I've been reading Oxtoby's Measure and Category and in chapter 6 he discusses the game of Banach-Mazur for two players $A$ and $B$ on the unit interval $I_0 \subset \mathbb{R}$. Towards the end of the chapter he gives the following explanation for why Bernstein sets are not determined:
Let A be the intersection of $I_0$ with a Bernstein set. Then neither $A$ nor $B=A^C$ contains an uncountable $G_\delta$ set (Lemma 5.1). Consequently, for any interval $I\subset I_0$, neither of the sets $A\cap I$ or $B\cap I$ is of first category. (For if one is of first category, the other is of second category having the property of Baire (*). By theorem 4.4 any such set contains an uncountable $G_\delta$ set.) Consequently, this game $\langle A,B\rangle$ is not determined in favor of either player.
What I don't understand is (*). If for example $A\cap I$ was of first category, then why would that result in $B\cap I$ being of second category and having the property of Baire?
Thank you for your help!
If both $A\cap I$ and $B\cap I$ were of first category, so would be $I=(A\cap I)\cup(B\cap I)$, since every countable union of first category sets is of first category (this is easy from the definition of first category).
But a Baire space is never of first category in itself: indeed suppose that $X$ is a topological space which can be written as $X=\bigcup_{n\in\Bbb N} F_n$ for closed nowhere dense sets $F_n$. Then $A_n=X\setminus F_n$ are dense open sets, but $\bigcap_{n\in\Bbb N} A_n=\varnothing$, showing that $X$ is not Baire.
In particular if one of $A\cap I$ and $B\cap I$ is of first category then the other one cannot be, which is exactly the same as saying that the other one is of second category.