Arbitrary union of meagre open sets 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.
 A: 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
A: I'll post this answer CW, because I don't have time to work out the details. Hopefully, someone can fill these in or shoot down the strategy.
On wikipedia, there is a cute remark about characterizing meagre sets using a Banach-Mazur game. Basically, you have two players who take turns to build a nested sequence of open sets $O_n$. If $U=\cap_{n=1}^\infty O_n$ is the resulting intersection, one of the players aims to have $U \cap X =\varnothing$ and the other player aims to have a point from $X$ in $U$.
Then, $X$ is meagre iff the player who wants the empty intersection has a winning strategy.
Couldn't this characterization be used in this problem? As I mentioned, I haven't been able to make the details work right...
