}
+      | c = [[De Morgan's Laws (Predicate Logic)]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists i \in I: x
+      | o = \in
+      | r = S \setminus T_i
+      | c = {{Defof|Set Difference}}: note $x \in S$ from above
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}
+      | c = {{Defof|Union of Family}}
+}}
+{{end-eqn}}
+Therefore:
+:$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Union}
+Tags: De Morgan's Laws, Indexed Families
+
+\begin{theorem}
+:$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
+where:
+:$\displaystyle \bigcap_{i \mathop \in I} T_i := \set {x: \forall i \in I: x \in T_i}$
+that is, the [[Definition:Intersection of Family|intersection]] of $\family {T_i}_{i \mathop \in I}$.
+\end{theorem}
+
+\begin{proof}
+Suppose:
+:$\displaystyle x \in S \setminus \bigcup_{i \mathop \in I} T_i$
+Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = S \setminus \bigcup_{i \mathop \in I} T_i
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \notin
+      | r = \bigcup_{i \mathop \in I} T_i
+      | c = {{Defof|Set Difference}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \neg \leftparen {\exists i \in I: \leftparen x}
+      | o = \in
+      | r = \rightparen {\rightparen {T_i} }
+      | c = {{Defof|Union of Family}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall i \in I: \neg \leftparen x
+      | o = \in
+      | r = \rightparen {T_i}
+      | c = [[De Morgan's Laws (Predicate Logic)]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall i \in I: x
+      | o = \in
+      | r = S \setminus T_i
+      | c = {{Defof|Set Difference}}: note $x \in S$ from above
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}
+      | c = {{Defof|Intersection of Family}}
+}}
+{{end-eqn}}
+Therefore:
+:$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle \complement \paren {\bigcap \mathbb T} = \bigcup_{H \mathop \in \mathbb T} \complement \paren H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \complement \paren {\bigcap \mathbb T}
+      | r = \mathbb U \setminus \paren {\bigcap \mathbb T}
+      | c = {{Defof|Set Complement}}
+}}
+{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = \bigcup_{H \mathop \in \mathbb T} \complement \paren H
+      | c = {{Defof|Set Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/General Case/Complement of Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle \complement \paren {\bigcup \mathbb T} = \bigcap_{H \mathop \in \mathbb T} \complement \paren H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \complement \paren {\bigcup \mathbb T}
+      | r = \mathbb U \setminus \paren {\bigcup \mathbb T}
+      | c = {{Defof|Set Complement}}
+}}
+{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws for Set Difference: Difference with Union]]
+}}
+{{eqn | r = \bigcap_{H \mathop \in \mathbb T} \complement \paren H
+      | c = {{Defof|Set Complement}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\overline {T_1 \cap T_2} = \overline T_1 \cup \overline T_2$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/Complement of Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\overline {T_1 \cup T_2} = \overline T_1 \cap \overline T_2$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a [[Definition:Set|set]] and $I$ is some [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
+Then:
+:$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Cardinality|cardinality]] $\card I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
+Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
+:$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
+as:
+:$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$
+The proof of this is more amenable to proof by [[Principle of Mathematical Induction]].
+For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
+:$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$
+$\map P 1$ is true, as this just says $S \setminus T_1 = S \setminus T_1$.
+=== Base Case ===
+$\map P 2$ is the case:
+:$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$
+which has been [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|proved]]. 
+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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle S \setminus \bigcap_{i \mathop = 1}^k T_i = \bigcup_{i \mathop = 1}^k \paren {S \setminus T_i}$
+=== Induction Step ===
+Now we need to show:
+:$\displaystyle S \setminus \bigcap_{i \mathop = 1}^{k + 1} T_i = \bigcup_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = S \setminus \bigcap_{i \mathop = 1}^{k + 1} T_i
+      | r = S \setminus \paren {\bigcap_{i \mathop = 1}^k T_i \cap T_{k + 1} }
+      | c = [[Intersection is Associative]]
+}}
+{{eqn | r = \paren {S \setminus \bigcap_{i \mathop = 1}^k T_i} \cup \paren {S \setminus T_{k + 1} }
+      | c = [[De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof#Base Case|Base case]]
+}}
+{{eqn | r = \paren {\bigcup_{i \mathop = 1}^k \paren {S \setminus T_i} } \cup \paren {S \setminus T_{k + 1} }
+      | c = [[De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection/Proof#Induction Hypothesis|Induction hypothesis]]
+}}
+{{eqn | r = \bigcup_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}
+      | c = [[Union is Associative]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$
+that is:
+:$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+: $\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \left({S \setminus T_i}\right)$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a [[Definition:Set|set]] and $I$ is some [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
+Then:
+: $\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Cardinality|cardinality]] $\size I$ of the [[Definition:Indexing Set|indexing set]] $I$ be $n$.
+Then by the definition of [[Definition:Cardinality|cardinality]], it follows that $I \cong \N^*_n$ and we can express the proposition:
+:$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
+as:
+:$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$
+The proof of this is more amenable to proof by [[Principle of Mathematical Induction]].
+For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
+:$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$.
+$\map P 1$ is true, as this just says $S \setminus T_1 = S \setminus T_1$.
+=== Base Case ===
+$\map P 2$ is the case:
+:$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$
+which has been [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Union|proved]].
+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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle S \setminus \bigcup_{i \mathop = 1}^k T_i = \bigcap_{i \mathop = 1}^k \paren {S \setminus T_i}$
+=== Induction Step ===
+Now we need to show:
+:$\displaystyle S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i = \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i
+      | r = S \setminus \paren {\bigcup_{i \mathop = 1}^k T_i \cup T_{k + 1} }
+      | c = [[Union is Associative]]
+}}
+{{eqn | r = \paren {S \setminus \bigcup_{i \mathop = 1}^k T_i} \cap \paren {S \setminus T_{k + 1} }
+      | c = [[De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof#Base Case|Base case]]
+}}
+{{eqn | r = \paren {\bigcap_{i \mathop = 1}^k \paren {S \setminus T_i} } \cap \paren {S \setminus T_{k + 1} }
+      | c = [[De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union/Proof#Induction Hypothesis|Induction hypothesis]]
+}}
+{{eqn | r = \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}
+      | c = [[Intersection is Associative]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$
+that is:
+:$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+: $\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \left({S \setminus T_i}\right)$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle S \setminus \bigcap \mathbb T = \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$
+where:
+:$\displaystyle \bigcap \mathbb T := \set {x: \forall T' \in \mathbb T: x \in T'}$
+that is, the [[Definition:Intersection of Set of Sets|intersection]] of $\mathbb T$
+\end{theorem}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$\displaystyle S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$
+where:
+:$\displaystyle \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$
+that is, the [[Definition:Union of Set of Sets|union]] of $\mathbb T$.
+\end{theorem}
+
+\begin{proof}
+Suppose:
+:$\displaystyle x \in S \setminus \bigcup \mathbb T$
+Note that by [[Set Difference is Subset]] we have that $x \in S$ (we need this later).
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = S \setminus \bigcup \mathbb T
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \notin
+      | r = \bigcup \mathbb T
+      | c = {{Defof|Set Difference}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \neg (\exists T' \in  \mathbb T: (x
+      | o = \in
+      | r = T'))
+      | c = {{Defof|Union of Set of Sets}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall T' \in  \mathbb T: \neg (x
+      | o = \in
+      | r = T')
+      | c = [[De Morgan's Laws (Predicate Logic)]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall T' \in  \mathbb T: x
+      | o = \in
+      | r = S \setminus T'
+      | c = {{Defof|Set Difference}}: note $x \in S$ from above
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}
+      | c = {{Defof|Intersection of Set of Sets}}
+}}
+{{end-eqn}}
+Therefore:
+: $\displaystyle S \setminus \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \left({S \setminus T'}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x \in S \setminus \paren {T_1 \cap T_2}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cap T_2} }
+      | c = {{Defof|Set Difference}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 \land x \in T_2} }
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S} \land \paren {x \notin T_1 \lor x \notin T_2}
+      | c = [[De Morgan's Laws (Logic)/Disjunction of Negations|De Morgan's Laws: Disjunction of Negations]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S \land x \notin T_1}) \lor \paren {x \in S \land x \notin T_2}
+      | c = [[Rule of Distribution]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = x \in \paren {S \setminus T_1} \cup \paren {S \setminus T_2}
+      | c = {{Defof|Set Union}} and {{Defof|Set Difference}}
+}}
+{{end-eqn}}
+By definition of [[Definition:Set Equality/Definition 1|set equality]]:
+: $S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Difference/Difference with Union}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x \in S \setminus \paren {T_1 \cup T_2}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cup T_2} }
+      | c = {{Defof|Set Difference}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 \lor x \in T_2} }
+      | c = {{Defof|Set Union}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S} \land \paren {x \notin T_1 \land x \notin T_2}
+      | c = [[De Morgan's Laws (Logic)/Conjunction of Negations|De Morgan's Laws: Conjunction of Negations]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \paren {x \in S \land x \notin T_1} \land \paren {x \in S \land x \notin T_2}
+      | c = Rules of [[Rule of Idempotence|Idempotence]], [[Rule of Commutation|Commutation]] and [[Rule of Association|Association]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = x \in \paren {S \setminus T_1} \cap \paren {S \setminus T_2}
+      | c = {{Defof|Set Intersection}} and {{Defof|Set Difference}}
+}}
+{{end-eqn}}
+By definition of [[Definition:Set Equality/Definition 1|set equality]]:
+:$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$
+{{qed}}
+{{Namedfor|Augustus De Morgan}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Ring of Integers Modulo 2 and Parity Ring}
+Tags: Ring of Integers Modulo m, Parity Ring, Ring Isomorphisms, Field Isomorphisms
+
+\begin{theorem}
+The [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]] and the [[Definition:Parity Ring|parity ring]] are [[Definition:Isomorphic Algebraic Structures|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+To simplify the notation, let the elements of $\Z_2$ be identified as $0$ for $\eqclass 0 2$ and $1$ for $\eqclass 1 2$.
+Let $f$ be the [[Definition:Mapping|mapping]] from the [[Definition:Parity Ring|parity ring]] $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ and the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]] $\struct {\Z_2, +_2, \times_2}$:
+:$f: \struct {\set {\text{even}, \text{odd} }, +, \times} \to \struct {\Z_2, +_2, \times_2}$:
+::$\forall x \in R: \map f x = \begin{cases}
+0 & : x = \text{even} \\
+1 & : x = \text{odd}  \end{cases}$
+The [[Definition:Bijection|bijective]] nature of $f$ is apparent:
+:$f^{-1}: \struct {\Z_2, +_2, \times_2} \to \struct {\set {\text{even}, \text{odd} }, +, \times}$:
+::$\forall x \in \Z_2: \map {f^{-1} } x = \begin{cases}
+\text{even} & : x = 0 \\
+\text{odd}  & : x = 1 \end{cases}$
+Thus the following equations can be checked:
+{{begin-eqn}}
+{{eqn | ll= 0 +_2 0 =
+      | l = \map f {\text{even} } +_2 \map f {\text{even} }
+      | r = \map f {\text{even} + \text{even} }
+      | rr= = 0
+}}
+{{eqn | ll= 0 +_2 1 =
+      | l = \map f {\text{even} } +_2 \map f {\text{odd} }
+      | r = \map f {\text{even} + \text{odd} }
+      | rr= = 1
+}}
+{{eqn | ll= 1 +_2 0 =
+      | l = \map f {\text{odd} } +_2 \map f {\text{even} }
+      | r = \map f {\text{odd} + \text{even} }
+      | rr= = 1
+}}
+{{eqn | ll= 1 +_2 1 =
+      | l = \map f {\text{odd} } +_2 \map f {\text{odd} }
+      | r = \map f {\text{odd} + \text{odd} }
+      | rr= = 0
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | ll= 0 \times_2 0 =
+      | l = \map f {\text{even} } \times_2 \map f {\text{even} }
+      | r = \map f {\text{even} \times \text{even} }
+      | rr= = 0
+}}
+{{eqn | ll= 0 \times_2 1 =
+      | l = \map f {\text{even} } \times_2 \map f {\text{odd} }
+      | r = \map f {\text{even} \times \text{odd} }
+      | rr= = 0
+}}
+{{eqn | ll= 1 \times_2 0 =
+      | l = \map f {\text{odd} } \times_2 \map f {\text{even} }
+      | r = \map f {\text{odd} \times \text{even} }
+      | rr= = 0
+}}
+{{eqn | ll= 1 \times_2 1 =
+      | l = \map f {\text{odd} } \times_2 \map f {\text{odd} }
+      | r = \map f {\text{odd} \times \text{odd} }
+      | rr= = 1
+}}
+{{end-eqn}}
+{{qed}}
+These results can be determined from their [[Definition:Cayley Table|Cayley tables]]:
+=== [[Parity Ring/Cayley Tables|Cayley Tables for Parity Ring]] ===
+{{:Parity Ring/Cayley Tables}}
+=== [[Ring of Integers Modulo 2/Cayley Tables|Cayley Tables for $\Z_2$]] ===
+{{:Ring of Integers Modulo 2/Cayley Tables}}
+\end{proof}<|endoftext|>
+\section{Parity Addition is Associative}
+Tags: Parity Ring, Parity Addition is Associative
+
+\begin{theorem}
+Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
+The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Associative Operation|associative]]:
+:$\forall a, b, c \in R: \paren {a + b} + c = a + \paren {b + c}$
+\end{theorem}
+
+\begin{proof}
+From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
+:$\struct {\set {\text {even}, \text {odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
+the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
+The result follows from:
+:[[Modulo Addition is Associative]]
+and:
+:[[Isomorphism Preserves Associativity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a, b, c \in R$.
+That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
+By definition of [[Definition:Odd Integer|odd]]:
+:$\text{odd} = 2 m + 1$
+for some $m \in \Z$.
+By definition of [[Definition:Even Integer|even]]:
+:$\text{even} = 2 n + 0$
+for some $n \in \Z$.
+Thus we can define the [[Definition:Mapping|mapping]] $f: R \to \Z$ as:
+:$\forall x \in R: f \left({x}\right) := \begin{cases}
+0 & : x \text { is even} \\
+1 & : x \text { is odd} \end{cases}$
+Thus an element of $R$ can be expressed as an arbitrary [[Definition:Integer|integer]] of the form:
+:$x = 2 k + f \left({x}\right)$
+where:
+:$k \in \Z$ is an [[Definition:Integer|integer]]
+:$f \left({x}\right)$ is either $0$ or $1$ according to whether $x$ is [[Definition:Even Integer|even]] or [[Definition:Odd Integer|odd]].
+Let $+_2$ be used to denote the [[Definition:Binary Operation|operation]] of $+$ in $R$, that is, the addition of two [[Definition:Parity of Integer|parities]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \left({a +_2 b}\right) +_2 c
+      | r = \left({2 r + f \left({a}\right) + 2 s + f \left({b}\right)}\right) + 2 t + f \left({c}\right)
+      | c = where $r, s, t \in \Z$
+}}
+{{eqn | r = 2 r + f \left({a}\right) + \left({2 s + f \left({b}\right) + 2 t + f \left({c}\right)}\right)
+      | c = [[Integer Addition is Associative]]
+}}
+{{eqn | r = a +_2 \left({b +_2 c}\right)
+      | c = Definition of [[Definition:Odd Integer|odd]] and [[Definition:Even Integer|even]]
+}}
+{{end-eqn}}
+The result follows by the identification of $+_2$ to be used to denote the [[Definition:Binary Operation|operation]] of $+$ in $R$:
+:$\left({a + b}\right) + c = a + \left({b + c}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parity Addition is Commutative}
+Tags: Parity Ring, Parity Addition is Commutative
+
+\begin{theorem}
+Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
+The [[Definition:Binary Operation|operation]] $+$ is [[Definition:Commutative Operation|commutative]]:
+:$\forall a, b \in R: a + b = b + a$
+\end{theorem}
+
+\begin{proof}
+From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
+:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
+the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
+The result follows from:
+:[[Modulo Addition is Commutative]]
+and:
+:[[Isomorphism Preserves Commutativity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a, b \in R$.
+That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
+By definition of [[Definition:Odd Integer|odd]]:
+:$\text{odd} = 2 m + 1$
+for some $m \in \Z$.
+By definition of [[Definition:Even Integer|even]]:
+:$\text{even} = 2 n + 0$
+for some $n \in \Z$.
+Thus we can define the [[Definition:Mapping|mapping]] $f: R \to \Z$ as:
+:$\forall x \in R: \map f x := \begin{cases}
+0 & : x \text { is even} \\
+1 & : x \text { is odd} \end{cases}$
+Thus an element of $R$ can be expressed as an arbitrary [[Definition:Integer|integer]] of the form:
+:$x = 2 k + \map f x$
+where:
+:$k \in \Z$ is an [[Definition:Integer|integer]]
+:$\map f x$ is either $0$ or $1$ according to whether $x$ is [[Definition:Even Integer|even]] or [[Definition:Odd Integer|odd]].
+Let $+_2$ be used to denote the [[Definition:Binary Operation|operation]] of $+$ in $R$, that is, the addition of two [[Definition:Parity of Integer|parities]].
+Then:
+{{begin-eqn}}
+{{eqn | l = a +_2 b
+      | r = 2 r + \map f a + 2 s + \map f b
+      | c = where $r, s \in \Z$
+}}
+{{eqn | r = 2 s + \map f b + 2 r + \map f a
+      | c = [[Integer Addition is Commutative]]
+}}
+{{eqn | r = b +_2 a
+      | c = Definition of [[Definition:Odd Integer|odd]] and [[Definition:Even Integer|even]]
+}}
+{{end-eqn}}
+The result follows by the identification of $+_2$ to be used to denote the [[Definition:Binary Operation|operation]] of $+$ in $R$:
+:$a + b = b + a$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parity Multiplication is Associative}
+Tags: Parity Ring, Parity Multiplication is Associative
+
+\begin{theorem}
+Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
+The [[Definition:Binary Operation|operation]] $\times$ is [[Definition:Associative Operation|associative]]:
+:$\forall a, b, c \in R: \paren {a \times b} \times c = a \times \paren {b \times c}$
+\end{theorem}
+
+\begin{proof}
+From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
+:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
+the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
+The result follows from:
+:[[Modulo Multiplication is Associative]]
+and:
+:[[Isomorphism Preserves Associativity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a, b, c \in R$.
+That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$.
+By definition of [[Definition:Odd Integer|odd]]:
+:$\text{odd} = 2 m + 1$
+for some $m \in \Z$.
+By definition of [[Definition:Even Integer|even]]:
+:$\text{even} = 2 n + 0$
+for some $n \in \Z$.
+Thus we can define the [[Definition:Mapping|mapping]] $f: R \to \Z$ as:
+:$\forall x \in R: \map f x := \begin{cases}
+0 & : x \text { is even} \\
+1 & : x \text { is odd} \end{cases}$
+Thus an element of $R$ can be expressed as an arbitrary [[Definition:Integer|integer]] of the form:
+:$x = 2 k + \map f x$
+where:
+:$k \in \Z$ is an [[Definition:Integer|integer]]
+:$\map f x$ is either $0$ or $1$ according to whether $x$ is [[Definition:Even Integer|even]] or [[Definition:Odd Integer|odd]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a \times b} \times c
+      | r = \paren {\paren {2 r + \map f a} \paren {2 s + \map f b} } \paren {2 t + \map f c}
+      | c = where $r, s, t \in \Z$
+}}
+{{eqn | r = \paren {2 r + \map f a} \paren {\paren {2 s + \map f b} \paren {2 t + \map f c} }
+      | c = [[Integer Multiplication is Associative]]
+}}
+{{eqn | r = a \times \paren {b \times c}
+      | c = Definition of [[Definition:Odd Integer|odd]] and [[Definition:Even Integer|even]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parity Multiplication is Commutative}
+Tags: Parity Ring, Parity Multiplication is Commutative
+
+\begin{theorem}
+Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the [[Definition:Parity Ring|parity ring]].
+The [[Definition:Binary Operation|operation]] $\times$ is [[Definition:Commutative Operation|commutative]]:
+:$\forall a, b \in R: a \times b = b \times a$
+\end{theorem}
+
+\begin{proof}
+From [[Isomorphism between Ring of Integers Modulo 2 and Parity Ring]]:
+:$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] with $\struct {\Z_2, +_2, \times_2}$
+the [[Definition:Ring of Integers Modulo m|ring of integers modulo $2$]].
+The result follows from:
+:[[Modulo Multiplication is Commutative]]
+and:
+:[[Isomorphism Preserves Associativity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a, b \in R$.
+That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.
+By definition of [[Definition:Odd Integer|odd]]:
+:$\text{odd} = 2 m + 1$
+for some $m \in \Z$.
+By definition of [[Definition:Even Integer|even]]:
+:$\text{even} = 2 n + 0$
+for some $n \in \Z$.
+Thus we can define the [[Definition:Mapping|mapping]] $f: R \to \Z$ as:
+:$\forall x \in R: \map f x := \begin{cases}
+0 & : x \text { is even} \\
+1 & : x \text { is odd} \end{cases}$
+Thus an element of $R$ can be expressed as an arbitrary [[Definition:Integer|integer]] of the form:
+:$x = 2 k + \map f x$
+where:
+:$k \in \Z$ is an [[Definition:Integer|integer]]
+:$\map f x$ is either $0$ or $1$ according to whether $x$ is [[Definition:Even Integer|even]] or [[Definition:Odd Integer|odd]].
+Then:
+{{begin-eqn}}
+{{eqn | l = a \times b
+      | r = \paren {2 r + \map f a} \paren {2 s + \map f b}
+      | c = where $r, s \in \Z$
+}}
+{{eqn | r = \paren {2 s + \map f b} \paren {2 r + \map f a}
+      | c = [[Integer Multiplication is Commutative]]
+}}
+{{eqn | r = b \times a
+      | c = Definition of [[Definition:Odd Integer|odd]] and [[Definition:Even Integer|even]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition}
+Tags: Examples of Group Isomorphisms, Roots of Unity, Additive Groups of Integers Modulo m
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\struct {R_n, \times}$ be the [[Definition:Complex Roots of Unity|complex $n$th roots of unity]] under [[Definition:Complex Multiplication|complex multiplication]].
+Let $\struct {\Z_n, +_n}$ be the [[Definition:Integers Modulo m|integers modulo $n$]] under [[Definition:Modulo Addition|modulo addition]].
+Then $\struct {R_n, \times}$ and $\struct {\Z_n, +_n}$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Integers Modulo m|set of integers modulo $n$]] is the [[Definition:Set|set]] exemplified by the [[Definition:Integer|integers]]:
+:$\Z_n = \set {0, 1, \ldots, n - 1}$
+The [[Definition:Complex Roots of Unity|complex $n$th roots of unity]] is the [[Definition:Set|set]]:
+:$R_n = \set {z \in \C: z^n = 1}$
+From [[Complex Roots of Unity in Exponential Form]]:
+:$R_n = \set {1, e^{\theta / n}, e^{2 \theta / n}, \ldots, e^{\left({n - 1}\right) \theta / n} }$
+where $\theta = 2 i \pi$.
+Let $z, w, \in R_n$.
+Then:
+: $\paren {z w}^n = z^n w^n = 1$
+and so $z w \in R_n$.
+Thus $\struct {R_n, \times}$ is a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+Consider the [[Definition:Mapping|mapping]] $f: \Z_n \to R_n$ defined as:
+:$\forall r \in \Z_n: \map f r = e^{r \theta / n}$
+which can be seen to be a [[Definition:Bijection|bijection]] by inspection.
+Let $j, k \in \Z_n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f j \map f k
+      | r = e^{j \theta / n} e^{k \theta / n}
+      | c = 
+}}
+{{eqn | r = e^{j \theta / n + k \theta / n}
+      | c = 
+}}
+{{eqn | r = e^{\paren {j + k} \theta / n}
+      | c = 
+}}
+{{eqn | r = \map f {j +_n k}
+      | c = 
+}}
+{{end-eqn}}
+Thus $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition}
+Tags: Examples of Group Isomorphisms: Order 4, Additive Groups of Integers Modulo m, Group of Gaussian Integer Units, Isomorphism between Gaussian Integer Units and Integers Modulo 4 under Addition
+
+\begin{theorem}
+Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
+Let $\struct {\Z_n, +_4}$ be the [[Definition:Integers Modulo m|integers modulo $4$]] under [[Definition:Modulo Addition|modulo addition]].
+Then $\struct {U_\C, \times}$ and $\struct {\Z_4, +_4}$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Mapping|mapping]] $f: \Z_4 \to U_\C$ be defined as:
+{{begin-eqn}}
+{{eqn | l = \map f 0
+      | r = 1
+}}
+{{eqn | l = \map f 1
+      | r = i
+}}
+{{eqn | l = \map f 2
+      | r = -1
+}}
+{{eqn | l = \map f 3
+      | r = -i
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Modulo Addition/Cayley Table/Modulo 4|Cayley Table of Integers Modulo $4$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $\struct {\Z_4, +_4}$ is as follows:
+{{:Modulo Addition/Cayley Table/Modulo 4}}
+=== [[Group of Gaussian Integer Units/Cayley Table|Cayley Table of Gaussian Integer Units]] ===
+The [[Definition:Cayley Table|Cayley table]] for $\struct {U_\C, \times}$ is as follows:
+{{:Group of Gaussian Integer Units/Cayley Table}}
+\end{proof}
+
+\begin{proof}
+From [[Gaussian Integer Units are 4th Roots of Unity]]:
+:$U_\C$ is the [[Definition:Set|set]] consisting of the [[Definition:Complex Roots of Unity|(complex) $4$th roots of $1$]].
+The result follows from [[Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition]].
+\end{proof}<|endoftext|>
+\section{Gaussian Integer Units are 4th Roots of Unity}
+Tags: Gaussian Integers, Complex Roots of Unity
+
+\begin{theorem}
+The [[Definition:Unit of Ring|units]] of the [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
+:$\set {1, i, -1, -i}$
+are the [[Definition:Complex Roots of Unity|(complex) $4$th roots of $1$]].
+\end{theorem}
+
+\begin{proof}
+We have that $i = \sqrt {-1}$ is the [[Definition:Imaginary Unit|imaginary unit]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = 1^4
+      | o = 
+      | rr= = 1
+}}
+{{eqn | l = i^4
+      | r = \paren {-1}^2
+      | rr= = 1
+}}
+{{eqn | l = \paren {-1}^4
+      | r = 1^2
+      | rr= = 1
+}}
+{{eqn | l = \paren {-i}^4
+      | r = \paren {-1}^2 \cdot \paren {-1}^2
+      | rr= = 1
+}}
+{{end-eqn}}
+So $\set {1, i, -1, -i}$ constitutes the [[Definition:Set|set]] of the [[Definition:Complex Roots of Unity|$4$th roots of unity]].
+{{qed}}
+[[Category:Gaussian Integers]]
+[[Category:Complex Roots of Unity]]
+pwj7xje8bvnpqyul3894062zu76x0j9
+\end{proof}<|endoftext|>
+\section{Isomorphism by Cayley Table}
+Tags: Cayley Tables, Isomorphisms
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]] whose [[Definition:Underlying Set of Structure|underlying sets]] are both [[Definition:Finite Set|finite]].
+Then $\struct {S, \circ}$ and $\struct {T, *}$ are [[Definition:Isomorphic Algebraic Structures|isomorphic]] {{iff}}:
+:a [[Definition:Bijection|bijection]] $f: S \to T$ can be found such that:
+::the [[Definition:Cayley Table|Cayley table]] of $\struct {T, *}$ can be generated from the [[Definition:Cayley Table|Cayley table]] of $\struct {S, \circ}$ by replacing each [[Definition:Entry of Cayley Table|entry]] of $S$ with its [[Definition:Image of Element under Mapping|image]] under $f$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $S$ and $T$ be [[Definition:Isomorphic Algebraic Structures|isomorphic]].
+Then by definition there exists an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $f: S \to T$.
+An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Bijection|bijection]] by definition.
+Thus the existence of the posited [[Definition:Bijection|bijection]] has been demonstrated.
+By the definition of [[Definition:Set Equivalence|set equivalence]], $S$ and $T$ have the same [[Definition:Cardinality|cardinality]].
+Let $\card S = \card T = n$.
+Let:
+:$S = \set {s_1, s_2, \ldots, s_n}$
+and:
+:$T = \set {t_1, t_2, \ldots, t_n}$
+be such that:
+:$\forall i \in \set {1, 2, \ldots, n}: \map f {s_i} = t_i$
+Let $S \sqbrk {j, k}$ denote the [[Definition:Entry of Cayley Table|entry]] of $S$ in row $j$ and column $k$.
+Thus:
+:$S \sqbrk {j, k} = s_j \circ s_k$
+for some $s_j, s_k \in S$.
+We have that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Thus:
+:$\map f {S \sqbrk {j, k} } = \map f {s_j} * \map f {s_k}$
+But:
+:$\map f {S \sqbrk {j, k} } = T \sqbrk {j, k}$
+:$\map f {s_j} = t_j$
+:$\map f {s_k} = t_k$
+Thus the [[Definition:Cayley Table|Cayley table]] of $\struct {T, *}$ has been generated from the [[Definition:Cayley Table|Cayley table]] of $\struct {S, \circ}$ by replacing each [[Definition:Entry of Cayley Table|entry]] of $S$ with its [[Definition:Image of Element under Mapping|image]] under $f$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let there exist a [[Definition:Bijection|bijection]] $f: S \to T$ such that the [[Definition:Cayley Table|Cayley table]] of $\struct {T, *}$ can be generated from the [[Definition:Cayley Table|Cayley table]] of $\struct {S, \circ}$ by replacing each [[Definition:Entry of Cayley Table|entry]] of $S$ with its [[Definition:Image of Element under Mapping|image]] under $f$.
