Proposition 2.7.4 (Arzelà–Ascoli)2963E35FB0D3

Proof. We may assume that $K\neq \varnothing $. Let $\varepsilon >0$. By equicontinuity, every $x\in K$ is the center of an open ball on which $|f-f(x)|<\varepsilon $ for all $f\in \mathcal F$; by compactness, $K$ is covered by finitely many such balls $B_ l$ with centers $x_ l$, $1\leq l\leq N$. Then every $f\in \mathcal F$ is bounded on $K$ by $M=\varepsilon +\max _ l\sup _{g\in \mathcal F}|g(x_ l)|$, which is finite by pointwise boundedness. Let $G$ be a finite set of scalars such that every scalar of absolute value at most $M$ lies at distance less than $\varepsilon $ from some element of $G$. For each map $\gamma \colon \{ 1,\ldots ,N\} \to G$, the functions $f\in \mathcal F$ with $|f(x_ l)-\gamma (l)|<\varepsilon $ for all $l$ form a set $\mathcal F_\gamma $, and these finitely many sets cover $\mathcal F$. If $f,g\in \mathcal F_\gamma $ and $y\in B_ l$, then

Choosing one function in each nonempty $\mathcal F_\gamma $, we obtain a finite set within uniform distance $4\varepsilon $ of $\mathcal F$, and hence within $5\varepsilon $ of its closure $\overline{\mathcal F}$ in $C(K)$. Thus $\overline{\mathcal F}$ is totally bounded. It is also complete, because $C(K)$ is complete for the uniform metric: a uniform Cauchy sequence converges pointwise, and the limit is continuous since the convergence is uniform.

A complete and totally bounded metric space $Y$ is sequentially compact and compact. Given a sequence in $Y$, cover $Y$ by finitely many sets of diameter less than $1$ and keep one containing infinitely many terms. Cover it by finitely many sets of diameter less than $1/2$, keep one containing infinitely many of those terms, and so on. A diagonal choice gives a Cauchy subsequence, which converges because $Y$ is complete.

Suppose now that an open cover of $Y$ had no finite subcover, and put $S_0=Y$. Covering $S_{n-1}$ by finitely many subsets of diameter less than $1/n$, we see that one of them, $S_ n$, is not covered by finitely many members of the cover, since otherwise $S_{n-1}$ would be. This gives nested nonempty sets $S_1\supset S_2\supset \cdots $ with $\operatorname {diam}S_ n<1/n$, none of which is covered by finitely many members of the cover. Points $y_ n\in S_ n$ form a Cauchy sequence with a limit $y$. If the open ball of radius $s$ about $y$ in $Y$ lies in a member $U$ of the cover, then $S_ n$ lies in this ball, and hence in $U$, as soon as $1/n+\operatorname {dist}(y_ n,y)<s$, a contradiction. Hence $\overline{\mathcal F}$ is compact, and every sequence in $\mathcal F$ has a uniformly convergent subsequence. $\square$