Timeline for Is there a measure zero set which isn't meagre?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 9, 2021 at 19:36 | history | suggested | Yngve Moe | CC BY-SA 4.0 |
Fixed dead link
|
| Sep 9, 2021 at 16:49 | review | Suggested edits | |||
| S Sep 9, 2021 at 19:36 | |||||
| S Dec 21, 2019 at 1:13 | history | suggested | user136906 | CC BY-SA 4.0 |
Fix formatting
|
| Dec 20, 2019 at 23:39 | review | Suggested edits | |||
| S Dec 21, 2019 at 1:13 | |||||
| Oct 19, 2015 at 16:47 | comment | added | Stefan Geschke | @The User: Yes, you are absolutely right. Thanks for pointing this out. | |
| May 20, 2013 at 20:24 | comment | added | The User | In fact Shelah proved that “ZF+CC+all sets of real numbers are measurable” is equiconsistent to the existence of an inaccessible cardinal. Without countable choice you could simply choose a model of ZF where $\mathbb{R}$ is a countable union of countable sets. | |
| Oct 29, 2010 at 2:33 | vote | accept | Anton Geraschenko | ||
| Oct 29, 2010 at 2:32 | comment | added | Anton Geraschenko | Even though this isn't strictly an answer to the question, I'm going to accept it on the reasoning that it'll probably be most useful to someone visiting this thread in the future. I don't like the thought of somebody reading just the first couple of answers and missing this gem. | |
| Oct 26, 2010 at 6:49 | history | edited | Stefan Geschke | CC BY-SA 2.5 |
added 15 characters in body
|
| Oct 25, 2010 at 9:48 | history | answered | Stefan Geschke | CC BY-SA 2.5 |