# Baire space definitions and equivalents

I have seen some equivalents of Baire space definitions,i.e. Countable intersections of open dense sets is dense and countable union of closed nowhere dense sets has empty interior. In Dugundji's topology, he states this is equivalent to the entire space is of second category in itself, i.e. It is not a countable union of nowhere dense sets.

I can see this is necessary for a Baire space (nonempty), but is it sufficient as Dugundji claims? I can't seem to prove this. Thanks

• Consider $X = [0,1] \cup \mathbb{Q}$, with the subspace topology induced by the standard topology on $\mathbb{R}$. – Daniel Fischer Apr 22 '17 at 21:14

Thus spaces like $[0,1] \cup \mathbb{Q}$, or the topological sum of a Baire space and a non-Baire space are non-Baire spaces that are of the second category in itself. (See below)
A correct condition would be that every nonempty open subset of $X$ is of the second category (in itself or in $X$, for open subspaces this is equivalent).
Let $X$ a topological space, and $U \subset X$ open. If $N\subset X$ is nowhere dense in $X$, then $N\cap U$ is nowhere dense in $U$. For suppose $V := \operatorname{int}_U(\operatorname{cl}_U(N\cap U)) \neq \varnothing$. Since $U$ is open, $V$ is also open in $X$, and $V \subset \operatorname{cl}_U(N\cap U) = \operatorname{cl}_X(N) \cap U \subset \operatorname{cl}_X(N)$, so if $N\cap U$ is not nowhere dense in $U$, then $N$ is not nowhere dense in $X$. And if $N \subset U$ is nowhere dense in $U$, then $N$ is nowhere dense in $X$. For if $V := \operatorname{int}_X(\operatorname{cl}_X(N)) \neq \varnothing$, then $V\cap U \neq \varnothing$ (since $\operatorname{cl}_X(N) \subset \operatorname{cl}_X(U)$), and $V\cap U \subset \operatorname{cl}_U(N)$, so then $N$ is not nowhere dense in $U$.