Every space is “almost” Baire?

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?

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@KCd: The result mentioned in the first sentence is indeed called the Banach category theorem. See e.g. Otxoby's book, Theorem 16.1, page 62. – t.b. Dec 5 '11 at 7:11
Every locally compact Hausdorff space is a Baire space. Take a non-metrizable one, e.g. an uncountable product of $\{0,1\}$ with itself. – t.b. Dec 5 '11 at 7:12
So: three main cases for the Baire category theorem: (a) complete metric space, (b) locally compact Hausdorff space, (c) locally countably compact regular space. OK, an interesting example would be a Baire space that is none of these. – GEdgar Dec 5 '11 at 14:01
@GEdgar: Take $p\in\beta\omega\setminus\omega$ and let $X=\{p\}\cup\omega$ as a subspace of $\beta\omega$. The topology at $p$ ensures that $X$ is in none of your three classes, but it’s certainly Baire: any intersection of dense open subsets must include $\omega$. – Brian M. Scott Dec 5 '11 at 18:18
@GEdgar: the Sorgenfrey line is not one of them and still Baire. – Henno Brandsma Dec 5 '11 at 19:01

On the very same page in Oxtoby: Measure and Category (GTM2), that was linked in t.b.'s comment, we can read this note, which follows after Banach category theorem.

It follows that any topological space is the union of an open (or closed) Baire subspace and a set of first category.

Banach Category Theorem: In a topological space $X$, the union of any family of open sets of first category is of first category.

There are several equivalent definitions of Baire space, see the article at Wikipedia.

This is taken from my notes here. I am giving the link here for the case that I made some mistake when expanding the macros.

Maybe the proof from the paper I am linking will be more readable - I guess I've included here too many details which are unnecessary for people which are experienced enough in working with meager, comeager sets and similar stuff.

Every space is union of open (closed) Baire subspace and a set of first category. (See also [BK,Lemma 4.1])

Let $\mathcal G$ be a system of all meager open subsets of $X$. Then $G=\bigcup\mathcal G$ is an open subset of first category an $Y:=X\setminus G$ is a Baire space.

(Since $Y$ is closed, we have $\overline{A}\mathstrut^Y=\overline{A}\mathstrut^X$ for any $A\subseteq Y$. Also $\operatorname{Int}_X(A)\subseteq\operatorname{Int}_Y(A)$. Hence every nowhere dense subset of $Y$ is nowhere dense in $X$ and every meager subset of $Y$ is meager in $X$. If $U\ne\emptyset$ would be an open meager subset of $Y$, then $U\cup G$ would be an open meager in $X$, contradicting the definition of $G$.)

Now if we put $G'=G\cup\partial G=G\cup (\overline G\cap \overline{X\setminus G})$, then $G'$ is meager (since boundary of any open set is nowhere-dense) and closed (since for open set $\partial G=\overline G\setminus\operatorname{Int} G=\overline G\setminus G$ and $G\cup\partial G=\overline G$). The subspace $X\setminus G'$ is a Baire space.

(Now $Y:=X\setminus G'$ is open, thus we have $\operatorname{Int}_X(A)=\operatorname{Int}_Y(A)$ and $\overline{A}\mathstrut^Y=\overline{A}\mathstrut^X\cap X \subseteq \overline{A}\mathstrut^X$ for any $A\subseteq Y$. If $A$ is nowhere dense in $Y$, then $\operatorname{Int}_Y \overline{A}\mathstrut^Y = \operatorname{Int}_X (\overline{A}\mathstrut^X\cap Y)=\emptyset$ $\Rightarrow$ $A\cup G \subseteq A\cup G' \subseteq \overline{A}\mathstrut^X \cup G'$ is meager in $X$ $\Rightarrow$ $A\cup G$ is meager in $X$. Consequently, if $A$ is meager in $Y$, then $A\cup G$ is meager in $X$. Thus for any open meager subset $U$ of $Y$ the set $U\cap G$ is meager in $X$. Again, $U\ne\emptyset$ would contradict the maximality of $G$.)

• [BK] M. Balcerzak and A. Kharazishvili. On uncountable unions and intersections of measurable sets. Georgian Mathematical Journal, 6(3):201--212, 1999. link
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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. $\Box$

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