Timeline for Is every Baire metric space a complete metric space in disguise?
Current License: CC BY-SA 4.0
8 events
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| Sep 23, 2020 at 18:11 | comment | added | Benjamin Steinberg | Seems the original question was asked on mo mathoverflow.net/questions/150472/… | |
| Sep 23, 2020 at 15:54 | comment | added | YCor | Feel free to edit, providing the MSE post as context and asking the question from your comment. | |
| Sep 23, 2020 at 15:44 | comment | added | Will Brian | @fedja: Let me address your comment. I think it's consistent with ZF that every metrizable Baire space contains a dense, completely metrizable $G_\delta$. (I'd have to think some more to be sure, though.) But using the Axiom of Choice, one may construct a Bernstein set -- a subset $X$ of $\mathbb R$ such that neither $X$ nor $\mathbb R \setminus X$ contains a homeomorphic copy of the Cantor space. Then $X$ is Baire, but no uncountable subset of $X$ is completely metrizable. | |
| Sep 23, 2020 at 15:30 | comment | added | fedja | @YCor Great, thank you! Of course, it just pushes the question one level up to "Does every metrizable Baire topological space contain a dense $G_\delta$ subset homeomorphic to a complete metric space?" (for some reason I understood that going down from a complete metric space won't change anything but the idea to add some dust back did not occur to me). However, as posed, the question is answered completely, indeed :-) | |
| Sep 23, 2020 at 14:44 | comment | added | YCor | This is answered negatively here at MathSE. | |
| Sep 23, 2020 at 14:43 | comment | added | YCor | An obvious restatement: is every metrizable Baire topological space homeomorphic to a complete metric space? | |
| Sep 23, 2020 at 14:42 | history | edited | YCor |
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| Sep 23, 2020 at 14:38 | history | asked | fedja | CC BY-SA 4.0 |