Timeline for Partition of R into midpoint convex sets
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2019 at 21:57 | comment | added | Dylan Thurston | I believe Sierpinski's result that every midpoint-convex, measurable function is continuous is essentially equivalent to this answer. | |
| Apr 24, 2010 at 18:49 | comment | added | Joel David Hamkins | François, very nice. | |
| Apr 24, 2010 at 8:17 | vote | accept | filipm | ||
| Apr 24, 2010 at 8:17 | |||||
| Apr 23, 2010 at 19:08 | comment | added | François G. Dorais | You can avoid the inaccessible cardinal by using Baire category as in this answer - mathoverflow.net/questions/16666/… | |
| Apr 23, 2010 at 13:20 | comment | added | Joel David Hamkins | Thomas, yes, I see. | |
| Apr 23, 2010 at 13:04 | comment | added | Thomas Kragh | @Joel David Hamkins: If such a set contains a ray it is a ray by trivial arguments. | |
| Apr 23, 2010 at 12:59 | comment | added | Wadim Zudilin | This solution answers my question(s). | |
| Apr 23, 2010 at 12:55 | comment | added | Joel David Hamkins | Keivan, but don't you need to work a little bit harder? You showed that A or B contains a ray, but you really want that they are both rays. But I think you can just repeat your argument again on the complement of the ray that was found. | |
| Apr 23, 2010 at 12:52 | comment | added | Joel David Hamkins | Solovay showed that if the existence of an inaccessible cardinal is consistent with ZFC, then the assertion that every set of reals is Lebesgue measurable is consistent with ZF+DC. So if large cardinals are consistent, then this argument shows that even DC is insufficient to make the desired partition. | |
| Apr 23, 2010 at 12:52 | comment | added | Steven Gubkin | doh ${ }$ | |
| Apr 23, 2010 at 12:50 | comment | added | Thomas Kragh | The assumptions on $A$ is $(A+A)/2 \subset A$. | |
| Apr 23, 2010 at 12:47 | comment | added | Steven Gubkin | I think this is close to an answer but I do not understand why A+A containing an interval implies that A. | |
| Apr 23, 2010 at 12:41 | history | answered | Keivan Karai | CC BY-SA 2.5 |