nyuuzyou/wikis huggingface dataset:
Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual

== Theorem ==
Let $\struct {R, +, \times}$ be a commutative ring with unity.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the double dual of $G$.
For each $x \in G$, let $x^\wedge: G^* \to R$ be defined as:
:$\forall t \in G^*: \map {x^\wedge} t = \map t x$
Then:
:$x^\wedge \in G^{**}$
== Proof ==
We have that $x^\wedge$ is a mapping from $G^* \to R$.
It remains to be demonstrates that $x^\wedge$ is in fact a linear transformation.
Hence we need to show that:
:$(1): \quad \forall u, v \in G^*: \map {x^\wedge} {u + v} = \map {x^\wedge} u + \map {x^\wedge} v$
:$(2): \quad \forall u \in G^*: \forall \lambda \in R: \map {x^\wedge} {\lambda \times u} = \lambda \times \map {x^\wedge} u$
Hence:
and:
Hence the result.
== Also see ==
* Definition:Evaluation Linear Transformation/Module Theory, which uses $x^\wedge$ in its definition
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Category:Evaluation Linear Transformations (Module Theory)
Underlying Set of Topological Space is Clopen

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is both open and closed in $T$.
== Proof ==
From the definition of topology, $S$ is open in $T$.
From Underlying Set of Topological Space is Closed $S$ is closed in $T$.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.3 \ \text{(b)}$
Category:Clopen Sets
Underlying Set of Topological Space is Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is closed in $T$.
== Proof ==
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ $S \setminus U$ is closed in $T$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Set Difference with Empty Set is Self:
:$S \setminus \O = S$
Hence $S$ is closed in $T$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $4$
Category:Closed Sets
Underlying Set of Topological Space is Everywhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is everywhere dense in $T$.
== Proof ==
From Underlying Set of Topological Space is Closed, $S$ is closed in $T$.
From Closed Set Equals its Closure, $S = S^-$.
The result follows from definition of everywhere dense.
Category:Denseness
Uniform Absolute Convergence of Infinite Product of Complex Functions

== Theorem ==
Let $X$ be a compact metric space.
Let $\sequence {f_n}$ be a sequence of continuous functions $X \to \C$.
Let $\ds \sum_{n \mathop = 1}^\infty f_n$ converge uniformly absolutely on $X$.
Then:
:$\map f x = \ds \prod_{n \mathop = 1}^\infty \paren {1 + \map {f_n} x}$ converges uniformly absolutely on $X$
:$f$ is continuous
:there exists $n_0 \in \N$ such that $\ds \prod_{n \mathop = n_0}^\infty \paren {1 + \map {f_n} x}$ has no zeroes.
== Proof ==
=== Lemma ===
Because $\ds \sum_{n \mathop = 1}^\infty f_n$ converges uniformly, there exists $n_0 > 0$ such that $\size {\map {f_n} x} < \dfrac 1 2$ for $n \ge n_0$ and $x \in X$.
Then:
:$\map \Re {1 + \map {f_n} x} > 0$
By Bounds for Complex Logarithm:
:$\size {\map \log {1 + \map {f_n} x} } \le \dfrac 3 2 \size {\map {f_n} x}$
:$\size {\map \log {1 + \size {\map {f_n} x} } } \le \dfrac 3 2 \size {\map {f_n} x}$
for $n \ge n_0$.
By Comparison Test for Uniformly Convergent Series:
:$g = \ds \sum_{n \mathop = n_0}^\infty \map \log {1 + f_n}$
and:
:$h = \ds \sum_{n \mathop = n_0}^\infty \map \log {1 + \size {f_n} }$
converge uniformly.
By Uniform Limit Theorem, $g$ and $h$ are continuous.
By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $g$ and $h$ are bounded
By Exponential of Series Equals Infinite Product:
:$\map \exp g = \ds \prod_{n \mathop = n_0}^\infty \paren {1 + f_n}$
:$\map \exp h = \ds \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$
By the lemma, the products converge uniformly.
By Uniform Limit Theorem, $\map \exp g$ is continuous.
Because the second product converges uniformly, the first converges uniformly absolutely.
By Image of Complex Exponential Function, $\map \exp g$ has no zeroes.
Category:Infinite Products
Uniform Antiprism is Uniform Polyhedron

