Timeline for Is there a form of choice that can elude Kunen's inconsistency theorem?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2023 at 8:58 | comment | added | Robert Furber | @JamesHanson For any large cardinal axiom A, one cannot prove con(ZF) => con(ZF + A) unless ZF is inconsistent, so that doesn't say much. | |
| Jan 13, 2023 at 3:24 | history | became hot network question | |||
| Jan 12, 2023 at 21:05 | vote | accept | Zuhair Al-Johar | ||
| Jan 12, 2023 at 20:34 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
deleted 1 character in body
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| Jan 12, 2023 at 20:33 | answer | added | Gabe Goldberg | timeline score: 23 | |
| Jan 12, 2023 at 20:06 | comment | added | Zuhair Al-Johar | @JamesHanson, the question is about if it is an open question whether Reinhardt cardinals are consistent with ZF + some form of choice? Or is it the case that it is proved that all known forms of choice are inconsistent with having a Reinhardt's cardinal. Put it the other way, what is the minimal known form of choice needed for Kunen's inconsistency theorem to work? | |
| Jan 12, 2023 at 20:02 | comment | added | James E Hanson | I'm confused as to why that doesn't answer your question. | |
| Jan 12, 2023 at 20:01 | comment | added | Zuhair Al-Johar | @JamesHanson, AFAIK, Yes! | |
| Jan 12, 2023 at 19:58 | comment | added | James E Hanson | Isn't it open whether Rheinhardt cardinals are consistent with $\mathsf{ZF}$ at all? | |
| Jan 12, 2023 at 19:23 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |