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.

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 $N \in \N$ meet $N^{-1} < \dfrac \epsilon 3$.

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 $n, m \ge N_p$:


 * $\map d {\map {g_} p, \map {g_m} p} < \dfrac \epsilon 3$

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.