# Question about proof of Arzelà-Ascoli

(Arzelà-Ascoli, $\Longleftarrow$) Let $K$ be a compact metric space. Let $S \subset (C(K), \|\cdot\|_\infty)$ be closed, bounded and equicontinuous. Then $S$ is compact, that is, for a sequence $f_n$ in $S$ we can find a convergent subsequence (conv. in $\|\cdot\|_\infty)$.

Proof: Note that a compact metric space is separable. Let $D$ be a dense, countable subset of $K$. By assumption, $S$ is bounded, hence there exists $M$, such that $\|f\|_\infty \leq M$ for all $f$ in $S$, in particular, for $f_n$.

Now consider the space of all functions from $D$ to $[-M,M]$. By Tychonoff, $[-M,M]^D$ is compact. Define $g_n = f_n\mid_D$. Then $g_n$ is a sequence in $[-M,M]^D$, hence has a convergent subsequence $g_{n_k}$.

This is what it says in my notes. Now I've been thinking about what exactly "convergent" means in the last sentence. I'm quite sure it means pointwise. But theoretically, I can endow $[-M,M]^D$ with a norm or metric (that induces the product topology). So "convergent" could mean convergent with respect to that metric, no? Or does that not make sense?

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You even need a metric to conclude a convergent subsequence. And the metric $e$ for the product topology is obvious as $D$ is countable, i.e. $e(f,g) = \sum_{d\in D} |f(d)-g(d)|$. – Vobo Jul 12 '12 at 16:31
Ooh, of course. Thank you! Why don't you post it as an answer, then I can upvote and accept. – Rudy the Reindeer Jul 12 '12 at 16:35
@Vobo How do we know $\sum_{d\in D} |f(d)-g(d)| < \infty$? – Rudy the Reindeer Jul 12 '12 at 16:36
I forgot the converging factor, see the answer below. – Vobo Jul 12 '12 at 16:42
I am quite sure that if you read the rest of the proof in your notes, the answer to your question will be become clear. – wildildildlife Jul 12 '12 at 18:44

You even need a metric to conclude a convergent sub*sequence*. And the metric d for the product topology is obvious for e.g. $D=\{ x_n| n\in N\}$, i.e. \begin{align} d(f,g)=\sum_{n=1}^\infty 2^{-n}|f(x_n)−g(x_n)|. \end{align}

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For this to be finite, each term in the sum would have to be at least smaller than $1$, that is, $|f(d) - g(d)| < 2^n$. Why do we know that this is the case? – Rudy the Reindeer Jul 12 '12 at 16:48
You just need a uniform bound. Since $S$ is bounded, this is true. – copper.hat Jul 12 '12 at 16:51
You don’t actually need a metric to conclude that there’s a convergent subsequence: first countability is sufficient. Of course you have a metric, so you might as well use it! @Matt: $|f(x_n)-g(x_n)|\le 2M$ for each $n$, so $d(f,g)\le 2M$. – Brian M. Scott Jul 12 '12 at 20:40
Nice, thank you everyone, Vobo, @BrianM.Scott and @copper.hat! – Rudy the Reindeer Jul 13 '12 at 6:02