Timeline for Is every convex subset of a Borel-linearly ordered space measurable?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2015 at 20:00 | vote | accept | Julian Newman | ||
| Sep 21, 2015 at 19:24 | answer | added | Nate Eldredge | timeline score: 2 | |
| Sep 20, 2015 at 17:24 | comment | added | Nate Eldredge | A thought: is it possible for a Borel order to have a set isomorphic to $\omega_1$? It seems unlikely. If not, then take a cofinal well-ordered subset $C$ of $A$, which must then be countable. Fix some $x_0 \in A$ and then you have $A \cap [x,\infty) = \bigcup_{y \in C} [x,y]$ which is Borel. Applying the same argument to the reverse order, $A \cap (-\infty, x]$ is also Borel. | |
| Sep 20, 2015 at 16:39 | history | asked | Julian Newman | CC BY-SA 3.0 |