Timeline for Axiom of Countable Choice and meager sets
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2020 at 16:30 | comment | added | Martin Väth | Asaf, yes that's what I meant exactly: It is consistent (with ZF) that UMM is unprovable. That's why I wrote "quite the opposite" (to the assertion that UMM is provable in ZF). | |
| Sep 13, 2020 at 20:03 | comment | added | Asaf Karagila♦ | Martin, ZF proves that $\Bbb R$ is not meager, it does not prove that it is a countable union of countable sets. Therefore it is consistent that the countable union of meager sets is not meager. | |
| Sep 13, 2020 at 16:19 | comment | added | Martin Väth | @D.S. Lipham. Quite the opposite. From GabeGoldberg's (correct) observation, it follows even in ZF that the real line is not meager. Hence the argument in the original question shows indeed that UMM is unprovable in ZF. | |
| Sep 13, 2020 at 16:05 | comment | added | D.S. Lipham | @GabeGoldberg But wouldn't that prove UMM in ZF? | |
| Sep 13, 2020 at 15:54 | comment | added | Gabe Goldberg | The Baire category theorem for separable spaces is provable in ZF since the dependent choices one makes can be restricted to a countable dense subset, which is wellordered at the outset so that one can choose canonically. | |
| Sep 13, 2020 at 15:38 | history | edited | Martin Väth | CC BY-SA 4.0 |
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| Sep 13, 2020 at 13:49 | history | undeleted | Martin Väth | ||
| Sep 13, 2020 at 13:48 | history | deleted | Martin Väth | via Vote | |
| Sep 13, 2020 at 13:46 | review | First posts | |||
| Sep 13, 2020 at 14:42 | |||||
| Sep 13, 2020 at 13:42 | history | answered | Martin Väth | CC BY-SA 4.0 |