# 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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