Timeline for A question about Cantor's Power Set theorem without the Axiom of Choice
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| May 4, 2015 at 22:11 | comment | added | Asaf Karagila♦ | I noticed a minor mistake in the answer; it is possible that $\operatorname{Card}(X)^2\nleq \operatorname{Card}(2^X)$; but it's always provable that if $X$ has more than four elements, then $\operatorname{Card}(2^X)\nleq\operatorname{Card}(X)^2$. We only need the one direction, as I point out in that Math.SE answer; so the second one has no business being here. | |
| May 4, 2015 at 22:09 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
It is possible that X^2\nleq 2^X, so I had to correct that.
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| May 1, 2015 at 19:39 | comment | added | Asaf Karagila♦ | Tell me about it! :-) | |
| May 1, 2015 at 19:38 | comment | added | Garabed Gulbenkian | You are right. I was finally able to digest all the steps of your proof. Proving theorems about all infinite cardinal numbers can be quite tricky when the Axiom of Choice is not available. | |
| Apr 30, 2015 at 21:40 | comment | added | Asaf Karagila♦ | Does the last paragraph uses anywhere that the set is Dedekind finite? Only that it has five different elements. | |
| Apr 30, 2015 at 19:09 | comment | added | Garabed Gulbenkian | In ZF set theory without the Axiom of Choice, there exist infinite sets X which are neither Alephs nor Dedekind-finite. Is it still true for such sets that CARD(2^X) is greater than 2*CARD(X)? | |
| Apr 29, 2015 at 20:52 | comment | added | Garabed Gulbenkian | Thanks for the neat proof-in ZF without the Axiom of Choice-that CARD(X(I)) is greater than CARD(X) when X is infinite. | |
| Apr 29, 2015 at 20:37 | vote | accept | Garabed Gulbenkian | ||
| Apr 28, 2015 at 19:16 | history | answered | Asaf Karagila♦ | CC BY-SA 3.0 |