TITLE: Doubt in Kechris's Classical Descriptive Set Theory
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.