
Re: Problem understanding proof of Jordan Curve Theorem
Posted:
Jul 10, 2013 11:49 AM


On Wed, 10 Jul 2013 05:17:36 0700 (PDT), Rupert <rupertmccallum@yahoo.com> wrote:
>On Wednesday, July 10, 2013 1:53:07 PM UTC+2, Rupert wrote: >> On Wednesday, July 10, 2013 1:40:08 PM UTC+2, Paul wrote: >> >> > On Wednesday, July 10, 2013 9:41:05 AM UTC+1, Rupert wrote: >> >> > >> >> > > 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: > > > On Tuesday, July 9, 2013 11:56:54 PM UTC+2, peps...@gmail.com wrote: > > > > > > > On Tuesday, July 9, 2013 10:53:04 PM UTC+1, Paul 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. > > > > > > > > > > > > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Thank you very much for your help, > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > 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. > > > > > > > > > > > > > > > > > > > > > > > > > > > > Paul Epstein > > > > > > > > > > > > There exists a disc D whose boundary circle meets the Jordan polygon in exactly one point. (So in that case ab=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 ab is a continuous realvalued >function defined on this space. A continuous realvalued 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 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. >> >> > >> >> > >> >> > >> >> > Good news/ bad news. The bad news is that I still fail to understand your proof. It seems that the discs D_n are regarded as points in a compact topology. I don't understand how this construction works or what this particular topology looks like. >> >> > >> >> > >> >> > >> >> > The good news is that I have been able to use some of the same ideas to prove it for myself as follows: >> >> > >> >> > >> >> > >> >> > Lemma: If a and b are distinct points on the Jordan polygon, then the perpendicular bisector of the line ab intersects with the bounded component of the Jordan polygon. >> >> > >> >> > >> >> > >> >> > Proof of above lemma: >> >> > >> >> > The idea is to use a proof of the Jordan curve theorem for polygons. Consider Tverberg's proof which uses vertical rays. We adapt this proof for rays which have the same gradient as the perpendicular bisector. Tverberg uses a coordinate change to cover the case where two vertices of the Jordan polygon share the same x coordinate. This coordinatechange idea doesn't work very well for this lemma since we may need to keep changing coordinates infinitely often. So we use a different proof of the Jordan polygon theorem  the proof here: mizar.org/trybulec65/4.pdf >> >> > >> >> > In this proof, when the ray hits a vertex, it is argued that if a parallel displacement of the ray so that it misses the vertex is small enough, then all small displacements give the same parity for each point p outside the polygon. >> >> > >> >> > >> >> > >> >> > So we use this argument. It is certainly possible that the perpendicular bisector meets a vertex of the polygon. If so, displace the perpendicular bisector slightly. Then, using either the perpendicular bisector, or the displaced bisector, the points on the bisector change parity when they meet the Jordon polygon. Hence the proof of the Jordan polygon theorem leads to the proof of the lemma. >> >> > >> >> > >> >> > >> >> > Now I can supply the answer to my own question. Let N be larger than the diameter of the Jordan polygon. Now consider the triplets (x, y, d) which have the property that x and y are both on the Jordan curve and that either (x = y and d = 0) >> >> > >> >> > >> >> > >> >> > Actually, I still struggle to finish this argument. I was going to place there center of the circles on the perpendicular bisector and let d represent the diameter and argue that the triplets were closed. >> >> > >> >> > >> >> > >> >> > Surely, I'm missing something really easy. >> >> > >> >> > >> >> > >> >> > Perhaps you can clarify this >> >> > >> >> > >> >> > >> >> > " 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". >> >> > >> >> > >> >> > >> >> > Paul Epstein >> >> >> >> My idea was that you could topologize the set of all discs in R^2 of positive radius, by identifying it with R^2 x R+ (map a disc to the ordered pair consisting of its centre and its radius), and then the set of discs I described can be given the subspace topology. And then one can prove that it is compact. >> >> >> >> I haven't checked your proof carefully, but if it works it's probably a nicer proof. > >Sorry. To make the space compact you need to include the zero radius as a possibility.
No prooblem. Since we're trying to maximize ab there is a delta>0 such that we can consider only disks where ab > delta. This means there is r_0 > 0 such that we can consider only disks of radius >= r_0. And now there's your compactness.
Why can we restrict to the radius >= r_0? Lest I have the notation in the paper wrong, here I'm assuming that gamma: S_1 > R^2 is our mapping, with image Gamma subset R^2. Now the inverse of gamma, mapping Gamma to S_1, is continuous...

