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show/hide this revision's text 3 Figured out the full case; added 22 characters in body

Here is an answer for

First, consider Gerhard's easier special case, where the open sets are

Perhaps a proof for the

A similar idea works in generalcase can be made , by putting 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 $C_n^i$ together more carefullyU_\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 $$C_n=\bigcup_\alpha [C_\alpha^n\capU_\alpha-\bigcup_{\beta\lt\alpha}U_\beta].$$ Note that these $C_n$cover the union $U=\bigcup_\alpha U_\alpha$, since any $a\in U$ isin some least $U_\alpha$ and I should like so it gets into some $C_\alpha^n\capU_\alpha$ 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 onsome nonempty set $V$, then there is some least $\alpha$containing members of $V$, and so we reduce to see 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 membersof this set in $C_n$ are contributed by $C_\alpha^n$, which isnowhere dense. So the closure of $C_n$ is nowhere dense, and so$U$ is meager, as desired. QED
show/hide this revision's text 2 added 31 characters in body

Here is an answer for 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

Perhaps a proof for the general case can be made by putting the sets $C_n^i$ together more carefully, and I should like to see this.
show/hide this revision's text 1

Here is an answer for 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 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

Perhaps a proof for the general case can be made by putting the sets $C_n^i$ together more carefully.