We have proved Arzela-Ascoli Theorem as a family of functions in $C(X)$ where $X$ is a compact metric space is relative compact iff it is equibounded and equicontinuous.

We’ve given a corollary for this. Every equicontinuous and equibounded sequence of $C(X)$ has an uniformly convergent subsequence.

I’ve seen the Rudin’s proof but there is some details like first countablity. Is there any basic and short way to obtain this corollary from the theorem we proved?

I’ve seen some proofs more and I’ve tried to do it but I’m not able to. If someone illuminate me on this shortly, I appreciate.

Thanks in advance

  • $\begingroup$ That’s just the “easy definition” of compactness (any sequence has a convergent subsequence) because $C(X)$ is a metric space. $\endgroup$ – Mindlack Nov 4 '19 at 22:33
  • $\begingroup$ @Mindlack closure of $f_n$ is compact thus all of them has a convergent subsequence. Equicontinuity and pointwise convergence implies uniform convergence. Is it a right approach? $\endgroup$ – user519955 Nov 4 '19 at 22:37
  • $\begingroup$ No, I mean that $C(X)$ is a metric space with uniform convergence. $\endgroup$ – Mindlack Nov 4 '19 at 22:39
  • $\begingroup$ @Mindlack it has uniform continuity. Why should it have uniform convergence? $\endgroup$ – user519955 Nov 4 '19 at 22:40
  • $\begingroup$ No. I mean that in the topology of $C(X)$ (which is metrizable), a sequence $f_n$ converges to $f$ iff $f_n$ converges uniformly to $f$. $\endgroup$ – Mindlack Nov 4 '19 at 22:45

Note the followings:

  1. $C(X)$ is equipped with the metric $d_{\sup}(f,g):=\sup_{x\in X}|f(x)-g(x)|$, and so, a sequence $(f_n)$ in $C(X)$ converges under the metric $d_{\sup}$ if and only if $(f_n)$ converges uniformly.

  2. Arzela-Ascoli theorem tells that a subset $K$ of the metric space $(C(X),d_{\sup})$ is relatively compact if and only if $K$ is equibounded and equicontinuous.

  3. A relatively compact sequence in a metric space has a convergent subsequence.

The the desired corollary is simply a combination of all these observations. Indeed, let $(f_n)$ be a sequence in $C(X)$. Then

\begin{align*} &\text{$(f_n)$ equibounded and equicontinuous} \\ &\Rightarrow \text{$(f_n)$ relative compact in $(C(X),d_{\sup})$} \tag{by A-A} \\ &\Rightarrow \text{$(f_n)$ has a convergent subsequence in $(C(X),d_{\sup})$} \tag{by 3}\\ &\Rightarrow \text{$(f_n)$ has a uniformly convergent subsequence.} \tag{by 1} \end{align*}


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