== Theorem ==
Let $P$ be a uniform antiprism.
Then $P$ is a uniform polyhedron.
== Proof ==
From Uniform Antiprism is Isogonal, $P$ is isogonal.
Hence the result.
== Sources ==
* 
* 
Category:Uniform Prisms
Category:Uniform Polyhedra
Uniform Continuity on Metric Space does not imply Compactness

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a uniformly continuous mapping on $A_1$.
Then $M_1$ does not necessarily have to be a compact metric space.
== Proof ==
Let $M_1 = \struct {A_1, d_1}$ be any metric space which is not compact.
Let $I_{M_1}: M_1 \to M_1$ be the identity mapping.
From Identity Mapping is Uniformly Continuous, $I_{M_1}$ is uniformly continuous on $M_1$.
Hence the result.
== Sources ==
* : $5$: Compact spaces: $5.8$: Compactness and Uniform Continuity: Remark $5.8.3$
Category:Metric Spaces
Category:Compact Spaces
Category:Uniformly Continuous Mappings
Uniform Contraction Mapping Theorem

== Theorem ==
Let $M$ and $N$ be metric spaces.
Let $M$ be complete.
Let $f : M \times N \to M$ be a continuous uniform contraction.
Then for all $t \in N$ there exists a unique $\map g t \in M$ such that $\map f {\map g t, t} = \map g t$, and the mapping $g: N \to M$ is continuous.
== Proof ==
For every $t\in N$, the mapping:
:$f_t: M \to M : x \mapsto \map f {x, t}$ is a contraction.
By the Banach Fixed-Point Theorem, there exists a unique $\map g t \in M$ such that $\map {f_t} {\map g t} = \map g t$.
We show that $g$ is continuous.
Let $K < 1$ be a uniform Lipschitz constant for $f$.
Let $s, t \in N$.
Then
and thus:
:$\map d {\map g s, \map g t} \le \dfrac 1 {1 - K} \map d {\map f {\map g t, s}, \map f {\map g t, t} }$
The continuity of $g$ now follows from that of $f$ using the definition of product metric
== Also see ==
* Implicit Function Theorem for Lipschitz Contractions
Category:Fixed Point Theorems
Category:Implicit Functions
Category:Metric Spaces
Category:Named Theorems
Uniform Convergence is Hereditary

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {f_n}$ be a sequence of mappings defined on $A$.
Let $\sequence {f_n}$ be uniformly convergent on $S \subseteq A$.
Then $\sequence {f_n}$ is uniformly convergent on every metric subspace of $S$.
That is, uniform convergence is a hereditary property of a metric space.
== Proof ==
Category:Metric Spaces
Uniform Convergence of General Dirichlet Series

== Theorem ==
Let $\map \arg z$ denote the argument of the complex number $z \in \C$.
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converges uniformly for all $s$ such that:
:$\cmod {\map \arg {s - s_0} } \le a < \dfrac \pi 2$
== Proof ==
Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges. 
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at $0$ by writing:
Thus it suffices to show $\map g s$ converges uniformly for for $\cmod {\map \arg s} \le a < \frac \pi 2$
By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon > 0$ there exists an $N$ independent of $s$ such that for all $m, n > N$:
:$\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} } < \epsilon$
By Abel's Lemma: Formulation 2 we may write:
Because $\map S {k, j}$ is the difference of two terms of a convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j > N$:
:$\ds \cmod {\map S {k, j} } < \frac {\epsilon \cos a} 3$
which gives us:
We see that:
From Shape of Secant Function, we have that on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
:$\cmod {\map \arg s} \le a \implies \map \sec {\map \arg s} \le \sec a$
which gives us:
:$\map \sec {\map \arg s} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} } \le \sec a \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }$
Hence:
Because $\sigma>0$, we have that $ - \lambda_k \sigma <0$ and hence:
:$e^{-\lambda_k \sigma} < 1 \le \sec a$
which gives us:
== Sources ==
* : $\text {II}$: Elementary Theory of the Convergence of Dirichlet's series $\S 2$
Category:General Dirichlet Series
Uniform Limit Theorem

