11
$\begingroup$

I am currently giving lectures in real analysis and a student asked an interesting question I couldn't answer, so I'm posting it here:

Let's say that a metric space $X$ is Baire if every countable intersection of dense open sets is dense. Then, by the Baire category theorem, every complete metric space is Baire. The converse doesn't hold literally, of course, because the Baire property is preserved under topological isomorphisms while the completeness is not. The question is whether it is the only obstacle, i.e.,

Is it true that every Baire metric space $(X,d)$ admits a metric $D$ generating the same topology such that $(X,D)$ is complete?

$\endgroup$
6
  • 3
    $\begingroup$ An obvious restatement: is every metrizable Baire topological space homeomorphic to a complete metric space? $\endgroup$
    – YCor
    Commented Sep 23, 2020 at 14:43
  • 9
    $\begingroup$ This is answered negatively here at MathSE. $\endgroup$
    – YCor
    Commented Sep 23, 2020 at 14:44
  • 3
    $\begingroup$ @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 :-) $\endgroup$
    – fedja
    Commented Sep 23, 2020 at 15:30
  • 4
    $\begingroup$ @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. $\endgroup$
    – Will Brian
    Commented Sep 23, 2020 at 15:44
  • 2
    $\begingroup$ Feel free to edit, providing the MSE post as context and asking the question from your comment. $\endgroup$
    – YCor
    Commented Sep 23, 2020 at 15:54

0

You must log in to answer this question.

Browse other questions tagged .