+By the definition of [[Definition:Set Equivalence|set equivalence]], $S$ and $T$ have the same [[Definition:Cardinality|cardinality]].
+Let $\card S = \card T = n$.
+Let the rows and columns of the [[Definition:Cayley Table|Cayley table]] of $\left({S, \circ}\right)$ both be numbered from $1$ to $n$.
+Let the entries heading the rows and columns be denoted $s_1, s_2, \ldots, s_n$.
+Let $S \sqbrk {j, k}$ denote the [[Definition:Entry of Cayley Table|entry]] of $S$ in row $j$ and column $k$.
+By the method of construction:
+:$S \sqbrk {j, k} = s_j \circ s_k$
+for some $s_j, s_k \in S$.
+Let $f$ be used to generate the [[Definition:Cayley Table|Cayley table]] of $\struct {T, *}$ from the [[Definition:Cayley Table|Cayley table]] of $\struct {S, \circ}$.
+Then for all $i, j \in \set {1, 2, \ldots, n}$:
+:$\map f {S \sqbrk {j, k} } = T \sqbrk {j, k}$
+:$\map f {s_j} = t_j$
+:$\map f {s_k} = t_k$
+But by definition of the [[Definition:Cayley Table|Cayley table]] of $\struct {T, *}$:
+:$T \sqbrk {j, k} = t_j * t_k$
+Thus:
+:$\forall j, k \in \set {1, 2, \ldots, n}: \map f {s_j \circ s_k} = \map f {s_j} * \map f {s_k}$
+demonstrating that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+{{qed}}
+[[Category:Cayley Tables]]
+[[Category:Isomorphisms]]
+pr241m4nkfod3fojbk28b02aad0395f
+\end{proof}<|endoftext|>
+\section{Bijection between Integers and Even Integers}
+Tags: Examples of Bijections, Integers, Even Integers
+
+\begin{theorem}
+Let $\Z$ be the [[Definition:Set|set]] of [[Definition:Integer|integers]].
+Let $2 \Z$ be the [[Definition:Set|set]] of [[Definition:Even Integer|even integers]].
+Then there exists a [[Definition:Bijection|bijection]] $f: \Z \to 2 \Z$ between the two.
+\end{theorem}
+
+\begin{proof}
+Let $f: \Z \to 2 \Z$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in \Z: \map f n = 2 n$
+Let $m, n \in \Z$ such that $\map f m = \map f n$.
+{{begin-eqn}}
+{{eqn | l = \map f m
+      | r = \map f n
+}}
+{{eqn | ll= \leadsto
+      | l = 2 m
+      | r = 2 n
+      | c = Definition of $f$
+}}
+{{eqn | ll= \leadsto
+      | l = m
+      | r = n
+      | c = dividing both sides by $2$
+}}
+{{end-eqn}}
+Thus $f$ is an [[Definition:Injection|injection]].
+Let $n \in 2 Z$.
+Then by definition:
+:$n = 2 r$
+for some $r \in \Z$
+That is:
+:$n = \map f r$
+and so $f$ is a [[Definition:Surjection|surjection]].
+So $f$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
+Hence by definition, $f$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+[[Category:Examples of Bijections]]
+[[Category:Integers]]
+[[Category:Even Integers]]
+1guf8fayzhy4b82sb3b08kplc5wrbl6
+\end{proof}<|endoftext|>
+\section{Set of Integers under Addition is Isomorphic to Set of Even Integers under Addition}
+Tags: Integers, Even Integers, Additive Group of Integers
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Algebraic Structure|algebraic structure]] formed by the [[Definition:Set|set]] of [[Definition:Integer|integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Addition|addition]].
+Let $\struct {2 \Z, +}$ be the [[Definition:Algebraic Structure|algebraic structure]] formed by the [[Definition:Set|set]] of [[Definition:Even Integer|even integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Addition|addition]].
+Then $\struct {\Z, +}$ and $\struct {2 \Z, +}$ are [[Definition:Isomorphic Algebraic Structures|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+Let $f: \Z \to 2 \Z$ be the [[Definition:Mapping|mapping]]:
+:$\forall n \in \Z: \map f n = 2 n$
+From [[Bijection between Integers and Even Integers]], $f$ is a [[Definition:Bijection|bijection]].
+Let $m, n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f {m + n}
+      | r = 2 \paren {m + n}
+      | c = Definition of $f$
+}}
+{{eqn | r = 2 m + 2 n
+      | c = [[Integer Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \map f m + \map f n
+      | c = [[Integer Multiplication Distributes over Addition]]
+}}
+{{end-eqn}}
+Thus $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers under Addition do not form Group}
+Tags: Natural Numbers
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, +}$ consisting of the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ under [[Definition:Natural Number Addition|addition]] $+$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+From [[Natural Numbers under Addition form Commutative Monoid]], $\struct {\N, +}$ has an [[Definition:Identity Element|identity element]] $0$.
+However, for any $x \in \N$ such that $x \ne 0$ there exists no $y \in \N$ such that $x + y = 0$.
+Thus the general element of $\struct {\N, +}$ has no [[Definition:Inverse Element|inverse]].
+Hence the result by definition of [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers under Multiplication do not form Group}
+Tags: Natural Number Multiplication, Examples of Groups, Natural Numbers under Multiplication do not form Group
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ under [[Definition:Natural Number Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]].
+\end{theorem}<|endoftext|>
+\section{Rational Numbers under Multiplication do not form Group}
+Tags: Rational Multiplication
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Q, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ under [[Definition:Rational Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $\struct {\Q, \times}$ is a [[Definition:Group|group]].
+By the definition of the [[Definition:Zero (Number)|number $0 \in \Q$]]:
+:$\forall x \in \Q: x \times 0 = 0 = 0 \times x$
+Thus $0$ is a [[Definition:Zero Element|zero]] in the [[Definition:Abstract Algebra|abstract algebraic sense]].
+From [[Group with Zero Element is Trivial]], $\struct {\Q, \times}$ is the [[Definition:Trivial Group|trivial group]].
+But $\Q$ contains other [[Definition:Element|elements]] besides $0$.
+From this [[Proof by Contradiction|contradiction]] it follows that $\struct {\Q, \times}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers under Multiplication do not form Group}
+Tags: Real Multiplication
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\R, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ under [[Definition:Real Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $\struct {\R, \times}$ is a [[Definition:Group|group]].
+By the definition of the [[Definition:Zero (Number)|number $0 \in \R$]]:
+:$\forall x \in \R: x \times 0 = 0 = 0 \times x$
+Thus $0$ is a [[Definition:Zero Element|zero]] in the [[Definition:Abstract Algebra|abstract algebraic sense]].
+From [[Group with Zero Element is Trivial]], $\struct {\R, \times}$ is the [[Definition:Trivial Group|trivial group]].
+But $\R$ contains other [[Definition:Element|elements]] besides $0$.
+From this [[Proof by Contradiction|contradiction]] it follows that $\struct {\R, \times}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers under Multiplication do not form Group}
+Tags: Complex Multiplication
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\C, \times}$ consisting of the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ under [[Definition:Complex Multiplication|multiplication]] $\times$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $\struct {\C, \times}$ is a [[Definition:Group|group]].
+By the definition of the [[Definition:Zero (Number)|number $0 \in \C$]]:
+:$\forall x \in \C: x \times 0 = 0 = 0 \times x$
+Thus $0$ is a [[Definition:Zero Element|zero]] in the [[Definition:Abstract Algebra|abstract algebraic sense]].
+From [[Group with Zero Element is Trivial]], $\struct {\C, \times}$ is the [[Definition:Trivial Group|trivial group]].
+But $\C$ contains other [[Definition:Element|elements]] besides $0$.
+From this [[Proof by Contradiction|contradiction]] it follows that $\struct {\C, \times}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetry Group of Square is Group}
+Tags: Symmetry Group of Square
+
+\begin{theorem}
+The [[Definition:Symmetry Group of Square|symmetry group of the square]] is a [[Definition:Abelian Group|non-abelian]] [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Let us refer to this group as $D_4$.
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== G0: Closure ===
+From the [[Symmetry Group of Square/Cayley Table|Cayley table]] it is seen directly that $D_4$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== G1: Associativity ===
+[[Composition of Mappings is Associative]].
+{{qed|lemma}}
+=== G2: Identity ===
+The [[Definition:Identity Element|identity]] is $e$ as defined.
+{{qed|lemma}}
+=== G3: Inverses ===
+Each element can be seen to have an [[Definition:Inverse Element|inverse]]:
+:$r^{-1} = r^3$ and so $\left({r^3}\right)^{-1} = r$
+:$r^2$, $t_{AC}$, $t_{BD}$, $t_x$ and $t_y$ are all [[Definition:Self-Inverse Element|self-inverse]].
+{{qed|lemma}}
+Thus $D_4$ is seen to be a [[Definition:Group|group]].
+Note that from the [[Symmetry Group of Square/Cayley Table|Cayley table]] it can be observed directly that:
+:$r \circ t_x = t_{BD}$
+:$t_x \circ r = t_{AC}$
+thus illustrating by [[Proof by Counterexample|counterexample]] that $D_4$ is not [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dihedral Group is Group}
+Tags: Dihedral Groups
+
+\begin{theorem}
+Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order $2 n$]].
+Then $D_n$ is indeed a [[Definition:Group|group]].
+\end{theorem}<|endoftext|>
+\section{Symmetric Group is Subgroup of Monoid of Self-Maps}
+Tags: Symmetric Groups, Permutation Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself
+Let $\struct {\Gamma \paren S, \circ}$ denote the [[Definition:Symmetric Group|symmetric group on $S$]].
+Let $\struct {S^S, \circ}$ be the [[Definition:Monoid|monoid]] of [[Definition:Self-Map|self-maps]] under [[Definition:Composition of Mappings|composition of mappings]].
+Then $\struct {\Gamma \paren S, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {S^S, \circ}$.
+\end{theorem}
+
+\begin{proof}
+By [[Symmetric Group is Group]], $\struct {\Gamma \paren S, \circ}$ is a [[Definition:Group|group]].
+Let $\phi \in \Gamma \paren S$ be a [[Definition:Permutation|permutation]] on $S$.
+As a [[Definition:Permutation|permutation]] is a [[Definition:Self-Map|self-map]], it follows that $\phi \in S^S$.
+Thus by definition $\Gamma \paren S$ is a [[Definition:Subset|subset]] of $S^S$.
+So by definition, $\Gamma \paren S$, is a [[Definition:Subgroup|subgroup]] of $\struct {S^S, \circ}$.
+{{qed}}
+[[Category:Symmetric Groups]]
+[[Category:Permutation Theory]]
+7qa78uey4kkvp6bapsfmznyb8xgrmxv
+\end{proof}<|endoftext|>
+\section{Cancellability by Cayley Table}
+Tags: Cayley Tables
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Finite Algebraic Structure|finite algebraic structure]].
+Let $\mathcal T$ be the [[Definition:Cayley Table|Cayley table]] for $\left({S, \circ}\right)$.
+Let $a \in S$ be an [[Definition:Element|element]] of $S$.
+Then $a$ is [[Definition:Cancellable Element|cancellable]] for $\circ$ {{iff}}:
+:$(1): \quad$ no [[Definition:Element|element]] of $S$ is repeated in $\mathcal T$ in the row headed by $a$
+and:
+:$(2): \quad$ no [[Definition:Element|element]] of $S$ is repeated in $\mathcal T$ in the column headed by $a$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $a \in S$ be [[Definition:Cancellable Element|cancellable]] for $\circ$.
+Suppose there exists $x \in S$ which appears twice in a row in $\mathcal T$ headed by $a$.
+Thus by definition of the structure of a [[Definition:Cayley Table|Cayley table]]:
+:$\exists y_1, y_2 \in S: a \circ y_1 = x = a \circ y_2$
+such that $y_1 \ne y_2$.
+That [[Proof by Contradiction|contradicts]] the stipulation that $a$ is [[Definition:Cancellable Element|cancellable]] for $\circ$.
+So no [[Definition:Element|element]] of $S$ is repeated in $\mathcal T$ in the row headed by $a$.
+Similarly, suppose there exists $x \in S$ which appears twice in a column in $\mathcal T$ headed by $a$.
+Thus by definition of the structure of a [[Definition:Cayley Table|Cayley table]]:
+:$\exists y_1, y_2 \in S: y_1 \circ a = x = y_2 \circ a$
+such that $y_1 \ne y_2$.
+That [[Proof by Contradiction|contradicts]] the stipulation that $a$ is [[Definition:Cancellable Element|cancellable]] for $\circ$.
+So no [[Definition:Element|element]] of $S$ is repeated in $\mathcal T$ in the column headed by $a$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $a$ be such that:
+:$(1): \quad$ no [[Definition:Element|element]] of $S$ is repeated in $\mathcal T$ in the row headed by $a$
+and:
+:$(2): \quad$ no [[Definition:Element|element]] of $S$ is repeated in $\mathcal T$ in the column headed by $a$.
+Thus from $(1)$:
+:$\forall y_1, y_2 \in S: y_1 \ne y_2 \implies a \circ y_1 \ne a \circ y_2$
+So by the [[Rule of Transposition]]:
+:$\forall y_1, y_2 \in S: a \circ y_1 = a \circ y_2 \implies y_1 = y_2$
+Also from $(2)$:
+:$\forall y_1, y_2 \in S: y_1 \ne y_2 \implies y_1 \circ a \ne y_2 \circ a$
+So by the [[Rule of Transposition]]:
+:$\forall y_1, y_2 \in S: y_1 \circ a = y_2 \circ a \implies y_1 = y_2$
+Thus it follows that $a$ is [[Definition:Cancellable Element|cancellable]] for $\circ$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset not necessarily Submagma}
+Tags: Magmas
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
+Let $T \subseteq S$.
+Then it is not necessarily the case that:
+: $\struct {T, \circ} \subseteq \struct {S, \circ}$
+That is, it does not always follow that $\struct {T, \circ}$ is a [[Definition:Submagma|submagma]] of $\struct {S, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {\Z, -}$ be the [[Definition:Magma|magma]] which is the [[Definition:Set|set]] of [[Definition:Integer|integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Subtraction|subtraction]].
+We have that the [[Definition:Natural Numbers|natural numbers]] $\N$ are a [[Definition:Subset|subset]] of the [[Definition:Integer|integers]].
+Consider $\struct {\N, -}$, the [[Definition:Natural Numbers|natural numbers]] under [[Definition:Natural Number Subtraction|subtraction]].
+We have that [[Natural Number Subtraction is not Closed]].
+For example:
+: $1 - 2 = -1 \notin \N$
+Thus $\struct {\N, -}$ is not [[Definition:Closed Algebraic Structure|closed]].
+So $\struct {\N, -}$ is not a [[Definition:Submagma|submagma]] of $\struct {\Z, -}$
+Hence it is not true to write $\struct {\N, -} \subseteq \struct {\Z, -}$, despite the fact that $\N \subseteq \Z$.
+Thus $\struct {\N, -}$ is not a [[Definition:Submagma|submagma]] of $\struct {\Z, -}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Subsets of Symmetry Group of Square}
+Tags: Symmetry Groups, Symmetry Group of Square
+
+\begin{theorem}
+Recall the [[Definition:Symmetry Group of Square|symmetry group of the square]]:
+\end{theorem}
+
+\begin{proof}
+Recall that a [[Definition:Submagma|submagma]] of an [[Definition:Algebraic Structure|algebraic structure]] $\mathcal S$ is a [[Definition:Subset|subsets]] of $\mathcal S$ which is [[Definition:Closed Algebraic Structure|closed]].
+Let $\mathcal X$ be the [[Definition:Set|set]] of all [[Definition:Submagma|submagmas]] of $\mathcal S$.
+From [[Empty Set is Submagma of Magma]]:
+:$\O \in \mathcal X$
+From [[Magma is Submagma of Itself]]:
+:$\mathcal S \in \mathcal X$
+From [[Idempotent Magma Element forms Singleton Submagma]]:
+:$\set e \in \mathcal X$
+Let us refer to the [[Definition:Cayley Table|Cayley table]]:
+=== [[Symmetry Group of Square/Cayley Table|Cayley Table of Symmetry Group of Square]] ===
+{{:Symmetry Group of Square/Cayley Table}}
+Taking each of the elements of $\mathcal S$ in order:
+{{begin-eqn}}
+{{eqn | l = t_x \circ t^x
+      | r = e
+}}
+{{eqn | l = t_y \circ t^y
+      | r = e
+}}
+{{eqn | l = t_{AC} \circ t^{AC}
+      | r = e
+}}
+{{eqn | l = t_{BD} \circ t^{BD}
+      | r = e
+}}
+{{eqn | l = r^2 \circ r^2
+      | r = e
+}}
+{{eqn | l = r \circ r
+      | r = r^2
+}}
+{{eqn | l = r \circ r^2
+      | r = r^3
+}}
+{{eqn | l = r \circ r^3
+      | r = e
+}}
+{{eqn | l = r^3 \circ r^3
+      | r = r^2
+}}
+{{eqn | l = r^3 \circ r^2
+      | r = r
+}}
+{{eqn | l = r^3 \circ r
+      | r = e
+}}
+{{end-eqn}}
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = \set {e, r^2}
+      | o = \in
+      | r = \mathcal X
+}}
+{{eqn | l = \set {e, t_x}
+      | o = \in
+      | r = \mathcal X
+}}
+{{eqn | l = \set {e, t_y}
+      | o = \in
+      | r = \mathcal X
+}}
+{{eqn | l = \set {e, t_{AC} }
+      | o = \in
+      | r = \mathcal X
+}}
+{{eqn | l = \set {e, t_{BD} }
+      | o = \in
+      | r = \mathcal X
+}}
+{{eqn | l = \set {e, r, r^2, r^3}
+      | o = \in
+      | r = \mathcal X
+}}
+{{end-eqn}}
+Next note by inspection that:
+:$\set {e, r^2, t_x, t_y} \in \mathcal X$
+and:
+:$\set {e, r^2, t_{AC}, t_{BD} } \in \mathcal X$
+{{begin-eqn}}
+{{eqn | l = \set {e, r^2, t_x, t_y}
+      | o = \in
+      | r = \mathcal X
+}}
+{{eqn | l = \set {e, r^2, t_{AC}, t_{BD} }
+      | o = \in
+      | r = \mathcal X
+}}
+{{end-eqn}}
+Finally note by inspection that:
+: any [[Definition:Closed Algebraic Structure|closed]] [[Definition:Subset|subset]] of $\mathcal S$ which contains both $r$ and any of the reflections contains all the elements of $\mathcal S$
+: any [[Definition:Closed Algebraic Structure|closed]] [[Definition:Subset|subset]] of $\mathcal S$ which contains both $r^3$ and any of the reflections contains all the elements of $\mathcal S$.
+Thus there are no more [[Definition:Proper Subset|proper subsets]] of $\mathcal S$ which are [[Definition:Submagma|submagmas]] of $\mathcal S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Piecewise Continuous Function does not necessarily have Improper Integrals}
+Tags: Piecewise Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$, $a < b$. 
+Let $f$ be a [[Definition:Piecewise Continuous Function|piecewise continuous function]]:
+{{:Definition:Piecewise Continuous Function}}
+Then it is not necessarily the case that $f$ is a [[Definition:Piecewise Continuous Function with Improper Integrals|piecewise continuous function with improper integrals]]:
+{{:Definition:Piecewise Continuous Function with Improper Integrals}}
+\end{theorem}
+
+\begin{proof}
+Consider the function:
+:$f \left({x}\right) = \begin{cases}
+0 & : x = a \\
+\dfrac 1 {x - a} & : x \in \left({a \,.\,.\, b}\right]
+\end{cases}$
+Since $\dfrac 1 {x - a}$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\left({a \,.\,.\, b}\right)$, $f$ is [[Definition:Continuous Real Function on Open Interval|continuous]] on $\left({a \,.\,.\, b}\right)$.
+Therefore, $f$ is a [[Definition:Piecewise Continuous Function|piecewise continuous function]] for the [[Definition:Finite Subdivision|(finite) subdivision]] $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.
+We now consider whether the [[Definition:Improper Integral over Open Interval|improper integral]] $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ exists.
+Let $c$ be a point in $\left({a \,.\,.\, b}\right)$.
+From the definition of [[Definition:Improper Integral over Open Interval|improper integral]], the existence of $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ requires that $\displaystyle \lim_{\gamma \mathop \to a^+} \int_\gamma^c \dfrac 1 {x - a} \rd x$ exists.
+We have
+{{begin-eqn}}
+{{eqn | l = \int_\gamma^c \frac 1 {x - a} \rd x
+      | r = \ln \left({x - a}\right) \vert _\gamma^c
+}}
+{{eqn | r = \ln \left({c - a}\right) - \ln \left({\gamma - a}\right)
+}}
+{{eqn | r = \ln \left({c - a}\right) + \ln \left({\frac 1 {\gamma - a} }\right)
+}}
+{{end-eqn}}
+The last {{RHS}} approaches $\infty$ as $\gamma$ approaches $a$ from above.
+So $\displaystyle \int_{a^+}^c \dfrac 1 {x - a} \rd x$ does not exist.
+Therefore, $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ does not exist either.
+Accordingly, $f$ is not a [[Definition:Piecewise Continuous Function with Improper Integrals|piecewise continuous function with improper integrals]].
+{{qed}}
+[[Category:Piecewise Continuous Functions]]
+7t74sablucz0x8v3jocui9occt1lzso
+\end{proof}<|endoftext|>
+\section{Right Cancellable Element is Right Cancellable in Subset}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $\left ({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $\left ({T, \circ}\right) \subseteq \left ({S, \circ}\right)$.
+Let $x \in T$ be [[Definition:Right Cancellable Element|right cancellable]] in $S$.
+Then $x$ is also [[Definition:Right Cancellable Element|right cancellable]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in T$ be [[Definition:Right Cancellable Element|right cancellable]] in $S$.
+That is:
+: $\forall a, b \in S: a \circ x = b \circ x \implies a = b$
+Therefore:
+: $\forall c, d \in T: c \circ x = d \circ x \implies c = d$
+Thus $x$ is [[Definition:Right Cancellable Element|right cancellable]] in $T$.
+{{qed}}
+[[Category:Abstract Algebra]]
+hzjs6p195emmj6ujwsi0gz9wpdsg06q
+\end{proof}<|endoftext|>
+\section{Left Cancellable Element is Left Cancellable in Subset}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
+Let $x \in T$ be [[Definition:Left Cancellable Element|left cancellable]] in $S$.
+Then $x$ is also [[Definition:Left Cancellable Element|left cancellable]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in T$ be [[Definition:Left Cancellable Element|left cancellable]] in $S$.
+That is:
+: $\forall a, b \in S: x \circ a = x \circ b \implies a = b$
+Therefore:
+: $\forall c, d \in T: x \circ c = x \circ d \implies c = d$
+Thus $x$ is [[Definition:Left Cancellable Element|left cancellable]] in $T$.
+{{qed}}
+[[Category:Abstract Algebra]]
+msfv3rbqh4ysjgyrf5b5uzs8ty9ymm3
+\end{proof}<|endoftext|>
+\section{Left Cancellable Elements of Semigroup form Subsemigroup}
+Tags: Subsemigroups, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
+Let $C_\lambda$ be the set of [[Definition:Left Cancellable Element|left cancellable elements]] of $\struct {S, \circ}$.
+Then $\struct {C_\lambda, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $C_\lambda$ be the set of [[Definition:Left Cancellable Element|left cancellable elements]] of $\struct {S, \circ}$:
+:$C_\lambda = \set {x \in S: \forall a, b \in S: x \circ a = x \circ b \implies a = b}$
+Let $x, y \in C_\lambda$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {x \circ y} \circ a
+      | r = \paren {x \circ y} \circ b
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ \paren {y \circ a}
+      | r = x \circ \paren {y \circ b}
+      | c = by [[Definition:Associative|associativity]] of $\circ$
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ a
+      | r = x \circ b
+      | c = as $y \in C_\lambda$
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = b
+      | c = as $x \in C_\lambda$
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ y
+      | o = \in
+      | r = C_\lambda
+      | c = 
+}}
+{{end-eqn}}
+Thus $\struct {C_\lambda, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+Therefore by the [[Subsemigroup Closure Test]] $\struct {C_\lambda, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
+{{qed}}
+[[Category:Subsemigroups]]
+[[Category:Cancellability]]
+b9eqfcllj7pbuvrl4sw72qy0mq7828r
+\end{proof}<|endoftext|>
+\section{Right Cancellable Elements of Semigroup form Subsemigroup}
+Tags: Subsemigroups, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
+Let $C_\rho$ be the set of [[Definition:Right Cancellable Element|right cancellable elements]] of $\struct {S, \circ}$.
+Then $\struct {C_\rho, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $C_\rho$ be the set of [[Definition:Right Cancellable Element|right cancellable elements]] of $\struct {S, \circ}$:
+:$C_\rho = \set {x \in S: \forall a, b \in S: a \circ x = b \circ x \implies a = b}$
+Let $x, y \in C_\rho$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \circ \paren {x \circ y}
+      | r = b \circ \paren {x \circ y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a \circ x} \circ y
+      | r = \paren {b \circ x} \circ y
+      | c = by [[Definition:Associative|associativity]] of $\circ$
+}}
+{{eqn | ll= \leadsto
+      | l = a \circ x
+      | r = b \circ x
+      | c = as $y \in C_\rho$
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = b
+      | c = as $x \in C_\rho$
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ y
+      | o = \in
+      | r = C_\rho
+      | c = 
+}}
+{{end-eqn}}
+Thus $\struct {C_\rho, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+Therefore by the [[Subsemigroup Closure Test]] $\struct {C_\rho, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
+{{qed}}
+[[Category:Subsemigroups]]
+[[Category:Cancellability]]
+mv9ktdrtnrrawipd59a65a8rjr13i30
+\end{proof}<|endoftext|>
+\section{Surjection iff Right Cancellable/Necessary Condition}
+Tags: Surjection iff Right Cancellable
+
+\begin{theorem}
+Let $f$ be a [[Definition:Surjection|surjection]].
+Then $f$ is [[Definition:Right Cancellable Mapping|right cancellable]].
+\end{theorem}<|endoftext|>
+\section{Surjection iff Right Cancellable/Sufficient Condition}
+Tags: Surjection iff Right Cancellable
+
+\begin{theorem}
+Let $f$ be a [[Definition:Mapping|mapping]] which is [[Definition:Right Cancellable Mapping|right cancellable]].
+Then $f$ is a [[Definition:Surjection|surjection]].
+\end{theorem}<|endoftext|>
+\section{Absolute Value Function on Integers induces Equivalence Relation}
+Tags: Equivalence Relations, Integers, Absolute Value Function
+
+\begin{theorem}
+Let $\Z$ be the [[Definition:Integer|set of integers]].
+Let $\mathcal R$ be the [[Definition:Relation|relation]] on $\Z$ defined as:
+:$\forall x, y \in \Z: \left({x, y}\right) \in \mathcal R \iff \left\vert{x}\right\vert = \left\vert{y}\right\vert$
+where $\left\vert{x}\right\vert$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
+Then $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+$\mathcal R$ is shown to be an [[Definition:Equivalence Relation|equivalence relation]] thus:
+=== Reflexivity ===
+: $\forall x \in \Z: \left\vert{x}\right\vert = \left\vert{x}\right\vert$
+Thus $\mathcal R$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+: $\forall x, y \in \Z: \left\vert{x}\right\vert = \left\vert{y}\right\vert \implies \left\vert{y}\right\vert = \left\vert{x}\right\vert$
+Thus $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+: $\forall x, y, z \in \Z: \left\vert{x}\right\vert = \left\vert{y}\right\vert \land \left\vert{y}\right\vert = \left\vert{z}\right\vert \implies \left\vert{x}\right\vert = \left\vert{z}\right\vert$
+Thus $\mathcal R$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+Thus, by definition, $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value induces Equivalence not Compatible with Integer Addition}
+Tags: Congruence Relations, Integer Addition, Absolute Value Function
+
+\begin{theorem}
+Let $\Z$ be the [[Definition:Integer|set of integers]].
+Let $\RR$ be the [[Definition:Relation|relation]] on $\Z$ defined as:
+:$\forall x, y \in \Z: \tuple {x, y} \in \RR \iff \size x = \size y$
+where $\size x$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
+Then $\RR$ is not a [[Definition:Congruence Relation|congruence relation]] for [[Definition:Integer Addition|integer addition]].
+\end{theorem}
+
+\begin{proof}
+From [[Absolute Value Function on Integers induces Equivalence Relation]], $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]].
+However, consider that:
+{{begin-eqn}}
+{{eqn | l = \size {-1} = \size 1
+      | o = \leadsto
+      | r = -1 \mathop \RR 1
+      | c = 
+}}
+{{eqn | l = \size 2 = \size 2
+      | o = \leadsto
+      | r = 2 \mathop \RR 2
+      | c = 
+}}
+{{end-eqn}}
+By conventional [[Definition:Integer Addition|integer addition]]:
+:$-1 + 2 = 1$
+while:
+:$1 + 2 = 3$
+But it does not hold that:
+:$\size 1 = \size 3$
+Therefore $\RR$ is not a [[Definition:Congruence Relation|congruence relation]] for [[Definition:Integer Addition|integer addition]].
+Hence the result, by [[Proof by Counterexample]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence Relation is Congruence for Right Operation}
+Tags: Left and Right Operations, Equivalence Relations
+
+\begin{theorem}
+Every [[Definition:Equivalence Relation|equivalence]] is a [[Definition:Congruence Relation|congruence]] for the [[Definition:Right Operation|right operation]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence relation]] on the structure $\left({S, \rightarrow}\right)$.
