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
Indeed it is a mistake. If a topological space has a nonempty open Baire subspace, it is of the second category in itself. But if it also has a nonempty meagre open subset, it is not a Baire space.
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 $U$.
From that it follows that every open subspace of a Baire space is itself a Baire space, and that every space that has a nonempty open Baire subspace is of the second category in itself.
