Timeline for Uncountable chain of nested sets without choice
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 1, 2019 at 21:33 | comment | added | Anindya | Thanks for the answers in the comments above. Why the votes to close, though ? It's not a super-advanced question, but from the comments , it looks like there are some non-trivial issues here. | |
| Sep 30, 2019 at 12:25 | comment | added | Andreas Blass | If, as previous comments indicate, $\kappa$ is supposed to be a well-ordered cardinal, then the answer is yes, because it's provable in ZF (without choice) that every infinite ordinal $\alpha$ admits a bijection to $\alpha\times\alpha$. Indeed, if you know this with choice, then it follows without choice, because you can find the required bijection in Gödel's constructible universe $L$. | |
| Sep 30, 2019 at 8:09 | comment | added | Wojowu | Note that if we intend for $\kappa$ to not necessarily be well-orderable (and taking into account the other comments to make the statement make sense) this is definitely false in general, as $S$ may not have any proper subsets of the same cardinality. | |
| Sep 30, 2019 at 5:27 | comment | added | Asaf Karagila♦ | @Nate: That is not the usual definition. However, in order to make sense of $\lambda\in\kappa$ we pretty much have to assume that $\kappa$ is indeed a well-ordered cardinal. | |
| Sep 30, 2019 at 4:35 | review | Close votes | |||
| Oct 6, 2019 at 3:05 | |||||
| Sep 30, 2019 at 4:18 | history | edited | Martin Sleziak |
added the (axiom-of-choice) tag
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| Sep 30, 2019 at 4:10 | history | edited | LeechLattice |
edited tags
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| Sep 30, 2019 at 3:05 | comment | added | Nate Eldredge | Are we using here the usual ZF definition of "cardinal" as "an ordinal not in bijection with any lesser ordinal"? If so, then $\kappa$ automatically has a well ordering. | |
| Sep 30, 2019 at 2:47 | history | asked | Anindya | CC BY-SA 4.0 |