+Then:
+: $x_1 \rightarrow y_1 = y_1$
+: $x_2 \rightarrow y_2 = y_2$
+Suppose $x_1 \mathop {\mathcal R} x_2 \land y_1 \mathop {\mathcal R} y_2$.
+It follows directly that:
+: $\left({x_1 \rightarrow y_1}\right) \mathop {\mathcal R} \left({x_2 \rightarrow y_2}\right)$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence Relation is Congruence for Left Operation}
+Tags: Left and Right Operations, Equivalence Relations
+
+\begin{theorem}
+Every [[Definition:Equivalence Relation|equivalence]] is a [[Definition:Congruence Relation|congruence]] for the [[Definition:Left Operation|left operation]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal R$ be an [[Definition:Equivalence Relation|equivalence relation]] on the structure $\left({S, \leftarrow}\right)$.
+Then:
+: $x_1 \leftarrow y_1 = x_1$
+: $x_2 \leftarrow y_2 = x_2$
+Suppose $x_1 \mathop {\mathcal R} x_2 \land y_1 \mathop {\mathcal R} y_2$.
+It follows directly that:
+: $\left({x_1 \leftarrow y_1}\right) \mathop {\mathcal R} \left({x_2 \leftarrow y_2}\right)$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Left Coset Space forms Partition}
+Tags: Coset Space forms Partition
+
+\begin{theorem}
+The [[Definition:Left Coset Space|left coset space]] of $H$ forms a [[Definition:Set Partition|partition]] of its [[Definition:Group|group]] $G$, and hence:
+{{begin-eqn}}
+{{eqn | l = x \equiv^l y \pmod H
+      | o = \iff
+      | r = x H = y H
+}}
+{{eqn | l = \neg \paren {x \equiv^l y} \pmod H
+      | o = \iff
+      | r = x H \cap y H = \O
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+: [[Left Congruence Modulo Subgroup is Equivalence Relation]]
+: [[Relation Partitions Set iff Equivalence]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Coset Space forms Partition}
+Tags: Coset Space forms Partition
+
+\begin{theorem}
+The [[Definition:Right Coset Space|right coset space]] of $H$ forms a [[Definition:Set Partition|partition]] of its [[Definition:Group|group]] $G$:
+{{begin-eqn}}
+{{eqn | l = x \equiv^r y \pmod H
+      | o = \iff
+      | r = H x = H y
+}}
+{{eqn | l = \neg \paren {x \equiv^r y} \pmod H
+      | o = \iff
+      | r = H x \cap H y = \O
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+: [[Right Congruence Modulo Subgroup is Equivalence Relation]]
+: [[Relation Partitions Set iff Equivalence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Cosets are Equal iff Product with Inverse in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+Let $H x$ denote the [[Definition:Right Coset|right coset]] of $H$ by $x$.
+Then:
+:$H x = H y \iff x y^{-1} \in H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = H x
+      | r = H y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \equiv^r
+      | r = y \bmod H
+      | c = [[Right Coset Space forms Partition]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x y^{-1}
+      | o = \in
+      | r = H
+      | c = [[Equivalent Statements for Congruence Modulo Subgroup]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Left Cosets are Equal iff Product with Inverse in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+Let $x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$.
+Then:
+:$x H = y H \iff x^{-1} y \in H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x H
+      | r = y H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \equiv^l
+      | r = y \pmod H
+      | c = [[Left Coset Space forms Partition]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x^{-1} y
+      | o = \in
+      | r = H
+      | c = [[Equivalent Statements for Congruence Modulo Subgroup]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Left Congruence Class Modulo Subgroup is Left Coset}
+Tags: Cosets
+
+\begin{theorem}
+Let $\mathcal R^l_H$ be the [[Definition:Equivalence Relation|equivalence]] defined as [[Definition:Left Congruence Modulo Subgroup|left congruence modulo $H$]].
+The [[Definition:Equivalence Class|equivalence class]] $\eqclass g {\mathcal R^l_H}$ of an element $g \in G$ is the [[Definition:Left Coset|left coset]] $g H$.
+This is known as the '''left congruence class of $g \bmod H$'''.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \eqclass g {\mathcal R^l_H}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \eqclass g {\mathcal R^l_H}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists h \in H:
+      | l = g^{-1} x
+      | r = h
+      | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists h \in H:
+      | l = x
+      | r = g h
+      | c = [[Definition:Group|Group properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = g H
+      | c = {{Defof|Left Coset}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass g {\mathcal R^l_H}
+      | o = \subseteq
+      | r = g H
+      | c = {{Defof|Subgroup}}
+}}
+{{end-eqn}}
+Now let $x \in g H$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = g H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists h \in H:
+      | l = x
+      | r = g h
+      | c = {{Defof|Left Coset}}
+}}
+{{eqn | ll= \leadsto
+      | l = g^{-1} x
+      | r = h \in H
+      | c = [[Definition:Group|Group properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \eqclass g {\mathcal R^l_H}
+      | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | r = g H \subseteq \eqclass g {\mathcal R^l_H}
+      | c = {{Defof|Subgroup}}
+}}
+{{end-eqn}}
+Thus:
+:$\eqclass g {\mathcal R^l_H} = g H$
+that is, the [[Definition:Equivalence Class|equivalence class]] $\eqclass g {\mathcal R^l_H}$ of an element $g \in G$ equals the [[Definition:Left Coset|left coset]] $g H$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Right Congruence Class Modulo Subgroup is Right Coset}
+Tags: Cosets
+
+\begin{theorem}
+Let $\mathcal R^r_H$ be the [[Definition:Equivalence Relation|equivalence]] defined as [[Definition:Right Congruence Modulo Subgroup|right congruence modulo $H$]].
+The [[Definition:Equivalence Class|equivalence class]] $\eqclass g {\mathcal R^r_H}$ of an element $g \in G$ is the [[Definition:Right Coset|right coset]] $H g$.
+This is known as the '''right congruence class of $g \bmod H$'''.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \eqclass g {\mathcal R^r_H}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \eqclass g {\mathcal R^r_H}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists h \in H:
+      | l = x g^{-1}
+      | r = h
+      | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists h \in H:
+      | l = x
+      | r = h g
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = H g
+      | c = {{Defof|Right Coset}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass g {\mathcal R^r_H}
+      | o = \subseteq
+      | r = H g
+      | c = {{Defof|Subgroup}}
+}}
+{{end-eqn}}
+Now let $x \in g H$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = H g
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists h \in H:
+      | l = x
+      | r = h g
+      | c = {{Defof|Right Coset}}
+}}
+{{eqn | ll= \leadsto
+      | l = x g^{-1}
+      | r = h \in H
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \eqclass g {\mathcal R^r_H}
+      | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | l = H g
+      | o = \subseteq
+      | r = \eqclass g {\mathcal R^r_H}
+      | c = {{Defof|Subgroup}}
+}}
+{{end-eqn}}
+Thus:
+:$\eqclass g {\mathcal R^r_H} = H g$
+That is, the [[Definition:Equivalence Class|equivalence class]] $\eqclass g {\mathcal R^r_H}$ of an element $g \in G$ equals the [[Definition:Right Coset|right coset]] $g H$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Left Congruence Modulo Subgroup is Equivalence Relation}
+Tags: Congruence Modulo Subgroup
+
+\begin{theorem}
+Let $x \equiv^l y \pmod H$ denote the [[Definition:Relation|relation]] that $x$ is [[Definition:Left Congruence Modulo Subgroup|left congruent modulo $H$]] to $y$.
+Then the [[Definition:Relation|relation]] $\equiv^l$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+For clarity of expression, we will use the notation:
+:$\tuple {x, y} \in \mathcal R^l_H$
+for:
+:$x \equiv^l y \pmod H$
+From the definition of [[Definition:Left Congruence Modulo Subgroup|left congruence modulo a subgroup]], we have:
+:$\mathcal R^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$
+We show that $\mathcal R^l_H$ is an [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+We have that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+From [[Identity of Subgroup]]:
+:$e \in H$
+Hence:
+:$\forall x \in G: x^{-1} x = e \in H \implies \tuple {x, x} \in \mathcal R^l_H$
+and so $\mathcal R^l_H$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+{{begin-eqn}}
+{{eqn | l = \tuple {x, y}
+      | o = \in
+      | r = \mathcal R^l_H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^{-1} y
+      | o = \in
+      | r = H
+      | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x^{-1} y}^{-1}
+      | o = \in
+      | r = H
+      | c = {{GroupAxiom|0}}
+}}
+{{end-eqn}}
+But then:
+: $\tuple {x^{-1} y}^{-1} = y^{-1} x \implies \tuple {y, x} \in \mathcal R^l_H$
+Thus $\mathcal R^l_H$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+{{begin-eqn}}
+{{eqn | l = \tuple {x, y}, \tuple {y, z}
+      | o = \in
+      | r = \mathcal R^l_H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^{-1} y
+      | o = \in
+      | r = H
+      | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
+}}
+{{eqn | lo= \land
+      | l = y^{-1} z
+      | o = \in
+      | r = H
+      | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x^{-1} y} \tuple {y^{-1} z} = x^{-1} z
+      | o = \in
+      | r = H
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x, z}
+      | o = \in
+      | r = R^l_H
+      | c = {{Defof|Left Congruence Modulo Subgroup|Left Congruence Modulo $H$}}
+}}
+{{end-eqn}}
+Thus $\mathcal R^l_H$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+So $\mathcal R^l_H$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Congruence Modulo Subgroup is Equivalence Relation}
+Tags: Congruence Modulo Subgroup
+
+\begin{theorem}
+Let $x \equiv^r y \pmod H$ denote the [[Definition:Relation|relation]] that $x$ is [[Definition:Right Congruence Modulo Subgroup|right congruent modulo $H$]] to $y$
+Then the [[Definition:Relation|relation]] $\equiv^r$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+For clarity of expression, we will use the notation:
+:$\tuple {x, y} \in \mathcal R^l_H$
+for:
+:$x \equiv^l y \pmod H$
+From the definition of [[Definition:Left Congruence Modulo Subgroup|left congruence modulo a subgroup]], we have:
+:$\mathcal R^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$
+We show that $\mathcal R^l_H$ is an [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+We have that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+From [[Identity of Subgroup]]:
+:$e \in H$
+Hence:
+:$\forall x \in G: x x^{-1} = e \in H \implies \tuple {x, x} \in \mathcal R^r_H$
+and so $\mathcal R^r_H$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+{{begin-eqn}}
+{{eqn | l = \tuple {x, y}
+      | o = \in
+      | r = \mathcal R^r_H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y^{-1}
+      | o = \in
+      | r = H
+      | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x y^{-1} }^{-1}
+      | o = \in
+      | r = H
+      | c = {{GroupAxiom|0}}
+}}
+{{end-eqn}}
+But then:
+: $\tuple {x y^{-1} }^{-1} = y x^{-1} \implies \tuple {y, x} \in \mathcal R^r_H$
+Thus $\mathcal R^r_H$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+{{begin-eqn}}
+{{eqn | l = \tuple {x, y}, \tuple {y, z}
+      | o = \in
+      | r = \mathcal R^r_H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y^{-1}
+      | o = \in
+      | r = H
+      | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
+}}
+{{eqn | lo= \land
+      | l = y z^{-1}
+      | o = \in
+      | r = H
+      | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x y^{-1} } \tuple {y z^{-1} } = x z^{-1}
+      | o = \in
+      | r = H
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x, z}
+      | o = \in
+      | r = R^r_H
+      | c = {{Defof|Right Congruence Modulo Subgroup|Right Congruence Modulo $H$}}
+}}
+{{end-eqn}}
+Thus $\mathcal R^r_H$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+So $\mathcal R^r_H$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Epimorphism is Isomorphism iff Kernel is Trivial}
+Tags: Quotient Groups, Group Epimorphisms, Group Isomorphisms, Group Epimorphism is Isomorphism iff Kernel is Trivial
+
+\begin{theorem}
+Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be [[Definition:Group|groups]].
+Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a [[Definition:Group Epimorphism|group epimorphism]].
+Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identities]] of $G$ and $H$ respectively.
+Let $K = \map \ker \phi$ be the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
+Then:
+:the [[Definition:Group Epimorphism|epimorphism]] $\phi$ is an [[Definition:Group Isomorphism|isomorphism]]
+{{iff}}
+:$K = \set {e_G}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\phi$ be an [[Definition:Group Isomorphism|isomorphism]].
+Then by definition $\phi$ is a [[Definition:Bijection|bijective]] [[Definition:Homomorphism|homomorphism]].
+Thus by definition of [[Definition:Bijection|bijection]], $\phi$ is an [[Definition:Injection|injection]].
+By definition of [[Definition:Injection|injection]], there exists exactly one [[Definition:Element|element]] $x$ of $G$ such that $\map \phi x = e_H$.
+From [[Epimorphism Preserves Identity]], that element $x$ is $e_G$:
+:$\map \phi {e_G} = e_H$
+Thus by definition of [[Definition:Kernel of Group Homomorphism|kernel]]:
+:$\map \ker \phi = \set {e_G}$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $K := \map \ker \phi = \set {e_G}$.
+From the [[Quotient Theorem for Epimorphisms]]:
+:$\mathcal R_\phi$ is compatible with $\oplus$
+and thus from [[Kernel is Normal Subgroup of Domain]]:
+:$K \lhd G$
+From [[Congruence Relation induces Normal Subgroup]], $\mathcal R_\phi$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence defined by $K$]].
+Let $\mathcal R_K$ be the [[Definition:Congruence Modulo Subgroup|congruence modulo $K$]] induced by $K$.
+Suppose $\map \phi x = \map \phi y$.
+Then:
+:$x \mathop {\mathcal R_K} y$
+as $\mathcal R_\phi = \mathcal R_K$ from [[Congruence Modulo Subgroup is Equivalence Relation]].
+Thus by [[Congruence Class Modulo Subgroup is Coset]]:
+:$x \oplus y^{-1} \in K$
+Hence:
+:$x \oplus y^{-1} = e_G$
+and so:
+:$x = y$
+Thus $\phi$ is [[Definition:Injection|injective]].
+By definition, an [[Definition:Injection|injective]] [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Kernel is Trivial iff Monomorphism/Group|Kernel is Trivial iff Monomorphism]], $\phi$ is a [[Definition:Group Monomorphism|monomorphism]] {{iff}} $K = \left\{{e_G}\right\}$.
+By definition, a [[Definition:Group|group]] $G$ is an [[Definition:Group Epimorphism|epimorphism]] is an [[Definition:Group Isomorphism|isomorphism]] {{iff}} $G$ is also a [[Definition:Group Monomorphism|monomorphism]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Quotient Group is Isomorphic to Quotient Group by Preimage under Quotient Mapping}
+Tags: Group Isomorphisms, Group Epimorphisms, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H \lhd G$ where $\lhd$ denotes that $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $K \lhd G/H$ and $L = q_H^{-1} \left[{K}\right]$, where:
+:$q_H: G \to G/H$ is the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to the [[Definition:Quotient Group|quotient group]] $G/H$
+:$q_H^{-1} \left[{K}\right]$ is the [[Definition:Preimage of Subset under Mapping|preimage of $K$ under $q_H$]].
+Then there exists a [[Definition:Group Isomorphism|group isomorphism]] $\phi: \left({G / H}\right) / K \to G / L$ defined as:
+::$\phi \circ q_K \circ q_H = q_L$
+\end{theorem}
+
+\begin{proof}
+By [[Quotient Mapping on Structure is Canonical Epimorphism]], both $q_K$ and $q_H$ are [[Definition:Group Epimorphism|epimorphisms]].
+From [[Composite of Group Epimorphisms is Epimorphism]] we have that $q_K \circ q_H: G \to \left({G / H}\right) / K$ is also an [[Definition:Group Epimorphism|epimorphism]].
+From [[Preimage of Normal Subgroup of Quotient Group under Quotient Mapping is Normal]]:
+:$L \lhd G$
+By [[Quotient Theorem for Group Epimorphisms]]:
+:there exists a [[Definition:Group Isomorphism|group isomorphism]] $\psi: G / L \to \left({G / H}\right) / K$ satisfying:
+::$\psi \circ q_L = q_K \circ q_L$
+Let $\phi = \psi^{-1}$.
+Then $\phi$ is a [[Definition:Group Isomorphism|group isomorphism]] from $\left({G / H}\right) / K$ to $G / L$:
+:$\phi \circ q_k \circ q_H = \phi \circ \psi \circ q_L = q_L$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers are not Well-Ordered under Conventional Ordering}
+Tags: Real Numbers, Well-Orderings
+
+\begin{theorem}
+Let $\left({\R, \leqslant}\right)$ be the [[Definition:Ordered Structure|ordered structure]] consisting of the [[Definition:Real Number|real numbers]] under the [[Definition:Conventional Ordering|conventional ordering]].
+Then $\left({\R, \leqslant}\right)$ is not a [[Definition:Well-Ordered Set|well-ordered set]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]
+Consider the [[Definition:Set|set]]:
+: $A := \left\{ {x \in \R: x > 1}\right\}$
+Suppose $a$ were the [[Definition:Smallest Element|smallest element]] of $A$.
+Then $a > 1$.
+But then:
+:$1 < \dfrac {a + 1} 2 < a$
+and so $a$ has been shown not to be the [[Definition:Smallest Element|smallest element]] of $A$ after all.
+Hence $A$ has no [[Definition:Smallest Element|smallest element]].
+Thus there exists a [[Definition:Subset|subset]] of $\R$ which has no [[Definition:Smallest Element|smallest element]].
+Thus $\R$ is not a [[Definition:Well-Ordered Set|well-ordered]] under $\leqslant$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum is Unique}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $T$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then $T$ has at most one [[Definition:Supremum of Set|supremum]] in $S$.
+\end{theorem}
+
+\begin{proof}
+Let $c$ and $c'$ both be [[Definition:Supremum of Set|suprema]] of $T$ in $S$.
+From the definition of [[Definition:Supremum of Set|supremum]], $c$ and $c'$ are [[Definition:Upper Bound of Set|upper bounds]] of $T$ in $S$.
+By that definition:
+:$c$ is an [[Definition:Upper Bound of Set|upper bound]] of $T$ in $S$ and $c'$ is a [[Definition:Supremum of Set|supremum]] of $T$ in $S$ implies that $c' \preceq c$
+:$c'$ is an [[Definition:Upper Bound of Set|upper bound]] of $T$ in $S$ and $c$ is a [[Definition:Supremum of Set|supremum]] of $T$ in $S$ implies that $c \preceq c'$.
+So:
+:$c' \preceq c \land c \preceq c'$
+and thus by the [[Definition:Antisymmetric Relation|antisymmetry]] of the [[Definition:Ordering|ordering]] $\preceq$:
+:$c = c'$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Infimum is Unique}
+Tags: Infima
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $T$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then $T$ has at most one [[Definition:Infimum of Set|infimum]] in $S$.
+\end{theorem}
+
+\begin{proof}
+Let $c$ and $c'$ both be [[Definition:Infimum of Set|infima]] of $T$ in $S$.
+From the definition of [[Definition:Infimum of Set|infimum]], $c$ and $c'$ are [[Definition:Lower Bound of Set|lower bounds]] of $T$ in $S$.
+By that definition:
+: $c$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$ in $S$ and $c'$ is an [[Definition:Infimum of Set|infimum]] of $T$ in $S$ implies that $c \preceq c'$
+: $c'$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$ in $S$ and $c$ is an [[Definition:Infimum of Set|infimum]] of $T$ in $S$ implies that $c' \preceq c$.
+So:
+:$c' \preceq c \land c \preceq c'$
+and thus by the [[Definition:Antisymmetric Relation|antisymmetry]] of the [[Definition:Ordering|ordering]] $\preceq$:
+:$c = c'$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Subgroups in Lattice}
+Tags: Lattice Theory, Subgroups
+
+\begin{theorem}
+Then:
+:$\inf \set {H, K} = H \cap K$
+\end{theorem}
+
+\begin{proof}
+Let $H, K \in \mathbb G$.
+From [[Set of Subgroups forms Complete Lattice]]:
+:$\inf \set {H, K} = H \cap K$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Subgroups in Lattice}
+Tags: Lattice Theory, Subgroups
+
+\begin{theorem}
+Let either $H$ or $K$ be [[Definition:Normal Subgroup|normal]] in $G$.
+Then:
+:$\sup \left\{{H, K}\right\} = H \circ K$
+where $H \circ K$ denotes [[Definition:Subset Product|subset product]].
+\end{theorem}
+
+\begin{proof}
+Recall that [[Set of Subgroups forms Complete Lattice]].
+Let $L = \sup \left\{{H, K}\right\}$.
+Let either $H$ or $K$ be [[Definition:Normal Subgroup|normal]] in $G$.
+Since $L$ contains $H$ and $K$, then $L$ contains $H \circ K$.
+The smallest [[Definition:Subgroup|subgroup]] of $G$ containing $H$ and $K$ is:
+: $\left \langle {H, K} \right \rangle$
+the [[Definition:Generator of Subgroup|subgroup generated by $H$ and $K$]].
+From [[Subset Product with Normal Subgroup as Generator]]:
+:$\left \langle {H, K} \right \rangle = H \circ K$
+when either $H$ or $K$ is [[Definition:Normal Subgroup|normal]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lexicographic Order on Totally Ordered Sets is Total Ordering}
+Tags: Lexicographic Order, Total Orderings
+
+\begin{theorem}
+Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
+Let $\preccurlyeq$ be the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
+:$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2} \iff \paren {x_1 \prec_1 y_1} \lor \paren {x_1 = y_1 \land x_2 \preceq_2 y_2}$
+Then:
+:$\preccurlyeq$ is a [[Definition:Total Ordering|total ordering]] on $S_1 \times S_2$
+{{iff}}
+:both $\preceq_1$ and $\preceq_2$ are [[Definition:Total Ordering|total orderings]].
+\end{theorem}
+
+\begin{proof}
+Note that from [[Lexicographic Order is Ordering]] we have that $\struct {S_1 \times S_2, \preccurlyeq}$ is an [[Definition:Ordered Set|ordered set]].
+=== Necessary Condition ===
+Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be [[Definition:Totally Ordered Set|totally ordered sets]].
+Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$.
+The following cases are to be examined:
+$(1): \quad x_1 \ne y_1$
+As $\preceq_1$ is a [[Definition:Total Ordering|total ordering]] it follows that either $x_1 \preceq_1 y_1$ or $y_1 \preceq_1 x_1$.
+Thus by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:either $\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ or $\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$.
+$(2): \quad x_1 = y_1$
+As $\preceq_2$ is a [[Definition:Total Ordering|total ordering]] it follows that one of the following holds:
+:$x_2 \preceq_2 y_2$ in which case $\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$
+:$y_2 \preceq_2 x_2$ in which case $\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$
+:$x = y$ in which case $\tuple {x_1, x_2} = \tuple {y_1, y_2}$
+by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]],
+In all cases $\tuple {x_1, x_2}$ is [[Definition:Comparable|comparable]] with $\tuple {y_1, y_2}$.
+That is, $\preccurlyeq$ is a [[Definition:Total Ordering|total ordering]] on $S_1 \times S_2$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\left({S_1 \times S_2, \preccurlyeq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+There are two cases to address:
+$(1): \quad$ Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ such that $x_1 = x_2$.
+As $\preccurlyeq$ is a [[Definition:Total Ordering|total ordering]] it follows that one of the following holds:
+:$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ in which case $x_2 \preceq_2 y_2$
+:$\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$ in which case $y_2 \preceq_2 x_2$
+:$\tuple {x_1, x_2} = \tuple {y_1, y_2}$ in which case $x = y$
+by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]].
+As the above holds for all $x_2, y_2 \in S_2$, it follows that $\preceq_2$ is a [[Definition:Total Ordering|total ordering]] on $S_2$.
+$(2): \quad$ Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \times S_2$ such that $x_1 \ne x_2$.
+As $\preccurlyeq$ is a [[Definition:Total Ordering|total ordering]] it follows that one of the following holds:
+:$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2}$ in which case $x_1 \preceq_1 y_1$
+:$\tuple {y_1, y_2} \preccurlyeq \tuple {x_1, x_2}$ in which case $y_1 \preceq_1 x_1$
+by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]].
+As the above holds for all $x_1, y_1 \in S_1$, it follows that $\preceq_1$ is a [[Definition:Total Ordering|total ordering]] on $S_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lexicographic Order is Ordering}
+Tags: Lexicographic Order
+
+\begin{theorem}
+Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $\preccurlyeq$ be the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
+:$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right) \iff \left({x_1 \prec_1 y_1}\right) \lor \left({x_1 = y_1 \land x_2 \preceq_2 y_2}\right)$
+Then $\preccurlyeq$ is an [[Definition:Ordering|ordering]] on $S_1 \times S_2$.
+\end{theorem}
+
+\begin{proof}
+In the following, $\left({x_1, x_2}\right), \left({y_1, y_2}\right), \left({z_1, z_2}\right) \in S_1 \times S_2$.
+Checking in turn each of the criteria for an [[Definition:Ordering|ordering]]:
+=== Reflexivity ===
+From [[Equality of Ordered Pairs]]:
+:$x_1 = x_2 \land y_1 = y_2 \iff \left({x_1, x_2}\right) = \left({y_1, y_2}\right)$
+Thus:
+:$\left({x_1, x_2}\right) = \left({x_1, x_2}\right)$
+and so:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({x_1, x_2}\right)$
+by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]].
+So $\preccurlyeq$ has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Transitivity ===
+Let:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$
+and:
+:$\left({y_1, y_2}\right) \preccurlyeq \left({z_1, z_2}\right)$
+$(1): \quad$ Let $x_1 = y_1 = z_1$.
+Then by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$x_2 \preceq_2 y_2$
+and:
+:$y_2 \preceq_2 z_2$
+As $\preceq_2$ is an [[Definition:Ordering|ordering]], it is [[Definition:Transitive Relation|transitive]].
+Thus:
+:$x_2 \preceq z_2$
+and it follows by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]] that:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$
+$(2): \quad$ Let $x_1 = y_1$ and $y_1 \ne z_1$.
+Then by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$y_1 \prec z_1$
+But as $x_1 = y_1$:
+:$x_1 \prec z_1$
+and so by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$
+$(3): \quad$ Let $x_1 \ne y_1$ and $y_1 = z_1$.
+Then by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$x_1 \prec y_1$
+But as $y_1 = z_1$:
+:$x_1 \prec z_1$
+and so by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$
+$(4): \quad$ Let $x_1 \ne y_1$ and $y_1 \ne z_1$.
+Then by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$x_1 \prec y_1$
+and:
+:$y_1 \prec z_1$
+As $\preceq_1$ is an [[Definition:Ordering|ordering]], it is [[Definition:Transitive Relation|transitive]].
+:$x_1 \prec z_1$
+So by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$
+Thus in all cases it can be seen that:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$
+So $\preccurlyeq$ has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+=== Antisymmetry ===
+Suppose that:
+:$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$
+and:
+:$\left({y_1, y_2}\right) \preccurlyeq \left({x_1, x_2}\right)$
+Suppose $x_1 \ne y_1$.
+Then by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$x_1 \preceq_1 y_1$
+and:
+:$y_1 \preceq_1 x_1$
+But $\preceq_1$ is an [[Definition:Ordering|ordering]] and so $x_1 = y_1$.
+From that contradiction it follows that $x_1 = y_1$.
+Then by definition of [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$x_2 \preceq_2 y_2$
+and:
+:$y_2 \preceq_2 x_2$
+As $\preceq_2$ is an [[Definition:Ordering|ordering]], it is [[Definition:Antisymmetric Relation|antisymmetric]].
+Therefore:
+:$x_2 = y_2$
+and so:
+:$\left({x_1, x_2}\right) = \left({y_1, y_2}\right)$
+So $\preccurlyeq$ has been shown to be [[Definition:Antisymmetric Relation|antisymmetric]].
+{{qed|lemma}}
+$\preccurlyeq$ has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
+Hence by definition it is an [[Definition:Ordering|ordering]].
+{{qed}}
+[[Category:Lexicographic Order]]
+20j34o6ba1mr2xw3oudisfdaldqfog1
+\end{proof}<|endoftext|>
+\section{Lexicographic Order on Pair of Well-Ordered Sets is Well-Ordering}
+Tags: Lexicographic Order, Well-Orderings
+
+\begin{theorem}
+Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]].
+Let $\preccurlyeq$ be the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]''':
+:$\tuple {x_1, x_2} \preccurlyeq \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preceq_2 y_2}$
+Then:
+:$\preccurlyeq$ is a [[Definition:Well-Ordering|well-ordering]] on $S_1 \times S_2$
+{{iff}}
+:both $\preceq_1$ and $\preceq_2$ are [[Definition:Well-Ordering|well-orderings]].
+\end{theorem}
+
+\begin{proof}
+From [[Lexicographic Order on Totally Ordered Sets is Total Ordering]] we have that $\struct {S_1 \times S_2, \preccurlyeq}$ is a [[Definition:Ordered Set|ordered set]].
+=== Necessary Condition ===
+Let $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ be [[Definition:Well-Ordered Set|well-ordered sets]].
+By definition of [[Definition:Well-Ordering|well-ordering]] we have that $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are [[Definition:Totally Ordered Set|totally ordered sets]].
+From [[Lexicographic Order on Totally Ordered Sets is Total Ordering]] it follows that $\struct {S_1 \times S_2, \preccurlyeq}$ is a [[Definition:Totally Ordered Set|totally ordered set]].
+It remains to be shown that the [[Definition:Cartesian Product|Cartesian product]] of arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] of $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ has a [[Definition:Smallest Element|smallest element]] under $\preccurlyeq$.
+Let $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$.
+Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in T_1 \times T_2$.
