TITLE: Every space is "almost" Baire?
QUESTION [6 upvotes]: There is this theorem called the Banach category theorem which states that in every topological space any union of open sets of first category is of first category.
Now doesn't this imply that every topological space X (let's say which is not of first category) is the union of a set of first category (the union of all open sets of first category of X) and a subspace which is Baire (the union of all open sets of second category)?
Doesn't this mean that every space is "almost" Baire if we understand sets of first category as negligible?
A side question: Is there an easy example of a Baire space which is not a completely metrizable space?

REPLY [2 votes]: I thought a more complete treatment of these beautiful and perhaps surprising results (Thms. 2 and 3) might be welcome.
I have not provided proofs of Prop. 1 - 3, as they are standard exercises. (I will be glad to supply proofs upon request.)
First, some definitions:
Suppose $X$ is a topological space and $E \subset X$.
$E^c := $the complement of $E$.
int$(E) := \overline {E^c}^c$.
bd$(E) := \overline E \cap \overline {E^c}$.
$E$ is dense if $\overline E = X$.
$E$ is nowhere dense (nwd) if int($\overline E$) is empty. 
$E$ is meager (aka 'first category') if it is a countable union of nwd sets.
$X$ is Baire provided that if $\{O_i\}$ is any collection of dense open subsets of $X$, then $\cap O_i$ is dense.
Prop. 1. If $E$ is open or closed, then bd($E$) is nwd.
Prop. 2. The following are equivalent: 
(a) $E$ is nwd. 
(b) If $O$ is open and non-empty, there exists an open non-empty set $P \subset O$ disjoint from $E$.
(c) If $O$ is open and $O \cap E$ is non-empty, there exists an open non-empty set $P \subset O$ disjoint from $E$.
Prop. 3. Suppose $A \subset B \subset X$.
(a) If $B$ is meager in $X$, then $A$ is meager in $X$. 
(b) If $A$ is meager in $B$, then $A$ is meager in $X$.
Prop. 4. The following are equivalent: 
(a) $X$ is Baire.
(b) If $O$ is open and non-empty, then $O$ is not meager.
Proof: This is a standard result. 
For a fairly sketchy proof, 
see this article. $\Box$
Prop. 5. Any open subset of a Baire space is Baire in the induced topology.
Proof: Suppose $X$ is Baire, $O$ is open in $X, P$ is open in $O$, and $P$ is non-empty. 
Then $P$ is open in $X$, so by Prop. 4(b), $P$ is not meager in $X$. 
Therefore by Prop. 3(b), $P$ is not meager in $O$. Again by Prop. 4(b), $O$ is Baire. $\Box$
Thm. 1 (Banach Category Theorem). The union of any family of open meager subsets of $X$ is meager.
Proof: (I follow Oxtoby here, with some minor modifications and more details.)
Let $M_0$ be such a family. $M := \cup M_0.$ 
Suppose $G_0 = \{U_\alpha \ | \ \alpha \in A\}$ is a maximal family of disjoint open non-empty sets 
with the property that each is contained in some member of $M_0. \ G := \cup G_0.$ 
We claim that $H := M \cap G^c$ is nwd. Otherwise, there exists $O$ open and non-empty 
with $O \subset \overline H \subset \overline M \cap G^c. 
\ P := O \cap M \subset G^c$, so $P$ is open and disjoint from all members of $G_0$. 
Since $\emptyset \neq O \subset \overline M, P$ is a non-empty subset of $M$. 
Therefore, there exists $S \in M_0$ with $T := P \cap S \neq \emptyset$. 
Then $T \cup G_0$ has the required properties and is a larger family than $G_0$ (#), establishing our claim.
For any $U_\alpha \in G_0$, we note that $U_\alpha$, being a subset of some set in $M_0$ is meager by Prop. 3(a). 
Suppose $U_\alpha = \cup N_{\alpha i}$ is a countable union of nwd sets. $N_i := \cup_ {\alpha \in A} N_{\alpha i}.$ 
We claim that $N_i$ is nwd. Suppose $O$ is open and $O \cap N_i \neq \emptyset$. 
Then there exists $\alpha$ such that $O \cap N_{\alpha i} \neq \emptyset. \ P := O \cap U_\alpha \ $.
Since $U_\alpha$ is open, so is $P$. Observe that $O \cap N_{\alpha i} \subset O \cap U_\alpha = P$, whence $P \neq \emptyset$.
Since $N_{\alpha i}$ is nwd., by Prop. 2(b), 
there exists $Q$ open with $\emptyset \neq Q \subset P$ and $Q \cap N_{\alpha i} = \emptyset$. 
Since $Q \subset P \subset U_a$ and the $U_a$'s are disjoint, $Q \cap U_\beta = \emptyset$ for $\beta \neq \alpha$, 
whence $Q \cap N_{\beta i} = \emptyset \ $.
It follows that $Q \cap N_i = \emptyset \ $. 
Clearly $Q \subset P \subset O$, so Prop. 2(c) is satisfied by $O, Q,$ and $N_i$, establishing our claim.
Since $G = \cup G_0 = \cup N_i$, it is meager. Now $G \subset M$, so $M = (M \cap G^c) \cup G$. 
Thus $M$, being the union of a nwd subset and a meager one, is also meager. $\Box$
Thm. 2. $X$ contains an open meager subset whose complement is Baire in the induced topology.
Proof: Let $M_0$ consist of all open meager subsets of X. $\ M := \cup M_0.$ 
By Thm. 1, $M$ is meager. We will show that $B := M^c$ is Baire in the topology induced on $B$ by $X$. 
By Prop. 4(b), to prove this it suffices to show that if $O$ is open and non-empty in $B$, then $O$ is not meager in $B$. 
Suppose such an $O$ is meager in $B$. By Prop. 3(b), $O$ is meager in $X$. 
Now $O = P \cap B$, where $P$ is open in $X. \ P \cap M$, being a subset of $M$, is meager in $X$ by Prop. 3(a). 
Therefore, $P = (P \cap B) \cup (P \cap M) = O \cup (P \cap M)$ is open and meager in X, whence $P \in M_0$. 
It follows that $P \subset M$, whence $O \subset M$. Since $O \subset B$, we must have $O = \emptyset$ (#). $\Box$
Thm. 3. $X$ contains a closed meager subset whose complement is Baire in the induced topology.
Proof: Let $M$ and $B$ be as in the proof of Thm. 2. $F := \overline M = M \cup \textrm {bd}(M). 
\ M$ is open and meager, and by Prop. 1, bd($M$) is nwd. Therefore, $F$ is meager. 
$O := F^c \subset M^c = B. \ O$ is open in $X$, therefore in $B$. 
In the proof of Thm. 2 we showed that $B$ is Baire. By Prop. 5, $O$ is Baire. $\Box$