Timeline for Is there a measure zero set which isn't meagre?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2010 at 23:25 | comment | added | George Lowther | So, the set of normal numbers is meagre and has full Lebesgue measure. | |
| Oct 25, 2010 at 18:33 | comment | added | Bjørn Kjos-Hanssen | @Greg Kuperberg: well, as long as you have a co-meager set of measure 0, such as in the classical example from Oxtoby's book, then you have that. What's neat about Andreas Blass' example, though, is that it is very intuitive that it should have measure 0 and be comeager. | |
| Oct 25, 2010 at 12:39 | comment | added | Greg Kuperberg | This is really ideal, expressing $[0,1]$ as a union of a meager set and a measure 0 set. That kills two birds with one stone. | |
| Oct 25, 2010 at 12:34 | history | answered | Andreas Blass | CC BY-SA 2.5 |