# Corollary to Arzelà–Ascoli Theorem

As I know the Arzelà–Ascoli Theorem for functions between metric spaces, it is the following:

Theorem: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces with $X$ compact and $Y$ complete. Also, Let $A \subset C(X \to Y)$. Then $A$ is compact if and only if $A$ is closed, equicontinuous, and pointwise compact.

I understand the proof to this reasonably well. What I need help in doing is proving the following corollary:

Corollary: Let $(X,d)$ be a compact metric space and let $A \subset C(X \to \mathbb{R}^n)$. If $A$ is equicontinuous and pointwise bounded, then every sequence in $A$ has a convergent subsequence.

Could anyone give me some assistance in proving this?

• What exactly are you having difficulty with? Seems pretty straightforward to try and show compactness of $\overline{A}$ using Arzela-Ascoli. Commented Jun 13, 2015 at 2:08
• I am unclear of how we can apply the theorem when only pointwise bounded is present, and not pointwise compact and closed. Commented Jun 13, 2015 at 2:13
• I don't think $A$ is compact, by $A$ pointwise bounded you can show it's closure is closed and pointwise compact, then apply the theorem. The corollary you are proving should be that every sequence in $A$ has a subsequence converging in $C(X\to\mathbb{R}^n)$
– Qidi
Commented Jun 13, 2015 at 2:26
• Yes, that is correct. I believe that every sequence has a convergent subsequence, not necessarily having the limit contained in $A$. I guess that is simple enough then. Thank you. Commented Jun 13, 2015 at 2:29
• You can post your own answer to this question, explaining what you have learned. Otherwise it will remain listed as "unanswered". Commented Jun 13, 2015 at 3:18

Because $A$ is pointwise bounded, its closure $\overline{A}$ can be shown to be pointwise compact, so that $\overline{A}$ will satisfy the hypotheses for Arzelà–Ascoli.