Timeline for Is every set being cardinal definable consistent with ZF + negation of Choice?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 27, 2021 at 18:09 | vote | accept | Zuhair Al-Johar | ||
| Dec 27, 2021 at 9:13 | answer | added | Elliot Glazer | timeline score: 12 | |
| Dec 26, 2021 at 19:59 | comment | added | Zuhair Al-Johar | @Gro-Tsen, I see your point. I actually tried to provide some context by the linked sites, but I'll add that point to the question. Thanks. | |
| Dec 26, 2021 at 19:57 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
added 192 characters in body
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| Dec 26, 2021 at 19:15 | comment | added | Dmytro Taranovsky | This is a good question. I expect the answer to be yes, but one essentially has to cook up a model of ZF with cardinal structure as rich as its set structure (at least where choice fails). | |
| Dec 26, 2021 at 18:51 | comment | added | Gro-Tsen | @Wojowu Actually I knew that, but my point was mostly to remind the poster that when asking a question it is in good taste to give motivation and context, including the answer to very obviously related questions when possible. | |
| Dec 26, 2021 at 17:54 | comment | added | Wojowu | @Gro-Tsen The answer is no in that case. The class of ordinal-definable sets has a definable well-ordering over ZF, so if all sets are ordinal-definable, AC holds. | |
| Dec 26, 2021 at 16:30 | comment | added | Gro-Tsen | Do you know the answer if “cardinal” is replaced by “ordinal” in your question? If so, it would be seemly to recall what it is as part of the context for your question. If not, you might also say so. | |
| Dec 26, 2021 at 12:45 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |