Timeline for Pair of curves joining opposite corners of a square must intersect---proof?
Current License: CC BY-SA 2.5
11 events
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| Aug 14, 2010 at 18:48 | comment | added | Will Jagy | Piecewise $C^1,$ mutual vertex avoidance and transversality should also be workable. | |
| Aug 14, 2010 at 18:27 | comment | added | Will Jagy | Peter, I think you ought to have the path avoid vertices, for the same reason you don't want it to coincide with an edge of the polygon. I imagine this can be worked out for the path between your original $x$ and $x_0$ also piecewise linear, where no vertex of either figure (polygon or path) lies on an edge of the other, and all edge to edge intersections are transverse. $$ $$ Goodness, Chelsea 6, West Bromwich Albion 0. | |
| Aug 14, 2010 at 18:13 | comment | added | Pierre-Yves Gaillard | Peter: Thanks for your answer. If I understand your comments (please correct me if I'm wrong), what you have in mind is a short selfcontained presentation of John's argument, based on a weakened version of the polygonal JCT. In this case, it'd be great, I think, if you wrote it, with the necessary details, as a separate answer, complementary to John's. | |
| Aug 14, 2010 at 17:34 | comment | added | Peter LeFanu Lumsdaine | @Pierre: thanks, yes, I did mean $\mathbb{R}^2$; and while the segment can't cross $C$ infinitely often (a polygon has finitely many edges by definition, at least in the conventions I know) it could contain an edge of $C$ as a subsegment, in which case we do have to look at what directions the adjacent edges of $C$ go off in. @Per: you're right, of course; this doesn't establish the full Jordan curve theorem; I was just thinking of what was necessary for the application at hand (which just needs disconnectedness plus the fact that the second curve's endpoints are in opposite components). | |
| Aug 14, 2010 at 14:49 | comment | added | Per Vognsen | Peter: That proof goes through even when the polygon isn't simple, e.g. a polygonal figure-eight, where the parity function disconnects the plane into more than two components. So, there's a bit more work to do. | |
| Aug 14, 2010 at 13:54 | comment | added | Pierre-Yves Gaillard | There is an abundant literature about the JCT for polygons. To search Google for the words: jordan curve theorem for polygons, click here: google.com/… | |
| Aug 14, 2010 at 11:18 | comment | added | Pierre-Yves Gaillard | Dear Peter: Your line segment may cross $C$ infinitely many times. [You probably mean $\mathbb{R}^2\setminus C$, not $\mathbb{R}\setminus C$.] | |
| Aug 14, 2010 at 6:47 | comment | added | Peter LeFanu Lumsdaine | This is gorgeous! It's possibly worth pointing out quite how elementary the Jordan curve theorem for polygons is, to show how little is being black-boxed here. Fix some point $x_0$ not on $C$, and for any (other) point $x$ not on $C$, look at the line segment from $x$ to $x_0$; count the parity of how many times it crosses $C$ (counting double/none if it hits vertices of $C$). This is well-defined and locally constant on $\mathbb{R}\setminus C$ (this is where we use that $C$ is a polygon); so as a locally constant, surjective function to $\{0,1\}$, it disconnects $\mathbb{R}^2$. | |
| Aug 14, 2010 at 6:14 | comment | added | Victor Protsak | This is a great argument! | |
| Aug 14, 2010 at 4:15 | history | edited | John Stillwell | CC BY-SA 2.5 |
"polygonal paths" instead of "polygons" at first
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| Aug 14, 2010 at 3:09 | history | answered | John Stillwell | CC BY-SA 2.5 |