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.


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    $\begingroup$ I don't know. You might start with the assumption that the open sets are pairwise disjoint, and see if that helps. Gerhard "Ask Me About System Design" Paseman, 2012.10.09 $\endgroup$ Oct 9, 2012 at 16:55

2 Answers 2


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

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    $\begingroup$ The statement of the question is called the Banach category theorem (see e.g. Oxtoby, Measure and category, Theorem 16.1). The general case can be reduced to your claim by considering the union $V = \bigcup V_j$ of an arbitrary family of meager open sets and taking a maximal family of pairwise disjoint open sets $U_i$ such that each $U_i$ is contained in some $V_j$. Then it follows that $\overline{V} \setminus \bigcup U_i$ is meager (otherwise the family $U_i$ wouldn't have been maximal) and your claim does the rest. $\endgroup$ Oct 9, 2012 at 18:07
  • $\begingroup$ Theo, thanks. Meanwhile, I also found a way to do it by enumerating the family in a well-ordered sequence. But perhaps this may amount to essentially the same thing as the maximal family argument... $\endgroup$ Oct 9, 2012 at 18:22
  • $\begingroup$ Since Joel's proof uses well-ordering and Theo's maximality one may ask whether the result itself depends on the axiom of choice. $\endgroup$ Oct 10, 2012 at 6:03
  • $\begingroup$ Jochen, good question, and I had had the same thought. It appears that Theo's maximality argument may need a little less choice than my argument, since to construct a maximal set of his sort it suffices to well-order the basic open sets, but in my argument, I need to well-order the family of open meager sets, which might be much larger. For example, one can find maximal sets without any AC when there are only countably many basic open sets (but in this case the main theorem is immediate by countable unions). $\endgroup$ Oct 10, 2012 at 12:11
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    $\begingroup$ There is also use of AC in both arguments just to pick the covers $C_n^i$, and this is not just countable AC or DC, since one needs to choose a cover of $U_i$ for each $i$. $\endgroup$ Oct 10, 2012 at 12:44

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...

  • $\begingroup$ It would be pretty cool if AD and this characterization could prove the result without choice. I still think it helps to assume disjoint open sets to start. Gerhard "Thanks For The Hat Tip" Paseman, 2012.10.10 $\endgroup$ Oct 10, 2012 at 15:32

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