Timeline for A possible ${\sf (ZF)}$-theorem in the spirit of the $3$-set-lemma
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 23 at 17:36 | vote | accept | Dominic van der Zypen | ||
| Feb 23 at 9:27 | comment | added | Emil Jeřábek | If $\kappa_0$ is meant to be a well-ordered cardinal, then the statement is equivalent to the axiom of choice for families of finite sets, just like (S$_3$). This is shown in my answer to the linked MO question. | |
| Feb 23 at 8:55 | answer | added | bof | timeline score: 4 | |
| Feb 23 at 7:09 | comment | added | Dominic van der Zypen | @bof Thanks for your argument for the special case - it's nice to think about it in coloring terms! If you want to elaborate this in an answer, that would be great. | |
| Feb 22 at 20:22 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
slight change in the final paragraph
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| Feb 22 at 17:20 | comment | added | Dominic van der Zypen | Thanks @AndreasBlass for your comment; I made the question more general in the spirit of ${\sf S}_3$. | |
| Feb 22 at 17:15 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
generalized the question and put original question as remark
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| Feb 22 at 15:46 | comment | added | Andreas Blass | Your question has $\mathbb N$ where $(S_3)$ has an arbitrary nonempty $X$. Doesn't that change make AC unnecessary in the proof? | |
| Feb 22 at 15:29 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
deleted 378 characters in body
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| Feb 22 at 15:00 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
added 536 characters in body
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| Feb 22 at 14:52 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |