Timeline for Can we have all classes that can be partitioned into non-singleton finite sets, be sets?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 10, 2022 at 23:42 | comment | added | Noah Schweber | @AlecRhea Whoops, fixed! | |
| Aug 10, 2022 at 23:42 | history | edited | Noah Schweber | CC BY-SA 4.0 |
deleted 4 characters in body
|
| Aug 10, 2022 at 23:39 | comment | added | Alec Rhea | Is the last condition supposed to be $m\approx n$? | |
| Aug 10, 2022 at 20:52 | comment | added | Zuhair Al-Johar | I'll post that as a new question. | |
| Aug 10, 2022 at 20:38 | comment | added | Zuhair Al-Johar | Yes! I had the feel that well ordered sets can violate what I'm saying, I think we need to require $X$ being non-well orderable, for this to be possible. So, the idea is that all non-well orderable classes of pairwise disjoint larger than singleton finite sets is a set. | |
| Aug 10, 2022 at 20:32 | vote | accept | Zuhair Al-Johar | ||
| Aug 10, 2022 at 19:26 | history | answered | Noah Schweber | CC BY-SA 4.0 |