I am doing problem 3 from section 45 in Munkres. The problem is Prove Arzela's Theorem, which states: Let $X$ be compact: let $f_n \in \mathcal{C}(X,\mathbb{R}^k)$. If the collection $\{f_n\}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence. Here is a sketch of my proof:

Let X is compact and $\{f_n\}\subseteq \mathcal{C}(X,\mathbb{R}^k)$. Since $\{f_n\}$ is pointwise bounded and equicontinuous by Ascoli's Theorem $\overline{\{f_n\}}$ is compact. Since $\overline{\{f_n\}}$ is a compact subset of a complete metric space it's complete. Since $\overline{\{f_n\}}$ is compact then the sequence $\{f_n\}$ has a convergent subsequence $\{f_{n_i}\} \to f$. Since we are in the uniform metric, this subsequence converges uniformly.

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    $\begingroup$ Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof. $\endgroup$ – Song Dec 9 '18 at 0:55
  • $\begingroup$ I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent. $\endgroup$ – Issacg628496 Dec 9 '18 at 1:08
  • $\begingroup$ So, should I instead show that Arzela's Theorem implies Ascoli's? $\endgroup$ – Issacg628496 Dec 9 '18 at 1:17
  • $\begingroup$ @Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here. $\endgroup$ – Chill2Macht Dec 9 '18 at 1:18
  • $\begingroup$ @Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $\mathcal{F}\subset C(X,\mathbb{R}^k)$, and the problem requires the sequence version. $\endgroup$ – Song Dec 9 '18 at 1:30

The argument should say: as we have a compact subset of a metric space, the subset is sequentially compact and so we have a convergent subsequence for the sequence $(f_n)_n$ (completeness is irrelevant as we don't have a Cauchy sequence). The rest seems correct.


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