+As both $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are [[Definition:Well-Ordered Set|well-ordered sets]] it follows that both $\struct {T_1, \preceq_1}$ and $\struct {T_2, \preceq_2}$ have [[Definition:Smallest Element|smallest elements]].
+Let those [[Definition:Smallest Element|smallest elements]] be $m_1$ and $m_2$ respectively.
+Consider the [[Definition:Element|element]] $\tuple {m_1, m_2} \in T_1 \times T_2$.
+Let $\tuple {x_1, x_2} \in T_1 \times T_2$.
+We have that:
+:$m_1 \preceq_1 x_1$
+and:
+:$m_2 \preceq_2 x_2$
+The following cases are to be examined:
+$(1): \quad m_1 \ne x_1$
+By defiinition of $m_1$ as the [[Definition:Smallest Element|smallest element]] of $T_1$:
+:$m_1 \preceq_1 x_1$
+Thus by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$\tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$
+$(2): \quad m_1 = x_1$
+By definition of $m_2$ as the [[Definition:Smallest Element|smallest element]] of $T_2$:
+:$m_2 \preceq_2 x_2$
+Thus by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$\tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$
+Thus it follows that $T_1 \times T_2$ has a [[Definition:Smallest Element|smallest element]] $\tuple {m_1, m_2}$.
+As $T_1 \times T_2$ is the [[Definition:Cartesian Product|Cartesian product]] of arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] of of $S_1$ and $S_2$, it follows that $\struct {S_1 \times S_2, \preccurlyeq}$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\struct {S_1 \times S_2, \preccurlyeq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
+By definition of [[Definition:Well-Ordering|well-ordering]] we have that $\struct {S_1 \times S_2, \preccurlyeq}$ is a [[Definition:Totally Ordered Set|totally ordered set]].
+From [[Lexicographic Order on Totally Ordered Sets is Total Ordering]] $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ are [[Definition:Totally Ordered Set|totally ordered sets]].
+It remains to be shown that arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] of $\struct {S_1, \preceq_1}$ and $\struct {S_2, \preceq_2}$ have a [[Definition:Smallest Element|smallest elements]].
+Let $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$.
+Let $\tuple {m_1, m_2} \in T_1 \times T_2$ be the [[Definition:Smallest Element|smallest element]] of $T_1 \times T_2$.
+Thus:
+:$\forall \tuple {x_1, x_2} \in T_1 \times T_2: \tuple {m_1, m_2} \preccurlyeq \tuple {x_1, x_2}$
+There are two cases to address:
+$(1): \quad$ Let $m_1 = x_1$.
+Thus by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$m_2 \preceq_2 x_2$
+This is the case for all $\tuple {x_1, x_2} \in T_1 \times T_2$ such that $m_1 = x_1$.
+Therefore:
+:$\forall x_2 \in T_2: m_2 \preceq_2 x_2$
+Thus by definition, $m_2$ is the [[Definition:Smallest Element|smallest element]] of $T_2$.
+We have that $T_2$ is an arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $\struct {S_2, \preceq_2}$.
+Therefore by definition $T_2$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+$(2): \quad$ Let $m_1 \ne x_1$.
+Thus by definition of the [[Definition:Lexicographic Order|lexicographic order on $S_1 \times S_2$]]:
+:$m_1 \prec_1 x_1$
+This is the case for all $\tuple {x_1, x_2} \in T_1 \times T_2$ such that $m_1 \ne x_1$.
+Therefore:
+:$\forall x_1 \in T_1: m_1 \preceq_1 x_1$
+Thus by definition, $m_1$ is the [[Definition:Smallest Element|smallest element]] of $T_1$.
+We have that $T_1$ is an arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $\struct {S_1, \preceq_1}$.
+Therefore by definition $T_1$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Commuting Elements of Monoid Commute}
+Tags: Monoids, Commutativity, Powers (Abstract Algebra)
+
+\begin{theorem}
+:$\forall m, n \in \N: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
+\end{theorem}
+
+\begin{proof}
+Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is [[Definition:A Fortiori|a fortiori]] also a [[Definition:Semigroup|semigroup]].
+From [[Powers of Commuting Elements of Semigroup Commute]]:
+:$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
+That is:
+:$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
+It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.
+Let $n \in \N$:
+{{begin-eqn}}
+{{eqn | l = a^m \circ b^0
+      | r = a^m \circ e
+      | c = Definition of $b^0$
+}}
+{{eqn | r = a^m
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = e \circ a^m
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = b^0 \circ a^m
+      | c = Definition of $b^0$
+}}
+{{end-eqn}}
+Similarly, let $m \in \N$:
+{{begin-eqn}}
+{{eqn | l = a^0 \circ b^n
+      | r = e \circ b^n
+      | c = Definition of $a^0$
+}}
+{{eqn | r = b^n
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = b^n \circ e
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = b^n \circ a^0
+      | c = Definition of $a^0$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = a^0 \circ b^0
+      | r = e \circ e
+      | c = Definition of $a^0$ and $b^0$
+}}
+{{eqn | r = b^0 \circ a^0
+      | c = Definition of $a^0$ and $b^0$
+}}
+{{end-eqn}}
+Thus:
+:$a^m \circ b^n = b^n \circ a^m$
+holds for $n = 0$ and $m = 0$.
+Thus:
+:$\forall m, n \in \N: a^m \circ b^n = b^n \circ a^m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Product of Commuting Elements in Monoid equals Product of Powers}
+Tags: Monoids, Commutativity, Powers (Abstract Algebra)
+
+\begin{theorem}
+:$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
+\end{theorem}
+
+\begin{proof}
+Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is [[Definition:A Fortiori|a fortiori]] also a [[Definition:Semigroup|semigroup]].
+From [[Power of Product of Commuting Elements in Semigroup equals Product of Powers]]:
+:$\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
+That is:
+:$\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n \circ b^n$
+It remains to be shown that the result holds for the cases where $n = 0$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {a \circ b}\^0
+      | r = \circ^0 \paren {a \circ b}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = e
+      | c = Definition of $\circ^0$
+}}
+{{eqn | r = e \circ e
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \paren {\circ^0 a} \circ \paren {\circ^0 b}
+      | c = Definition of $\circ^0$
+}}
+{{eqn | r = a^0 \circ b^0
+      | c = Definition of $\circ$
+}}
+{{end-eqn}}
+Thus:
+:$\paren {a \circ b}^n = a^n \circ b^n$
+holds for $n = 0$.
+Thus:
+:$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Index Laws/Product of Indices/Monoid}
+Tags: Index Laws, Monoids
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity element]] is $e$.
+For $a \in S$, let $\circ^n a = a^n$ be the [[Definition:Power of Element of Magma with Identity|$n$th power of $a$]].
+Then:
+:$\forall m, n \in \N: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
+\end{theorem}
+
+\begin{proof}
+Because $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], it is [[Definition:A Fortiori|a fortiori]] a [[Definition:Semigroup|semigroup]].
+Hence, from [[Index Laws for Semigroup/Product of Indices|Index Laws for Semigroup: Product of Indices]]:
+:$\forall m, n \in \N_{>0}: \circ^{n m} a = \circ^m \paren {\circ^n a} = \circ^n \paren {\circ^m a}$
+That is:
+:$\forall m, n \in \N_{>0}: a^{n m} = \paren {a^n}^m = \paren {a^m}^n$
+It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.
+{{begin-eqn}}
+{{eqn | l = n \times 0
+      | r = 0
+      | c = 
+}}
+{{eqn | r = 0 \times  m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \circ^{n \times 0} a
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | r = \circ^{0 \times m} a
+      | c = 
+}}
+{{end-eqn}}
+so the condition holds for either $n = 0$ or $m = 0$.
+Finally, we also have:
+:$\circ^n \paren {\circ^0 a} = e = \circ^0 \paren {\circ^m a}$
+:$\circ^0 \paren {\circ^n a}\= e = \circ^m \paren {\circ^0 a}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Proper Subset of Finite Set}
+Tags: Set Theory, Subsets, Cardinality
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Finite Set|finite sets]] such that $A \subsetneqq B$.
+Let $\card B = n$, where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinality]].
+Then $\card A < n$.
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by the [[Principle of Mathematical Induction]] on $n$, the [[Definition:Cardinality|cardinality]] of $B$.
+Let $S$ be the set of $n \in \N$ such that every [[Definition:Proper Subset|proper subset]] of any [[Definition:Set|set]] with $m$ [[Definition:Element|elements]] is [[Definition:Finite Set|finite]] and has (strictly) fewer than $n$ [[Definition:Element|elements]].
+=== Basis for the Induction ===
+The case $n = 0$ is verified as follows:
+Suppose $\card B = 0$.
+From [[Cardinality of Empty Set]], $B = \O$. 
+There are no sets $A$ such that $A \subsetneq \O$.
+Therefore $0 \in S$ by [[Definition:Vacuous Truth|vacuous truth]].
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $k \in \N$ with $k \ge 0$.
+Suppose that $k \in S$.
+That is:
+:every [[Definition:Proper Subset|proper subset]] of any [[Definition:Set|set]] with $k$ [[Definition:Element|elements]] is [[Definition:Finite Set|finite]] and has (strictly) fewer than $k$ [[Definition:Element|elements]].
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+It remains to be proved that $k + 1 \in S$:
+:every [[Definition:Proper Subset|proper subset]] of any [[Definition:Set|set]] with $k + 1$ [[Definition:Element|elements]] is [[Definition:Finite Set|finite]] and has (strictly) fewer than $k + 1$ [[Definition:Element|elements]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Suppose we have any set $B$ such that $\card B = k + 1$.
+Let $A \subsetneqq B$.
+From the definition of $\subsetneqq$:
+:$\exists b \in B: b \notin A$
+Therefore:
+:$A \subseteq B \setminus \set b$
+From [[Cardinality Less One]]:
+:$\card {B - \set b} = k$
+Suppose $A = B - \set b$.
+Then $\card A = k < k + 1$ and so has fewer than $k + 1$ [[Definition:Element|elements]] and is [[Definition:Finite Set|finite]] by definition.
+Otherwise:
+:$A \subsetneqq B - \set b$
+So by the [[Cardinality of Proper Subset of Finite Set#Induction Hypothesis|induction hypothesis]]:
+:$A$ is [[Definition:Finite Set|finite]] and has fewer than $k$ [[Definition:Element|elements]].
+But as $k < k + 1$ it also follows that $A$ is [[Definition:Finite Set|finite]] and has fewer than $k + 1$ [[Definition:Element|elements]].
+Thus $A \in S$.
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First-Countable Space is Hausdorff iff All Convergent Sequences have Unique Limit}
+Tags: First-Countable Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
+Then $T$ is [[Definition:Hausdorff Space|Hausdorff]] {{iff}} all [[Definition:Convergent Sequence|convergent sequences]] on $T$ have a [[Definition:Unique|unique]] [[Definition:Limit Point of Sequence|limit point]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+This is shown in [[Convergent Sequence in Hausdorff Space has Unique Limit]].
+Note that it does not require [[Definition:First-Countable Space|first-countability]].
+{{qed|lemma}}
+=== Necessary Condition ===
+We prove the [[Definition:Contrapositive Statement|contrapositive]].
+Suppose $T$ is not [[Definition:Hausdorff Space|Hausdorff]].
+Then by definition:
+:$\exists x, y \in S, x \ne y: \forall U, V \in \tau: x \in U, y \in V: U \cap V \ne \O$
+By [[Definition:First-Countable Space|first-countability]], $x$ and $y$ has [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+Let $\set {X_n: n \in \N}$ and $\set {Y_n: n \in \N}$ be [[Definition:Countable Set|countable]] [[Definition:Local Basis|local bases]] for $x$ and $y$ respectively.
+Let $U_n = \displaystyle \bigcap_{k \mathop = 1}^n X_n$ and $V_n = \displaystyle \bigcap_{k \mathop = 1}^n Y_n$.
+We have for all $n \in \N$:
+:$x \in U_n, y \in V_n$
+:$U_m \subseteq U_n, V_m \subseteq V_n$ for all $m \ge n$
+We have that:
+:$\forall n \in \N: U_n \cap V_n \ne \O$
+hence we can pick any $z_n \in U_n \cap V_n$.
+We show that $\sequence {z_n}$ [[Definition:Convergent Sequence|converges]] to both $x$ and $y$.
+Let $A$ be an [[Definition:Open Set (Topology)|open set]] such that $x \in A$.
+By definition of a [[Definition:Local Basis|local basis]]:
+:$\exists N \in \N: X_N \subseteq A$
+We also have:
+:$\forall n \ge N: U_n \subseteq U_N \subseteq X_N \subseteq A$
+Therefore for any $n \ge N$:
+:$z_n \in U_n \cap V_n \subseteq U_n \subseteq A$
+By definition of [[Definition:Convergent Sequence|convergence]], $\sequence {z_n}$ [[Definition:Convergent Sequence|converges]] to $x$.
+Similarly $\sequence {z_n}$ [[Definition:Convergent Sequence|converges]] to $y$.
+Hence $\sequence {z_n}$ does not have [[Definition:Unique|unique]] [[Definition:Limit Point of Sequence|limit point]].
+{{qed}}
+[[Category:First-Countable Spaces]]
+[[Category:Hausdorff Spaces]]
+hjqs2yokgfjt4z0o78tp2r10yrx7q1e
+\end{proof}<|endoftext|>
+\section{Topologically Distinguishable Points are Distinct}
+Tags: Separation Axioms
+
+\begin{theorem}
+Let $T = \left({X, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x, y \in X$ be [[Definition:Topologically Distinguishable|topologically distinguishable]].
+Then the [[Definition:Singleton|singleton sets]] $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are [[Definition:Disjoint Sets|disjoint]] and so:
+: $x \ne y$
+\end{theorem}
+
+\begin{proof}
+Let $x$ and $y$ be '''topologically distinguishable'''.
+Then either:
+:$\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$
+or:
+:$\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$
+Suppose $x = y$.
+Then:
+:$\neg \left({\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x}\right)$
+and:
+:$\neg \left({\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y}\right)$
+Hence the result by [[Proof by Contradiction]]:
+:$x \ne y$
+and so by [[Singleton Equality]]:
+:$\left\{{x}\right\} \ne \left\{{y}\right\}$
+{{qed}}
+[[Category:Separation Axioms]]
+fibtb9jvxk7u2j8s651wkn5xs3chb5l
+\end{proof}<|endoftext|>
+\section{Restriction of Strict Total Ordering is Strict Total Ordering}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\left({S, \prec}\right)$ be a [[Definition:Strict Total Ordering|strict total ordering]].
+Let $T \subseteq S$.
+Let $\prec \restriction_T$ be the [[Definition:Restriction of Ordering|restriction]] of $\prec$ to $T$.
+Then $\prec \restriction_T$ is a [[Definition:Strict Total Ordering|strict total ordering]] of $T$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Strict Total Ordering|strict total ordering]], $\prec$ is:
+: $(1): \quad$ a [[Definition:Relation|relation]] which is [[Definition:Transitive|transitive]] and [[Definition:Antireflexive Relation|antireflexive]] 
+: $(2): \quad$ a [[Definition:Relation|relation]] which is [[Definition:Connected Relation|connected]].
+By [[Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering]]:
+: $\prec \restriction_T$ is a [[Definition:Strict Ordering|strict ordering]].
+It follows from [[Restriction of Connected Relation is Connected]] that:
+:$\prec \restriction_T$ is [[Definition:Connected Relation|connected]].
+Thus $\prec \restriction_T$ is a [[Definition:Strict Ordering|strict ordering]] which is [[Definition:Connected Relation|connected]].
+So by definition $\prec \restriction_T$ is a [[Definition:Strict Total Ordering|strict total ordering]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Recursive Definition}
+Tags: Named Theorems, Mapping Theory, Natural Numbers, Recursive Definitions, Principle of Recursive Definition
+
+\begin{theorem}
+Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
+Let $T$ be a [[Definition:Set|set]].
+Let $a \in T$.
+Let $g: T \to T$ be a [[Definition:Mapping|mapping]].
+Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]] $f: \N \to T$ such that:
+:$\forall x \in \N: \map f x = \begin{cases}
+a & : x = 0 \\
+\map g {\map f n} & : x = n + 1
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+Consider $\N$ defined as a [[Definition:Peano Structure|Peano structure]] $\struct {P, 0, s}$.
+The result follows from [[Principle of Recursive Definition for Peano Structure]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider $\N$ defined as the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction of the natural numbers]].
+The result follows from [[Principle of Recursive Definition for Minimal Infinite Successor Set]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider $\N$, defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
+Let the [[Definition:Mapping|mapping]] $f$ be defined as:
+:$\map f x = \begin{cases}
+a & : x = 0 \\
+\map s {\map f n} & : x = n \circ 1 \end{cases}$
+if $\map f n$ is defined.
+Let $S' = \set {n \in S: \map f n \text{ is defined} }$.
+Then:
+:$0 \in S'$
+and:
+:$n + 1 \in S'$
+so by [[Principle of Mathematical Induction for Naturally Ordered Semigroup]]:
+:$S' = S$
+Thus the [[Definition:Domain|domain]] of $f$ is $S$.
+Consequently, $f$ is a [[Definition:Mapping|mapping]] from $S$ into $T$ which satisfies:
+:$\map f 0 = a$
+and:
+:$\map f {n \circ 1} = \map s {\map f n}$
+for all $n \in S$.
+{{qed}}
+=== Objections ===
+$(1):$
+In the above argument, $S'$ is not precisely defined.
+In order for a [[Definition:Set|set]] to be defined, that definition should be able to be expressed solely in terms of [[Definition:Logic|logic]] and definitions from [[Definition:Set Theory|set theory]].
+In this case, the expression "is defined" does not meet that criterion.
+$(2):$
+The [[Definition:Mapping|mapping]] $f$ is not defined properly.
+A [[Definition:Mapping|mapping]] needs to be defined as a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]] which needs to satisfy a [[Definition:Propositional Function|condition]].
+In the above, it is indeed specified that
+:$\tuple {0, a} \in f$
+and:
+:$\tuple {n \circ 1, \map s x} \in f$ whenever $\tuple {n, x} \in f$
+Thus it appears either that:
+:$f$ itself is used to define $f$
+or else that:
+:$f$ itself changes during the process in which it is being defined.
+Neither of these possibilities can be accepted.
+$(3):$
+The only property of $\struct {S, \circ, \preceq}$ used in the argument is that it satisfies [[Principle of Mathematical Induction for Naturally Ordered Semigroup]].
+However, consider the [[Definition:Commutative Semigroup|commutative semigroup]] $\struct {D, +}$ which has [[Definition:Element|elements]] $0$ and $1$, such that:
+:$D$ is the only [[Definition:Subset|subset]] of $D$:
+::containing $0$
+:and
+::containing $x + 1$ whenever it contains $x$.
+Then:
+:for any set $T$
+:any [[Definition:Mapping|mapping]] $g: T \to T$
+:any [[Definition:Element|element]] $a \in T$
+there exists a [[Definition:Mapping|mapping]] $f: D \to T$ such that:
+:$\map f y = \begin{cases}
+a & : y = 0 \\
+\map s {\map f x} & : y = x + 1 \end{cases}$
+But consider the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $2$]]:
+:$\struct {\Z_2, +_2}$
+which is indeed a [[Definition:Commutative Semigroup|commutative semigroup]] containing $0$ and $1$ which satisfies the hypothesis.
+But if $g: \N \to \N$ is the [[Definition:Mapping|mapping]] defined as:
+:$\map g n = n + 1$
+there is no [[Definition:Mapping|mapping]] $f: \Z_2 \to \N$ which satisfies:
+:$\map f y = \begin{cases}
+0 & : y = 0 \\
+\map s {\map f x} & : y = x +_2 1 \end{cases}$
+for all $x \in \Z_2$.
+Hence the argument is [[Definition:Invalid Argument|invalid]].
+\end{proof}<|endoftext|>
+\section{Principle of Recursive Definition/General Result}
+Tags: Mapping Theory, Natural Numbers, Principle of Recursive Definition
+
+\begin{theorem}
+Let $p \in \N$.
+Let $p^\ge$ be the [[Definition:Upper Closure of Element|upper closure]] of $p$ in $\N$:
+:$p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$
+Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Mapping|mapping]] $f: p^\ge \to T$ such that:
+:$\forall x \in p^\ge: \map f x = \begin{cases}
+a & : x = p \\
+\map g {\map f n} & : x = n + 1
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+Consider $\N$, defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
+For simplicity, let $S' = p^\ge$.
+First an '''admissible mapping''' is defined.
+Let $n \in S'$.
+The [[Definition:Mapping|mapping]] $h: \closedint p n \to T$ is defined as an '''admissible mapping''' for $n$ {{iff}}:
+:$\forall r \in \closedint p n: \map h r = \begin{cases}
+a & : r = p \\
+\map g {\map h y} & : r = y \circ 1
+\end{cases}$
+where $\closedint p n$ denotes an [[Definition:Integer Interval|integer interval]] of $S$:
+:$\closedint p n = \set {r \in S: p \le r \le n}$
+Define:
+:$A = \set {x \in S': x \text { has exactly one admissible mapping} }$
+We now use the [[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction for a Naturally Ordered Semigroup]] to prove that $A = S'$.
+=== Basis for the Induction ===
+Consider the [[Definition:Integer Interval|integer interval]] $\closedint p p = \set p$.
+Let the [[Definition:Mapping|mapping]] $h_p: \set p \to T$ be defined as:
+:$\map {h_p} p = a$
+By definition, $h_p$ is the [[Definition:Unique|unique]] '''admissible mapping''' for $p$.
+Thus $p \in A$.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Let $n \in A$ such that $p \le n$.
+That is, there exists a [[Definition:Unique|unique]] '''admissible mapping''' for $n$.
+This is the [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+From the [[Principle of Recursive Definition/General Result#Induction Hypothesis|induction hypothesis]]:
+:Let $h$ be the [[Definition:Unique|unique]] '''admissible mapping''' for $n$.
+By [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]:
+:$n \circ 1 \notin \closedint p n$
+So by [[Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor]]:
+:$\closedint p {n \circ 1} = \closedint p n \cup \set {n \circ 1}$
+So we can define a [[Definition:Mapping|mapping]] $\phi: \closedint p {n \circ 1} \to T$ by:
+:$\forall r \in \closedint p {n \circ 1}: \map \phi r = \begin{cases}
+\map h r & : r \in \closedint p n \\
+\map g {\map h n} & : r = n \circ 1
+\end{cases}$
+By definition, $\phi$ is an '''admissible mapping''' for $n \circ 1$.
+Let $\phi'$ be an '''admissible mapping''' for $n \circ 1$.
+[[Definition:Unique|Uniqueness]] of $\phi$ is then proved by showing that $\phi' = \phi$, as follows.
+The [[Definition:Restriction of Mapping|restriction]] of $\phi'$ to $\closedint p n $ is an '''admissible mapping''' for $n$
+Thus by the [[Principle of Recursive Definition/General Result#Induction Hypothesis|induction hypothesis]]:
+:$\phi' {\restriction_{\closedint p n} } = \phi {\restriction_{\closedint p n} }$
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \map {\phi'} {n \circ 1}
+      | r = \map g {\map {\phi'} n}
+      | c = 
+}}
+{{eqn | r = \map g {\map h n}
+      | c = 
+}}
+{{eqn | r = \map \phi {n \circ 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus as:
+:$\phi' {\restriction_{\closedint p n} } = \phi {\restriction_{\closedint p n} }$
+and:
+:$\map {\phi'} {n \circ 1} = \map \phi {n \circ 1}$
+it follows by [[Equality of Mappings]] that:
+:$\phi' = \phi$
+Thus:
+:$n \circ 1 \in A$
+Thus by the [[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction for a Naturally Ordered Semigroup]] it follows that:
+:$A = S'$
+For all $n \in S'$, let $h_n$ be the unique '''admissible mapping''' for $n$.
+Let $n \ge p$.
+The [[Definition:Restriction of Mapping|restriction]] of $h_{n \circ 1}$ to $\closedint p n$ is an '''admissible mapping''' for $n$.
+Therefore:
+:$h_{n \circ 1} {\restriction_{\closedint p n} } = h_n$
+So:
+:$\forall r \in \closedint p n: \map {h_{n \circ 1} } r = \map {h_n} r$
+and in particular:
+:$\map {h_{n \circ 1} } n = \map {h_n} n$
+Let $f: S' \to T$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in S': \map f n = \map {h_n} n$
+Then:
+:$\map f p = \map {h_p} p = a$
+and, for all $n \in S'$:
+{{begin-eqn}}
+{{eqn | l = \map f {n \circ 1}
+      | r = \map {h_{n \circ 1} } {n \circ 1}
+}}
+{{eqn | r = \map g {\map {h_{n \circ 1} } n}
+}}
+{{eqn | r = \map g {\map {h_n} n}
+}}
+{{eqn | r = \map g {\map f n}
+}}
+{{end-eqn}}
+Thus $f: S' \to T$ is a [[Definition:Mapping|mapping]] with the desired properties.
+ 
+Let $f'$ be any [[Definition:Mapping|mapping]] satisfying the desired properties. 
+It remains to be proved that:
+:$f' = f$
+that is, that $f$ is the only [[Definition:Mapping|mapping]] with the desired properties.
+By the foregoing, for all $n \in S'$, the [[Definition:Restriction of Mapping|restriction]] of $f'$ to $\closedint p n$ is an '''admissible mapping''' for $n$.
+Therefore:
+:$f' = h_n$
+So:
+:$\forall n \in S': \map {f'} n = \map {h_n} n = \map f n$
+and the proof is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers are not Close Packed}
+Tags: Integers
+
+\begin{theorem}
+The [[Definition:Integer|integers]] $\Z$ are not [[Definition:Close Packed|close packed]].
+That is:
+:$\forall n \in \Z: \not \exists m \in \Z: n < m < n + 1$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Immediate Successor Element|immediate successor element]], this is equivalent to the statement:
+:$\forall n \in \Z: n + 1$ is the [[Definition:Immediate Successor Element|immediate successor]] to $n$
+We have that [[Integers form Ordered Integral Domain]].
+From [[One Succeeds Zero in Well-Ordered Integral Domain]]:
+:$\not \exists r \in \Z: 0 < r < 1$
+From [[Properties of Ordered Ring]]:
+:$a < b \implies n + a < n + b$
+Putting $a = 0, b = 1, m = n + r$:
+:$\not \exists m \in \Z: n + 0 < m < n + 1$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Zero is Zero Element}
+Tags: Real Numbers
+
+\begin{theorem}
+:$\forall x \in \R: 0 \times x = 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 0 \times x
+      | r = \paren {0 + 0} \times x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 3$]]
+}}
+{{eqn | r = 0 \times x + 0 \times x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R D$]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 \times x
+      | r = 0
+      | c = [[Real Addition Identity is Zero/Corollary|Real Addition Identity is Zero: Corollary]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Real Zero equals Zero}
+Tags: Real Numbers
+
+\begin{theorem}
+Let $0$ denote the [[Real Addition Identity is Zero|identity for addition]] in the [[Definition:Real Number|real numbers]] $\R$.
