# Arzelà - Ascoli Theorem for an arbitrary family of functions

I have just completed an exercise in Abott's Understanding analysis concerning the Arzelà - Ascoli theorem for functions on $\Bbb{R}$. The statement of the exercise is as follows:

Let $\{f_n\}$ be an equicontinuous family of functions on $[0,1]$ that is uniformly bounded on the same interval, viz. there exists a constant $M$ such that $|f_n(x)| \leq M$ for all $n \in \Bbb{N}$ and for all $x \in [0,1]$. Then $\{f_n\}$ has a uniformly convergent subsequence.

Now this exercise generalises nicely to the following statement for functions on an arbitrary compact metric space nicely, namely:

Let $(X,d)$ be a compact metric space and $C(X;\Bbb{R}^n)$ the space of all continuous functions from $X$ to $\Bbb{R}^n$ with the sup metric derived from $d$. If $\{f_n\}$ is an equicontinuous and uniformly bounded family on $C(X;\Bbb{R}^n)$ then $\{f_n\}$ has a uniformly convergent subsequence.

This generalises directly from the exercise in Abbott because a compact metric space $X$ is separable (has a countable dense subset $E$) and everything else follows through from the proof for functions on $[0,1]$.

Now I am trying to apply the statement of the Arzelà - Ascoli theorem above for metric spaces to prove the following problem below:

Let $(X,d)$ bee a compact metric space and let $C(X;\Bbb{R}^n)$ be the space of all continuous functions from $X$ to $\Bbb{R}^n$. Let $\mathcal{F}$ be any subfamily of $C(X;\Bbb{R}^n)$ which is closed, uniformly bounded and uniformly equicontinuous. Then $\mathcal{F}$ is compact in the sup metric on $C(X;\Bbb{R}^n)$ derived from $d$.

It seems to me that as long as I can choose a countable subset $\{g_n\}$ of $\mathcal{F}$, the problem should be done by the Arzelà - Ascoli theorem. This is because we already know that $\{g_n\}$ is uniformly bounded and uniformly equicontinuous. Why do we need the hypothesis that $\mathcal{F}$ is closed?

Thanks.

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A compact subspace of a Hausdorff space is always closed. You are probably forgetting that the limit of the convergence subsequence must be in $\mathcal F$ to conclude that $\mathcal F$ is sequentially compact. –  azarel Jun 6 '12 at 1:36
@azarel OMG facepalm you are completely right. –  user38268 Jun 6 '12 at 1:37
@azarel One more thing, how can I just choose arbitrarily a countable subset of $\mathcal{F}$ as I did above with the $\{g_n\}$'s? –  user38268 Jun 6 '12 at 1:38
It suffices to show that $\mathcal F$ is sequentially compact i.e. every sequence $\{g_n\}\subset \mathcal F$ has a convergent subsequence. –  azarel Jun 6 '12 at 1:45
@azarel Thanks! –  user38268 Jun 6 '12 at 1:46