Timeline for A Problem about partitioning $S^2$
Current License: CC BY-SA 3.0
12 events
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| Jan 13, 2013 at 8:11 | comment | added | woodbass | @Alexandre Eremenko:Does not "A the set of points with argument commensurable with $\pi$" imply that A is the ray $(-\infty,0)$ in $\mathbb{R}$? $B=\mathbb{C}-A-\lbrace{0\lbrace}$? My understanding is correct? | |
| Jan 12, 2013 at 21:43 | comment | added | Alexandre Eremenko | @woodbass: I do not understand your objection: there is no "line containing A" in my example; A is a dense set in my example. | |
| Jan 12, 2013 at 18:45 | comment | added | woodbass | @Alexandre Eremenko: There is a mistake in your construction. The line containing C and A passes through both B and D. | |
| Dec 30, 2012 at 8:49 | history | edited | Angelo | CC BY-SA 3.0 |
added 4 characters in body
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| Dec 29, 2012 at 1:02 | comment | added | Gerhard Paseman | Actually, my last example allows for monochromatic circles, regardless of whether they are great or not. Lee's example specifically prohibits monochromatic great circles. I suspect there is no coloring that has every circle with exactly three of the four colors, and certainly no such coloring must have a monchromatic open subset. Your example is not likely to be bested soon. Gerhard "Making More Words To Eat" Paseman, 2012.12.28 | |
| Dec 28, 2012 at 23:06 | comment | added | Gerhard Paseman | And, of course, this last example is similar to Lee Mosher's example. Time for me to stop. Gerhard "Dizzy From Coming Full Circle" Paseman, 2012.12.28 | |
| Dec 28, 2012 at 22:56 | comment | added | Gerhard Paseman | Time to eat my words. Pick a sphere colored A, then pick a circle on it and color it B. Then pick three points on the circle and color two of them C and one of them D. Any circle that has D, C, and B points must avoid the A color. This example can be generalized, but I don't see any set but A having measure greater than zero (with respect to the measure of spherical area). Gerhard "Please Pass The Salt Shaker" Paseman, 2012.12.28 | |
| Dec 28, 2012 at 22:40 | comment | added | Gerhard Paseman | With Alexandre's example in mind, I think a painting argument shows that any such coloring must have two of the colors limited to singleton sets. Looking at the circles passing through D and C, they determine a pencil of pairs of arcs which must be monochromatic off of the C and D regions, and then D and C' (C' also having color of C) determines a distinct pencil from the first (excepting one pair of arcs). It then becomes clear that any connected open subset not colored by C or D is monochromatic. Gerhard "Some Details Remain, Of Course" Paseman, 2012.12.28 | |
| Dec 28, 2012 at 22:16 | comment | added | Gerhard Paseman | Further, if C had more than one point, I suspect there is no nontrivial two coloring of the remainder that would avoid a circle with four colors. Gerhard "Not A Geometric Ramsey Theorist" Paseman, 2012.12.28 | |
| Dec 28, 2012 at 22:09 | comment | added | Yuichiro Fujiwara | You can do the curly braces by \lbrace and \rbrace in math mode. I learned this by right-clicking math in others' posts. You can literally see how they typed their math this way. | |
| Dec 28, 2012 at 22:07 | comment | added | Gerhard Paseman | For the non projected version, let C and D be (singleton sets of) antipodal points, and then partition the (pairs of) great circle arcs between them into two sets. Gerhard "Wish I Thought Of That" Paseman, 2012.12.28 | |
| Dec 28, 2012 at 21:56 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |