# Proof of equicontinuous and pointwise bounded implies compact

I tried to prove the Arzela-Ascoli theorem:

Let $X$ be a compact Hausdorff space and let $C(X)$ denote the space of continuous functions $f: X \to \mathbb R$ endowed with the sup norm $\|\cdot \|_\infty$. Then $S \subseteq C(X)$ is relatively compact if and only if it is pointwise bounded and equicontinuous.

Proof:

(this direction caused no problems)

$\implies$: Let $S\subseteq C(X)$ be relatively compact. Since $C(X)$ is a complete metric space (a Banach space, in fact)we may use the fact that a set is totally bounded if and only if it is relatively compact. Hence $S$ is totally bounded and hence bounded which implies pointwise bounded.

It remains to be shown that $S$ is equicontinuous. To this end, let $\varepsilon > 0$. Since $S$ it totally bounded it can be covered by a finite number of $\varepsilon/3$ balls. Let $f_1, \dots , f_n$ denote the centres of these balls. Since $f \in C(X)$ are uniformly continuous, for every $f_i$ there exists a $\delta_i >0$ such that $|x-y|<\delta_i$ implies $|f_i(x) - f_i(y)|<{\varepsilon \over 3}$. Let $\delta = \min_i \delta_i$. Then for $|x-y|<\delta$, $$|f(x)-f(y)| \le |f(x) - f_i(x)| + |f_i(x) - f_i (y)| + |f_i(y) -f(y)| < \varepsilon$$ where $f_i$ is such that $f$ is contained in the $\varepsilon/3$ ball with centre $f_i$.

This direction is causing the problem:

$\Longleftarrow$: Let $S \subseteq C(X)$ be equicontinuous and pointwise bounded. The goal is to show that $\overline{S}$ is compact. By the general Heine-Borel theorem for metric spaces a subset of a metric space is compact if and only if it is complete and totally bounded. Furthermore, closed subsets of complete metric spaces are complete hence it is enough to show that $\overline{S}$ is totally bounded.

To this end let $\varepsilon > 0$. Since $\overline{S}$ is equicontinuous there exists $\delta > 0$ such that $|x-y|<\delta$ implies $|f(x) - f(y) |< \varepsilon$. Since $X$ is compact it may be covered by finitely many $\delta$ balls. Let $x_1, \dots , x_n$ denote the centres of these balls. Since $\overline{S}$ is pointwise bounded for every $x_i$ there exists $K_i$ such that $\sup_{f \in \overline{S}} |f(x_i)| \le K_i$.

How to construct the finite collection of $\varepsilon$ balls that cover $\overline{S}$?

• Note that bounded and totally bounded are two completely different things in infinite dimensional Banach spaces (it is perhaps thus better to tell the latter property precompactness). In fact, the closed unit ball of an infinite dimensional Banach space is always bounded but not precompact. – Jochen Apr 17 '14 at 10:31
• @Jochen Thank you for your comment but I don't understand why you mention it. In my proof I use that (in any metric space) bounded implies totally bounded. Please could you elaborate. – Student Apr 17 '14 at 11:57
• It is not true that bounded implies totally bounded in a metric space. Consider the sequence $(1,0,\dots)$, $(0,1,0,\dots)$, $\dots$ in $\ell^\infty$. This sequence is bounded by $1$ in norm but is not totally bounded because any open ball of radius $1/2$ contains at most one of the terms. – user137301 Apr 17 '14 at 12:01
• @user137301 Good point. I will try to fix my proof. – Student Apr 17 '14 at 15:30
• I can't seem to fix my proof. – Student Apr 18 '14 at 12:06

Hint: Suppose that $f,g \in \bar{S}$ are such that for each $i$, we have $|f(x_i) - g(x_i)| < \epsilon$. Show that $\|f-g\| < \epsilon$. Now think about a finite covering of $\prod_{i=1}^n [-K_i, K_i]$ by boxes measuring $\epsilon$ in each dimension.
• Then I can finish by showing that these $A_{j_1, \dots , j_n}$ cover $\overline{S}$? – Student Apr 18 '14 at 12:38