1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.