Timeline for A Baire subset of reals that is not Suslin measurable
Current License: CC BY-SA 4.0
12 events
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Jan 4, 2022 at 0:51 | history | undeleted | Gabe Goldberg | ||
Jan 4, 2022 at 0:48 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
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Dec 20, 2021 at 14:22 | history | deleted | Gabe Goldberg | via Vote | |
Dec 20, 2021 at 14:22 | comment | added | Gabe Goldberg | I think defining a surjection from $\omega^\omega$ onto the result of applying the Suslin operation to all countable sequences of open sets is all a bit more subtle than I thought, so I'm going to delete this answer... One shouldn't try to work in ZF at 5am I guess. | |
Dec 20, 2021 at 14:07 | comment | added | Gabe Goldberg | Also I think Kechris's book Classical Descriptive Set Theory says a lot about analytic and coanalytic sets as does Moschovakis's Descriptive Set Theory. | |
Dec 20, 2021 at 13:56 | comment | added | Gabe Goldberg | I made some edits, hope that clears things up | |
Dec 20, 2021 at 13:54 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
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Dec 20, 2021 at 13:49 | comment | added | Clement Yung | Sorry, I'm not sure what you're saying. Can you elaborate? | |
Dec 20, 2021 at 13:46 | comment | added | Gabe Goldberg | I guess was considering all sets obtained applying your Suslin operation to schemes of open sets, which may not itself be closed under your Suslin operation in ZF. | |
Dec 20, 2021 at 13:44 | comment | added | Clement Yung | Also, do you suggest any references that I can refer to for an exploration on the theory of coanalytic sets? | |
Dec 20, 2021 at 13:43 | comment | added | Clement Yung | Thank you very much! However, it appears that as Gerald Edgar suggested, in a model of $\mathsf{ZF}$ which every subset of reals is a countable union of countable sets, so trivially every subset of reals is Suslin measurable (and hence there doesn't exist a Baire set that is not Suslin measurable). Why does that not contradict your proof? | |
Dec 20, 2021 at 13:33 | history | answered | Gabe Goldberg | CC BY-SA 4.0 |