== Theorem ==
Let $\struct {M, d_M}$ and $\struct {N, d_N}$ be metric spaces.
Let $\sequence {f_n}$ be a sequence of mappings from $M$ to $N$ such that:
:$(1): \quad \forall n \in \N: f_n$ is continuous at every point of $M$
:$(2): \quad \sequence {f_n}$ converges uniformly to $f$
Then:
:$f$ is continuous at every point of $M$.
== Proof ==
Let $a \in M$.
We are given that $d_N$ is a metric on $N$.
By applying twice:
Let $\epsilon \in \R_{>0}$.
Since $\sequence {f_n}$ converges uniformly to $f$:
We are given that $\forall n \in \N: f_n$ is continuous.
Hence:
Combining $(3)$, $\text {(4 a)}$, $\text {(4 b)}$ and $(5)$:
As $a$ and $\epsilon$ are arbitrary, it follows by Universal Instantiation of $n$ that:
Hence, $f$ is continuous at every point of $M$.
== Sources ==
* : $\S 5.2$, Theorem $5.10$
Category:Functional Analysis
Category:Continuous Mappings
Category:Metric Spaces
Category:Named Theorems
Uniform Limit of Analytic Functions is Analytic

== Theorem ==
Let $U$ be an open subset of $\C$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of analytic functions $f_n : U \to \C$.
Let $\sequence {f_n}$ converge locally uniformly to $f$ on $U$.
Then $f$ is analytic.
== Proof ==
Let $z \in U$.
By , there is an $\epsilon > 0$ so that:
:$\map {B_\epsilon} z \subset U$
and $f_n$ converges uniformly on $\map {B_\epsilon} z$.
Let $\gamma$ be any simple closed curve in $\map {B_\epsilon} z$.
Then $f_n \to f$ uniformly on $\gamma$, because $\gamma \subset \map {B_\epsilon} z$.
Thus by Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions, we have:
:$\ds \lim_{n \mathop \to \infty} \int_\gamma \map {f_n} z \rd z = \int_\gamma \map f z \rd z$
Since each $f_n$ is analytic, we have that:
:$\ds \forall n \in \N: \int_\gamma \map {f_n} z \rd z = 0$
So we conclude also that:
:$\ds \int_\gamma \map f z \rd z = 0$
Since $\gamma$ was arbitrary, we have by Morera's Theorem that $f$ is analytic in $\map {B_\epsilon} z$.
Since $z$ was arbitrary, $f$ is analytic on all of $U$.
== Also see ==
* Derivative of Uniform Limit of Analytic Functions
Category:Complex Analysis
Category:Uniform Convergence
Uniform Limit of Sequence of Continuous Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of functions $f_n : X \to \C$ that vanishes at infinity.
Let $f : X \to \C$ be a function such that:
:$f_n - f$ is bounded for each $n \in \N$
and:
:$\ds \sup_{x \mathop \in X} \cmod {\map {f_n} x - \map f x} \to 0$ as $n \to \infty$.
Then $f$ vanishes at infinity.
== Proof ==
Let $\epsilon > 0$. 
Since:
:$\ds \sup_{x \mathop \in X} \cmod {\map {f_n} x - \map f x} \to 0$ as $n \to \infty$
we can take $N \in \N$ such that:
:$\ds \sup_{x \mathop \in X} \cmod {\map {f_N} x - \map f x} < \frac \epsilon 2$
Further since $f_N$ vanishes at infinity, there exists a compact set $F \subseteq X$ such that:
:$\cmod {\map {f_N} x} < \dfrac \epsilon 2$ for each $x \in X \setminus F$.
Since:
:$\cmod {\map {f_N} x - \map f x} < \dfrac \epsilon 2$ for each $x \in X$
we in particular have:
:$\cmod {\map {f_N} x - \map f x} < \dfrac \epsilon 2$ for each $x \in X \setminus F$
Then from Triangle Inequality for Complex Numbers, we have:
for each $x \in X \setminus F$.
So $f$ vanishes at infinity.
Category:Complex-Valued Functions Vanishing at Infinity
Uniform Matroid is Matroid

