Timeline for Pair of curves joining opposite corners of a square must intersect---proof?
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| Aug 14, 2010 at 19:34 | comment | added | Will Jagy | That's my sense as well. I'm quite happy with an all-continuous discussion, I did not PL as an undergraduate. There does seem to be an entirely PL discussion that works. The parallel discussions remind me of a book review on circle packing in the July 2009 Bulletin, lovely pictures, see if I can make a working link: ams.org/journals/bull/2009-46-03/S0273-0979-09-01245-2/… | |
| Aug 14, 2010 at 18:26 | comment | added | HJRW | It seems to me that this approach is converging to John Stillwell's answer below. | |
| Aug 14, 2010 at 6:30 | comment | added | HJRW | Will, unfortunately I had misspelled the author's name. Apologies! The reference is: Thomassen, Carsten, The Jordan-Schönflies theorem and the classification of surfaces. Amer. Math. Monthly 99 (1992), no. 2, 116--130. | |
| Aug 14, 2010 at 5:36 | comment | added | Will Jagy | Henry, what is the Thomason reference? I see two books, $$ $$ Other: Bollobás, Béla. / Thomason, Andrew / Erdîos, Paul, 1913- Title: Combinatorics, geometry, and probability : a tribute to Paul Erdîos / edited by Béla Bollobás, Andrew Thomason. Publ: Cambridge University Press Year: 1997 $$ $$ Other: Brightwell, Graham / Leader, Imre / Scott, Alexander / Thomason, Andrew / Bollobás, Béla. Title: Combinatorics and probability : celebrating Béla Bollabás's 60th birthday / [edited by] Graham Brightwell [et al.] Publ: Cambridge University Press Year: 2007 | |
| Aug 14, 2010 at 5:10 | comment | added | HJRW | Will, yes, that sounds right. Regarding getting between $K_{3,3}$ and $K_5$, there's a graph $X$ that contains a $K_{3,3}$ subdivision and that can be reduced by a single edge-contraction to $K_5$. So it follows that if $K_5$ is planar then so is $K_{3,3}$. I'm not sure about the other way around, off the top of my head, but presumably one can do something similar. | |
| Aug 14, 2010 at 5:07 | comment | added | HJRW | Just to clarify slightly, Thomason quickly observes that if a graph is planar then it has a polygonal embedding in the plane; so you can talk about faces before you've proved the JCT. | |
| Aug 14, 2010 at 4:22 | comment | added | Will Jagy | Thierry, thank you. $$ $$ jmoy, in my answer I say "Note that the concept of inside for the square uses elementary ideas such as convexity." You can demand that the square be in the $x,y$ plane with vertices at $(0,0), \; (1,0) \; (1,1) \; (0,1).$ A point $(x,y)$ is strictly inside when $ 0 < x < 1$ and $0 < y < 1.$ Then, place the fifth point at $(2,2),$ so that three of the connecting arcs are line segments and the final one, to $(0,0)$ is a semicircle with center at $(1,1).$ If you need any more explanation you might email me. | |
| Aug 14, 2010 at 4:10 | comment | added | Thierry Zell | Is the JCT used to define faces? I'd say yes, because a typical proof of Euler's formula relies on the idea that adding an edge will split a face in two. Even an exotic proof of the formula ought to use a similar argument, I believe. | |
| Aug 14, 2010 at 4:05 | comment | added | Jyotirmoy Bhattacharya | Thanks! This was something new to me. But if I tried to formalize the idea of "inside" and "outside" and give explicit constructions for the arcs wouldn't it all turn messy again? | |
| Aug 14, 2010 at 3:25 | history | edited | Will Jagy | CC BY-SA 2.5 |
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| Aug 14, 2010 at 2:45 | comment | added | Will Jagy | Henry, I don't think I know how to sort out first principles here. The Munkres book, same page, in exercise 4, has the reader use JCT to show that $K{3,3}$ is non-planar. I'm going to guess that the three facts are roughly equivalent in the sense of quick proofs in both directions for any pair. So the questions become, does the nonplanarity of the complete graph on five vertices imply JCT quickly, and is there some trickery where each nonplanar graph gives the other, all "quickly." | |
| Aug 14, 2010 at 2:24 | comment | added | Kerry | very nice proof! | |
| Aug 14, 2010 at 1:41 | comment | added | HJRW | Do you really need the Jordan Curve Theorem to see that $K_5$ is non-planar? I ask because Thomason gives a proof of the JCT using the fact that $K_{3,3}$ is non-planar. | |
| Aug 13, 2010 at 18:43 | comment | added | Somnath Basu | That's a neat idea! | |
| Aug 13, 2010 at 18:27 | history | answered | Will Jagy | CC BY-SA 2.5 |