Timeline for A question about Cantor's Power Set theorem without the Axiom of Choice
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 29, 2015 at 20:37 | vote | accept | Garabed Gulbenkian | ||
| Apr 28, 2015 at 20:02 | comment | added | Joel David Hamkins | One can also argue like this: if $X$ is infinite Dedekind finite, then $X$ is strictly smaller than $X(f)$, since it injects into $X(f)$ by the map $x\mapsto\{x\}$, and so if they were bijective than $X$ would be bijective with a strictly smaller set, a contradiction. | |
| Apr 28, 2015 at 19:18 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
added 126 characters in body; edited tags
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| Apr 28, 2015 at 19:16 | answer | added | Asaf Karagila♦ | timeline score: 5 | |
| Apr 28, 2015 at 19:15 | comment | added | Emil Jeřábek | Unless I’m missing something, $|X|=|X(f)|=|X(I)|$ implies $|P(X)|\le2|X|$, and it is easy to see that $|X(f)|\le|X|$ implies $2|X|\le|X|$, so this is not possible. | |
| Apr 28, 2015 at 18:29 | history | asked | Garabed Gulbenkian | CC BY-SA 3.0 |