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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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Rupert

Posts: 3,809
Registered: 12/6/04
Re: Problem understanding proof of Jordan Curve Theorem
Posted: Jul 10, 2013 4:41 AM
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On Wednesday, July 10, 2013 10:31:29 AM UTC+2, peps...@gmail.com wrote:
> On Wednesday, July 10, 2013 9:06:49 AM UTC+1, Rupert wrote:
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> > 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.
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> 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.
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> I know why it's a subset of S^1 x S^1 but I don't know why it's a closed subset.
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Let's say you have a convergent sequence of points (a_n,b_n) in S^1 x S^1 such that, for each n, gamma(a_n) and gamma(b_n) lie on the boundary circle of some disc contained in the bounded component of R^2 \ Gamma. Then there is a sequence of discs D_n, such that each disc is an open disc contained in the bounded component of R^2\Gamma, and for all n, gamma(a_n) and gamma(b_n) lie on the boundary circle of D_n. Now, the set of all discs whose closure is contained in the closure of the bounded component of R^2\Gamma can be viewed as a compact space. The sequence (D_n) is a sequence in this compact space and must have a convergent subsequence which converges to some disc D. Prove that the interior of the disc D is contained in the bounded component of R^2\Gamma and that its boundary circle contains gamma(a) and gamma(b) where a and b are the limits of the respective sequences (a_n) and (b_n). This shows you that the set in question is closed under taking limits of convergent sequences and is therefore closed. This is one way to argue the point. There may be more elegant ways, perhaps.



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