# Proof of the Arzelà–Ascoli Theorem

I'm stuck on a particular line of the proof of The Arzelà–Ascoli Theorem.

In lectures, we have:

$$1.$$ Defined equicontinuous as:

Let $$X$$ be a metric space, $$C(X) = \{f: X \rightarrow \mathbb{R}\text{ continuous} \}$$ the space of continuous functions, $$S \subset C(X)$$. Let $$x \in X$$ be a point. Then $$S$$ is equicontinuous at $$x$$ if $$\forall \varepsilon > 0$$, $$\exists \delta > 0$$ such that $$y \in B(x, \delta)$$, $$f \in S$$ $$\implies$$ $$|f(x) - f(y)| < \varepsilon$$.

And obviously $$S$$ is equicontinuous if it is equicontinuous at all points of $$X$$.

$$2.$$ Stated The Arzelà–Ascoli Theorem as:

Suppose that $$X$$ is a compact metric space and $$S \subset C(X)$$ is a subspace.

Then, $$S$$ is compact $$\iff$$ $$S$$ is closed, bounded, and equicontinuous.

In the proof of the forward direction:

Closed and bounded are clear, so it remains to show that $$S$$ is equicontinuous. We already know that $$S$$ is totally bounded, so let $$\varepsilon > 0$$ and fix $$x \in X$$. Then $$\exists F \subset S$$ finite such that $$S \subset \bigcup_{f \in F}B(f, \frac{\varepsilon}{3})$$.

Since $$F$$ is equicontinuous...

This is the line that I get stuck at, why is $$F$$ equicontinuous?

• Because it is finite. A finite subset of $C(X)$ is equicontinuous (in general this is not true, but since $X$ is compact, for all $f \in C(X)$, $f$ is uniformly continuous). Commented Mar 30, 2016 at 10:16

Recall that for all $f \in C(X)$, $f$ is uniformly continuous because $X$ is compact.
Given $F \subset C(X)$ a finite set, we show that it is equicontinuous. Fix $\varepsilon > 0$. We need to find som $\delta > 0$ such that some condition is satisfied.
For all $f \in F$, by uniform continuity of $f$, there exist $\delta_f$ such that etcetera. Your required $\delta$ is $\delta = \min_{f \in F} \delta_f$.