+Then:
+:$-0 = 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = -0 + 0
+      | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$]]
+}}
+{{eqn | ll= \leadsto
+      | l = -0
+      | r = 0
+      | c = [[Real Addition Identity is Zero/Corollary|Real Addition Identity is Zero: Corollary]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Negative Real Number}
+Tags: Real Numbers
+
+\begin{theorem}
+:$\forall x \in \R: -\paren {-x} = x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \paren {-x} + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$: Inverse for Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {-x} + 0
+      | r = -\paren {-x} + \paren {-x} + x
+      | c = adding $-\paren {-x}$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {-x} + 0
+      | r = \paren {-\paren {-x} + \paren {-x} } + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {-x} + 0
+      | r = 0 + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$: Inverse for Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {-x}
+      | r = x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 3$: Identity for Addition]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplication by Negative Real Number}
+Tags: Real Multiplication
+
+\begin{theorem}
+:$\forall x, y \in \R: x \times \paren {-y} = -\paren {x \times y} = \paren {-x} \times y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x \times \paren {\paren {-y} + y}
+      | r = x \times 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$: Inverse for Addition]]
+}}
+{{eqn | r = 0
+      | c = [[Real Zero is Zero Element]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x \times \paren {-y} } + \paren {x \times y}
+      | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R D$: Distributivity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {x \times \paren {-y} } + \paren {x \times y} } + \paren {-\paren {x \times y} }
+      | r = 0 + \paren {-\paren {x \times y} }
+      | c = adding $-\paren {x \times y}$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x \times \paren {-y} } + \paren {\paren {x \times y} + \paren {-\paren {x \times y} } }
+      | r = 0 + \paren {-\paren {x \times y} }
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x \times \paren {-y} } + 0
+      | r = 0 + \paren {-\paren {x \times y} }
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$: Inverse for Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = x \times \paren {-y}
+      | r = -\paren {x \times y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 3$: Identity for Addition]]
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \paren {x + \paren {-x} } \times y
+      | r = 0 \times y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$: Inverse for Addition]]
+}}
+{{eqn | r = 0
+      | c = [[Real Zero is Zero Element]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x \times y} + \paren {\paren {-x} \times y}
+      | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R D$: Distributivity]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {x \times y} + \paren {\paren {x \times y} + \paren {\paren {-x} \times y} }
+      | r = -\left({x \times y}\right) + 0
+      | c = adding $-\left({x \times y}\right)$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-\paren {x \times y} + \paren {x \times y} } + \paren {\paren {-x} \times y}
+      | r = -\paren {x \times y} + 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + \paren {\paren {-x} \times y}
+      | r = -\paren {x \times y} + 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 4$: Inverse for Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-x} \times y
+      | r = -\paren {x \times y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R A 3$: Identity for Addition]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Sum of Real Numbers}
+Tags: Real Addition, Real Subtraction
+
+\begin{theorem}
+:$\forall x, y \in \R: -\paren {x + y} = -x - y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = -\paren {x + y}
+      | r = \paren {-1} \times \paren {x + y}
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = \paren {\paren {-1} \times x} + \paren {\paren {-1} \times y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R D$: Distributivity]]
+}}
+{{eqn | r = \paren {-x} + \paren {-y}
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = -x - y
+      | c = {{Defof|Real Subtraction}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Sum of Real Numbers/Corollary}
+Tags: Real Addition, Real Subtraction
+
+\begin{theorem}
+:$\forall x, y \in \R: -\paren {x - y} = -x + y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = -\paren {x - y}
+      | r = -\paren {x + \paren {-y} }
+      | c = {{Defof|Real Subtraction}}
+}}
+{{eqn | r = -x - \paren {-y}
+      | c = [[Negative of Sum of Real Numbers]]
+}}
+{{eqn | r = -x + \paren {-\paren {-y} }
+      | c = {{Defof|Real Subtraction}}
+}}
+{{eqn | r = -x + y
+      | c = [[Negative of Negative Real Number]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Multiplication Identity is One/Corollary}
+Tags: Real Multiplication
+
+\begin{theorem}
+:$\forall x \in \R_{\ne 0}: x \times y = x \implies y = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x \times y
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 x \times \paren {x \times y}
+      | r = \frac 1 x \times x
+      | c = as long as $x \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\frac 1 x \times x} \times y
+      | r = \frac 1 x \times x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1 \times y
+      | r = 1
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = 1
+      | c = [[Real Multiplication Identity is One]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Divided by Itself}
+Tags: Real Division
+
+\begin{theorem}
+:$\forall x \in \R_{\ne 0}: \dfrac x x = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | lo= \forall x \ne 0:
+      | l = \frac x x
+      | r = x \times \frac 1 x
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = 1
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 4$: Inverses]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Division by One}
+Tags: Real Division
+
+\begin{theorem}
+:$\forall x, y \in \R: \dfrac x 1 = x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac x 1
+      | r = x \times \frac 1 1
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = x \times 1
+      | c = [[Real Number Divided by Itself]]
+}}
+{{eqn | r = x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Reciprocals of Real Numbers}
+Tags: Real Multiplication, Reciprocals
+
+\begin{theorem}
+:$\forall x, y \in \R_{\ne 0}: \dfrac 1 x \times \dfrac 1 y = \dfrac 1 {x \times y}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac 1 {x \times y} \times \paren {x \times y}
+      | r = 1
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x \times y} \times \paren {x \times y} \times \frac 1 y
+      | r = 1 \times \frac 1 y
+      | c = as $y \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\frac 1 {x \times y} \times x} \times \paren {y \times \frac 1 y}
+      | r = 1 \times \frac 1 y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\frac 1 {x \times y} \times x} \times 1
+      | r = 1 \times \frac 1 y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverse]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x \times y} \times x
+      | r = \frac 1 y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\frac 1 {x \times y} \times x} \times \frac 1 x
+      | r = \frac 1 y \times \frac 1 x
+      | c = as $x \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x \times y} \times \paren {x \times \frac 1 x}
+      | r = \frac 1 y \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x \times y} \times 1
+      | r = \frac 1 y \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverse]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x \times y}
+      | r = \frac 1 y \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x \times y}
+      | r = \frac 1 x \times \frac 1 y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 2$: Commutativity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Quotients of Real Numbers}
+Tags: Real Multiplication, Real Division
+
+\begin{theorem}
+:$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y \times \dfrac w z = \dfrac {x \times w} {y \times z}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac x y \times \frac w z
+      | r = x \times \frac 1 y \times w \times \frac 1 z
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = x \times w \times \frac 1 y \times \frac 1 z
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 2$: Commutativity]]
+}}
+{{eqn | r = \paren {x \times w} \times \paren {\frac 1 y \times \frac 1 z}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity]]
+}}
+{{eqn | r = \paren {x \times w} \times \frac 1 {y \times z}
+      | c = [[Product of Reciprocals of Real Numbers]]
+}}
+{{eqn | r = \dfrac {x \times w} {y \times z}
+      | c = {{Defof|Real Division}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Quotients of Real Numbers}
+Tags: Real Addition, Real Division
+
+\begin{theorem}
+:$\forall x, w \in \R, y, z \in \R_{\ne 0}: \dfrac x y + \dfrac w z = \dfrac {\paren {x \times z} + \paren {y \times w} } {y \times z}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac x y + \frac w z
+      | r = \paren {x \times \frac 1 y} + \paren {w \times \frac 1 z}
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = \paren {x \times \frac 1 y \times 1} + \paren {1 \times w \times \frac 1 z}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 3$: Identity]]
+}}
+{{eqn | r = \paren {x \times \frac 1 y \times z \times \frac 1 z} + \paren {y \times \frac 1 y \times w \times \frac 1 z}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 4$: Inverses]]
+}}
+{{eqn | r = \paren {x \times z \times \frac 1 y \times \frac 1 z} + \paren {y \times w \times \frac 1 y \times \frac 1 z}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 2$: Commutativity]]
+}}
+{{eqn | r = \paren {\paren {x \times z} \times \paren {\frac 1 y \times \frac 1 z} } + \paren {\paren {y \times w} \times \paren {\frac 1 y \times \frac 1 z} }
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M 1$: Associativity]]
+}}
+{{eqn | r = \paren {\paren {x \times z} + \paren {y \times w} } \times \paren {\frac 1 y \times \frac 1 z}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R D$: Distributivity]]
+}}
+{{eqn | r = \paren {\paren {x \times z} + \paren {y \times w} } \times \frac 1 {y z}
+      | c = [[Product of Reciprocals of Real Numbers]]
+}}
+{{eqn | r = \frac {\paren {x \times z} + \paren {y \times w} } {y \times z}
+      | c = {{Defof|Real Division}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reciprocal of Real Number is Non-Zero}
+Tags: Real Numbers, Reciprocals
+
+\begin{theorem}
+:$\forall x \in \R: x \ne 0 \implies \dfrac 1 x \ne 0$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that:
+:$\exists x \in \R_{\ne 0}: \dfrac 1 x = 0$
+From [[Real Zero is Zero Element]]
+:$\dfrac 1 x \times x = 0$
+But from [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverses]]:
+:$\dfrac 1 x \times x = 1$
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reciprocal of Quotient of Real Numbers}
+Tags: Real Division, Reciprocals
+
+\begin{theorem}
+:$\forall x, y \in \R_{\ne 0}: \dfrac 1 {x / y} = \dfrac y x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {x / y}
+      | r = \frac 1 {x \times \dfrac 1 y}
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = 1 \times \frac 1 {x \times \dfrac 1 y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{eqn | r = \paren {y \times \frac 1 y} \times \frac 1 {x \times \dfrac 1 y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverses]]
+}}
+{{eqn | r = y \times \paren {\frac 1 y \times \frac 1 {x \times \dfrac 1 y} }
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity]]
+}}
+{{eqn | r = y \times \frac 1 {y \times \paren {x \times \dfrac 1 y} }
+      | c = [[Product of Reciprocals of Real Numbers]]
+}}
+{{eqn | r = y \times \frac 1 {x \times \paren {y \times \dfrac 1 y} }
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity and $\R \text M 2$: Commutativity]]
+}}
+{{eqn | r = y \times \frac 1 {x \times 1}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverses]]
+}}
+{{eqn | r = y \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{eqn | r = \frac y x
+      | c = {{Defof|Real Division}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Quotients of Real Numbers}
+Tags: Real Division
+
+\begin{theorem}
+:$\forall x \in \R, y, w, z \in \R_{\ne 0}: \dfrac {x / y} {w / z} = \dfrac {x \times z} {y \times w}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {x / y} {w / z}
+      | r = \frac x y \times \frac 1 {w / z}
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = \frac x y \times \frac z w
+      | c = [[Reciprocal of Quotient of Real Numbers]]
+}}
+{{eqn | r = \dfrac {x \times z} {y \times w}
+      | c = [[Product of Quotients of Real Numbers]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Real Number with Quotient}
+Tags: Real Multiplication, Real Division
+
+\begin{theorem}
+:$\forall a, x \in \R, y \in \R_{\ne 0}: \dfrac {a \times x} y = a \times \dfrac x y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {a \times x} y
+      | r = \paren {a \times x} \times \frac 1 y
+      | c = {{Defof|Real Division}}
+}}
+{{eqn | r = a \times \paren {x \times \frac 1 y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 1$: Associativity]]
+}}
+{{eqn | r = a \times \frac x y
+      | c = {{Defof|Real Division}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood in Topological Subspace}
+Tags: Topology, Neighborhoods
+
+\begin{theorem}
+Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $S \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
+Let $\tau_S$ denote the [[Definition:Topological Subspace|subspace topology]] on $S$.
+Let $x \in S$ be an arbitrary point of $S$.
+Let $E \subseteq S$.
+Then:
+:$E$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $\struct {S, \tau_S}$
+{{iff}}:
+:$\exists D \subseteq X$ such that:
+::$D$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $X$
+::$E = D \cap S$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $E$ be a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $\struct {S, \tau_S}$. 
+By the definition of [[Definition:Neighborhood of Point|neighborhood]]:
+:$\exists U \in \tau_S : x \in U \subseteq E$
+Now, by the definition of the [[Definition:Topological Subspace|subspace topology]]:
+:$\exists V \in \tau: U = V \cap S$
+Take $D := V \cup E$.
+We have that:
+:$V \subseteq D$
+and:
+:$V \in \tau$
+Thus $D$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $X$.
+Therefore it holds true that:
+:$D \cap S = \paren {V \cap S} \cup \paren {E \cap S} = U \cup E = E$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\exists D \subseteq X$ such that $D$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $\struct {X, \tau}$.
+Let $E = D \cap S$
+Then, by the definition of [[Definition:Neighborhood of Point|neighborhood]]:
+:$\exists V \in \tau: x \in V \subseteq D$
+By the definition of the [[Definition:Topological Subspace|subspace topology]]:
+:$V \cap S \in \tau_S$
+As a consequence:
+:$E = D \cap S \supseteq V \cap S \in \tau_S$
+That is:
+:$E$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $\struct {S, \tau_S}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Quotient of Real Numbers}
+Tags: Real Division
+
+\begin{theorem}
+:$\forall x \in \R, y \in \R_{\ne 0}: \dfrac {-x} y = -\dfrac x y = \dfrac x {-y}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {-x} y
+      | r = \frac {\paren {-1} \times x} y
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = \paren {-1} \times \frac x y
+      | c = [[Product of Real Number with Quotient]]
+}}
+{{eqn | r = - \frac x y
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = \paren {-1} \times \frac x y
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = \paren {-1} \times 1 \times \frac x y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{eqn | r = \paren {-1} \times \frac {-1} {-1} \times \frac x y
+      | c = [[Real Number Divided by Itself]]
+}}
+{{eqn | r = \frac {-1 \times -1} {-1} \times \frac x y
+      | c = [[Product of Real Number with Quotient]]
+}}
+{{eqn | r = \frac {-\paren {-1} } {-1} \times \frac x y
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = \frac 1 {-1} \times \frac x y
+      | c = [[Negative of Negative Real Number]]
+}}
+{{eqn | r = \frac {1 \times x} {\paren {-1} \times y}
+      | c = [[Product of Quotients of Real Numbers]]
+}}
+{{eqn | r = \frac {1 \times x} {- y}
+      | c = [[Multiplication by Negative Real Number/Corollary|Multiplication by Negative Real Number: Corollary]]
+}}
+{{eqn | r = \frac x {- y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Hausdorff Space is Locally Compact}
+Tags: Compact Spaces, Hausdorff Spaces, Locally Compact Hausdorff Spaces, Compact Hausdorff Space is Locally Compact
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Hausdorff Space|Hausdorff space]] which is [[Definition:Compact Topological Space|compact]].
+Then $T$ is [[Definition:Locally Compact Hausdorff Space|locally compact]].
+\end{theorem}
+
+\begin{proof}
+By [[Compact Space is Weakly Locally Compact]], $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+Thus $T$ is a [[Definition:Locally Compact Hausdorff Space|locally compact Hausdorff space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Strictly Positive Real Numbers is Strictly Positive}
+Tags: Real Addition
+
+\begin{theorem}
+:$x, y \in \R_{>0} \implies x + y \in \R_{>0}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = 0
+      | c =
+}}
+{{eqn | lo= \land
+      | l = y
+      | o = >
+      | r = 0
+      | c = [[Real Number Ordering is Compatible with Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = x + y
+      | o = >
+      | r = 0 + 0
+      | c = [[Real Number Inequalities can be Added]]
+}}
+{{eqn | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 3$: Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Strictly Positive Real Numbers is Strictly Positive}
+Tags: Real Multiplication
+
+\begin{theorem}
+:$x, y \in \R_{>0} \implies x \times y \in \R_{>0}$
+\end{theorem}
+
+\begin{proof}
+Let $y > 0$.
+From [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$]]: [[Definition:Relation Compatible with Operation|compatible]] with [[Definition:Real Multiplication|multiplication]]:
+:$x > z \implies x \times y > z \times y$
+Thus setting $z = 0$:
+:$x > 0 \implies x \times y > 0 \times y$
+But from [[Real Zero is Zero Element]]:
+:$0 \times y = 0$
+Hence the result:
+:$x, y > 0 \implies x \times y > 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number is Greater than Zero iff its Negative is Less than Zero}
+Tags: Real Numbers
+
+\begin{theorem}
+:$\forall x \in \R: x > 0 \iff \paren {-x} < 0$
+\end{theorem}
+
+\begin{proof}
+Let $x > 0$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + \paren {-x}
+      | o = >
+      | r = 0 + \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = >
+      | r = 0 + \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = >
+      | r = \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 3$: Identity]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $\paren {-x} < 0$.
+{{begin-eqn}}
+{{eqn | l = \paren {-x}
+      | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + \paren {-x}
+      | o = <
+      | r = x + 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = x + 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 3$: Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Real Numbers is Dual of Order of their Negatives}
+Tags: Real Numbers, Inequalities, Order of Real Numbers is Dual of Order of their Negatives
+
+\begin{theorem}
+:$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$
+\end{theorem}
+
+\begin{proof}
+Let $x > y$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + \paren {-x}
+      | o = >
+      | r = y + \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 1$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = >
+      | r = y + \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + \paren {-y}
+      | o = >
+      | r = y + \paren {-x} + \paren {-y}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 1$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + \paren {-y}
+      | o = >
+      | r = \paren {y + \paren {-y} } + \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 1$: Associativity and $\R \text A 2$: Commuativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + \paren {-y}
+      | o = >
+      | r = 0 + \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-y}
+      | o = >
+      | r = \paren {-x}
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 3$: Identity]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-x}
+      | o = <
+      | r = \paren {-y}
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $\paren {-x} < \paren {-y}$.
+{{begin-eqn}}
+{{eqn | l = \paren {-x}
+      | o = <
+      | r = \paren {-y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-x} + x
+      | o = <
+      | r = \paren {-y} + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 1$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \paren {-y} + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + y
+      | o = <
+      | r = \paren {-y} + x + y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 1$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + y
+      | o = <
+      | r = \paren {\paren {-y} + y} + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 1$: Associativity and $\R \text A 2$: Commuativity]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + y
+      | o = <
+      | r = 0 + x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = <
+      | r = x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 3$: Identity]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = >
+      | r = y
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = y
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y - x
+      | o = >
+      | r = 0
+      | c = [[Inequality iff Difference is Positive]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | lr = -x + y
+      | o = >
+      | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 2 $: Commutativity of Addition]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x + -\paren {-y}
+      | o = >
+      | r = 0
+      | c = [[Negative of Negative Real Number]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -x - \paren {-y}
+      | o = >
+      | r = 0
+      | c = {{Defof|Real Subtraction}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -y
+      | o = <
+      | r = -x
+      | c = [[Inequality iff Difference is Positive]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square of Non-Zero Real Number is Strictly Positive}
+Tags: Real Numbers
+
+\begin{theorem}
+:$\forall x \in \R: x \ne 0 \implies x^2 > 0$
+\end{theorem}
+
+\begin{proof}
+There are two cases to consider:
+:$(1): \quad x > 0$
+:$(2): \quad x < 0$
+Let $x > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \times x
+      | o = >
+      | r = 0
+      | c = [[Product of Strictly Positive Real Numbers is Strictly Positive]]
+}}
+{{end-eqn}}
+Let $x < 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \times x
+      | o = >
+      | r = x \times 0
+      | c = [[Real Number Ordering is Compatible with Multiplication/Negative Factor|Real Number Ordering is Compatible with Multiplication: Negative Factor]]
+}}
+{{eqn | r = 0
+      | c = [[Real Zero is Zero Element]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minus One is Less than Zero}
+Tags: Real Numbers
+
+\begin{theorem}
+:$-1 < 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = <
+      | r = 1
+      | c = [[Real Zero is Less than Real One]]
+}}
+{{eqn | ll= \leadsto
+      | l = -0
+      | o = >
+      | r = -1
+      | c = [[Order of Real Numbers is Dual of Order of their Negatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = >
+      | r = -1
+      | c = [[Negative of Real Zero equals Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = -1
+      | o = <
+      | r = 0
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Zero is Less than Real One}
+Tags: Real Numbers
+
+\begin{theorem}
+The [[Definition:Real Number|real number]] $0$ is less than the [[Definition:Real Number|real number]] $1$:
+:$0 < 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 1 \times 1
+      | o = >
+      | r = 0
+      | c = [[Square of Non-Zero Real Number is Strictly Positive]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = >
+      | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 3$: Identity]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = 1
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Real Numbers is Positive iff Numbers have Same Sign}
+Tags: Real Multiplication
+
+\begin{theorem}
+The [[Definition:Real Multiplication|product]] of two [[Definition:Real Number|real numbers]] is greater than $0$ {{iff}} either both are greater than $0$ or both are less than $0$.
+:$\forall x, y \in \R: x \times y > 0 \iff \paren {x, y \in \R_{>0} } \lor \paren {x, y \in \R_{<0} }$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \times y > 0$.
+{{AimForCont}} either $x = 0$ or $y = 0$.
+Then from [[Real Zero is Zero Element]]:
+:$x \times y = 0$
+Therefore by [[Proof by Contradiction]]:
+:$y \ne 0$ and $x \ne 0$
+{{qed|lemma}}
+Let $x > 0$.
+{{AimForCont}} $y < 0$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \times y
+      | o = <
+      | r = 0 \times 0
+      | c = [[Real Number Ordering is Compatible with Multiplication/Negative Factor|Real Number Ordering is Compatible with Multiplication: Negative Factor]]
+}}
+{{eqn | ll= \leadsto
+      | l = x \times y
+      | o = <
+      | r = 0
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+But [[Definition:By Hypothesis|by hypothesis]], $x \times y > 0$.
+Therefore by [[Proof by Contradiction]]:
+:$y > 0$
+{{qed|lemma}}
+Let $x < 0$.
+{{AimForCont}} $y > 0$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \times y
+      | o = <
+      | r = 0 \times 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Multiplication|multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = x \times y
+      | o = <
+      | r = 0
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+But [[Definition:By Hypothesis|by hypothesis]], $x \times y > 0$.
+Therefore by [[Proof by Contradiction]]:
+:$y > 0$
+{{qed|lemma}}
+Thus:
+:$x \times y > 0 \implies \paren {x > 0 \land y > 0} \lor \paren {x < 0 \land y < 0}$
+{{qed|lemma}}
+=== Necesssary Condition ===
+Let $x > 0$ and $y > 0$.
+Then from [[Product of Strictly Positive Real Numbers is Strictly Positive]]:
+:$x \times y > 0$
+{{qed|lemma}}
+Let $x < 0$ and $y < 0$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \times y
+      | o = >
+      | r = 0 \times 0
+      | c = [[Real Number Ordering is Compatible with Multiplication/Negative Factor|Real Number Ordering is Compatible with Multiplication: Negative Factor]]
+}}
+{{eqn | ll= \leadsto
+      | l = x \times y
+      | o = >
+      | r = 0
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Thus if either $x, y > 0$ or $x, y < 0$:
+:$x \times y > 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reciprocal of Strictly Positive Real Number is Strictly Positive}
+Tags: Reciprocals, Real Numbers, Inequalities
+
+\begin{theorem}
+:$\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$
+\end{theorem}
+
+\begin{proof}
+Let $x > 0$.
+{{AimForCont}} $\dfrac 1 x < 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \times \dfrac 1 x
+      | o = <
+      | r = 0 \times 0
+      | c = [[Real Number Ordering is Compatible with Multiplication/Negative Factor|Real Number Ordering is Compatible with Multiplication: Negative Factor]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = <
+      | r = 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text M 4$: Inverse]]
+}}
+{{end-eqn}}
+But from [[Real Zero is Less than Real One]]:
+:$1 > 0$
+Therefore by [[Proof by Contradiction]]:
+:$\dfrac 1 x > 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Strictly Positive Real Numbers is Dual of Order of their Reciprocals}
+Tags: Real Numbers
+
+\begin{theorem}
+:$\forall x, y \in \R: x > y > 0 \implies \dfrac 1 x < \dfrac 1 y$
+\end{theorem}
+
+\begin{proof}
+From [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]:
+:$(1): \quad x > 0 \implies \dfrac 1 x > 0$
+:$(2): \quad y > 0 \implies \dfrac 1 y > 0$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \times \frac 1 x
+      | o = >
+      | r = y \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R O2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Multiplication|multiplication]] and from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = x \times \frac 1 x \times \frac 1 y
+      | o = >
+      | r = y \times \frac 1 x \times \frac 1 y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R O2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Multiplication|multiplication]] and from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x \times \frac 1 x} \times \frac 1 y
+      | o = >
+      | r = \paren {y \times \frac 1 y} \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M1$ (Associativity) and $\R M2$ (Commutativity)]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1 \times \frac 1 y
+      | o = >
+      | r = 1 \times \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M4$ (Inverses)]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 y
+      | o = >
+      | r = \frac 1 x
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R M3$ (Identity)]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 x
+      | o = <
+      | r = \frac 1 y
+      | c = {{Defof|Dual Ordering}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mean of Unequal Real Numbers is Between them}
+Tags: Real Numbers
+
+\begin{theorem}
+:$\forall x, y \in \R: x < y \implies x < \dfrac {x + y} 2 < y$
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = <
+      | r = 1
+      | c = [[Real Zero is Less than Real One]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 + 0
+      | o = <
+      | r = 1 + 1
+      | c = [[Real Number Inequalities can be Added]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \frac 1 {1 + 1}
+      | c = [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \frac 1 2
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + x
+      | o = <
+      | r = x + y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R O1$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x + x} \times \frac 1 2
+      | o = <
+      | r = \paren {x + y} \times \frac 1 2
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R O2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Multiplication|multiplication]] and from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = <
+      | r = \frac {x + y} 2
+      | c = {{Defof|Real Division}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + y
+      | o = <
+      | r = y + y
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R O1$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Addition|addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x + y} \times \frac 1 2
+      | o = <
+      | r = \paren {y + y} \times \frac 1 2
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R O2$]]: [[Definition:Relation Compatible with Operation|compatibility]] with [[Definition:Real Multiplication|multiplication]] and from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x + y} 2
+      | o = <
+      | r = y
+      | c = {{Defof|Real Division}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Inductive Set as Subset of Real Numbers is Inductive Set}
+Tags: Inductive Sets
+
+\begin{theorem}
+Let $\AA$ be a [[Definition:Set of Sets|set]] of [[Definition:Inductive Set as Subset of Real Numbers|inductive sets defined as subsets of real numbers]].
+Then their [[Definition:Set Intersection|intersection]] is an [[Definition:Inductive Set as Subset of Real Numbers|inductive set]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Inductive Set as Subset of Real Numbers|inductive set]]:
+:$\forall S \in \AA: 1 \in S$
+Thus by definition of [[Definition:Intersection of Set of Sets|set intersection]]:
+:$\displaystyle 1 \in \bigcap_{S \mathop \in \AA} S$
+Also by definition of [[Definition:Inductive Set as Subset of Real Numbers|inductive set]]:
+:$\forall S \in \AA: x \in S \implies x + 1 \in S$
+So:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \bigcap_{S \mathop \in \AA} S
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall S \in \AA:
+      | l = x
+      | o = \in
+      | r = S
+      | c = {{Defof|Intersection of Set of Sets}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall S \in \AA:
+      | l = x + 1
+      | o = \in
+      | r = S
+      | c = {{Defof|Inductive Set as Subset of Real Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = x + 1
+      | o = \in
+      | r = \bigcap_{S \mathop \in \AA} S
+      | c = {{Defof|Intersection of Set of Sets}}
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Inductive Set as Subset of Real Numbers|inductive set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $n \in \N_{>0}$ be a non-zero [[Definition:Natural Number|natural number]].
+Let $\N^*_n$ denote the [[Definition:Initial Segment of One-Based Natural Numbers|Initial segment]] $\set {1, 2, \ldots, n}$ of the non-zero [[Definition:Natural Number|natural numbers]].
+Then every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N^*_n$ has a [[Definition:Greatest Element|greatest element]].
+\end{theorem}
+
+\begin{proof}
+The proof will proceed by the [[Principle of Mathematical Induction]] on $\N_{>0}$.
+Let $S$ be the [[Definition:Set|set]] defined as:
+:$S := \left\{ {n \in \N:}\right.$ Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N^*_n$ has a [[Definition:Greatest Element|greatest element]]$\left.{}\right\}$
+=== Basis for the Induction ===
+We have that:
+: $\N^*_1 = \set 1$
+The only [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]]  of $\set 1$ is $\set 1$ itself.
+This has a [[Definition:Greatest Element|greatest element]] $1$.
+So $1 \in S$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+It is to be shown that, if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.
+This is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N^*_k$ has a [[Definition:Greatest Element|greatest element]].
+It is to be demonstrated that it follows that:
+:Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N^*_{k + 1}$ has a [[Definition:Greatest Element|greatest element]].
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Let $T \subseteq \N^*_{k + 1}$.
+Either $k + 1 \in T$ or $k + 1 \notin T$.
+If $k + 1 \notin T$ then $T \subseteq \N^*_k$.
+In that case, then from the [[Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element#Induction Hypothesis|induction hypothesis]]:
+: $T$ has a [[Definition:Greatest Element|greatest element]].
+If $k + 1 \in T$ then:
+:$\forall x \in T: x \le k + 1$
+and so $k + 1$ is the [[Definition:Greatest Element|greatest element]] of $T$.
+So in both cases $T$ has a [[Definition:Greatest Element|greatest element]] and so $k + 1 \in S$.
+So $k \in S \implies k + 1 \in S$ and the result follows by the [[Principle of Mathematical Induction]]:
+:$\forall n \in \N$: every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N^*_n$ has a [[Definition:Greatest Element|greatest element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Not every Non-Empty Subset of Natural Numbers has Greatest Element}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $S \subseteq \N_{>0}$.
+Then, despite [[Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element]], it is not necessarily the case that $S$ has a [[Definition:Greatest Element|greatest element]].
+\end{theorem}
+
+\begin{proof}
+Let:
+:$S = \set {x \in \N_{>0}: x > 1}$
+Then $S \subseteq \N_{>0}$.
+{{AimForCont}} $S$ has a [[Definition:Greatest Element|greatest element]].
+Let the [[Definition:Greatest Element|greatest element]] of $S$ be $k$.
+But $\N_{>0}$ is an [[Definition:Inductive Set as Subset of Real Numbers|inductive set]].
+Therefore $k + 1 \in \N_{>0}$.
+By definition of $S$:
+:$k + 1 \in S$
+Therefore $k$ cannot be the [[Definition:Greatest Element|greatest element]] of $S$.
+By [[Proof by Contradiction]] it follows that $S$ has no [[Definition:Greatest Element|greatest element]].
+It is not possible to apply [[Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element]], because $S$ is not a subset of an [[Definition:Initial Segment of Natural Numbers|initial segment]] of $\N_{>0}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Integral on Union of Adjacent Intervals}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ where $a < b$.
+Let $c$ be a point in $\openint a b$.
+Then:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a c$ and $\closedint c b$
+{{iff}}:
+:$f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+We need to prove that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+To do this it suffices to show that for all $\epsilon > 0$, there exists a [[Definition:Subdivision (Real Analysis)|subdivision]] $S$ of $\closedint a b$ such that $\map U S – \map L S < \epsilon$.
+Here, $\map U S$ and $\map L S$ are, respectively, the [[Definition:Upper Sum|upper]] and [[Definition:Lower Sum|lower]] sums of $f$ on $\closedint a b$ with respect to the [[Definition:Subdivision (Real Analysis)|subdivision]] $S$.
+Let a [[Definition:Strictly Positive Real Number|strictly positive]] $\epsilon$ be given.
+Since $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a c$, we know that a [[Definition:Subdivision (Real Analysis)|subdivision]] $S_1$ of $\closedint a c$ exists such that $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$.
+Since $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint c b$, we know that a [[Definition:Subdivision (Real Analysis)|subdivision]] $S_2$ of $\closedint c b$ exists such that $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$.
+Define the [[Definition:Subdivision (Real Analysis)|subdivision]] $S = S_1 \cup S_2$.
+We observe that $S$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$.
