# Ascoli's theorem: A subspace F of C(X,R^n)

1. Nov 15, 2009

### symbol0

I just read Ascoli's theorem: A subspace F of C(X,R^n) has compact closure if and only if F is equicontinuous and pointwise bounded.

Then it says, As a corollary: If the collection {fn} of functions in C(X,R^k) is pointwise bounded and equicontinuous, then the sequence (fn) has a uniformly convergent subsequence.

Can anybody tell me why the corollary follows from the theorem?

2. Nov 16, 2009

### g_edgar

Re: corollary

in a metric space, when you have a compact subset A, then any sequence in A has a convergent subsequence. Now apply this to the appropriate metric space of functions, so that convergence in the metric is uniform convergence of functions.

3. Nov 16, 2009

### symbol0

Re: corollary

Thanks Edgar,
The problem I see with your reply is that here the closure of the sequence is compact (by Ascoli's theorem), so as you say, this closure has a convergent subsequence. But we don't know that the sequence itself is compact, so how do we know that it has a convergent subsequence?

4. Nov 16, 2009

### quasar987

Re: corollary

I think that here, we are taking F as the subspace generated by the f_n's. Then F has compact closure. Now, the f_n are in F and so in particular, they are in the closure of F. Therefor the sequence {f_n} has a convergent subsequence (in C(X,R^k)).

AFAIK, it does not make sense to say that a sequence is compact.

5. Nov 16, 2009

### symbol0

Re: corollary

Of course!!! how could I miss that? ohh, I know.
For some reason, instead of thinking that every sequence of a compact metric space has a convergent subsequence, I was thinking: every compact metric space has a convergent subsequence. I guess I was very tired.

Thank you for clarifying.