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I came across this overview of a talk given by a CalTech student which claims that the proof of Arzela-Ascoli given in Carothers' "Real Analysis" is incorrect. I've been trying to find the flaw for some time now but I don't see it. Maybe there isn't one?

The usual proofs I've seen involve obtaining a countably dense subset (via separability) and making a diagonal argument to find a subsequence, but this one does not do that.

Here is the proof (of the just the converse direction because the forward direction is simple):

Arzela-Ascoli Theorem 11.8 : Let $X$ be a compact metric space, and let $\mathcal{F}$ be a subset of $C(X)$. Then $\mathcal{F}$ is compact if and only if $\mathcal{F}$ is closed, uniformly bounded, and equicontinuous.

Proof. ($\impliedby$) $ \ $ Suppose $\mathcal{F}$ is closed, uniformly bounded, and equicontinuous and let $(f_n)\in\mathcal{F}.$ We need to show that $(f_n)$ has a uniformly convergent subsequence.

First note that $(f_n)$ is equicontinuous. Thus given $\epsilon>0, \exists \delta >0: d(x,y)<\delta\implies |f_n(x)-f_n(y)|<\epsilon/3, \forall n$.

Next since $X$ is totally bounded, $X$ has a finite $\delta$-net, i.e, there exists $x_1,\ldots, x_k\in X$ such that each $x\in X$ satisfies $d(x,x_i)<\delta$. Now since $(f_n)$ is also uniformly bounded, each of the sequences $(f_n(x_i))_{n=1}^{\infty}$ is bounded in $\mathbb{R}$ for $i=1,2,\ldots, k$.

Thus by passing to a subsequence of the $f_n$ (and relabeling), we may suppose that $(f_n(x_i))_{n=1}^{\infty}$ converges for each $i=1,2,\ldots, k$.

In particular, we can find some $N$ such that $|f_m(x_i)-f_n(x_i)|<\epsilon/3$ for any $m,n\geq N$ and any $i=1,2,\ldots, k$.

And now we are done! Given $x\in X$, first find $i$ such that $d(x,x_i)<\delta$ and then whenever $m,n\geq N$, we will have \begin{align*} &|f_m(x)-f_n(x)|\\ &\leq|f_m(x)-f_m(x_i)|+|f_m(x_i)-f_n(x_i)|+|f_n(x_i)-f_n(x)|\\ &<\epsilon/3+\epsilon/3+\epsilon/3\\ &=\epsilon \end{align*}

That is, $f_n$ is uniformy Cauchy, since our choice of $N$ does not depend on $x$. Since $\mathcal{F}$ is closed in $C(X)$ by assumption, it follows that $f_n(x)$ converges to some $f\in\mathcal{F}$, uniformly.

stoic-santiago
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rosemary 2.0
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    In this proof the sequence $f_n$ does not remain "stable", for each $\epsilon$ he generates a new sub-sequence (in the step that $f_n(x_i)$ is assumed to converge), so the step "for every $\epsilon$ there is an $N$ so that for $n,m>N$ you have $|f_n(x)-f_m(x)|<\epsilon$" does not work. To complete the proof a diagonal argument is needed so you can assume the $f_n$ are always the same sequence. – s.harp Apr 06 '21 at 19:26
  • @s.harp $f_n(x_i)$ are never assumed to converge, they are just bounded in $\mathbb R$. Thereafter, the application of Bolzano-Weierstrass theorem gives a convergent subsequence. Isn't that right? – stoic-santiago Apr 24 '21 at 10:03
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    @epsilon-emperor At a certain point the author picks a sub-sequence of the $f_n$ so that $f_n(x_i)$ converges and re-assigns names so that now $f_n$ denotes this sub-sequence. This step may be done for any $\delta$-net (and that works), but the author uses this result for a net that is simultaneously $\delta$-fine for all $\delta>0$. That can also be done (diagonal argument) but requires further justification. – s.harp Apr 24 '21 at 22:03

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