Let $X$ be a topological space such that
$\forall x\in X$ $\exists U_x$ open neighborhood of $x$ such that $U_x$ is a Baire space. I have to prove that $X$ is a Baire space.
Proof :
Let $\{A_n\}_{n\in\mathbb{N}}$ a sequence of open dense subsets of $X$. I have to prove that the intersection $I:=\bigcap_{n\in\mathbb{N}} A_n$ is dense in $X$. My idea is to prove that $I\cap U\ne\emptyset$ for all open not-empty subset $U$ of $X$. Can anyone help me please?
Best Answer
If $Y$ is a Baire space and if $W$ is a non-empty open subspace of $Y$ then $W$ is a Baire space. Proof: Let $F$ be any non-empty countable family of dense open subsets of $W.$ Then $G=\{f\cup (Y\setminus \overline W):f\in F\}$ is a non-empty countable family of dense open subsets of $Y,$ so $\cap G$ is dense in $Y,$ so $\cap F=(\cap G)\cap W$ is dense in the open set $W.$
Let $A$ be a non-empty countable family of dense open subsets of $X $ and let $U$ be any non-empty open subset of $X.$ We need to show $\emptyset\ne U\cap (\cap A).$ Take $x\in U$ and let $x\in Y$ where $Y$ is an open Baire subspace of $X.$ Let $W=U\cap Y.$ And let $F=\{a\cap W:a\in A\}.$ Then $U\cap (\cap A)\supseteq \cap F\ne\emptyset$ by 1.