== Theorem ==
Let $S$ be a finite set of cardinality $n$.
Let $0 \le k \le n$.
Let $U_{k, n} = \struct{S, \mathscr I}$ be the uniform matroid of rank $k$.
Then $U_{k, n}$ is a matroid.
== Proof ==
It needs to be shown that $\mathscr I$ satisfies the matroid axioms $(I1)$, $(I2)$ and $(I3)$.
=== Matroid Axiom $(I1)$ ===
Now:
By definition of the uniform matroid of rank $k$:
:$\O \in \mathscr I$
It follows that $\mathscr I$ satisfies matroid axiom $(I1)$.
=== Matroid Axiom $(I2)$ ===
Let $X \in \mathscr I$.
Let $Y \subseteq X$.
Then:
By definition of the uniform matroid of rank $k$:
:$Y \in \mathscr I$
It follows that $\mathscr I$ satisfies matroid axiom $(I2)$.
=== Matroid Axiom $(I3)$ ===
Let $U, V \in \mathscr I$.
Let $\size V < \size U$.
From Intersection is Subset:
:$U \cap V \subseteq V$
:$U \cap V \subseteq U$
Then:
Then:
:$U \cap V \ne U$
It follows that:
:$U \cap V \subsetneqq U$
Then:
Then:
By definition of the uniform matroid of rank $k$:
:$V \cup \set x \in \mathscr I$
It follows that $\mathscr I$ satisfies matroid axiom $(I3)$.
== Sources ==
* Chapter $1.$ $\S 3.$ Examples of Matroids
Category:Matroid Theory
Uniform Mean Ergodic Theorem

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\mathbb F$.
Let $U : \HH \to \HH$ be a bounded linear operator such that:
:$\forall f \in \HH : \norm {\map U f} \le \norm f$
Then for each $f \in \HH$:
:$\ds \lim_{N - M \mathop \to \infty} \dfrac 1 {N - M} \sum_{n \mathop = M}^{N - 1} \map {U^n} f = \map P f$
i.e.
:$\ds \forall \epsilon \in \R_{>0}: \exists K \in \N: \forall M,N \in \N: N - M \ge K \implies \norm {\frac 1 {N - M} \sum_{n \mathop = M}^{N - 1} \map {U^n} f - \map P f} \le \epsilon$
where:
:$U^n$ denotes the $n$ times composition of $U$
:$I := \set {f \in \HH : \map U f = f}$
:$P : \HH \to I$ denotes the orthogonal projection on $I$
== Proof ==
== Sources ==
* $2.5$ The Mean Ergodic Theorem
Category:Mean Ergodic Theorem
Uniform Prism is Semiregular Polyhedron

== Theorem ==
A uniform prism is a semiregular polyhedron.
== Proof ==
Let $P$ be a uniform prism.
By definition, $P$ is regular.
Hence the bases of $P$ are regular polygons.
Also by definition, the lateral faces of $P$ are square so regular polygons.
It remains to be shown that $P$ is isogonal.
== Sources ==
* : $13$
* : $13$
* 
* 
Category:Prisms
Category:Semiregular Polyhedra
Uniform Prism is Uniform Polyhedron

== Theorem ==
Let $P$ be a uniform prism.
Then $P$ is a uniform polyhedron.
== Proof ==
That all the faces of $P$ are regular polygons follows from the definition of regular prism.
From Regular Prism is Isogonal, $P$ is isogonal.
Hence the result.
== Sources ==
* 
* 
Category:Uniform Prisms
Category:Uniform Polyhedra
Uniform Prisms are Countably Infinite

== Theorem ==
There is a countably infinite different varieties of uniform prisms.
== Proof ==
Hence a uniform prism is made of:
:$2$ bases which are regular polygons
:as many lateral faces as there are sides of one of the bases.
Hence for each type of regular polygon there exists a corresponding uniform prism.
There exists a type of regular polygon for each natural number greater than or equal to $3$.
There exists countably infinite set of natural numbers greater than or equal to $3$.
Hence the result.
Category:Uniform Prisms
Category:Countably Infinite
Uniform Product of Continuous Functions is Continuous

== Theorem ==
Let $X$ be a metric space.
Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.
Let $\sequence {f_n}$ be a sequence of bounded continuous mappings $f_n: X \to \mathbb K$.
Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly to $f$.
Then $f$ is continuous.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Infinite Product of Analytic Functions is Analytic
Category:Uniform Convergence
Category:Infinite Products
Category:Uniform Product of Continuous Functions is Continuous
Uniform Product of Continuous Functions is Continuous/Proof 1

== Theorem ==
== Proof ==
Let $n_0 \in \N$ be such that the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty f_n$ converges uniformly.
By the Uniform Limit Theorem, $\ds \prod_{n \mathop = n_0}^\infty f_n$ is continuous.
Because $f_1, \ldots, f_{n_0 - 1}$ are continuous, so is $\ds \prod_{n \mathop = 1}^\infty f_n$.
Category:Uniform Product of Continuous Functions is Continuous
Uniform Space whose Topology is Metrizable is not necessarily Metrizable

== Theorem ==
Let $\UU$ be a uniformity on a set $S$.
Let $\struct {\struct {S, \UU}, \tau}$ be the uniform space generated from $\UU$.
Let $T = \struct {S, \tau}$ be the uniformizable space yielded by $\struct {\struct {S, \UU}, \tau}$.
Let $T$ be a metrizable space.
Then it is not necessarily the case that $\UU$ is itself a metrizable uniformity.
== Proof ==
Let $T = \struct {S, \tau}$ be an uncountable discrete ordinal space.
From Uncountable Discrete Ordinal Space is Metrizable, $T$ is a metrizable space.
However, from Uncountable Discrete Ordinal Space has Unmetrizable Uniformity, there exists a uniformity $\UU$ which yields the uniformizable space $T = \struct {S, \tau}$ which is not itself a metrizable uniformity.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metric Uniformities
Category:Metrizable Topologies
Category:Uniformities
Uniform Tubular Neighborhood Theorem

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $S$ be a submanifold of $M$.
Suppose $S$ is compact.
Then $S$ has a uniform tubular neighborhood in $M$.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Tubular Neighborhoods and Fermi Coordinates
Category:Tubular Neighborhoods
Category:Named Theorems
Uniformity iff Quasiuniformity has Symmetric Basis

== Theorem ==
Let $S$ be a set.
Let $\UU$ be a quasiuniformity on $S$.
Then $\UU$ is a uniformity $\UU$ has a symmetric filter basis.
== Proof ==
Let $\UU$ be a quasiuniformity on $S$ which has a symmetric filter basis $\BB$.
From the definition of filter basis, all the elements of $\UU$ can be formed from intersections of elements of $\BB$.
But from Intersection of Symmetric Relations is Symmetric, it follows that all elements of $\UU$ are symmetric.
Now suppose $\UU$ is a uniformity.
If $\BB$ is a filter basis of $\UU$ then all the elements of $\BB$ are also elements of $\UU$.
Hence $\BB$ is a symmetric filter basis of $\UU$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Uniformities
Category:Uniformities
Uniformly Absolutely Convergent Product is Uniformly Convergent

== Theorem ==
Let $X$ be a set.
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let $\mathbb K$ be complete.
Let $\sequence {f_n}$ be a sequence of bounded mappings $f_n: X \to \mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly absolutely on $X$.
Then it converges uniformly.
== Proof ==
Category:Uniform Convergence
Category:Infinite Products
Uniformly Continuous Function Preserves Uniform Convergence

