Timeline for When does $A^A=2^A$ without the axiom of choice?
Current License: CC BY-SA 4.0
14 events
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| Aug 8, 2018 at 12:26 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added doi link (I did not find a free link - not behind a paywall - for this paper)
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| Aug 8, 2018 at 12:18 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added some links
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| Jul 2, 2014 at 16:26 | vote | accept | Asaf Karagila♦ | ||
| Jul 2, 2014 at 16:00 | comment | added | Ioanna | @Asaf, i just edited the answer to incorporate the comments, and try to dispel any confusion with these $\mathcal{F}$, upon which the construction of the model does not depend on. | |
| Jul 2, 2014 at 15:54 | history | edited | Ioanna | CC BY-SA 3.0 |
edited after the discussion in the comments, also 1 typo found.
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| Jul 2, 2014 at 15:42 | comment | added | Ioanna | In fact, according to [Pincus78], whose argument I found to be correct, in the Pincus model: for all infinite $x$, $x!=2^x=x^x=x^x-x!=|\mathcal{F}(x)|$ holds for all set valued operators $\mathcal{F}$ such that: (1) the predicate $y\in\mathcal{F}$ is absolute from $M$ to $V$, (2) $\mathsf{ZF}$ proves $|y|\leq x \Rightarrow |\mathcal{F}(y)|\leq |\mathcal{F}(y)|$ and $|2x|=|x|\Rightarrow 2^x\leq|\mathcal{F}(x)|$ for infinite $x$, and (3)$\mathsf{ZFC}$ proves $2^x=|\mathcal{F}(x)|$ for infinite $x$. The question now is, what other interesting $\mathcal{F}$ can we think of? :) | |
| Jul 2, 2014 at 15:26 | comment | added | Ioanna | In the Pincus model for all infinite $x$, $x!=2^x=x^x=x^x-x!$. | |
| Jul 2, 2014 at 15:21 | comment | added | Asaf Karagila♦ | @Andreas: It is easy to see that $x!\neq x^x$ in general, because in the case of a strongly amorphous set, $x!$ is a proper subset of $x^x$, which is a Dedekind-finite set. So there's that (basically, the basic Fraenkel model, yes). But in Pincus model it might be the case after all, which will answer my question. | |
| Jul 2, 2014 at 15:14 | comment | added | Andreas Blass | @AsafKaragila I don't think $x!=x^x$ in general. We always have an embedding of $x$ into $x^x$, using the constant functions, but I don't think there's always an embedding of $x$ into the set of permutations of $x$. Suppose, for example, that $x$ is the set of atoms in the basic Fraenkel model. | |
| Jul 2, 2014 at 14:47 | comment | added | Ioanna | oops, was writing the comment while you posted yours. I don't know whether $x!=x^x$ either. | |
| Jul 2, 2014 at 14:45 | comment | added | Ioanna | Well, the consequences dictionary is not the easiest book to search. I searched for this result a few times before I found it. About my first paragraph, I meant characterisation in the sense that for an infinite set $A$, it is not the case that $2^A=A^A \iff A=A\times A$, in particular $2^A=A^A \not\Rightarrow A=A\times A$. What did you mean by "characterise"? (And no problem at all, I'm not in any position to complain about slow email replies :) I also haven't forgotten about updating the file...) | |
| Jul 2, 2014 at 14:43 | comment | added | Asaf Karagila♦ | Okay, forget about the part I wrote I don't understand. Too much sleep deprivation. My brain is like cheese. I'm not 100% sure that $x!$ and $x^x$ must have the same cardinality, though. | |
| Jul 2, 2014 at 12:07 | comment | added | Asaf Karagila♦ | Strange. I know both papers, but I haven't read them thoroughly. I do recall searching the consequences dictionary and coming up short. I also don't quite understand your first paragraph; if it doesn't imply the axiom of choice, how does it mean it doesn't characterize? (And I haven't forgotten about the email. I'm writing my reply slowly...) | |
| Jul 2, 2014 at 11:17 | history | answered | Ioanna | CC BY-SA 3.0 |