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(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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    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.2012-07-12

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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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    Nice, thank you everyone, Vobo, @BrianM.Scott and @copper.hat!2012-07-13