Timeline for Existence of a certain set of 0/1-sequences without the Axiom of Choice
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 18, 2019 at 20:08 | comment | added | Andreas Blass | @AsafKaragila Avoiding large cardinals was my reason for using the Baire property here. An $\mathcal X$ as in the question can't be Lebesgue measurable either, but then I'd need an inaccessible cardinal, which somebody might not trust. | |
| Aug 18, 2019 at 20:05 | comment | added | Asaf Karagila♦ | Indeed that was my point. Even more so, the Shelah model shows that one does not even need to trust large cardinals to have such a model. | |
| Aug 18, 2019 at 20:03 | comment | added | Andreas Blass | @AsafKaragila As far as I"m concerned, it doesn't truly matter in this case, because the usual models for "all sets of reals have the Baire property" satisfy DC, so the problem disappears. (That's how I justified my laziness in the first place.) | |
| Aug 18, 2019 at 20:00 | comment | added | Asaf Karagila♦ | Indeed. That seems to be in line of what I remember on the topic. The question is, why does it truly matter in this case... :-) | |
| Aug 18, 2019 at 19:56 | comment | added | Andreas Blass | @AsafKaragila If one defines, as usual, "meager" to mean "covered by countably many closed sets with empty interior", then I think there's no danger of all sets (in Cantor space, say) being meager even without choice --- even when the reals are the union of countably many countable sets. The danger seems to be rather that the union of countably many meager sets need not be meager. | |
| Aug 18, 2019 at 12:33 | comment | added | Asaf Karagila♦ | But certainly the notion of having the Baire property doesn't make sense if every set is meager. So to that end, DC is certainly helpful. | |
| Aug 18, 2019 at 11:44 | comment | added | Andreas Blass | @AsafKaragila You're right that the separable case doesn't need any choice at all; I was just too lazy to check that and say it. | |
| Aug 18, 2019 at 8:34 | comment | added | Asaf Karagila♦ | While I agree that the statement of BCT makes more sense when assuming DC (which is equivalent to BCT, as we know), to make sense of BCT for separable spaces, such as the Cantor space, you don't need more than ZF. | |
| Aug 18, 2019 at 7:08 | vote | accept | M. Winter | ||
| Aug 17, 2019 at 23:20 | history | answered | Andreas Blass | CC BY-SA 4.0 |