# Doubt in Kechris's Classical Descriptive Set Theory

In Theorem 8.29 in the Kechris's book Classical Descriptive Set Theory, he writes that if $W=\bigcup_{i\in I} U_i$ where $U_i$ are pairwise disjoint and set $A$ is comeager in each $U_i$, then $A$ is comeager in $W$. $A$ is comeager in $U_i$ means $U_i \setminus A$ is meager. If index set is not countable then I don't see the claim.

-
I've had a look at Kechris' book and $U_i$'s in the proof of this theorem are open. I believe this might be important. – Martin Sleziak Jun 1 '11 at 14:07

Added: Note that I'm assuming that the sets $U_i$ are open, which is the situation that Kechris uses, as Martin pointed out in a comment.

For each $i$ there is a sequence of dense open sets $O_{n}^{(i)} \subset U_i$ such that $G_{i} = \bigcap_n O_{n}^{(i)}$ is contained in $A \cap U_{i}$ since $A \cap U_i$ is co-meager in $U_i$. Now each $O_n = \bigcup_{i} O_{n}^{(i)}$ is dense open in $W = \bigcup_i U_i$ and the intersection $G = \bigcap_n O_n$ is contained in $A$. Therefore $A$ is co-meager in $W$.

-

I do not claim that this proof is substantially different from Theo's one, for me this one was simpler. (Maybe they are even dual to each other. Probably a matter of taste, I have much more practice working with meager than with comeager sets.)

Suppose that $U_i\setminus A$ is meager for each $i\in I$, where $U_i$'s are pairwise disjoint open sets. This means that $U_i\setminus A=\bigcup_{n=1}^\infty N_{i,n}$, where each $N_{i,n}$ is nowhere dense. Denote

$$N_n=\bigcup_{i\in I} N_{i,n}.$$

Now

$$W\setminus A=\bigcup_{i\in I} (U_i\setminus A) = \bigcup_{n=1}^\infty N_n.$$

So it suffices to show that each $N_n$ is nowhere dense. This is equivalent to the claim that for each open set $U$ there exists an open subset $V\subseteq U$ such that $V\cap N_n=\emptyset$.

If $U\cap N_n=\emptyset$, then we can put $V=U$.

If $U\cap N_n\ne\emptyset$, then there exists $i\in I$ such that $U\cap N_{n,i}\ne\emptyset$. Since $N_{n,i}\subseteq U_i$, this implies that $(U\cap U_i)\cap N_{n,i}\ne\emptyset$. As $N_{n,i}$ is nowhere dense (and $U\cap U_i$ is open), there exists an open set $V\subseteq U\cap U_i$ such that $V\cap N_{n,i}=\emptyset$.

For any $j\ne i$ we have $V\cap N_{n,j} \subseteq U_i\cap U_j=\emptyset$. This implies that

$$V\cap N_n=V\cap (\bigcup_{j\in I} N_{n,j})=\emptyset.$$

EDIT In order to give credit, where credit is due, I should mention that this is very similar to the proof of Banach category theorem from Oxtoby's book, which I have been studying recently. Banach category theorem says that (arbitrary) union of open meager sets is again meager.

-
Well, a matter of taste and complements? Thanks for this complement to my answer! – t.b. Jun 1 '11 at 14:33