Timeline for Is the existence of a well-ordering on R independent of ZF?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2014 at 8:34 | comment | added | Lehs | @Ori Gurel-Gurevich: does this means that R is not a set according to ZF? | |
| Nov 12, 2009 at 0:27 | comment | added | Ori Gurel-Gurevich | Qiaochu: there's a classical argument for this. Let $X$ and $Y$ be two independent $U[0,1]$ RV. What is the probability that $X<Y$ in the well-ordering? | |
| Nov 12, 2009 at 0:25 | comment | added | Ori Gurel-Gurevich | Richard: you're right, I forgot about that. | |
| Nov 12, 2009 at 0:05 | comment | added | Richard Dore | One minor quibble is that Solovay's model requires the existence of an inaccessible. This is a minor assumption, but it is unnecessary here. | |
| Nov 11, 2009 at 23:25 | comment | added | Eric Wofsey | Well, you should also make it not differ from any Borel by a null set, and for this you need to also simultaneously diagonalize on the null Borel sets. But yeah, that's the idea. | |
| Nov 11, 2009 at 23:19 | vote | accept | Qiaochu Yuan | ||
| Nov 11, 2009 at 23:18 | comment | added | Qiaochu Yuan | Alright, so I think I know what the argument should be: a well-ordering of R induces a well-ordering of the Borel algebra, so we can bijectively assign to every r a Borel set B(r) and let S = {r | r \not \in B(r)}). This subset is non-empty because some B(r) is the empty set, and it is clearly non-measurable. Is this correct? | |
| Nov 11, 2009 at 23:02 | history | answered | Ori Gurel-Gurevich | CC BY-SA 2.5 |