== Theorem ==
Let $X$ be a set.
Let $M = \struct {A_1, d_1}$ and $N = \struct {A_2, d_2}$ be metric spaces.
Let $\sequence {g_n}$ be a sequence of mappings $g_n: X \to A_1$.
Let $g_n$ converge to $g$ uniformly on $X$ as $n \to \infty$.
Let $f: A_1 \to A_2$ be uniformly continuous on $A_1$.
Then the sequence $\sequence {f \circ g_n}$ converges to $g \circ f$ uniformly on $X$ as $n \to \infty$.
== Proof ==
Let $\epsilon \in \R_{>0}$.
Because $f$ is uniformly continuous on $A_1$, there exists $\delta \in \R_{>0}$ such that:
:$\forall x, y \in A_1: \map {d_2} {\map f x, \map f y} < \epsilon$ for $\map {d_1} {x, y} < \delta$
Because $\sequence {g_n}$ converges to $g$ uniformly on $X$ as $n \to \infty$, there exists $N > 0$ such that:
:$\map {d_2} {\map {g_n} x, \map g x} < \epsilon$ 
for all $n > N$ and all $x \in X$.
Thus:
:$\map {d_2} {\map f {\map {g_n} x}, \map f {\map g x} } < \epsilon$
for all $n > N$ and all $x \in X$.
Hence the sequence $\sequence {f \circ g_n}$ converges to $g \circ f$ uniformly on $X$ as $n \to \infty$.
== Also see ==
* Composition of Continuous Mapping on Compact Space Preserves Uniform Convergence
Category:Uniform Convergence
Category:Uniform Continuity
Uniformly Continuous Function is Continuous/Metric Space

== Theorem ==
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_1, d_1}\right)$ be metric spaces.
Let the mapping $f: M_1 \to M_2$ be uniformly continuous on $M_1$.
Then $f$ is continuous on $M_1$.
== Proof ==
Let $f$ be uniformly continuous on $M_1$.
Let $x \in M_1$.
Let $\epsilon > 0$.
As $f$ is uniformly continuous, $\exists \delta > 0$ such that:
:$\forall y \in M_1: d_1 \left({x, y}\right) < \delta: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$
Thus by definition $f$ is continuous at $x$.
Category:Metric Spaces
Category:Continuous Mappings on Metric Spaces
Category:Uniformly Continuous Mappings
Uniformly Continuous Function is Continuous/Real Function

== Theorem ==
Let $I$ be an interval of $\R$.
Let $f: I \to \R$ be a uniformly continuous real function on $I$.
Then $f$ is continuous on $I$.
== Proof 1 ==
== Proof 2 ==
Category:Real Analysis
Category:Continuous Real Functions
Category:Uniformly Continuous Real Functions
Category:Uniformly Continuous Real Function is Continuous
Uniformly Continuous Function is Continuous/Real Function/Proof 2

== Theorem ==
== Proof ==
Let $x \in I$.
Let $\epsilon \in \R_{>0}$.
As $f$ is uniformly continuous: 
:$\exists \delta \in \R_{>0}: \paren {x, y \in I, \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon}$
Then, for all $y \in I$ such that $\size {x - y} < \delta$:
:$\size {\map f x - \map f y} < \epsilon$
Thus by definition $f$ is continuous at $x$.
As $x$ was arbitrary, $f$ is continuous on all of $I$.
Category:Uniformly Continuous Real Function is Continuous
Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain/Lemma 1

== Theorem ==
Let $\tuple {X, d}$ be a metric space.
Let $\tuple {Y, d'}$ be a complete metric space.
Let $A \subseteq X$ be a set that is not closed.
Let $f : A \to Y$ be a uniformly continuous function.
Let $\sequence {a_n}$ be a sequence in $A$ convergent to $a \in A^-$. 
Then $\sequence {\map f {a_n} }$ converges. 
== Proof ==
From Set is Closed iff Equals Topological Closure, $A^- \setminus A$ is non-empty. 
Let $a \in A^- \setminus A$. 
From Point in Closure of Subset of Metric Space iff Limit of Sequence:
:there exists a sequence $\sequence {a_n}$ in $A$ converging to $a$.
Consider now the sequence $\sequence {\map f {a_n} }$ in $Y$.
Note that since $Y$ is complete, if we can show that $\sequence {\map f {a_n} }$ is a Cauchy sequence, then we know that it converges.
We therefore want to show that there exists an $N$ such that for $n, m > N$ we have:
:$\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$
Since $f$ is uniformly continuous, we can find $\delta > 0$ such that: 
:$\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$
whenever: 
:$\map d {a_n, a_m} < \delta$
Since $\sequence {a_n}$ is a Cauchy sequence, we can find $N$ such that for $n, m > N$ we have: 
:$\map d {a_n, a_m} < \delta$
giving:
:$\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$
for $n, m > N$. 
So $\sequence {\map f {a_n} }$ is a Cauchy sequence, and so is convergent.
Category:Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain