Timeline for Finite axiom of choice: how do you prove it from just ZF?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2013 at 14:19 | comment | added | David E Speyer | See math.stackexchange.com/questions/64237/finite-choice-without-ac | |
| Jan 20, 2013 at 12:38 | answer | added | Alexey Muranov | timeline score: 9 | |
| Jul 19, 2010 at 23:40 | answer | added | Andreas Blass | timeline score: 30 | |
| Jul 19, 2010 at 22:26 | history | edited | François G. Dorais |
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| Jul 19, 2010 at 21:51 | answer | added | Thomas Scanlon | timeline score: 29 | |
| Jul 19, 2010 at 21:51 | answer | added | François G. Dorais | timeline score: 22 | |
| Jul 19, 2010 at 21:48 | comment | added | Daniel Litt | Ah sorry; I think the OP's question boils down to: how, from the axioms of ZF, do we "explicitly write down an ordered pair"? Since if we can write down ordered pairs, we're obviously done. | |
| Jul 19, 2010 at 21:41 | comment | added | Charles Staats | The product is, by definition, the set of all ordered pairs. In this case, we can explicitly write down an ordered pair, so clearly the product is nonempty. But I'm not entirely sure I understand your comment, so my apologies if this comment does not address it. | |
| Jul 19, 2010 at 21:37 | comment | added | Daniel Litt | Well, saying the product is non-empty is equivalent to the axiom of choice. I assume the OP means, how do we show this claim from the axioms? | |
| Jul 19, 2010 at 21:33 | comment | added | Charles Staats | Sorry, that should be {(x, y)}. | |
| Jul 19, 2010 at 21:29 | comment | added | Charles Staats | By definition of "nonempty", x contains an element y, so let $f$ be given by the set of ordered pairs {({x}, y)}. | |
| Jul 19, 2010 at 21:18 | history | asked | user7758 | CC BY-SA 2.5 |