From: "Robin Chapman" <rjc@maths.ex.ac.uk>
Subject: Re: Ascoli's Theorem
Date: Tue, 25 Sep 2001 07:42:32 +0000 (UTC)
Newsgroups: sci.math
Summary: Proof of Ascoli's theorem (compact families of functions on compact fomains)

"Cengiz" <cengizbelentepe@hotmail.com> wrote in message
news:e8415163.0109241501.24046514@posting.google.com...

> I'm going through Simmon's Intro to Topology and Modern Analysis and
> am stumbling a bit on his proof of Ascoli's Theorem on pg 126 that "If
> X is a compact metric space, then a closed subspace of C(X,R) is
> compact <->it is bounded and equicontinuous"
> 
> My problem is going from bounded and equicontinuous -> compact.  In
> particular he uses a diagonalization which I can't seem to grasp.  I
> don't see how the subsequence is convergent.

Simmons defines a bunch of sequences:
S_i: f_{i 1}, {f_i 2}, f_{i 3} ....
such that S_{i+1} is a subsequence of S_i and for each i,
f_{i j}(x_i) -> a limit, a_i say, as j-> infinity.

Now he puts f_i = f_{i i}. The point about this is that f_i(x_k)
-> a limit for each k. To see this forget about the first few
of the f_i and consider the sequence
f_k, f_{k+1}, f_{k+2} ...
f_k lies in S_k. Now f_{k+1} = f_{k+1 k+1} lies in S_{k+1} which is
a subsequence of S_k. As it's the (k+1)-th term of S_{k+1},
it must be at least the (k+1)-th term of S_k.
Similarly f_{k+2} is at least the (k+2)-th term
of S_k and so on. Hence f_{k+i} = f_{k, s(i)} where s(i) >= k+i.
This is enough to show that f_{k+i}(x_k) tends to a_k as i -> infinity.
Hence f_i(x_k) -> a_k.

Now he finishes by showing that equicontinuity can be used to
prove the sequence of functions is uniformly convergent.

Robin Chapman
www.maths.ex.ac.uk/~rjc/rjc/html


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