Timeline for Partitioning a set of cardinality $\kappa$ into more than $\kappa$ disjoint subsets
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| 2 days ago | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
As the thread was bumped anyway, might as well http->https the link.
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| Dec 14, 2022 at 23:13 | comment | added | Andrés E. Caicedo | @Gabe Oops. Yes, of course, non-well-orderable. | |
| Dec 14, 2022 at 4:19 | history | edited | bof | CC BY-SA 4.0 |
corrected a typo
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| Dec 14, 2022 at 2:10 | comment | added | Gabe Goldberg | @AndrésE.Caicedo If $A$ is wellorderable, it cannot be partitioned into more than $|A|$ pieces. I guess you meant to ask whether a nonwellorderable uncountable set has such a "paradoxical" partition. I think it is not too hard to give a positive answer in the $L(\mathbb R)$ case for sets that are the surjective image of $\mathbb R$ using the fact that $\mathbb R$ injects into such a set and $\mathbb R$ surjects onto a strictly larger set. | |
| Dec 14, 2022 at 1:16 | comment | added | Timothy Chow | @AsafKaragila Got it. So I guess I'm proposing an anti-anti-anti-Banach-Tarski argument here? | |
| Dec 13, 2022 at 23:33 | comment | added | Asaf Karagila♦ | @TimothyChow: I was told that Stan was following my blog. And I am referring to a much earlier post, probably around 2014 or 2015. Specifically, this and that. | |
| Dec 13, 2022 at 22:59 | comment | added | Timothy Chow | @AsafKaragila What do you mean that "part of the motivation ... was my blog post"? Do you mean that Taylor & Wagon were motivated in part by the desire to respond to your blog post? But your blog post appears to be dated later than the Taylor & Wagon paper, if you're referring to the blog post linked in this answer of yours. | |
| Dec 13, 2022 at 22:49 | comment | added | Asaf Karagila♦ | @TimothyChow: To my understanding, part of the motivation behind the Taylor & Wagon paper was my blog post about how the Division Paradox is an anti-anti-Banach–Tarski-based-anti-AC argument. | |
| Dec 13, 2022 at 22:47 | comment | added | Timothy Chow | @SamHopkins Just to clarify, I'm not trying to demonstrate that there is something really paradoxical about denying AC. It's almost the opposite. I'm trying to undermine a specific argument that there is something really paradoxical about denying AC. | |
| Dec 13, 2022 at 22:38 | comment | added | Asaf Karagila♦ | @Sam: Presumably you want to refine that question, since any constant function will do the trick. Presumably you want to say that $f$ lifts to an injective function on $A/{\sim}$ somehow? (But probably that is "too strong" here.) | |
| Dec 13, 2022 at 22:30 | vote | accept | Timothy Chow | ||
| Dec 13, 2022 at 22:28 | comment | added | Andrés E. Caicedo | @Sam As far as I know, it is still open whether the failure of choice implies the existence of sets $A$ admitting a partition into more than $A$ pieces. | |
| Dec 13, 2022 at 22:27 | comment | added | Andrés E. Caicedo | But the real question should be: In Solovay's model, if $A$ is uncountable, is there a partition of $A$ into more than $A$ pieces? (Don't know whether this would be easier to answer if instead of Solovay's model one looks at $L(\mathbb R)$ assuming $\mathsf{AD}$. I would be curious to know, in any case.) | |
| Dec 13, 2022 at 21:45 | history | answered | Asaf Karagila♦ | CC BY-SA 4.0 |