Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(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?

share|improve this question
1  
You even need a metric to conclude a convergent sub*sequence*. 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. –  Matt N. Jul 12 '12 at 16:35
    
@Vobo How do we know $\sum_{d\in D} |f(d)-g(d)| < \infty$? –  Matt N. 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
add comment

1 Answer

up vote 2 down vote accepted

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}

share|improve this answer
    
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? –  Matt N. Jul 12 '12 at 16:48
1  
You just need a uniform bound. Since $S$ is bounded, this is true. –  copper.hat Jul 12 '12 at 16:51
1  
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! –  Matt N. Jul 13 '12 at 6:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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