Timeline for A Problem about partitioning $S^2$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2013 at 15:16 | comment | added | woodbass | @Alexandre Eremenko: I am sorry for misunderstanding your construction and take my objection back cordially. | |
| Jan 13, 2013 at 14:52 | comment | added | Alexandre Eremenko | @woodbass: the argument $\alpha$ is called commensurable with $\pi$ if $\alpha/\pi$ is rational. So the set $A$ consists of punctured lines (lines with $0$ deleted), not of rays. | |
| Jan 13, 2013 at 8:12 | comment | added | woodbass | @ Gerhard Paseman:If "A the set of points with argument commensurable with $\pi$" implies that A is the ray $(-\infty,0)$ in $\mathbb{R}$, then my understanding is correct and so is my objection. | |
| Jan 13, 2013 at 1:30 | comment | added | Gerhard Paseman | Woodbass, your comments are unclear to me. One of my suggestions is an example with monochromatic circles. Eremenko's example is a stereographic projection, with a coloring that has two singleton point classes. Any circle through the two points has (by construction) points of only one other color. Any line through C does not intersect both A and B, while any circle through C avoids D. I do not see any coloring yet in which all circles are tricolored. Gerhard "Ask Me About Color Counting" Paseman, 2013.01.12 | |
| Jan 12, 2013 at 18:46 | 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 6:28 | comment | added | Gerhard Paseman | Indeed,as I observed also in a comment to your solution, Alexandre. Your solution has the feature that at least two colors occur on every circle though, which is not the case for this last suggestion. Gerhard "Ask Me About System Design" Paseman, 2012.12.29 | |
| Dec 29, 2012 at 17:03 | comment | added | Alexandre Eremenko | Why not partition some circle S into 3 arbitrary non-empty sets, and the complement of S is the 4-th set. Let C be any round circle. It intersects S at at most 2 points, or coincides with S. So it intersects at most 3 sets. | |
| Dec 28, 2012 at 19:24 | comment | added | Lee Mosher | I took it to mean a great circle. If not, do you mean round circles? Topologically embedded circles? | |
| Dec 28, 2012 at 18:31 | comment | added | woodbass | Maybe you didn't understand my question. Under your partition, every circle in the the complement of the equator (i.e. the fourth set) passes through the fourth set only. | |
| Dec 28, 2012 at 18:23 | history | answered | Lee Mosher | CC BY-SA 3.0 |