Suppose $X=A\cup B$ is a separable metrizable space and $X$ is Baire. If $A$ is an open Polish subspace of $X$, $B$ is countable and closed nowhere dense in $X$, $A\cap B=\emptyset$, then is $X$ a Polish space?
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Isn't $A$ also closed since it's completely metrizable? Then $B$ would be empty as nowhere dense set. – user87690 Aug 26 '13 at 17:11
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$A$ must be $G_{\delta}$, but $A$ needs not to be closed. – aha Aug 26 '13 at 17:16
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It seems that $X$ needs not be Polish.
Take any perfect and nowhere dense closed set $E\subset\mathbb R$. Let $B$ be a countable dense subset of $E$ and $A:=\mathbb R\setminus E$. Then $X:=A\cup B$ is a Baire space since it is comeager in $\mathbb R$. The set $A$ is open in $\mathbb R$, hence Polish and open in $X$. The set $B$ is countable and nowhere dense in $X$ because $A$ is dense in $\mathbb R$ and hence in $X$. But $X$ cannot be Polish since $B$ is closed in $X$ and not Polish (being countable without isolated points).
I hope this works!
Etienne
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It works. Well, I am convinced your proof is correct. I looked for a hole but didn't find one. – Daniel Fischer Aug 28 '13 at 10:25
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And for a nice concrete example, take $B$ to be the set of ‘endpoints’ of the middle-thirds Cantor set $C$, and $A$ to be complement of the Cantor set (or for that matter $[0,1]\setminus C$). – Brian M. Scott Aug 28 '13 at 20:47
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Thank you! I think the proof is right. And I also want to know if $B$ could be made to be scattered of finite Cantor-Bendixson rank? – aha Aug 29 '13 at 11:00
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I would say no. It should be possible to prove by induction on the rank that if $B$ has finite Cantor-Bendixson rank then it is Polish. Once this is done, you conclude that $B$ is $G_\delta$ in the completion of $X$. Since $A$ is $G_\delta$ as well in the completion (being Polish), so is $X$, and hence $X$ is Polish. – Etienne Aug 29 '13 at 15:27