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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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Paul

Posts: 393
Registered: 7/12/10
Re: Problem understanding proof of Jordan Curve Theorem
Posted: Jul 10, 2013 4:31 AM
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On Wednesday, July 10, 2013 9:06:49 AM UTC+1, Rupert wrote:
> On Tuesday, July 9, 2013 11:56:54 PM UTC+2, peps...@gmail.com wrote:
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> > On Tuesday, July 9, 2013 10:53:04 PM UTC+1, Paul wrote:
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> > > I have been reading Tverberg's proof of the Jordan Curve Theorem:
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> > > http://www.maths.ed.ac.uk/~aar/jordan/tverberg.pdf
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> > > 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.
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> > > Thank you very much for your help,
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> > To clarify, when I say that I can't justify "that assertion", I mean the assertion about disk D. I _don't_ mean the assertion that continuous functions on compact sets attain their suprema.
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> > Paul Epstein
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> There exists a disc D whose boundary circle meets the Jordan polygon in exactly one point. (So in that case |a-b|=0). Now, the set of ordered pairs (a,b) such that there exists a disc D whose boundary circle meets the Jordan polygon at gamma(a) and gamma(b) is a closed subset of S^1 x S^1 (and is nonempty by the foregoing remark). As such it is compact with the subspace topology coming from S^1 x S^1, and the map sending (a,b) to |a-b| is a continuous real-valued function defined on this space. A continuous real-valued function defined on a nonempty compact set attains a maximum.


Thanks a lot for this reply. I tried a proof along very similar lines. But I don't know why the set of ordered pairs (a,b) such that there exists a disc D whose boundary circle meets the Jordan polygon at gamma(a) and gamma(b) is a closed subset of S^1 x S^1.

I know why it's a subset of S^1 x S^1 but I don't know why it's a closed subset.

Thanks again,

Paul Epstein



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