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Topic: Problem understanding proof of Jordan Curve Theorem
Replies: 13   Last Post: Jul 12, 2013 9:25 AM

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David C. Ullrich

Posts: 3,024
Registered: 12/13/04
Re: Problem understanding proof of Jordan Curve Theorem
Posted: Jul 10, 2013 12:02 PM
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On Tue, 9 Jul 2013 14:53:04 -0700 (PDT), Paul <pepstein5@gmail.com>
wrote:

>I have been reading Tverberg's proof of the Jordan Curve Theorem:
>http://www.maths.ed.ac.uk/~aar/jordan/tverberg.pdf
>
>I understand completely the proofs of lemmas 1 and 2, but I'm having trouble following the proof of lemma 3. In particular, it's not clear to me why there is a disc D with abs(a - b) maximal. This is probably a consequence of the fact that continuous functions on compact sets attain their maxima, but I can't justify that assertion.


It's clear.

The proof I'm about to give is "perfectly clear" in my book. It
may strike you as something that he should have written
out in detail because it's not obvious. I disagree - it
was immediately obvious to me, and the paper was
written with a certain audience in mind...

But regardless, before giving the proof, it really
_is_ clear, by a simple non-proof: Imagine some
disks with |a-b| increasing to the sup of |a-b|.
Some subsequence converges to a maximal
disk. At least that's clear from the video in
_my_ head.

Anyway, the actual proof is very simple. One
can easily fix what Rupert said, or one can
put the whole thing this way:

Say we have a sequence of disks such that [blah
blah], with |a_n - b_n| tending to the sup of all
possible |a-b|.

Passing to a subsequence, then a subsequuence
of that subsequence, etc, we can assume that
a_n -> a and b_n -> b.

Say the nth disk is D(p_n, r_n), with center p_n
and radius r_n. We can also assume that
p_n -> p and r_n -> r >= 0. IF we can show that
r > 0 then D(p,r) is a disk that maximizes |a-b|,
qed.

And it's clear that r > 0:

I didn't read the paper carefully; in case I don't
have the paper's notation straight, note that I'm
assuming we have a continuous 1-1 function
gamma: S_1 -> R^2, with image Gamma.
Say h : Gamma -> S_1 is the inverse of gamma.
Standard compact-hausdorff stuff shows that
h is continuous. Hence h is uniformly continuous.

Now note that

|gamma(a_n) - gamma(b_n)| <= 2 r_n.

So if we had r_n -> 0 that would imply
|gamma(a_n) - gamma(b_n)| -> 0.
Since h is uniformly continuous this would imply
that |a_n - b_n| -> 0, which is clearly false.

DU.


>Thank you very much for your help,
>
>Paul Epstein





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