# Proof of Arzela's Theorem

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.

• 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. – Song Dec 9 '18 at 0:55
• I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent. – Issacg628496 Dec 9 '18 at 1:08
• So, should I instead show that Arzela's Theorem implies Ascoli's? – Issacg628496 Dec 9 '18 at 1:17
• @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. – Chill2Macht Dec 9 '18 at 1:18
• @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. – 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.