On Tue, 9 Jul 2013 14:53:04 -0700 (PDT), Paul <firstname.lastname@example.org> 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.
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.
>Thank you very much for your help, > >Paul Epstein