Timeline for Decidability of the Axiom of Choice
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Feb 23, 2011 at 19:15 | answer | added | user1448 | timeline score: 2 | |
| Feb 10, 2010 at 11:18 | history | edited | Charles Stewart |
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| Feb 10, 2010 at 11:17 | history | edited | Charles Stewart |
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| Feb 6, 2010 at 6:35 | comment | added | Anton Geraschenko | @Johannes: I got confused by that too. I think that when Theo says ZF-without-C he means (ZG + ¬AC), in which AC is clearly decidable. | |
| Feb 6, 2010 at 3:19 | answer | added | Joel David Hamkins | timeline score: 7 | |
| Feb 6, 2010 at 3:00 | answer | added | Jeremy Shipley | timeline score: 2 | |
| Feb 6, 2010 at 1:21 | answer | added | François G. Dorais | timeline score: 8 | |
| Feb 6, 2010 at 0:41 | comment | added | Hailong Dao | Oh, my bad. ZFC is ZF+ axiom of choice, sorry! | |
| Feb 6, 2010 at 0:35 | comment | added | Pete L. Clark | And I agree with Zev and Theo: ZFC seems to be an acceptable (and certainly nontrivial; that's a huge theorem of Godel) answer to this question. [Normally I refer to posters by their initials, to save space. But (i) this is, or was, a short enough comment; and (ii) there is enough alphabet around already.] | |
| Feb 6, 2010 at 0:32 | comment | added | Pete L. Clark | A technical comment: instead of "consistent" one should say "relatively consistent", i.e., consistent if ZF is consistent. The point is that the consistency of ZF is not provable from ZF [nor from, as I understand it, by any acceptably "finitistic" formal system], by Godel's Second Incompleteness Theorem. | |
| Feb 6, 2010 at 0:31 | comment | added | Johannes Hahn | @Theo: Can you elaborate this? In what sense is the AC decidable in ZF ? | |
| Feb 6, 2010 at 0:29 | answer | added | Johannes Hahn | timeline score: 7 | |
| Feb 6, 2010 at 0:28 | comment | added | Theo Johnson-Freyd | Well, C is decidable in ZF-without-C too. And in my reading, the current post says "stronger than ZF". | |
| Feb 6, 2010 at 0:25 | comment | added | Harry Gindi | I've posted an answer that is much stronger than ZFC. You can quantify over anything! | |
| Feb 6, 2010 at 0:21 | comment | added | Hailong Dao | He did say "stronger than ZFC". | |
| Feb 6, 2010 at 0:17 | comment | added | Zev Chonoles | I don't know much about set theory, but I assume you're looking for an answer other than ZFC? | |
| Feb 6, 2010 at 0:06 | history | asked | Daniel Katz | CC BY-SA 2.5 |