# Baire's property iff first category has dense complement.

Show that $(S, d)$ has Baire's property iff every set of first category has a dense complement.

A set is of first category if it is a countable union of nowhere dense sets. First Category

Baire's Lemma: Let $(X, \rho)$ be a complete metric space and $\{U_n\}_{n=1}^{\infty}$ a sequence of open dense sets in $X$. Then the set $\bigcap_{n =1}^{\infty} U_n$ is also dense.

Note that A is nowhere dense iff $(\overline{A})^c$ is open and dense. Suppose $\forall A$, $A$ is first category. That is, $A$ is a countable union of nowhere dense sets. i.e., $A_i$ is nowhere dense. $\forall A, \; A = \bigcup_{n \in \mathbb{N}}A_i \implies (A)^c = \bigcap_{n \in \mathbb{N}}(A_{i})^{c}$. But we don't know if this is a countable dense set! We know that $(\overline{A}_i)^c$ would be, but not just $(A)^c$.

If $A$ is nowhere dense, what can we say about the compliment of $A$?

Can anyone clear this up or provide advice?

More scratch work:

In my proposed proof, I said let all $A \subset S$ be of first category. i.e., each $A_i$ is nowhere dense, i.e, $(\overline{A}_i)^c$ is open and dense. So, $\forall A, (\overline{A}_i)^c = \bigcup (\overline{A}_i)^c \implies ((\overline{A}_i)^c)^c = \bigcap (((\overline{A}_i)^c)^c) = \bigcap (\overline{A}_i)$. Still though, what can I say about $\overline{A}_i$?

$\newcommand{\cl}{\operatorname{cl}}$Suppose first that $\langle S,d\rangle$ has the property that the intersection of countably many dense open sets is dense in $S$, and let $A$ be a first category set in $S$; we want to show that $S\setminus A$ is dense in $S$. Since $A$ is first category, there are nowhere dense sets $A_k$ for $k\in\Bbb N$ such that $A=\bigcup_{k\in\Bbb N}A_k$. For $k\in\Bbb N$ let $U_k=S\setminus\cl A_k$; each $U_k$ is a dense open subset of $S$, so $\bigcap_{k\in\Bbb N}U_k$ is dense in $S$. But

$$\bigcap_{k\in\Bbb N}U_k=\bigcap_{k\in\Bbb N}(S\setminus\cl A_k)=S\setminus\bigcup_{k\in\Bbb N}\cl A_k\subseteq S\setminus\bigcup_{k\in\Bbb N}A_k=S\setminus A\;,$$

so $S\setminus A$ is also dense in $S$.

Now suppose that the complement of every first category subset of $S$ is dense in $S$, and let $\{U_k:k\in\Bbb N\}$ be a family of dense open subsets of $S$. For each $k\in\Bbb N$ let $A_k=S\setminus U_k$; $A_k$ is closed and nowhere dense in $S$, so $A=\bigcup_{k\in\Bbb N}A_k$ is first category in $S$. Finally,

$$\bigcap_{k\in\Bbb N}U_k=\bigcap_{k\in\Bbb N}(S\setminus A_k)=S\setminus\bigcup_{k\in\Bbb N}A_k=S\setminus A\;,$$

which is dense in $S$, as desired.

A good book with much information on such topics is John C. Oxtoby, Measure and Category: A Survey of the Analogies between Topological and Measure Spaces, 2nd edition. (The first edition is also good.)

• More detail than my answer so i'll put it as the 'Accepted' one. Question though, how did you find this post? It is a month old. – Ozera Jan 20 '15 at 22:08
• @Ozera: I’m a set-theoretic topologist, so from time to time I check the general-topology tag to see whether I’ve missed anything interesting. Apparently I missed your question when it was posted. – Brian M. Scott Jan 20 '15 at 22:19
• @Ozera: Retired now, so not always up to date on what’s going on, but in this case I do know: it’s not. (I don’t think that it ever has done.) I did find a list of all the REU math programs here; so far as I know, it’s current. – Brian M. Scott Jan 20 '15 at 22:33
• @Friedrich: You mean Suppose first and Now suppose? The problem is to prove that those two statements are equivalent, so I began by supposing the first and proving the second and concluded by supposing the second and proving the first. It isn’t a matter of considering different cases. – Brian M. Scott Jun 6 at 20:37
• @Friedrich: I’m glad that it helped. It does take a while to get comfortable with that cluster of concepts. – Brian M. Scott Jun 6 at 20:47

A nowhere dense set has empty interior. Therefore, the complement of a nowhere dense set is dense. Hence, I think my original solution is fine for one direction. If anyone else has comments to add, please do so.