Timeline for Is AC equivalent over ZF to 'every fibration can be equipped with a cleavage'?
Current License: CC BY-SA 4.0
14 events
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| Jan 20, 2021 at 18:28 | comment | added | Alec Rhea | @AsafKaragila Fair enough; apologies if my first comment came across as aggressive. | |
| Jan 20, 2021 at 18:24 | comment | added | Asaf Karagila♦ | @Alec: Will asked about a variant of AC. I merely pointed out that this variant is equivalent to full AC. Whatever Will may or may not saw in your question, you'll have to ask him. | |
| Jan 20, 2021 at 18:18 | comment | added | Alec Rhea | @AsafKaragila I'm trying to understand what connection Will saw between my question and the question he asked me; you answered his question with a yes, so I thought you sa the same connection he did. Your comment seems to be answering yes to the question "is choice for equipotent sets equivalent to full choice", but I never asked about choice for equipotent sets and I still don't see what connection Will saw without going through the construction given by Simon to go from your answer connecting equipotent choice to my question about fibrations. Am I missing something? | |
| Jan 20, 2021 at 18:01 | comment | added | Asaf Karagila♦ | @Alec: Did you read the first comment on your question? I'm not salty, I'm just trying to point out that once a discussion starts in the comments, it can take its own form. | |
| Jan 20, 2021 at 17:59 | comment | added | Alec Rhea | @AsafKaragila That is cool, but how is this related to my question? I'm not salty, I enjoyed reading your comment, but I don't see the connection Will made between my question and the question about choice for equipotent sets. | |
| Jan 20, 2021 at 16:34 | comment | added | Asaf Karagila♦ | @Will: Yes. Given any collection of non-empty sets, $\{A_i\mid i\i I\}$, let $\alpha$ be an ordinal such that $A_i\subseteq V_\alpha$ for all $i$, and consider $\{A_i\times V_\alpha\mid i\in I\}$. Since $V_\alpha$ is equipotent with its square, all those sets are of the same cardinality: $|V_\alpha|$. Now if $F$ is a choice function from this family, we easily get a choice function for the original family by composing with the left projection. | |
| Jan 20, 2021 at 14:35 | review | Suggested edits | |||
| Jan 20, 2021 at 14:50 | |||||
| Jan 20, 2021 at 8:16 | history | became hot network question | |||
| Jan 20, 2021 at 5:32 | vote | accept | Alec Rhea | ||
| Jan 20, 2021 at 3:31 | answer | added | Simon Henry | timeline score: 17 | |
| Jan 20, 2021 at 2:10 | history | edited | Alec Rhea | CC BY-SA 4.0 |
added 282 characters in body
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| Jan 20, 2021 at 2:00 | comment | added | Alec Rhea | @WillSawin Not that I can tell, but perhaps you're seeing something I'm not? The sets of Cartesian arrows parametrized by arrows in the base category and objects above their codomains in the overcategory need not have the same cardinality, unless I'm missing something. | |
| Jan 20, 2021 at 0:19 | comment | added | Will Sawin | Is this the same as asking if AC is equivalent to the axiom of choice for collections of sets which all have the same cardinality (i.e. for any two sets in the collection, there exists a bijection between them)? | |
| Jan 20, 2021 at 0:14 | history | asked | Alec Rhea | CC BY-SA 4.0 |