Let $X$ be a topological space. A subset $M$ of $X$ is called meagre (or of first category) if it is covered by the union of a countable family of closed subsets of $X$ with empty interior.


Can you help me to find a proof of the following theorem?


"Arbitrary union of meagre open subsets of $X$ is meagre."


The case of countable unions is easy because $\mathbb{N} \times \mathbb{N}$ is countable. The case in which $X$ is a Baire space (e.g. a complete metric space) is obvious because all open meagre sets are empty.


Thanks.



Arbitrary union of meagre open sets
First, consider Gerhard's easier special case, where the open sets are
disjoint.


**Claim.** The union of an arbitrary family of pairwise
disjoint open meager sets is meager.


Proof. Suppose that $U\_i$ are pairwise disjoint and meager, so
that $U\_i\subset\bigcup\_n C\_n^i$, where each $C\_n^i$ is closed and
nowhere dense. Let $C\_n=\bigcup\_i (C\_n^i\cap U\_i)$. This set is
not dense on any nonempty open set, because if it were dense on
some nonempty $V$, then it would be dense on some nonempty $V\cap
U\_i$, but that is impossible since by the disjointness hypothesis
only $C\_n^i$ contributes points to this set, and it is nowhere
dense. Thus, the closure of $C\_n$ is closed and nowhere dense, and
$\bigcup\_i U\_i$ is contained within $\bigcup\_n C\_n$, since each
$U\_i$ is contained within and is in fact equal to $\bigcup\_n
(C\_n^i\cap U\_i)$. So $\bigcup\_i U\_i$ is meager. QED


A similar idea works in general, by well-ordering the family of open sets.


**Theorem.** An arbitrary union of open meager sets is meager.


Proof. Suppose we have a family of open meager sets $U\_\alpha$,
indexed by ordinals $\alpha$, so that for each $\alpha$ we have
$U\_\alpha\subset\bigcup\_n C\_\alpha^n$ for some closed nowhere dense sets
$C\_\alpha^n$. Let $$C\_n=\bigcup\_\alpha [C\_\alpha^n\cap
U\_\alpha-\bigcup\_{\beta\lt\alpha}U\_\beta].$$ Note that these $C\_n$
cover the union $U=\bigcup\_\alpha U\_\alpha$, since any $a\in U$ is
in some least $U\_\alpha$ and so it gets into some $C\_\alpha^n\cap
U\_\alpha$ without being in $\bigcup\_{\beta\lt\alpha}U\_\beta$, and consequently is in $C\_n$. Also, each $C\_n$ is nowhere dense, because if $C\_n$ is dense on
some nonempty set $V$, then there is some least $\alpha$
containing members of $V$, and so we reduce to nonempty $V\subset U\_\alpha-\bigcup\_{\beta\lt\alpha}U\_\beta$; thus, $C\_n$ would be dense on
$V$ inside $U\_\alpha-\bigcup\_{\beta\lt\alpha}U\_\alpha$. But the only members
of this set in $C\_n$ are contributed by $C\_\alpha^n$, which is
nowhere dense. So the closure of $C\_n$ is nowhere dense, and so
$U$ is meager, as desired. QED


