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Figured out the full case; added 22 characters in body
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Joel David Hamkins
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Here is an answer forFirst, consider Gerhard's easier special case, where the open sets are disjoint disjoint.

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

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

Perhaps a proof for theA similar idea works in general case can be made, by puttingwell-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 carefully$U_\alpha$, 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 I should likeso 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 seenonempty $V\subset U_\alpha-\bigcup_{\beta\lt\alpha}U_\beta$; 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

Here is an answer for Gerhard's 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.

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

added 31 characters in body
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Joel David Hamkins
  • 236.5k
  • 44
  • 777
  • 1.4k

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.

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.

Here is an answer for Gerhard's 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.

Source Link
Joel David Hamkins
  • 236.5k
  • 44
  • 777
  • 1.4k

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.