Timeline for When does $A^A=2^A$ without the axiom of choice?
Current License: CC BY-SA 3.0
21 events
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| May 22 at 14:56 | history | protected | CommunityBot | ||
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Jul 2, 2014 at 16:26 | vote | accept | Asaf Karagila♦ | ||
| Jul 2, 2014 at 11:17 | answer | added | Ioanna | timeline score: 33 | |
| May 8, 2014 at 20:23 | comment | added | Asaf Karagila♦ | @Yair: I'm not quite clear on your idea there. Maybe you can explain it tomorrow? | |
| May 8, 2014 at 19:49 | comment | added | Yair Hayut | Maybe we can say something like this: take a model of $\neg AC$ in which $\forall A,\,|A|+|A|=|A|$. Now if you have in addition for every infinite set $A$ a set $B$ with $A\leq 2^B$ and $2^{A} = 2^{2^B}$ (sort of very weak $GCH$) then it would follow that $A^A = 2^A$ for every $A$: it is true for $X = P(B)$, since $|X \times X| = 2^{|B| + |B|} = 2^{|B|} = |X|$, and for $A\leq X$ with $|2^A| = |2^X|$, $2^X \cong 2^A \leq A^A \leq X^X \cong 2^X$. | |
| Dec 26, 2013 at 13:02 | comment | added | Asaf Karagila♦ | @Yair: That's an equivalent question to the second one I ask here. | |
| Dec 26, 2013 at 7:31 | comment | added | Yair Hayut | Does it known whether $\forall A\,2^{A\times A} = 2^A$ implies $AC$? | |
| Feb 26, 2013 at 6:09 | comment | added | Asaf Karagila♦ | Does anyone have an idea why this was downvoted? | |
| Feb 25, 2013 at 23:58 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
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| Feb 25, 2013 at 0:47 | comment | added | JRN | @Daniel Spector, also when $A=0$ (as some people define $0^0=1$). :) | |
| Feb 25, 2013 at 0:37 | comment | added | Asaf Karagila♦ | Thanks Andres, I didn't notice that typo (I can't type the accent on my keyboard, but that's another story!) | |
| Feb 25, 2013 at 0:36 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
Typo!
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| Feb 24, 2013 at 23:42 | comment | added | Andrés E. Caicedo | (Sierpiński.)${}$ | |
| Feb 24, 2013 at 21:19 | comment | added | Asaf Karagila♦ | I wasn't sure whether or not the additional question makes it eligible for [reference-request]. | |
| Feb 24, 2013 at 21:18 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
added 299 characters in body
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| Feb 24, 2013 at 9:54 | comment | added | Adam Epstein | Which for some reason is excluded from the initial statement :) | |
| Feb 24, 2013 at 9:10 | comment | added | Daniel Spector | Because of the question "When does $A^A=2^A$ without the axiom of choice?", I could't help but comment - when A=2... | |
| Feb 24, 2013 at 6:30 | comment | added | Asaf Karagila♦ | Eric, yes. That would be the first immediate consequence. | |
| Feb 24, 2013 at 6:21 | comment | added | Eric Wofsey | Given that $A\times A=A$ for all $A$ is equivalent to AC, you could also ask whether the same is true of $A^A=2^A$. | |
| Feb 24, 2013 at 5:43 | history | asked | Asaf Karagila♦ | CC BY-SA 3.0 |