Arzelà-Ascoli Theorem

== Theorem ==
Let $X$ be a compact Hausdorff space.
Let $Y$ be a metric space.
Let $\CC$ be the set of all continuous functions $X \to Y$.
Consider $\CC$ as a metric space with the supremum metric:
:$\ds \map {d_\infty} {f, g} := \sup_{x \mathop \in X} \map d {\map f x, \map g x}$
Then a set $F \subseteq \CC$ is relatively compact it is pointwise equicontinuous and pointwise relatively compact.
== Proof ==
=== Pointwise properties imply relative compactness ===
By Sequentially Compact Metric Space is Compact, it suffices to show that every sequence in $F$ has a uniformly convergent subsequence.
Note that the limit need not lie in $F$.
For the proof, we will first construct a set $P \subseteq X$ which we will use like a countable dense subset.
Then we will use $P$ to construct a subsequence of any given sequence in $F$.
This subsequence will yield the result.
==== Constructing $P$ ====
By pointwise equicontinuity, for every $n \in \N$, there is a neighbourhood $U_{n; x}$ around any $x$ such that for all $f \in F$ and all $y \in U$, we have:
:$\map {d_\infty} {\map f x, \map f y} < n^{-1}$
This gives, for every $n \in \N$, a cover $\set {U_{n; x}: x \in X}$ of $X$.
By compactness of $X$, each of these has a finite subcover.
Let $P_n$ consist of one point from each $U_{n; x}$ in this subcover.
Let $\ds P = \bigcup_{n \mathop = 1}^\infty P_n$.
As Countable Union of Finite Sets is Countable, $P$ is countable.
Therefore, we may let $P$ be enumerated by a sequence $\sequence {p_n}$.
==== Construction of subsequence ====
Let $\sequence {f_n}$ be a sequence in $F$.
Inductively construct sequences $\sequence {f_{n; k} }$ for each positive integer $k$ as follows.
We set $f_{n; 1} := f_n$.
If $\sequence {f_{n; k} }$ is constructed, consider $\sequence {\map {f_{n; k} } {p_k} }$.
As each $f \in F$ is pointwise relatively compact, the image of this sequence is relatively compact.
Therefore, by Countably Compact First-Countable Space is Sequentially Compact, there is a monotone $a: \N \to \N$ such that $\sequence {\map {f_{\map a n; k} } {p_k} }$ converges.
Let $f_{n; k + 1} := f_{\map a n; k}$.
Let $g_n := f_{n; n}$.
The sequence $\sequence {g_n}$ is a subsequence of $\sequence {f_n}$.
==== Convergence of subsequence ====
Let $\epsilon > 0$ be aribitrary.
Let $N \in \N$ such that $N^{-1} < \dfrac \epsilon 3$.
By construction of $\sequence {g_n}$, for any $p \in P_N$, $\map {g_n} p$ converges.
By Convergent Sequence in Metric Space is Cauchy Sequence, we can find $N_p$ such that for all $n, m \ge N_p$:
:$\map d {\map {g_n} p, \map {g_m} p} < \dfrac \epsilon 3$
By construction of $P_N$, for each $x \in X$, we can pick a $p_x \in P_N$ such that for all $n$:
:$\map d {\map {g_n} x, \map {g_n} {p_x} } < \dfrac \epsilon 3$
Let $K := \max \set {N_p: p \in P_N}$.
This means that for all $n, m \ge K$, we have:
Therefore, $\sequence {g_n}$ is a Cauchy sequence in $\CC$.
This implies that $\sequence {g_n}$ converges.
=== Relative compactness implies pointwise properties ===
By Evaluation Mapping is Continuous, the map $e: \CC \to Y; f \mapsto \map f x$ is continuous.
Hence if $F$ is relatively compact, so is $e \sqbrk F = \set {\map f x: f \in F}$, because Continuous Image of Compact Space is Compact.
That is, $F$ is then also pointwise relatively compact.
Now suppose by way of contradiction that $F$ is relatively compact, but not pointwise equicontinuous.
Then there is an $x \in X$ and an $\epsilon > 0$ such that for all neighbourhoods $U$ of $x$, there are $x_U \in U$ and $f_U \in F$ such that:
:$\map {d_\infty} {\map {f_U} x, \map {f_U} {x_U} } \ge \epsilon$
By Countably Compact First-Countable Space is Sequentially Compact, any sequence consisting of such $f_U$ has a convergent subsequence.
However, the terms of this subsequence are still in $f_U$. 
Therefore, by Limit Preserves Inequality, we have for all $x_U$:
:$\map {d_\infty} {\map f x, \map f {x_U} } \ge \epsilon$
This means that $f$ is not continuous.
This is in contradiction to the Uniform Limit Theorem.
Category:Hausdorff Spaces
Category:Metric Spaces