+We get by the definition of [[Definition:Upper Sum|upper sum]]:
+{{begin-eqn}}
+{{eqn | l = \map U S
+      | r = \map U {S_1} + \map U {S_2}
+      | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent
+}}
+{{end-eqn}}
+Also, by the definition of [[Definition:Lower Sum|lower sum]]:
+{{begin-eqn}}
+{{eqn | l = \map L S
+      | r = \map L {S_1} + \map L {S_2}
+      | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent
+}}
+{{end-eqn}}
+These two equations give:
+{{begin-eqn}}
+{{eqn | l = \map U S – \map L S
+      | r = \map U {S_1} + \map U {S_2} – \paren {\map L {S_1} + \map L {S_2} }
+}}
+{{eqn | r = \map U {S_1} – \map L {S_1} + \map U {S_2} – \map L {S_2}
+}}
+{{eqn | o = <
+      | r = \dfrac \epsilon 2 + \dfrac \epsilon 2
+      | c = since $\map U {S_1} – \map L {S_1} < \dfrac \epsilon 2$ and $\map U {S_2} – \map L {S_2} < \dfrac \epsilon 2$
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+This shows that $S$ satisfies $\map U S – \map L S < \epsilon$.
+We conclude from this that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$ since $S$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+We need to prove that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a c$ and $\closedint c b$.
+Let a [[Definition:Strictly Positive Real Number|strictly positive]] $\epsilon$ be given.
+The [[Definition:Darboux Integrable Function|Darboux integrability]] of $f$ on $\closedint a b$ implies that a [[Definition:Subdivision (Real Analysis)|subdivision]] $P$ of $\closedint a b$ exists such that $\map U P – \map L P < \epsilon$.
+Here, $\map U P$ and $\map L P$ are, respectively, the [[Definition:Upper Sum|upper]] and [[Definition:Lower Sum|lower]] sums of $f$ on $\closedint a b$ with respect to the [[Definition:Subdivision (Real Analysis)|subdivision]] $P$.
+Define the [[Definition:Subdivision (Real Analysis)|subdivision]] $S = P \cup \set c$.
+We observe that $S$ equals $P$ if $c$ is a point in $P$, otherwise $S$ is a finer [[Definition:Subdivision (Real Analysis)|subdivision]] than $P$.
+We have
+{{begin-eqn}}
+{{eqn | l = \epsilon
+      | o = >
+      | r = \map U P – \map L P 
+}}
+{{eqn | o = \ge
+      | r = \map U S – \map L P 
+      | c = as $\map U S \le \map U P $ by the definition of [[Definition:Upper Sum|upper sum]] and the fact that $S$ refines $P$
+}}
+{{eqn | o = \ge
+      | r = \map U S – \map L S 
+      | c = as $\map L S \ge \map L P $ by the definition of [[Definition:Lower Sum|lower sum]] and the fact that $S$ refines $P$
+}}
+{{end-eqn}}
+This shows that $S$ satisfies $\map U S – \map L S < \epsilon$.
+Define:
+: $S_1 = S \cap \set {x: x \le c}$
+: $S_2 = S \cap \set {x: x \ge c}$
+We observe:
+:$S_1$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a c$.
+:$S_2$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint c b$.
+:$S_1$ and $S_2$ are adjacent.
+:The union of $S_1$ and $S_2$ equals $S$.
+We get by the definition of [[Definition:Upper Sum|upper sum]]:
+{{begin-eqn}}
+{{eqn | l = \map U S
+      | r = \map U {S_1} + \map U {S_2}
+      | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent
+}}
+{{end-eqn}}
+Also, by the definition of [[Definition:Lower Sum|lower sum]]:
+{{begin-eqn}}
+{{eqn | l = \map L S
+      | r = \map L {S_1} + \map L {S_2}
+      | c = since $S = S_1 \cup S_2$ and $S_1$ and $S_2$ are adjacent
+}}
+{{end-eqn}}
+We have
+{{begin-eqn}}
+{{eqn | l = \epsilon
+      | o = >
+      | r = \map U S – \map L S
+}}
+{{eqn | r = \map U {S_1} + \map U {S_2} – \paren {\map L {S_1} + \map L {S_2} }
+      | c = as $\map U S = \map U {S_1} + \map U {S_2}$ and $\map L S = \map L {S_1} + \map L {S_2}$
+}}
+{{eqn | r = \map U {S_1} – \map L {S_1} + \map U {S_2} – \map L {S_2} 
+}}
+{{eqn | o = \ge
+      | r = \map U {S_1} – \map L {S_1} 
+      | c = as $\map U {S_2} – \map L {S_2} \ge 0$ by the definitions of [[Definition:Upper Sum|upper]] and [[Definition:Lower Sum|lower]] sums
+}}
+{{end-eqn}}
+This shows that $S_1$ satisfies $\map U {S_1} – \map L {S_1} < \epsilon$.
+A similar deduction focusing on $S_2$ instead of $S_1$ shows that $S_2$ satisfies $\map U {S_2} – \map L {S_2} < \epsilon$.
+$\map U {S_1} – \map L {S_1} < \epsilon$ gives that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a c$ since $S_1$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a c$.
+$\map U {S_2} – \map L {S_2} < \epsilon$ gives that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint c b$ since $S_2$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint c b$.
+{{qed}}
+[[Category:Integral Calculus]]
+3eh9p5joywrep39apfhisgn7y3kozau
+\end{proof}<|endoftext|>
+\section{Cartesian Product of Bijections is Bijection}
+Tags: Bijections, Cartesian Product
+
+\begin{theorem}
+Let $S_1 \times S_2$ be the [[Definition:Cartesian Product|Cartesian product]] of two [[Definition:Set|sets]] $S_1$ and $S_2$.
+Let $T_1 \times T_2$ be the [[Definition:Cartesian Product|Cartesian product]] of two [[Definition:Set|sets]] $T_1$ and $T_2$.
+Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be [[Definition:Bijection|bijections]].
+Let $f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$ be the [[Definition:Cartesian Product of Relations|Cartesian product]] of $f_1$ and $f_2$ defined as:
+:$\forall \left({s_1, s_2}\right) \in S_1 \times S_2: f_1 \times f_2 \left({s_1, s_2}\right) := \left({f_1 \left({s_1}\right), f_2 \left({s_2}\right)}\right)$
+Then $f_1 \times f_2$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+Because $f_1$ and $f_2$ are both [[Definition:Bijection|bijections]], it follows by definition that they are both [[Definition:Surjection|surjections]].
+Let $\left({t_1, t_2}\right) \in T_1 \times T_2$.
+Then:
+:as $f_1$ is a [[Definition:Surjection|surjection]], $\exists s_1 \in S_1: f_1 \left({s_1}\right) = t_1$
+:as $f_2$ is a [[Definition:Surjection|surjection]], $\exists s_2 \in S_2: f_2 \left({s_2}\right) = t_2$
+Thus:
+:$\exists \left({s_1, s_2}\right) \in S_1 \times S_2: f_1 \times f_2 \left({s_1, s_2}\right) = \left({t_1, t_2}\right)$
+So $f_1 \times f_2$ is a [[Definition:Surjection|surjection]].
+Because $f_1$ and $f_2$ are both [[Definition:Bijection|bijections]], it follows by definition that they are both [[Definition:Injection|injections]].
+Let:
+: $f_1 \times f_2 \left({a_1, a_2}\right) = \left({c_1, c_2}\right), f_1 \times f_2 \left({b_1, b_2}\right) = \left({d_1, d_2}\right)$
+for some $\left({a_1, a_2}\right), \left({b_1, b_2}\right) \in S_1 \times S_2$.
+Suppose $\left({c_1, c_2}\right) = \left({d_1, d_2}\right)$.
+Then by [[Equality of Ordered Pairs]]:
+:$c_1 = d_1$
+:$c_2 = d_2$
+By definition of $f_1 \times f_2$:
+:$f_1 \left({a_1}\right) = c_1$
+:$f_1 \left({b_1}\right) = d_1$
+and:
+:$f_2 \left({a_2}\right) = c_2$
+:$f_2 \left({b_2}\right) = d_2$
+As $f_1$ is an [[Definition:Injection|injection]]:
+:$c_1 = d_1 \implies a_1 = b_1$
+and as $f_2$ is an [[Definition:Injection|injection]]:
+:$c_2 = d_2 \implies a_2 = b_2$
+Thus it follows that:
+:$\left({c_1, c_2}\right) = \left({d_1, d_2}\right) \implies \left({a_1, a_2}\right) = \left({b_1, b_2}\right)$
+and so $f_1 \times f_2$ is an [[Definition:Injection|injection]].
+So $f_1 \times f_2$ is a [[Definition:Surjection|surjection]] and also an [[Definition:Injection|injection]].
+Hence by definition, $f_1 \times f_2$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+[[Category:Bijections]]
+[[Category:Cartesian Product]]
+a5gd6052hnmbkcqn4kezns4qy9nt7ci
+\end{proof}<|endoftext|>
+\section{First Order ODE/y' - f (y) phi' (x) over f' (y) = phi (x) phi' (x) over f' (y)}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+Let $\map f y$ and $\map \phi x$ be known [[Definition:Real Function|real functions]] of $y$ and $x$ respectively.
+The [[Definition:General Solution of Differential Equation|general solution]] of:
+:$(1): \quad \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x = \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}$
+is:
+:$\map f y = C e^{\map \phi x} - \map \phi x - 1$
+\end{theorem}
+
+\begin{proof}
+Let $u = \map f y$
+Then by the [[Chain Rule for Derivatives]]:
+:$\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x
+      | r = \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f'} y \dfrac {\d y} {\d x} - \map f y \, \map {\phi'} x
+      | r = \map \phi x \, \map {\phi'} x
+      | c = multiplying $(1)$ by $\map {f'} y$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d u} {\d x} - u \, \map {\phi'} x
+      | r = \map \phi x \, \map {\phi'} x
+      | c = substituting $u$ for $\map f y$
+}}
+{{end-eqn}}
+This is a [[Definition:Linear First Order Ordinary Differential Equation|linear first order ordinary differential equation]] in the form:
+:$\dfrac {\d u} {\d x} + \map P x u = \map Q x$
+whose [[Definition:General Solution of Differential Equation|general solution]] from [[Solution to Linear First Order Ordinary Differential Equation]] is:
+:$\displaystyle u e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$
+In this instance:
+:$\map P x = -\map {\phi'} x$
+and:
+:$\map Q x = \map \phi x \, \map {\phi'} x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = -\int \map {\phi'} x \rd x
+      | c = 
+}}
+{{eqn | r = -\map \phi x + A
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$\displaystyle u e^{-\map \phi x + A} = \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x + C$
+Let $v = \map \phi x$.
+Then by [[Integration by Substitution]]:
+{{begin-eqn}}
+{{eqn | l = \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x
+      | r = \int v e^{-v + A} \rd v
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int v e^{-v} e^A \rd v
+      | c = 
+}}
+{{eqn | r = e^A \int v e^{-v} \rd v
+      | c = 
+}}
+{{eqn | r = - e^A e^{-v} \paren {v + 1} + C'
+      | c = [[Primitive of x by Exponential of a x|Primitive of $x e^{a x}$]]
+}}
+{{end-eqn}}
+This leaves us with:
+:$u e^{-\map \phi x + A} = - e^A e^{-v} \paren {v + 1} + C$
+subsuming $C'$ into $C$.
+Substituting back in, the [[Definition:General Solution of Differential Equation|general solution]] is seen to be:
+:$\map f y e^{-\map \phi x + A} = - e^{- \map \phi x + A} \paren {\map \phi x + 1} + C$
+and so multiplying both sides by $e^{\map \phi x - A}$:
+{{begin-eqn}}
+{{eqn | l = \map f y 
+      | r = \paren {- \map \phi x - 1} + C e^{\map \phi x - A}
+}}
+{{eqn | r = C e^{- A} e^{\map \phi x} - \map \phi x - 1 
+}}
+{{eqn | r = C e^{\map \phi x} - \map \phi x - 1 
+      | c = By absorbing the constant factor $e^{- A}$ into $C$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of All Mappings/Finite Sets}
+Tags: Cardinality of Set of All Mappings, Finite Sets
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
+The [[Definition:Cardinality|cardinality]] of the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$ (that is, the total number of [[Definition:Mapping|mappings]] from $S$ to $T$) is:
+:$\card {T^S} = \card T^{\card S}$
+\end{theorem}
+
+\begin{proof}
+Let $\card S = n$ and $\card T = m$.
+First suppose that $n = 0$, that is, that $S = \O$.
+From [[Cardinality of Set of All Mappings from Empty Set]]:
+:$\card {T^\O} = 1 = m^0$
+and the result is seen to hold for $n = 0$.
+Next, suppose that $m = 0$, that is, that $T = \O$.
+From [[Cardinality of Set of All Mappings to Empty Set]]:
+:$\card {\O^S} = \begin{cases}
+1 & : S = \O \\ 
+0 & : S \ne \O
+\end{cases}$
+So if $n > 0$:
+:$\card {\O^S} = 0 = 0^n$
+and if $n = 0$:
+:$\card {T^S} = 1 = 0^0 = m^n$
+and the result holds.
+This fits in with the preferred definition of the value of $0^0$.
+Finally, suppose $m > 0$ and $n > 0$.
+Let $\sigma: \N_n \to S$ and $\tau: T \to \N_n$ be [[Definition:Bijection|bijections]].
+Then the mapping $\Phi: T^S \to \paren {\N_m}^{\paren {\N_n} }$ defined as:
+:$\forall f \in T^S: \Phi \paren f = \tau \circ f \circ \sigma$
+(where $\paren {\N_m}^{\paren {\N_n} }$ is the [[Definition:Set of All Mappings|set of all mappings]] from $\N_n$ to $\N_m$) is also a [[Definition:Bijection|bijection]].
+So we need only consider the case where $S = \N_n$ and $T = \N_m$.
+Let $m \in \N_{>0}$.
+For each $n \in \N$, let $\Bbb T \paren {n, m}$ be the [[Definition:Set of All Mappings|set of all mappings]] from $\N_n$ to $\N_m$.
+Let:
+:$\Bbb S = \set {n \in \N: \card {\Bbb T \tuple {n, m} } = m^n}$
+We have seen that $0 \in \Bbb S$.
+Let $n \in \Bbb S$.
+Let $\rho: \Bbb T \tuple {n + 1, m} \to \Bbb T \paren {n, m}$ defined by:
+:$\forall f \in \Bbb T \tuple {n + 1, m}: \rho \paren f =$ the [[Definition:Restriction of Mapping|restriction]] of $f$ to $\N_n$
+Given that $g \in \Bbb T \tuple {n, m}$, and $k \in \N_m$, let $g_k: \N_{n + 1} \to \N_m$ be defined by:
+:$\forall x \in \N_{n + 1}: g_k \paren x = \begin{cases}
+g \paren x & : x \in \N_n \\
+k & : x = n
+\end{cases}$
+Then:
+:$\rho^{-1} \paren {\set g} = \set {g_0, \ldots, g_{m-1} }$
+Thus $\rho^{-1} \paren {\set g}$ has $m$ elements.
+So clearly:
+:$\set {\rho^{-1} \paren {\set g}: g \in \Bbb T \tuple {n, m} }$
+is a [[Definition:Partition (Set Theory)|partition]] of $\Bbb T \tuple {n + 1, m}$.
+Hence, as $n \in \Bbb S$, the set $\Bbb T \tuple {n + 1, m}$ has $m \cdot m^n = m^{n + 1}$ elements by [[Number of Elements in Partition]].
+Thus $n + 1 \in \Bbb S$.
+By the [[Principle of Mathematical Induction]]:
+:$\Bbb S = \N$
+and the proof is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of All Mappings/Infinite Sets}
+Tags: Cardinality of Set of All Mappings, Infinite Sets
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]] such that either $S$ or $T$ is [[Definition:Infinite Set|infinite]].
+The [[Definition:Cardinality|cardinality]] of the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$ (that is, the total number of [[Definition:Mapping|mappings]] from $S$ to $T$) is:
+:$\card {T^S} = \card T^{\card S}$
+\end{theorem}<|endoftext|>
+\section{Cardinality of Set Union/General Case}
+Tags: Cardinality of Set Union
+
+\begin{theorem}
+Let $S_1, S_2, \ldots$ be [[Definition:Set|sets]].
+Then:
+{{begin-eqn}}     
+{{eqn | l = \card {\bigcup_{i \mathop = 1}^n S_i}
+      | r = \sum_{i \mathop = 1}^n \card {S_i}
+      | c =
+}}
+{{eqn | o = -
+      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \card {S_i \cap S_j}
+      | c =
+}}
+{{eqn | o = +
+      | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} \card {S_i \cap S_j \cap S_k}
+      | c =
+}}
+{{eqn | o = \cdots
+      | c =
+}}
+{{eqn | o = +
+      | r = \paren {-1}^{n - 1} \card {\bigcap_{i \mathop = 1}^n S_i}
+      | c =
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By [[Cardinality is Additive Function]], we can apply the [[Inclusion-Exclusion Principle]]:
+If $f: \mathcal S \to \R$ is an [[Definition:Additive Function (Measure Theory)|additive function]], then:
+{{begin-eqn}}     
+{{eqn | l = f \paren {\bigcup_{i \mathop = 1}^n S_i}
+      | r = \sum_{i \mathop = 1}^n f \paren {S_i}
+      | c =
+}}
+{{eqn | o = -
+      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} f \paren {S_i \cap S_j}
+      | c =
+}}
+{{eqn | o = +
+      | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} f \paren {S_i \cap S_j \cap S_k}
+      | c =
+}}
+{{eqn | o = \cdots
+      | c =
+}}
+{{eqn | o = +
+      | r = \paren {-1}^{n - 1} f \paren {\bigcap_{i \mathop = 1}^n S_i}
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Structure of Inverse Completion of Commutative Semigroup}
+Tags: Inverse Completions, Commutative Semigroups
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
+Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\left({S, \circ}\right)$.
+Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$.
+Then:
+:$T = S \circ' C^{-1}$
+where:
+:$C^{-1}$ is the [[Definition:Inverse of Subset of Monoid|inverse]] of $C$ in $T$
+:$S \circ' C^{-1}$ is the [[Definition:Subset Product|subset product]] of $S$ with $C^{-1}$.
+\end{theorem}
+
+\begin{proof}
+Let $a \in C$.
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = S
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = x \circ \left({a \circ' a^{-1} }\right)
+      | c = Definition of [[Definition:Invertible Element|Invertible Element]]
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = \left({x \circ a}\right) \circ' a^{-1}
+      | c = Definition of [[Definition:Associative|Associative]]
+}}
+{{eqn | ll= \implies
+      | l = x
+      | o = \in
+      | r = S \circ' C^{-1}
+      | c = Definition of [[Definition:Subset Product|Subset Product]]
+}}
+{{eqn | ll= \implies
+      | l = S
+      | o = \subseteq
+      | r = S \circ' C^{-1}
+      | c = Definition of [[Definition:Subset|Subset]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = C
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | o = \in
+      | r = C^{-1}
+      | c = Definition of [[Definition:Inverse of Subset of Monoid|Inverse of Subset]]
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | r = a \circ' a^{-1} \circ' y^{-1}
+      | c = Definition of [[Definition:Invertible Element|Invertible Element]]
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | r = a \circ' \left({y \circ' a}\right)^{-1}
+      | c = [[Inverse of Product]]
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | r = a \circ' \left({y \circ a}\right)^{-1}
+      | c = [[Definition:Extension of Operation|$\circ'$ extends $\circ$]]
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | o = \in
+      | r = S \circ' C^{-1}
+      | c = Definition of [[Definition:Subset Product|Subset Product]]
+}}
+{{eqn | ll= \implies
+      | l = C^{-1}
+      | o = \subseteq
+      | r = S \circ' C^{-1}
+      | c = Definition of [[Definition:Subset|Subset]]
+}}
+{{end-eqn}}
+Thus, as:
+: $C^{-1} \subseteq S \circ' C^{-1}$
+and:
+:$S \subseteq S \circ' C^{-1}$
+by [[Union is Smallest Superset]] it follows that:
+:$S \cup C^{-1} \subseteq S \circ' C^{-1}$
+From [[Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup]]:
+:$S \circ' C^{-1}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+So $S \circ' C^{-1}$ is a [[Definition:Semigroup|semigroup]] which contains $S \cup C^{-1}$.
+By definition of [[Definition:Generator of Semigroup|generator of semigroup]], it follows that:
+:$\left\langle{S \cup C^{-1} }\right\rangle \subseteq S \circ' C^{-1}$
+Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.
+Then by definition:
+:$x \in S$
+and:
+:$y^{-1} \in C^{-1}$
+and so by definition of [[Definition:Generator of Semigroup|generator of semigroup]]:
+:$x \circ' y^{-1} \in \left\langle{S \cup C^{-1} }\right\rangle$
+Thus:
+$S \circ' C^{-1} \subseteq \left\langle{S \cup C^{-1} }\right\rangle$
+By definition of [[Definition:Set Equality|set equality]]:
+:$S \circ' C^{-1} = \left\langle{S \cup C^{-1} }\right\rangle$
+and so by definition of the [[Definition:Inverse Completion|inverse completion]]:
+:$T = S \circ' C^{-1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse in Monoid is Unique}
+Tags: Monoids, Inverse Elements
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]].
+Then an [[Definition:Element|element]] $x \in S$ can have at most one [[Definition:Inverse Element|inverse]] for $\circ$.
+\end{theorem}
+
+\begin{proof}
+Let $e$ be the [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$.
+Suppose $x \in S$ has two [[Definition:Inverse Element|inverses]]: $y$ and $z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = y \circ e
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = y \circ \paren {x \circ z}
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | r = \paren {y \circ x} \circ z
+      | c = {{Defof|Associative Operation}}
+}}
+{{eqn | r = e \circ z
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | r = z
+      | c = {{Defof|Identity Element}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = e \circ y
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = \paren {z \circ x} \circ y
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | r = z \circ \paren {x \circ y}
+      | c = {{Defof|Associative Operation}}
+}}
+{{eqn | r = z \circ e
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | r = z
+      | c = {{Defof|Identity Element}}
+}}
+{{end-eqn}}
+So whichever way round you do it, $y = z$ and the [[Definition:Inverse Element|inverse]] of $x$ is [[Definition:Unique|unique]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Inverse/Monoid}
+Tags: Monoids, Inverse Elements
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]].
+Let $x \in S$ be [[Definition:Invertible Element|invertible]], and let its [[Definition:Inverse Element|inverse]] be $x^{-1}$.
+Then $x^{-1}$ is also [[Definition:Invertible Element|invertible]], and:
+:$\paren {x^{-1} }^{-1} = x$
+\end{theorem}
+
+\begin{proof}
+By [[Inverse in Monoid is Unique]], any [[Definition:Inverse Element|inverse]] of $x$ is [[Definition:Unique|unique]], and can be denoted $x^{-1}$.
+From [[Inverse of Inverse in General Algebraic Structure]]:
+:$x^{-1}$ is [[Definition:Invertible Element|invertible]] and its [[Definition:Inverse Element|inverse]] is $x$.
+That is:
+:$\paren {x^{-1} }^{-1} = x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Product/Monoid}
+Tags: Monoids, Inverse Elements, Inverse of Product
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $a, b \in S$ be [[Definition:Invertible Element|invertible]] for $\circ$, with [[Definition:Inverse Element|inverses]] $a^{-1}, b^{-1}$.
+Then $a \circ b$ is [[Definition:Invertible Element|invertible]] for $\circ$, and:
+: $\left({a \circ b}\right)^{-1} = b^{-1} \circ a^{-1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left({a \circ b}\right) \circ \left({b^{-1} \circ a^{-1} }\right)
+      | r = \left({\left({a \circ b}\right) \circ b^{-1} }\right) \circ a^{-1}
+      | c = [[Definition:Associative Operation|Associativity]]
+}}
+{{eqn | r = \left({a \circ \left({b \circ b^{-1} }\right)}\right) \circ a^{-1}
+      | c = [[Definition:Associative Operation|Associativity]]
+}}
+{{eqn | r = \left({a \circ e}\right) \circ a^{-1}
+      | c = Behaviour of [[Definition:Inverse Element|Inverse]]
+}}
+{{eqn | r = a \circ a^{-1}
+      | c = Behaviour of [[Definition:Identity Element|Identity]]
+}}
+{{eqn | r = e
+      | c = Behaviour of [[Definition:Inverse Element|Inverse]]
+}}
+{{end-eqn}}
+Similarly for $\left({b^{-1} \circ a^{-1}}\right) \circ \left({a \circ b}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Right Identity Element is Idempotent}
+Tags: Identity Elements, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $e_R \in S$ be a [[Definition:Right Identity|right identity]] with respect to $\circ$.
+Then $e_R$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Right Identity|right identity]]:
+:$\forall x \in S: x \circ e_R = x$
+Thus in particular:
+:$e_R \circ e_R = e_R$
+Therefore $e_R$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left Identity Element is Idempotent}
+Tags: Identity Elements, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $e_L \in S$ be a [[Definition:Left Identity|left identity]] with respect to $\circ$.
+Then $e_L$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Left Identity|left identity]]:
+:$\forall x \in S: e_L \circ x = x$
+Thus in particular:
+: $e_L \circ e_L = e_L$
+Therefore $e_L$ is [[Definition:Idempotent Element|idempotent]] under $\circ$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{More than one Right Identity then no Left Identity}
+Tags: Identity Elements
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]].
+If $\left({S, \circ}\right)$ has more than one [[Definition:Right Identity|right identity]], then it has no [[Definition:Left Identity|left identity]].
+\end{theorem}
+
+\begin{proof}
+Let $\left({S, \circ}\right)$ be an [[Definition:Algebraic Structure|algebraic structure]] with more than one [[Definition:Right Identity|right identity]].
+Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$.
+Suppose $\left({S, \circ}\right)$ has a [[Definition:Left Identity|left identity]].
+Call this [[Definition:Left Identity|left identity]] $e_L$.
+Then, by the behaviour of $e_L$, $e_{R_1}$ and $e_{R_2}$:
+: $e_{R_1} = e_{R_1} \circ e_L = e_L$
+: $e_{R_2} = e_{R_2} \circ e_L = e_L$
+So $e_{R_1} = e_L = e_{R_2}$, which contradicts the supposition that $e_{R_1}$ and $e_{R_2}$ are different.
+Therefore, in an algebraic structure with more than one [[Definition:Right Identity|right identity]], there can be no [[Definition:Left Identity|left identity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Completion is Commutative Monoid}
+Tags: Inverse Completions
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
+Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\left({S, \circ}\right)$.
+Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$.
+Then $\left({T, \circ'}\right)$ is a [[Definition:Commutative Monoid|commutative monoid]].
+\end{theorem}
+
+\begin{proof}
+From [[Inverse Completion is Commutative Semigroup]]:
+:$\left({T, \circ'}\right)$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+By definition of [[Definition:Inverse Completion|inverse completion]]:
+:$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$
+Thus by definition of [[Definition:Inverse Element|inverse element]]:
+:$e := x \circ x^{-1}$ is an [[Definition:Identity Element|identity element]] of $T$.
+By [[Identity is Unique]], $e$ is the only such [[Definition:Element|element]] of $T$.
+Thus $\left({T, \circ'}\right)$ is a [[Definition:Semigroup|semigroup]] with an [[Definition:Identity Element|identity element]].
+So, by definition, $\left({T, \circ'}\right)$ is a [[Definition:Commutative Monoid|commutative monoid]].
+{{qed}}
+[[Category:Inverse Completions]]
+92t9hf9kxhakzgw1zv7vnoqrizel2ts
+\end{proof}<|endoftext|>
+\section{Inverse Completion is Commutative Semigroup}
+Tags: Inverse Completions
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
+Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\left({S, \circ}\right)$.
+Let $\left({T, \circ'}\right)$ be an [[Definition:Inverse Completion|inverse completion]] of $\left({S, \circ}\right)$.
+Then $T = S \circ' C^{-1}$, and is a [[Definition:Commutative Semigroup|commutative semigroup]].
+\end{theorem}
+
+\begin{proof}
+From [[Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup]]:
+:$S \circ' C^{-1}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+From [[Structure of Inverse Completion of Commutative Semigroup]]:
+:$T = S \circ' C^{-1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complement of G-Delta Set is F-Sigma Set}
+Tags: F-Sigma Sets, G-Delta Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $X$ be an [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
+Then its [[Definition:Relative Complement|complement]] $S \setminus X$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $X$ be a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
+Let $X = \displaystyle \bigcap \UU$ where $\UU$ is a [[Definition:Countable Set|countable]] [[Definition:Set Intersection|intersection]] of [[Definition:Open Set (Topology)|open sets]] in $T$.
+Then from [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]] we have:
+:$\displaystyle S \setminus X = S \setminus \bigcap \UU = \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$
+By definition of [[Definition:Closed Set (Topology)|closed set]], each of the $S \setminus U$ are [[Definition:Closed Set (Topology)|closed sets]].
+So $\displaystyle \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$ is a [[Definition:Countable Set|countable]] [[Definition:Set Union|union]] of [[Definition:Closed Set (Topology)|closed sets]] in $T$.
+Hence $S \setminus X$ is, by definition, an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bounded Piecewise Continuous Function has Improper Integrals}
+Tags: Piecewise Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, $a < b$. 
+Let $f$ be [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
+Then $f$ is a [[Definition:Piecewise Continuous Function with Improper Integrals|piecewise continuous function with improper integrals]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]] on $\closedint a b$.
+It is sufficient to prove that the [[Definition:Improper Integral over Open Interval|improper integral]] $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
+Let $i \in \set {1, 2, \ldots, n}$.
+Let $c$ be a point in $\openint {x_{i − 1} } {x_i}$.
+By definition, the [[Definition:Improper Integral over Open Interval|improper integral]] $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists {{iff}}:
+:$\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$
+and:
+: $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$
+both exist.
+By [[Bounded Piecewise Continuous Function is Darboux Integrable]] we know that $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+Therefore, $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on every [[Definition:Closed Real Interval|closed subinterval]] of $\closedint a b$. 
+Accordingly, the following [[Definition:Darboux Integral|Darboux integrals]]:
+:$\displaystyle \int_{x_{i - 1} }^c \map f x \rd x$
+:$\displaystyle \int_{\gamma_1}^c \map f x \rd x$
+:$\displaystyle \int_c^{\gamma_2} \map f x \rd x$
+:$\displaystyle \int_c^{x_i} \map f x \rd x$
+all exist where $\gamma_1 \in \openint {x_{i − 1} } c$ and $\gamma_2 \in \openint c {x_i}$.
+Note that $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$ as $f$ is [[Definition:Bounded Piecewise Continuous Function|piecewise continuous and bounded]].
+Therefore, a [[Definition:Bound of Real-Valued Function|bound]] $B$ exists for $f$ on $\closedint a b$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \size {\int_{\gamma_1}^c \map f x \rd x - \int_{x_{i - 1} }^c \map f x \rd x}
+      | r = \size {\int_{x_{i - 1} }^{\gamma_1} \map f x \rd x}
+}}
+{{eqn | o = \le
+      | r = B \size {\gamma_1 - x_{i - 1} }
+      | c = [[Upper and Lower Bounds of Integral/Corollary|Corollary to Upper and Lower Bounds of Integral]]
+}}
+{{end-eqn}}
+which approaches $0$ as $\gamma_1$ approaches $x_{i − 1}$.
+This shows that $\displaystyle \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c \map f x \rd x$ equals $\displaystyle \int_{x_{i - 1} }^c \map f x \rd x$.
+From this we gather that $\displaystyle \lim_{\gamma \mathop \to  {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ exists.
+Also:
+{{begin-eqn}}
+{{eqn | l = \size {\int_c^{x_i} \map f x \rd x - \int_c^{\gamma_2} \map f x \rd x}
+      | r = \size {\int_{\gamma_2}^{x_i} \map f x \rd x}
+}}
+{{eqn | o = \le
+      | r = B \size {x_i - \gamma_2}
+      | c = [[Upper and Lower Bounds of Integral/Corollary|Corollary to Upper and Lower Bounds of Integral]]
+}}
+{{end-eqn}}
+which approaches $0$ as $\gamma_2$ approaches $x_i$.
+This shows that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma \map f x \rd x$ equals $\displaystyle \int_c^{x_i} \map f x \rd x$.
+From this we gather that $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exists.
+Since $i$ is arbitrary, we have shown that $\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ and $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exist for every $i \in \left\{{1, \ldots, n}\right\}$.
+Therefore, $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$.
+That is, $f$ is a [[Definition:Piecewise Continuous Function with Improper Integrals|piecewise continuous function with improper integrals]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{F-Sigma Set is not necessarily Closed Set}
+Tags: F-Sigma Sets, Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $X$ be an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
+Then it is not necessarily the case that $X$ is a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $X \subseteq S$ be an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
+From [[F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space|$F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space]]:
+:$X$ is neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]] in $T$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Not every Closed Set is G-Delta Set}
+Tags: G-Delta Sets, Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+Then it is not necessarily the case that $V$ is a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+From [[Closed Set of Uncountable Finite Complement Topology is not G-Delta|Closed Set of Uncountable Finite Complement Topology is not $G_\delta$]]:
+:$V$ is not a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Set of Uncountable Finite Complement Topology is not F-Sigma}
+Tags: F-Sigma Sets, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
+Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+Then $U$ is not an [[Definition:F-Sigma Set|$F_\sigma$ set]].
+\end{theorem}
+
+\begin{proof}
+Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+As $S$ is [[Definition:Uncountable Set|uncountable]], then so is $U$.
+By the definition of a [[Definition:Finite Complement Topology|finite complement topology]], all [[Definition:Closed Set (Topology)|closed sets]] of $T$ are [[Definition:Finite Set|finite]].
+From [[Countable Union of Countable Sets is Countable]], $U$ can not be expressed as the [[Definition:Set Union|union]] of a [[Definition:Countable Set|countable]] number of [[Definition:Closed Set (Topology)|closed sets]].
+So by definition $U$ is not an [[Definition:F-Sigma Set|$F_\sigma$ set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Set of Uncountable Finite Complement Topology is not G-Delta}
+Tags: G-Delta Sets, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
+Let $V \in \tau$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+Then $V$ is not a [[Definition:G-Delta Set|$G_\delta$ set]].
+\end{theorem}
+
+\begin{proof}
+Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+{{AimForCont}} $V$ is [[Definition:G-Delta Set|$G_\delta$ set]].
+Then by [[Complement of G-Delta Set is F-Sigma Set|Complement of $G_\delta$ Set is $F_\sigma$ Set]]:
+:$S \setminus V$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]].
+By definition of [[Definition:Closed Set (Topology)|closed set]], $S \setminus V$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+But by [[Open Set of Uncountable Finite Complement Topology is not F-Sigma|Open Set of Uncountable Finite Complement Topology is not $F_\sigma$]]:
+:$S \setminus V$ is not an [[Definition:F-Sigma Set|$F_\sigma$ set]].
+It follows by [[Proof by Contradiction]] that $V$ is not a [[Definition:G-Delta Set|$G_\delta$ set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{G-Delta Set is not necessarily Open Set}
+Tags: G-Delta Sets, Open Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $X$ be a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
+Then it is not necessarily the case that $X$ is a [[Definition:Open Set (Topology)|open set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $X \subseteq S$ be a [[Definition:G-Delta Set|$G_\delta$ set]] of $T$.
+From [[F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space|$F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space]]:
+:$X$ is neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]] in $T$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Not every Open Set is F-Sigma Set}
+Tags: F-Sigma Sets, Open Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $V$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+Then it is not necessarily the case that $V$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+From [[Open Set of Uncountable Finite Complement Topology is not F-Sigma|Closed Set of Uncountable Finite Complement Topology is not $F_\sigma$]]:
+:$U$ is not an [[Definition:F-Sigma Set|$F_\sigma$ set]] of $T$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Point of Sequence may only be Adherent Point of Range}
+Tags: Limit Points, Adherent Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $A$.
+Let $\alpha$ be a [[Definition:Limit Point of Sequence|limit point of $\sequence {x_n}$]].
+Then $\alpha$ may be only an [[Definition:Adherent Point|adherent point]] of $A$ and not a [[Definition:Limit Point of Set|limit point]] of $A$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$.
+Let $x \in S$.
+Then by definition of [[Definition:Discrete Space|discrete space]]:
+:$U = \set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+Consider the [[Definition:Sequence|sequence]] $\sequence {x_n}$ defined as:
+:$\forall n \in \N: x_n = x$
+That is:
+:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
+From [[Limit Point of Sequence in Discrete Space not always Limit Point of Open Set]]:
+:$x$ is not a [[Definition:Limit Point of Set|limit point]] of $U$.
+But from [[Limit Point of Sequence is Adherent Point of Range]]:
+:$x$ is an [[Definition:Adherent Point|adherent point]] of $\set x$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Point of Sequence in Discrete Space not always Limit Point of Open Set}
+Tags: Discrete Topology, Limit Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $U$.
+Let $x$ be the [[Definition:Limit Point of Sequence|limit point]] of $\sequence {x_n}$.
+Then $x$ is not always a [[Definition:Limit Point of Set|limit point]] of $U$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+By definition of [[Definition:Discrete Space|discrete space]]:
+:$U = \set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+Consider the [[Definition:Sequence|sequence]] $\sequence {x_n}$ defined as:
+:$\forall n \in \N: x_n = x$
+That is:
+:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
+Thus $x$ is the [[Definition:Limit Point of Sequence|limit point]] of $\sequence {x_n}$.
+But:
+:$U \setminus \set x = \O$
+and so $x$ is not a [[Definition:Limit Point of Set|limit point]] of $U$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range}
+Tags: Accumulation Points, Omega-Accumulation Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $S$.
+Let $\alpha$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$.
+Then $\alpha$ is also an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $\set {x_n: n \in \N}$.
+\end{theorem}
+
+\begin{proof}
+Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$ containing $\alpha$.
+By definition of [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$, $U$ contains [[Definition:Infinite Set|infinitely many]] [[Definition:Term of Sequence|terms]] of $\sequence {x_n}$.
+As $\sequence {x_n}$ is a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]]:
+$U$ contains [[Definition:Infinite Set|infinitely many]] [[Definition:Element|elements]] of $\set {x_n: n \in \N}$.
+Thus by definition $\alpha$ is an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $\set {x_n: n \in \N}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Point of Sequence is Adherent Point of Range}
+Tags: Limit Points, Adherent Points
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $\left \langle {x_n} \right \rangle$ be a [[Definition:Sequence|sequence]] in $S$.
+Let $\alpha$ be a [[Definition:Limit Point of Sequence|limit point of $\left \langle {x_n} \right \rangle$]].
+Then $\alpha$ is an [[Definition:Adherent Point|adherent point]] of $\left\{ {x_n: n \in \N}\right\}$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\left\{ {x_n: n \in \N}\right\}$ contains all the [[Definition:Term of Sequence|terms]] of $\left \langle {x_n} \right \rangle$.
+In particular, $\left\{ {x_n: n \in \N}\right\}$ contains $\alpha$.
+Thus by definition $\alpha$ is an [[Definition:Adherent Point|adherent point]] of $\left\{ {x_n: n \in \N}\right\}$.
+{{qed}}
+[[Category:Limit Points]]
+[[Category:Adherent Points]]
+ceqq0wwplri2wabadwh8wexrnj5lzps
+\end{proof}<|endoftext|>
+\section{Limit Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range}
+Tags: Limit Points, Omega-Accumulation Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $S$.
+Let $\alpha$ be a [[Definition:Limit Point of Set|limit point]] of $\sequence {x_n}$.
+Then $\alpha$ is also an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $\set {x_n: n \in \N}$.
+\end{theorem}
+
+\begin{proof}
+Let $\alpha$ be an [[Definition:Limit Point of Set|limit point]] of $\sequence {x_n}$.
+From [[Limit Point of Sequence is Accumulation Point]], $\alpha$ is an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {x_n}$.
+From [[Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range]], $\alpha$ is an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $\set {x_n: n \in \N}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Isolated Point}
+Tags: Isolated Points
+
+\begin{theorem}
+{{TFAE|def = Isolated Point of Subset|view = isolated point}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $x$ be an [[Definition:Isolated Point (Topology)/Subset/Definition 1|isolated point of $H$ by definition 1]].
+Then by definition:
+:$\exists U \in \tau: U \cap H = \set x$
+Thus we have an [[Definition:Open Set (Topology)|open set]] in $T$ such that $x \in U$ contains no other point of $H$ than $x$.
+Thus, by definition, $x$ is not a [[Definition:Limit Point of Set|limit point of $H$]].
+Thus $x$ is an [[Definition:Isolated Point (Topology)/Subset/Definition 2|isolated point of $H$ by definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $x$ be an [[Definition:Isolated Point (Topology)/Subset/Definition 2|isolated point of $H$ by definition 2]].
+{{AimForCont}} $x$ is a [[Definition:Limit Point of Set|limit point of $H$]].
+Then by definition every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
+That is:
+:$\forall U \in \tau: x \in U \implies \exists y \in S, y \ne x: y \in U \cap H$
+That is:
+:$\not \exists U \in \tau: U \cap H = \set x$
+because all $U$ with $x$ in them are such that there is at least one point in $U \cap H$ apart from $x$.
+Thus by [[Proof by Contradiction]] $x$ is not a [[Definition:Limit Point of Set|limit point of $H$]].
+That is, $x$ is an [[Definition:Isolated Point (Topology)/Subset/Definition 1|isolated point of $H$ by definition 1]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Closed Set}
+Tags: Closed Sets, Limit Points
+
+\begin{theorem}
+{{TFAE|def = Closed Set (Topology)|view = Closed Set|context = Topology (Mathematical Branch)|contextview = topology}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+=== Definition 1 implies Definition 2 ===
+Let $H$ be a [[Definition:Closed Set/Topology/Definition 1|closed set of $T$ by definition 1]].
+Let $H^\complement$ denote the [[Definition:Relative Complement|relative complement]] of $H$ in $S$.
+By definition of [[Definition:Closed Set (Topology)|closed set]] in $T$:
+:$H^\complement$ is [[Definition:Open Set (Topology)|open]] in $T$
+From [[Set is Open iff Neighborhood of all its Points]]:
+:$\forall x \in S: x \notin H \implies H^\complement$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+By definition of [[Definition:Limit Point of Set/Definition 4|limit point]]:
+:$\forall x \in S: x \notin H \implies x$ is not a [[Definition:Limit Point of Set|limit point]] of $H$
+Thus $H$ is a [[Definition:Closed Set/Topology/Definition 2|closed set of $T$ by definition 2]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $H$ be a [[Definition:Closed Set/Topology/Definition 2|closed set of $T$ by definition 2]].
+Then by definition:
+$\forall x \in S: x \notin H \implies x$ is not a [[Definition:Limit Point of Set|limit point]] of $H$
+By definition of [[Definition:Limit Point of Set/Definition 4|limit point]]:
+$\forall x \in S: x \notin H \implies H^\complement$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$
+By [[Set is Open iff Neighborhood of all its Points]]:
+$H^\complement$ is [[Definition:Open Set (Topology)|open]] in $T$
+Thus $H$ is a [[Definition:Closed Set/Topology/Definition 1|closed set of $T$ by definition 1]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem}
+Tags: Set Closures, Relative Complement, Kuratowski's Closure-Complement Problem
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $T$.
+By successive applications of the [[Definition:Unary Operation|operations]] of [[Definition:Relative Complement|complement relative to $S$]] and the [[Definition:Closure (Topology)|closure]], there can be as many as $14$ [[Definition:Distinct|distinct]] [[Definition:Subset|subsets]] of $S$ (including $A$ itself).
+\end{theorem}
+
+\begin{proof}
+That there can be as many as $14$ will be demonstrated by example.
+=== [[Kuratowski's Closure-Complement Problem/Complement|Complement]] ===
+{{:Kuratowski's Closure-Complement Problem/Complement}}
+=== [[Kuratowski's Closure-Complement Problem/Interior|Interior]] ===
+{{:Kuratowski's Closure-Complement Problem/Interior}}
+=== [[Kuratowski's Closure-Complement Problem/Closure|Closure]] ===
+{{:Kuratowski's Closure-Complement Problem/Closure}}
+=== [[Kuratowski's Closure-Complement Problem/Exterior|Exterior]] ===
+{{:Kuratowski's Closure-Complement Problem/Exterior}}
+=== [[Kuratowski's Closure-Complement Problem/Closure of Complement|Closure of Complement]] ===
+{{:Kuratowski's Closure-Complement Problem/Closure of Complement}}
+=== [[Kuratowski's Closure-Complement Problem/Closure of Interior|Closure of Interior]] ===
+{{:Kuratowski's Closure-Complement Problem/Closure of Interior}}
+=== [[Kuratowski's Closure-Complement Problem/Interior of Closure|Interior of Closure]] ===
+{{:Kuratowski's Closure-Complement Problem/Interior of Closure}}
+=== [[Kuratowski's Closure-Complement Problem/Interior of Closure of Interior|Interior of Closure of Interior]] ===
+{{:Kuratowski's Closure-Complement Problem/Interior of Closure of Interior}}
+=== [[Kuratowski's Closure-Complement Problem/Interior of Complement of Interior|Interior of Complement of Interior]] ===
+{{:Kuratowski's Closure-Complement Problem/Interior of Complement of Interior}}
+=== [[Kuratowski's Closure-Complement Problem/Closure of Interior of Complement|Closure of Interior of Complement]] ===
+{{:Kuratowski's Closure-Complement Problem/Closure of Interior of Complement}}
+=== [[Kuratowski's Closure-Complement Problem/Closure of Interior of Closure|Closure of Interior of Closure]] ===
+{{:Kuratowski's Closure-Complement Problem/Closure of Interior of Closure}}
+=== [[Kuratowski's Closure-Complement Problem/Interior of Complement of Interior of Closure|Interior of Complement of Interior of Closure]] ===
+{{:Kuratowski's Closure-Complement Problem/Interior of Complement of Interior of Closure}}
+=== [[Kuratowski's Closure-Complement Problem/Complement of Interior of Closure of Interior|Complement of Interior of Closure of Interior]] ===
+{{:Kuratowski's Closure-Complement Problem/Complement of Interior of Closure of Interior}}
+=== [[Kuratowski's Closure-Complement Problem/Proof of Maximum|Proof of Maximum]] ===
+It remains to be shown that there can be no more than $14$.
+{{:Kuratowski's Closure-Complement Problem/Proof of Maximum}}
+\end{proof}
+
+\begin{proof}
+Consider an arbitrary [[Definition:Subset|subset]] $A$ of a [[Definition:Topological Space|topological space]] $T = \left({S, \tau}\right)$.
+To simplify the presentation:
+:let $a$ be used to denote the [[Definition:Unary Operation|operation]] of taking the [[Definition:Relative Complement|complement of $A$ relative to $S$]]: $a \left({A}\right) = S \setminus A$
+:let $b$ be used to denote the [[Definition:Unary Operation|operation]] of taking the [[Definition:Topological Closure|closure of $A$ in $T$]]: $b \left({A}\right) = A^-$
+:let $I$ be used to denote the [[Definition:Identity Mapping|identity operation]] on $A$, that is: $I \left({A}\right) = A$.
+:let the [[Definition:Parenthesis|parentheses]] and the reference to $A$ be removed, so as to present, for example:
+::$a \left({b \left({a \left({A}\right)}\right)}\right)$
+:as:
+::$a b a$
+From [[Relative Complement of Relative Complement]]:
+:$a \left({a \left({A}\right)}\right) = A$
+or, using the compact notation defined above:
+:$(1): \quad a a = I$
+and from [[Closure of Topological Closure equals Closure]]:
+:$b \left({b \left({A}\right)}\right) = b \left({A}\right) = A^-$
+or, using the compact notation defined above:
+:$(2): \quad b b = b$
+Let $s$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Element|elements]] of $\left\{{a, b}\right\}$.
+By successive applications of $(1)$ and $(2)$, it is possible to eliminate all multiple consecutive instances of $a$ and $b$ in $s$, and so reduce $s$ to one of the following forms:
+:$\text{a)}: \quad a b a b \ldots a$
+:$\text{b)}: \quad b a b a \ldots a$
+:$\text{c)}: \quad a b a b \ldots b$
+:$\text{d)}: \quad b a b a \ldots b$
+From [[Closure of Complement of Closure is Regular Closed]]:
+:$b a b$ is [[Definition:Regular Closed Set|regular closed]].
+By [[Interior equals Complement of Closure of Complement]], the [[Definition:Interior (Topology)|interior]] of $A$ is:
+:$a b a$
+Recall the definition of [[Definition:Regular Closed Set|regular closed]]:
+:a set $A$ is [[Definition:Regular Closed Set|regular closed]] {{iff}} it equals the [[Definition:Topological Closure|closure]] of its [[Definition:Interior (Topology)|interior]].
+And so as $b a b$ is [[Definition:Regular Closed Set|regular closed]]:
+:$b a b = b a b a \left({b a b}\right)$
+So, adding an extra $b$ to either of $a b a b a b a$ or $b a b a b a$ will generate a string containing $b a b a b a b$ which can be reduced immediately to $b a b$.
+It follows that the possible different [[Definition:Subset|subsets]] of $S$ that can be obtained from $A$ by applying $a$ and $b$ can be generated by none other than:
+:$I$
+:$a$
+:$a b$
+:$a b a$
+:$a b a b$
+:$a b a b a$
+:$a b a b a b$
+:$a b a b a b a$
+:$b$
+:$b a$
+:$b a b$
+:$b a b a$
+:$b a b a b$
+:$b a b a b a$
+... a total of $14$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Complement}
+Tags: Kuratowski's Closure-Complement Problem
+
+\begin{theorem}
+The [[Definition:Relative Complement|complement of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A'
+      | r = \left({\gets \,.\,.\, 0}\right]
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left\{ {1} \right\}
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{2 \,.\,.\, 3}\right)
+      | c = {{Defof|Half-Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left({3 \,.\,.\, 4}\right]
+      | c = ... adjacent to [[Definition:Half-Open Real Interval|Half-Open Real Interval]]
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left({\R \setminus \Q \cap \left[{4 \,.\,.\, 5}\right]}\right)
+      | c = [[Definition:Irrational Number|Irrational Numbers]] from $4$ to $5$
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{5 \,.\,.\, \to}\right)
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-Comp.png|500px]]
+\end{theorem}
+
+\begin{proof}
+For ease of analysis, let:
+:$A_1 := \left({0 \,.\,.\, 1}\right)$
+:$A_2 := \left({1 \,.\,.\, 2}\right)$
+:$A_3 := \left\{ {3} \right\}$
+:$A_4 := \left({\Q \cap \left({4 \,.\,.\, 5}\right)}\right)$
+Thus:
+:$\displaystyle A = \bigcup_{i \mathop = 1}^4 A_i$
+By [[De Morgan's Laws (Set Theory)/Relative Complement/Family of Sets/Complement of Intersection|De Morgan's Laws]]:
+:$\displaystyle A' := \R \setminus A = \bigcap_{i \mathop = 1}^4 \left({\R \setminus A_i}\right)$
+{{begin-eqn}}
+{{eqn | l = \R \setminus A_1
+      | r = \left({\gets \,.\,.\, 0}\right] \cup \left[{1 \,.\,.\, \to}\right)
+      | c = 
+}}
+{{eqn | l = \R \setminus A_2
+      | r = \left({\gets \,.\,.\, 1}\right] \cup \left[{2 \,.\,.\, \to}\right)
+      | c = 
+}}
+{{eqn | l = \R \setminus A_3
+      | r = \left({\gets \,.\,.\, 3}\right) \cup \left({3 \,.\,.\, \to}\right)
+      | c = 
+}}
+{{eqn | l = \R \setminus A_4
+      | r = \left({\gets \,.\,.\, 4}\right] \cup \R \setminus \Q \cup \left[{5 \,.\,.\, \to}\right)
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws]]
+}}
+{{end-eqn}}
+from which the result follows by inspection.
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+d9oxzyl9w658n6mgvygx6400yaq9p0r
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Interior}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors
+
+\begin{theorem}
+The [[Definition:Interior (Topology)|interior of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^\circ
+      | r = \left({0 \,.\,.\, 1}\right) \cup \left({1 \,.\,.\, 2}\right)
+      | c = [[Definition:Union of Adjacent Open Intervals|Union of Adjacent Open Intervals]]
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-Int.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Interior equals Complement of Closure of Complement]]:
+:$A^\circ = A^{\prime \, - \, \prime}$
+From [[Kuratowski's Closure-Complement Problem/Closure of Complement|Kuratowski's Closure-Complement Problem: Closure of Complement]]:
+{{begin-eqn}}
+{{eqn | l = A^{\prime \, -}
+      | r = \left({\gets \,.\,.\, 0}\right]
+      | c = [[Definition:Unbounded Closed Real Interval|Unbounded Closed Real Interval]]
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left\{ {1} \right\}
+      | c = [[Definition:Singleton|Singleton]]
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{2 \,.\,.\, \to}\right)
+      | c = [[Definition:Unbounded Closed Real Interval|Unbounded Closed Real Interval]]
+}}
+{{end-eqn}}
+It can be determined by inspection that:
+:$A^{\prime \, - \, \prime} = \left({0 \,.\,.\, 1}\right) \cup \left({1 \,.\,.\, 2}\right)$
+Hence the result.
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+aohordjb80shikn1mcq1ohtd7rzl2q4
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Closure}
+Tags: Kuratowski's Closure-Complement Problem, Set Closures
+
+\begin{theorem}
+The [[Definition:Closure (Topology)|closure of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^-
+      | r = \closedint 0 2
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 3
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 4 5
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-Clos.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Closure of Union of Adjacent Open Intervals]]:
+:$\paren {\openint 0 1 \cup \openint 1 2}^- = \closedint 0 2$
+From [[Real Number is Closed in Real Number Line]]:
+:$\set 3$ is [[Definition:Closed Set (Topology)|closed]] in $\R$
+From [[Set is Closed iff Equals Topological Closure]]:
+:$\set 3^- = \set 3$
+From [[Closure of Rational Interval is Closed Real Interval]]:
+:$\paren {\Q \cap \openint 4 5 }^- = \closedint 4 5$
+The result follows from [[Closure of Finite Union equals Union of Closures]].
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Closures]]
+l9svs3wen8l6wzv0kj3p5en5vi0iw2t
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Exterior}
+Tags: Kuratowski's Closure-Complement Problem, Set Exteriors
+
+\begin{theorem}
+The [[Definition:Exterior (Topology)|exterior of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^e
+      | r = \openint \gets 0
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 2 3 \cup \openint 3 4
+      | c = {{Defof|Union of Adjacent Open Intervals}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 5 \to
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-Ext.png|500px]]
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Exterior (Topology)|exterior of $A$ in $\R$]] can be defined either as:
+:the [[Definition:Set Complement|complement]] of the [[Definition:Closure (Topology)|closure]] of $A$ in $\R$: $A^{- \, \prime}$
+or as:
+:the [[Definition:Interior (Topology)|interior]] of the [[Definition:Set Complement|complement]] of $A$ in $\R$: $A^{\prime \, \circ}$
+From [[Kuratowski's Closure-Complement Problem/Closure|Kuratowski's Closure-Complement Problem: Closure]]:
+{{begin-eqn}}
+{{eqn | l = A^-
+      | r = \closedint 0 2
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 3
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 4 5
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+It follows by inspection that:
+{{begin-eqn}}
+{{eqn | l = A^e = A^{- \, \prime}
+      | r = \openint \gets 0
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 2 3 \cup \openint 3 4
+      | c = {{Defof|Union of Adjacent Open Intervals}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 5 \to
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Exteriors]]
+6l2oj9k014ngv3qrsqeg832zewp7oal
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Closure of Complement}
+Tags: Kuratowski's Closure-Complement Problem
+
+\begin{theorem}
+The [[Definition:Closure (Topology)|closure]] of the [[Definition:Relative Complement|complement of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{\prime \, -}
+      | r = \hointl \gets 0
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 1
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointr 2 \to
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-ClosComp.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Complement|Kuratowski's Closure-Complement Problem: Complement]]:
+{{begin-eqn}}
+{{eqn | l = A'
+      | r = \hointl \gets 0
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 1
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointr 2 3
+      | c = {{Defof|Half-Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointl 3 4
+      | c = adjacent to [[Definition:Half-Open Real Interval|Half-Open Real Interval]]
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \paren {\R \setminus \Q \cap \closedint 4 5}
+      | c = [[Definition:Irrational Number|Irrational Numbers]] from $4$ to $5$
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointr 5 \to
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+From [[Real Number is Closed in Real Number Line]]:
+:$\set 3$ is [[Definition:Closed Set (Topology)|closed]] in $\R$
+and:
+:$\hointl \gets 0$ is [[Definition:Closed Set (Topology)|closed]] in $\R$
+and:
+:$\hointr 5 \to$ is [[Definition:Closed Set (Topology)|closed]] in $\R$
+Then from [[Set is Closed iff Equals Topological Closure]]:
+:$\set 3^- = \set 3$
+:$\hointl \gets 0^- = \hointl \gets 0$
+:$\hointr 5 \to^- = \hointr 5 \to$
+From [[Closure of Half-Open Real Interval is Closed Real Interval]]:
+:$\hointr 2 3 = \closedint 2 3$
+and:
+:$\hointl 3 4 = \closedint 3 4$
+From [[Closure of Irrational Interval is Closed Real Interval]]:
+:$\paren {\R \setminus \Q \cap \closedint 4 5}^- = \closedint 4 5$
+From [[Closure of Finite Union equals Union of Closures]]:
+{{begin-eqn}}
+{{eqn | l = A^{\prime \, -}
+      | r = \hointl \gets 0
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 1
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 2 3
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 3 4
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 4 5
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointr 5 \to
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+5j3uqggmkqu6tdln6ihka6gej6chaz3
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Closure of Interior}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors, Set Closures
+
+\begin{theorem}
+The [[Definition:Closure (Topology)|closure]] of the [[Definition:Interior (Topology)|interior of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{\circ \, -}
+      | r = \closedint 0 2
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-ClosInt.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Interior|Kuratowski's Closure-Complement Problem: Interior]]:
+:$A^\circ = \openint 0 1 \cup \openint 1 2$
+From [[Closure of Union of Adjacent Open Intervals]]:
+:$A^{\circ \, -} = \closedint 0 2$
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+[[Category:Set Closures]]
+1tlrjvqjiw9pxy8edfnepwds3ks1j02
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Interior of Closure}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors, Set Closures
+
+\begin{theorem}
+The [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{- \, \circ}
+      | r = \openint 0 2
+      | c = {{Defof|Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 4 5
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-IntClos.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Closure|Kuratowski's Closure-Complement Problem: Closure]]:
+{{begin-eqn}}
+{{eqn | l = A^-
+      | r = \closedint 0 2
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 3
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 4 5
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+From [[Interior of Closed Real Interval is Open Real Interval]]:
+:$\closedint 0 2^\circ = \openint 0 2$
+and:
+:$\closedint 4 5^\circ = \openint 4 5$
+From [[Interior of Singleton in Real Number Line is Empty]]:
+:$\set 3^\circ = \O$
+{{Finish|Get past this: [[Interior of Union is not necessarily Union of Interiors]]}}
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+[[Category:Set Closures]]
+ewc4ug707iz5wmjr2f1zooy69ikydy6
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Interior of Closure of Interior}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors, Set Closures
+
+\begin{theorem}
+The [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure]] of the [[Definition:Interior (Topology)|interior of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{\circ \, - \, \circ}
+      | r = \left({0 \,.\,.\, 2}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-IntClosInt.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Closure of Interior|Kuratowski's Closure-Complement Problem: Closure of Interior]]:
+{{begin-eqn}}
+{{eqn | l = A^{\circ \, -}
+      | r = \left[{0 \,.\,.\, 2}\right]
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+From [[Interior of Closed Real Interval is Open Real Interval]]:
+:$\left[{0 \,.\,.\, 2}\right]^\circ = \left({0 \,.\,.\, 2}\right)$
+Hence the result.
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+[[Category:Set Closures]]
+46s5m5z3ka0ul2mo5rjbcwvhbilhqig
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Interior of Complement of Interior}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors, Set Complement
+
+\begin{theorem}
+The [[Definition:Interior (Topology)|interior]] of the [[Definition:Relative Complement|complement]] of the [[Definition:Interior (Topology)|interior of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{\circ \, \prime \, \circ}
+      | r = \openint \gets 0
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 2 \to
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-IntCompInt.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Complement of Interior equals Closure of Complement]]:
+:$A^{\circ \, \prime} = A^{\prime \, -}$
+From [[Kuratowski's Closure-Complement Problem/Closure of Complement|Kuratowski's Closure-Complement Problem: Closure of Complement]]:
+{{begin-eqn}}
+{{eqn | l = A^{\prime \, -}
+      | r = \hointl \gets 0
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \set 1
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointr 2 \to
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+From [[Interior of Closed Real Interval is Open Real Interval]]:
+:$\hointl \gets 0^\circ = \openint \gets 0$
+and:
+:$\hointr 2 \to^\circ = \openint 2 \to$
+From [[Interior of Singleton in Real Number Line is Empty]]:
+:$\set 1^\circ = \O$
+{{Finish|Get past this: [[Interior of Union is not necessarily Union of Interiors]]}}
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+[[Category:Set Complement]]
+c2o9wneuuzcfipk8cgw6lb7l4ry38bx
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Closure of Interior of Complement}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors, Set Closures, Set Complement
+
+\begin{theorem}
+The [[Definition:Closure (Topology)|closure]] of the [[Definition:Interior (Topology)|interior]] of the [[Definition:Relative Complement|complement of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{\prime \, \circ \, -}
+      | r = \hointl \gets 0
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \closedint 2 4
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \hointr 5 \to
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-ClosIntComp.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Exterior|Kuratowski's Closure-Complement Problem: Exterior]]:
+{{begin-eqn}}
+{{eqn | l = A^e
+      | r = \openint \gets 0
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 2 3 \cup \openint 3 4
+      | c = {{Defof|Union of Adjacent Open Intervals}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \openint 5 \to
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{end-eqn}}
+From [[Closure of Open Real Interval is Closed Real Interval]]:
+:$\openint \gets 0^- = \hointl \gets 0$
+and:
+:$\openint 5 \to^- = \hointr 5 \to$
+From [[Closure of Union of Adjacent Open Intervals]]:
+:$\paren {\openint 2 3 \cup \openint 3 4}^- = \closedint 2 4$
+The result follows from [[Closure of Finite Union equals Union of Closures]].
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+[[Category:Set Closures]]
+[[Category:Set Complement]]
+1eih3f697wm4pv5pbxpfiymynq5qazp
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Closure of Interior of Closure}
+Tags: Kuratowski's Closure-Complement Problem, Set Interiors, Set Closures
+
+\begin{theorem}
+The [[Definition:Closure (Topology)|closure]] of the [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{- \, \circ \, -}
+      | r = \left[{0 \,.\,.\, 2}\right]
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{4 \,.\,.\, 5}\right]
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-ClosIntClos.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Interior of Closure|Kuratowski's Closure-Complement Problem: Interior of Closure]]:
+{{begin-eqn}}
+{{eqn | l = A^{- \, \circ}
+      | r = \left({0 \,.\,.\, 2}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left({4 \,.\,.\, 5}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+From [[Closure of Open Real Interval is Closed Real Interval]]:
+:$\left({0 \,.\,.\, 2}\right)^- = \left[{0 \,.\,.\, 2}\right]$
+and:
+:$\left({4 \,.\,.\, 5}\right)^- = \left[{4 \,.\,.\, 5}\right]$
+The result follows from [[Closure of Finite Union equals Union of Closures]].
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+[[Category:Set Interiors]]
+[[Category:Set Closures]]
+8m8djt5kb04phfl9rc9qu0tpwfb0ehx
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Interior of Complement of Interior of Closure}
+Tags: Kuratowski's Closure-Complement Problem
+
+\begin{theorem}
+The [[Definition:Interior (Topology)|interior]] of the [[Definition:Relative Complement|complement]] of the [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{- \, \circ \, \prime \, \circ}
+      | r = \left({\gets \,.\,.\, 0}\right)
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left({2 \,.\,.\, 4}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left({5 \,.\,.\, \to}\right)
+      | c = {{Defof|Unbounded Open Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-IntCompIntClos.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Interior of Closure|Kuratowski's Closure-Complement Problem: Interior of Closure]]:
+{{begin-eqn}}
+{{eqn | l = A^{- \, \circ}
+      | r = \left({0 \,.\,.\, 2}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left({4 \,.\,.\, 5}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+By inspection:
+{{begin-eqn}}
+{{eqn | l = A^{- \, \circ \, '}
+      | r = \left({\gets \,.\,.\, 0}\right]
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{2 \,.\,.\, 4}\right]
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{5 \,.\,.\, \to}\right)
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+From [[Interior of Closed Real Interval is Open Real Interval]]:
+:$\left({\gets \,.\,.\, 0}\right]^\circ = \left({\gets \,.\,.\, 0}\right)$
+and:
+:$\left[{2 \,.\,.\, 4}\right]^\circ = \left({2 \,.\,.\, 4}\right)$
+and:
+:$\left[{5 \,.\,.\, \to}\right)^\circ = \left({5 \,.\,.\, \to}\right)$
+{{Finish|Get past this: [[Interior of Union is not necessarily Union of Interiors]]}}
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+r98z9ozwhyltndvcnakqeb8bxhaa8d6
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Complement of Interior of Closure of Interior}
+Tags: Kuratowski's Closure-Complement Problem
+
+\begin{theorem}
+The [[Definition:Relative Complement|complement]] of the [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure]] of the [[Definition:Interior (Topology)|interior of $A$ in $\R$]] is given by:
+{{begin-eqn}}
+{{eqn | l = A^{\circ \, - \, \circ \, \prime}
+      | r = \left({\gets \,.\,.\, 0}\right]
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{eqn | o =
+      | ro= \cup
+      | r = \left[{2 \,.\,.\, \to}\right)
+      | c = {{Defof|Unbounded Closed Real Interval}}
+}}
+{{end-eqn}}
+:[[File:Kuratowski-Closure-Complement-Theorem-CompIntClosInt.png|500px]]
+\end{theorem}
+
+\begin{proof}
+From [[Kuratowski's Closure-Complement Problem/Interior of Closure of Interior|Kuratowski's Closure-Complement Problem: Interior of Closure of Interior]]:
+{{begin-eqn}}
+{{eqn | l = A^{\circ \, - \, \circ}
+      | r = \left({0 \,.\,.\, 2}\right)
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+The result follows by inspection.
+{{qed}}
+[[Category:Kuratowski's Closure-Complement Problem]]
+4paw14owtms8gr8v0mcwb31g4ih419q
+\end{proof}<|endoftext|>
+\section{Sum of Integrals on Adjacent Intervals for Integrable Functions}
+Tags: Definite Integrals, Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Darboux Integrable Function|Darboux integrable]] on any [[Definition:Closed Real Interval|closed interval]] $\mathbb I$.
+Let $a, b, c \in \mathbb I$.
+Then:
+:$\displaystyle \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, assume $a < b$.
+First let $a < c < b$.
+Let $P_1$ and $P_2$ be any [[Definition:Subdivision (Real Analysis)|subdivisions]] of $\closedint a c$ and $\closedint c b$ respectively.
+Then $P = P_1 \cup P_2$ is a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$.
+From the definitions of [[Definition:Upper Sum|upper sum]] and [[Definition:Lower Sum|lower sum]]:
+:$\map L {P_1} + \map L {P_2} = \map L P$
+:$\map U {P_1} + \map U {P_2} = \map U P$
+We consider the [[Definition:Lower Sum|lower sum]].
+The same conclusion can be obtained by investigating the [[Definition:Upper Sum|upper sum]].
+By definition of [[Definition:Darboux Integral|Darboux integral]]:
+:$\displaystyle \map L P \le \int_a^b \map f t \rd t$
+Thus, given the [[Definition:Subdivision (Real Analysis)|subdivisions]] $P_1$ and $P_2$, we have:
+:$\displaystyle \map L {P_1} + \map L {P_2} \le \int_a^b \map f t \rd t$
+and so:
+:$\displaystyle \map L {P_1} \le \int_a^b \map f t \rd t - \map L {P_2}$
+So, for any [[Definition:Subdivision (Real Analysis)|subdivision]] $P_2$ of $\closedint c b$, $\displaystyle \int_a^b \map f t \rd t - \map L {P_2}$ is an [[Definition:Upper Bound of Mapping|upper bound]] of $\map L {P_1}$.
+Thus:
+:$\displaystyle \map {\sup_{P_1} } {\map L {P_1} } \le \int_a^b \map f t \rd t - \map L {P_2}$
+where $\map {\sup_{P_1} } {\map L {P_1} }$ ranges over all [[Definition:Subdivision (Real Analysis)|subdivisions]] of $P_1$.
+Thus by definition of [[Definition:Definite Integral|definite integral]]:
+:$\displaystyle \int_a^c \map f t \rd t \le \int_a^b \map f t \rd t - \map L {P_2}$
+and so:
+:$\displaystyle \map L {P_2} \le \int_a^b \map f t \rd t - \int_a^c \map f t \rd t$
+Similarly, we find that:
+:$\displaystyle \int_c^b \map f t \rd t \le \int_a^b \map f t \rd t - \int_a^c \map f t \rd t$
+Therefore:
+:$\displaystyle \int_a^b \map f t \rd t \ge \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$
+Having established the above, we now need to put it into context.
+Let $P$ be ''any'' [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$, which may or may not include the point $c$.
+Let $Q = P \cup \set c$ be the subdivision of $\closedint a b$ obtained from $P$ by including with it, if necessary, the point $c$.
+It is easy to show that $\map L P \le \map L Q$.
+Let $Q_1$ be the [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$ which includes only the points of $Q$ that lie in $\closedint a c$.
+Let $Q_2$ be the [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$ which includes only the points of $Q$ that lie in $\closedint c b$.
+From the definition of [[Definition:Lower Sum|lower sum]]:
+:$\map L Q = \map L {Q_1} + \map L {Q_2}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map L P
+      | o = \le
+      | r = \map L Q
+      | c = 
+}}
+{{eqn | r = \map L {Q_1} + \map L {Q_2}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \int_a^c \map f t \rd t + \int_c^b \map f t \rd t
+      | c = 
+}}
+{{end-eqn}}
+So $\displaystyle \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$ is an [[Definition:Upper Bound of Mapping|upper bound]] for $\map L P$, where $P$ is any subdivision of $\closedint a b$.
+Thus:
+:$\displaystyle \map {\sup_P} {\map L P} \le \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$
+Thus, by [[Definition:Definite Integral|definition]]:
+:$\displaystyle \int_a^b \map f t \rd t \le \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$
+Combining this with the result:
+:$\displaystyle \int_a^b \map f t \rd t \ge \int_a^c \map f t \rd t + \int_c^b \map f t \rd t$
+the result follows.
+{{qed|lemma}}
+{{Explain|"It is easy to show that $\map L P \le \map L Q$". This needs a separate theorem, but really requires the definitions for "finer" and "coarser" mesh, which leads us off into areas of topology that I need to lay some more groundwork for.}}
+Now suppose $a < b < c$.
+Then from the definition of [[Definition:Darboux Integral|Darboux integral]]:
+:$\displaystyle \int_c^b \map f t \rd t := -\int_b^c \map f t \rd t$
+and it follows that:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \map f t \rd t
+      | r = \int_a^c \map f t \rd t - \int_b^c \map f t \rd t 
+      | c = main result
+}}
+{{eqn | r = \int_a^c \map f t \rd t + \int_c^b \map f t \rd t
+      | c = 
+}}
+{{end-eqn}}
+The case of $c < a < b$ is proved similarly.
+Finally, suppose $a = c < b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \map f t \rd t
+      | r = 0 + \int_c^b \map f t \rd t + \int_a^c \map f t \rd t
+      | c = as $a = c$
+}}
+{{eqn | r = \int_a^c \map f t \rd t + \int_c^b \map f t \rd t
+      | c = [[Integral on Zero Interval]], as $a = c$
+}}
+{{end-eqn}}
+The case of $a < c = b$ is proved similarly.
+Hence the result, from [[Proof by Cases]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Irrational Numbers is Real Numbers}
+Tags: Irrational Number Space, Real Number Line with Euclidean Topology, Set Closures
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Let $\struct {\R \setminus \Q, \tau_d}$ be the [[Definition:Irrational Number Space|irrational number space]] under the [[Definition:Euclidean Topology on Real Number Line|same topology]].
+Then:
+:$\paren {\R \setminus \Q}^- = \R$
+where $\paren {\R \setminus \Q}-$ denotes the [[Definition:Closure (Topology)|closure]] of $\R \setminus \Q$.
+\end{theorem}
+
+\begin{proof}
+From [[Irrationals are Everywhere Dense in Topological Space of Reals]], $\R \setminus \Q$ is [[Definition:Everywhere Dense|everywhere dense]] in $\R$.
+It follows by definition of [[Definition:Everywhere Dense|everywhere dense]] that $\paren {\R \setminus \Q}^- = \R$.
+{{qed}}
+[[Category:Irrational Number Space]]
+[[Category:Real Number Line with Euclidean Topology]]
+[[Category:Set Closures]]
+43wj1draqpd0swslfbj7dprl6gi9vm8
+\end{proof}<|endoftext|>
+\section{Closure of Union of Adjacent Open Intervals}
+Tags: Set Closures, Union of Adjacent Open Intervals
+
+\begin{theorem}
+Let $a, b, c \in R$ where $a < b < c$.
+Let $A$ be the [[Definition:Union of Adjacent Open Intervals|union of the two adjacent open intervals]]:
+:$A := \left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)$
+Then:
+:$A^- = \left[{a \,.\,.\, c}\right]$
+where:
+: $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = A^-
+      | r = \left({\left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)}\right)^-
+      | c = 
+}}
+{{eqn | r = \left({a \,.\,.\, b}\right)^- \cup \left({b \,.\,.\, c}\right)^-
+      | c = [[Closure of Finite Union equals Union of Closures]]
+}}
+{{eqn | r = \left[{a \,.\,.\, b}\right] \cup \left[{b \,.\,.\, c}\right]
+      | c = [[Closure of Open Ball in Metric Space]]
+}}
+{{eqn | r = \left[{a \,.\,.\, c}\right]
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Set Closures]]
+[[Category:Union of Adjacent Open Intervals]]
+p1ts9xym8w2qx2gim0b0f2bqkl6dw7t
+\end{proof}<|endoftext|>
+\section{Interior equals Complement of Closure of Complement}
+Tags: Set Closures, Set Interiors
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ and $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of $H$.
+Let $H^\prime$ denote the [[Definition:Relative Complement|complement of $H$ in $T$]]:
+:$H^\prime = T \setminus H$
+Then:
+:$H^\circ = H^{\prime \, - \, \prime}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = H^{\circ \, \prime}
+      | r = H^{\prime \, -}
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {H^{\circ \, \prime} }^\prime
+      | r = \paren {H^{\prime \, -} }^\prime
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = H^{\circ \, \prime \, \prime}
+      | r = H^{\prime \, - \, \prime}
+      | c = [[Composition of Mappings is Associative]]
+}}
+{{eqn | ll= \leadsto
+      | l = H^\circ
+      | r = H^{\prime \, - \, \prime}
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Set Closures]]
+[[Category:Set Interiors]]
+289qxln45jqaqop7asiju08vtc3a97r
+\end{proof}<|endoftext|>
+\section{Closure of Half-Open Real Interval is Closed Real Interval}
+Tags: Real Intervals, Set Closures
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+ 
+Let $H_1 = \hointl a b$ and $H_2 = \hointr a b$ be [[Definition:Half-Open Real Interval|half-open intervals]] of $\R$.
+Then the [[Definition:Closure (Topology)|closure]] of both $H_1$ and $H_2$ in $\R$ are the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+For an arbitrary $H \subseteq \R$, let $H^-$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $\R$.
+By the definition of [[Definition:Closure (Topology)/Definition 3|closure]]:
+:$H^-$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Closed Set (Topology)|closed set]] of $\struct {\R, \tau_d}$ containing $H$ as a [[Definition:Subset|subset]].
+Let $A := \openint a b$ be the [[Definition:Open Real Interval|open interval]] between $a$ and $b$.
+By definition of $H_1$ and $H_2$:
+:$A \subseteq H_1$
+and:
+:$A \subseteq H_2$
+From [[Closure of Open Real Interval is Closed Real Interval]]:
+:$\openint a b^- = \closedint a b$
+Thus as $\closedint a b$ is the [[Definition:Closure (Topology)|closure]] of $\openint a b$, it follows that:
+:$\closedint a b$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Closed Set (Topology)|closed set]] of $\struct {\R, \tau_d}$ containing $\openint a b$.
+But $\closedint a b$ also contains as [[Definition:Subset|subsets]] both $H_1$ and $H_2$.
+Thus we have:
+:$A \subseteq H_1 \subseteq \closedint a b$
+and:
+:$A \subseteq H_2 \subseteq \closedint a b$
+Thus it follows that $\closedint a b$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Closed Set (Topology)|closed set]] of $\struct {\R, \tau_d}$ containing $H_1$ and $H_2$.
+Hence the result.
+{{qed}}
+[[Category:Real Intervals]]
+[[Category:Set Closures]]
+e2fcgq4qmbpjesvpx91oj0d7xf7w5jz
+\end{proof}<|endoftext|>
+\section{Closure of Irrational Interval is Closed Real Interval}
+Tags: Real Intervals, Irrational Number Space, Set Closures
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number|real numbers]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $\struct{\R \setminus \Q, \tau_d}$ be the [[Definition:Irrational Number Space|irrational number space]] under the [[Definition:Euclidean Topology on Real Number Line|same topology]].
+Let $a, b \in \R$ such that $a < b$.
+Let $\Bbb I \subseteq \R$ be an [[Definition:Real Interval|interval]] of $\R$
+Then the [[Definition:Closure (Topology)|closure]] of the set:
+:$\Bbb I \cap \paren {\R \setminus \Q}$
+is the [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+Let $\Bbb I$ be an [[Definition:Open Real Interval|open real interval]].
+From [[Closure of Real Interval is Closed Real Interval]]:
+:$\Bbb I^- = \closedint a b$
+From [[Closure of Irrational Numbers is Real Numbers]]:
+:$\paren {\R \setminus \Q}^- = \R$
+From [[Closure of Intersection is Subset of Intersection of Closures]]:
+:$\paren{\Bbb I \cap \paren{\R \setminus \Q}}^- \subseteq \Bbb I^- \cap \paren {\R \setminus \Q}^-$
+From [[Intersection with Subset is Subset]]:
+:$\Bbb I^- \cap \paren {\R \setminus \Q}^- = \closedint a b$
+and so:
+:$\paren {\Bbb I \cap \paren {\R \setminus \Q} }^- \subseteq \closedint a b$
+From [[Irrationals are Everywhere Dense in Topological Space of Reals]]:
+:$\closedint a b \subseteq \paren {\Bbb I \cap \paren {\R \setminus \Q} }^-$
+and the result follows.
+{{qed}}
+[[Category:Real Intervals]]
+[[Category:Irrational Number Space]]
+[[Category:Set Closures]]
+2sgds1491k5cm9vkp7ei81psyz1rljc
+\end{proof}<|endoftext|>
+\section{Closure of Rational Interval is Closed Real Interval}
+Tags: Real Intervals, Irrational Number Space
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number|real numbers]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]] under the [[Definition:Euclidean Topology on Real Number Line|same topology]].
+Let $a, b \in \R$ such that $a < b$.
+Let $\Bbb I \subseteq \R$ be an [[Definition:Real Interval|interval]] of $\R$
+Then the [[Definition:Closure (Topology)|closure]] of the set :
+:$\Bbb I \cap \Q$
+is the [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+Let $\Bbb I$ be an [[Definition:Open Real Interval|open real interval]].
+From [[Closure of Real Interval is Closed Real Interval]]:
+:$\Bbb I^- = \closedint a b$
+From [[Closure of Rational Numbers is Real Numbers]]:
+:$\Q^- = \R$
+From [[Closure of Intersection is Subset of Intersection of Closures]]:
+:$\paren{\Bbb I \cap \Q}^- \subseteq \Bbb I^- \cap \Q^-$
+From [[Intersection with Subset is Subset]]:
+:$\Bbb I^- \cap \Q^- = \closedint a b$
+and so:
+:$\paren {\Bbb I \cap \Q}^- \subseteq \closedint a b$
+From [[Rationals are Everywhere Dense in Topological Space of Reals]]:
+:$\closedint a b \subseteq \paren {\Bbb I \cap \Q}^-$
+and the result follows.
+{{qed}}
+[[Category:Real Intervals]]
+[[Category:Irrational Number Space]]
+m64xrdzlhmu195l3hgeur2klv6xxdnb
+\end{proof}<|endoftext|>
+\section{Interior of Singleton in Real Number Line is Empty}
+Tags: Set Interiors, Singletons, Real Number Line with Euclidean Topology
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Let $a \in \R$ be a [[Definition:Real Number|real number]].
+Then:
+:$\set a^\circ = \O$
+where $\set a^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $\set a$ in $\R$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \set a^\circ
+      | r = \closedint a a^\circ
+      | c = {{Defof|Closed Real Interval}}
+}}
+{{eqn | r = \openint a a
+      | c = [[Interior of Closed Real Interval is Open Real Interval]]
+}}
+{{eqn | r = \set {x \in \R: a < x < a}
+      | c = {{Defof|Open Real Interval}}
+}}
+{{eqn | r = \O
+      | c = {{Defof|Empty Real Interval}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Set Interiors]]
+[[Category:Singletons]]
+[[Category:Real Number Line with Euclidean Topology]]
+9eq8s0dt9sqtlvbodatp8avtdc6tknh
+\end{proof}<|endoftext|>
+\section{Kuratowski's Closure-Complement Problem/Proof of Maximum}
+Tags: Set Closures, Relative Complement, Kuratowski's Closure-Complement Problem
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $T$.
+By successive applications of the [[Definition:Unary Operation|operations]] of [[Definition:Relative Complement|complement relative to $S$]] and the [[Definition:Topological Closure|closure]], there can be no more than $14$ [[Definition:Distinct|distinct]] [[Definition:Subset|subsets]] of $S$ (including $A$ itself).
+\end{theorem}
+
+\begin{proof}
+Consider an arbitrary [[Definition:Subset|subset]] $A$ of a [[Definition:Topological Space|topological space]] $T = \left({S, \tau}\right)$.
+To simplify the presentation:
+:let $a$ be used to denote the [[Definition:Unary Operation|operation]] of taking the [[Definition:Relative Complement|complement of $A$ relative to $S$]]: $a \left({A}\right) = S \setminus A$
+:let $b$ be used to denote the [[Definition:Unary Operation|operation]] of taking the [[Definition:Topological Closure|closure of $A$ in $T$]]: $b \left({A}\right) = A^-$
+:let $I$ be used to denote the [[Definition:Identity Mapping|identity operation]] on $A$, that is: $I \left({A}\right) = A$.
+:let the [[Definition:Parenthesis|parentheses]] and the reference to $A$ be removed, so as to present, for example:
+::$a \left({b \left({a \left({A}\right)}\right)}\right)$
+:as:
+::$a b a$
+From [[Relative Complement of Relative Complement]]:
+:$a \left({a \left({A}\right)}\right) = A$
+or, using the compact notation defined above:
+:$(1): \quad a a = I$
+and from [[Closure of Topological Closure equals Closure]]:
+:$b \left({b \left({A}\right)}\right) = b \left({A}\right) = A^-$
+or, using the compact notation defined above:
+:$(2): \quad b b = b$
+Let $s$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Element|elements]] of $\left\{{a, b}\right\}$.
+By successive applications of $(1)$ and $(2)$, it is possible to eliminate all multiple consecutive instances of $a$ and $b$ in $s$, and so reduce $s$ to one of the following forms:
+:$\text{a)}: \quad a b a b \ldots a$
+:$\text{b)}: \quad b a b a \ldots a$
+:$\text{c)}: \quad a b a b \ldots b$
+:$\text{d)}: \quad b a b a \ldots b$
+From [[Closure of Complement of Closure is Regular Closed]]:
+:$b a b$ is [[Definition:Regular Closed Set|regular closed]].
+By [[Interior equals Complement of Closure of Complement]], the [[Definition:Interior (Topology)|interior]] of $A$ is:
+:$a b a$
+Recall the definition of [[Definition:Regular Closed Set|regular closed]]:
+:a set $A$ is [[Definition:Regular Closed Set|regular closed]] {{iff}} it equals the [[Definition:Topological Closure|closure]] of its [[Definition:Interior (Topology)|interior]].
+And so as $b a b$ is [[Definition:Regular Closed Set|regular closed]]:
+:$b a b = b a b a \left({b a b}\right)$
+So, adding an extra $b$ to either of $a b a b a b a$ or $b a b a b a$ will generate a string containing $b a b a b a b$ which can be reduced immediately to $b a b$.
+It follows that the possible different [[Definition:Subset|subsets]] of $S$ that can be obtained from $A$ by applying $a$ and $b$ can be generated by none other than:
+:$I$
+:$a$
+:$a b$
+:$a b a$
+:$a b a b$
+:$a b a b a$
+:$a b a b a b$
+:$a b a b a b a$
+:$b$
+:$b a$
+:$b a b$
+:$b a b a$
+:$b a b a b$
+:$b a b a b a$
+... a total of $14$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Complement of Closure is Regular Closed}
+Tags: Regular Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $T$.
+Let $A^-$ denote the [[Definition:Topological Closure|closure]] of $A$ in $T$.
+Let $A'$ denote the [[Definition:Relative Complement|complement]] of $A$ in $S$: $A' = S \setminus A$.
+Then $A^{- ' -}$ is [[Definition:Regular Closed Set|regular closed]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Regular Closed Sets]]
+9u5q00j1eiexrxsc5qihc4py55jj1ea
+\end{proof}<|endoftext|>
+\section{Topology induced by Usual Metric on Positive Integers is Discrete}
+Tags: Discrete Topology, Metric Spaces
+
+\begin{theorem}
+Let $\Z_{>0}$ be the [[Definition:Set|set]] of [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]] on $\Z_{>0}$.
+Then the [[Definition:Topology Induced by Metric|metric topology]] for $d$ is a [[Definition:Discrete Topology|discrete topology]].
+\end{theorem}
+
+\begin{proof}
+Let $\tau_d$ denote the [[Definition:Topology Induced by Metric|metric topology]] for $d$.
+Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.
+Let $a \in \Z_{>0}$.
+Recall the definition of the [[Definition:Open Ball|open $\epsilon$-ball]] of $a$ in $\struct {\Z_{>0}, d}$:
+:$\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$
+But we have:
+:$\forall x \in \Z_{>0}, x \ne a: \map d {x, a} \ge 1$
+and so:
+:$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$
+It follows that:
+:$\map {B_\epsilon} a := \set a$
+Thus by definition of $\tau_d$:
+:$\forall a \in \Z_{>0}: \set a \in \tau_d$
+It follows from [[Basis for Discrete Topology]] that $\struct {\Z_{>0}, \tau_d}$ is a [[Definition:Discrete Space|discrete topological space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete}
+Tags: Discrete Topology, Metric Spaces
+
+\begin{theorem}
+Let $\Z_{>0}$ be the [[Definition:Set|set]] of [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the [[Definition:Metric|metric]] on $\Z_{>0}$ defined as:
+:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
+Then the [[Definition:Topology Induced by Metric|metric topology]] for $\delta$ is a [[Definition:Discrete Topology|discrete topology]].
+\end{theorem}
+
+\begin{proof}
+Let $\tau_\delta$ denote the [[Definition:Topology Induced by Metric|metric topology]] for $\delta$.
+In [[Scaled Euclidean Metric is Metric]] it is demonstrated that $\delta$ is indeed a [[Definition:Metric|metric]] on $\Z_{>0}$.
+Let $a \in \Z_{>0}$.
+Recall the definition of the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] of $a$ in $\struct {\Z_{>0}, \delta}$:
+:$\map {B_\epsilon} a := \set {x \in A: \map \delta {x, a} < \epsilon}$
+Let $x \in \R_{>0}$.
+Let $\epsilon \in \R_{>0}$ such that $\epsilon < \dfrac 1 {a \paren {a + 1} }$.
+But we have:
+{{begin-eqn}}
+{{eqn | ll= \forall x \in \Z_{>0}, x \ne a:
+      | l = \map \delta {x, a}
+      | r = \frac {\size {x - a} } {x a}
+      | c = 
+}}
+{{eqn | r = \size {\frac 1 x - \frac 1 a}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \size {\frac 1 {a + 1} - \frac 1 a}
+      | c = 
+}}
+{{eqn | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+and so:
+:$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$
+It follows that:
+:$\map {B_\epsilon} a := \set a$
+Thus by definition of $\tau_d$:
+:$\forall a \in \Z_{>0}: \set a \in \tau_\delta$
+It follows from [[Basis for Discrete Topology]] that $\struct {\Z_{>0}, \tau_\delta}$ is a [[Definition:Discrete Space|discrete topological space]].
+{{qed}}
+\end{proof}
\ No newline at end of file