Oxtoby - Thm 6.2 - There exists a strategy in which (A) wins iff $I_1 \cap B$ of first category

Question

I am reading the book Measure and Category of Oxtoby. In Theorem 6.2 it is stated that

There exists a strategy in which (A) wins iff $I_1 \cap B$ of first category

As in the picture below. My question is: Who is $I_1$ and how do we pick it.

Thank you, Shir

Answer 1

IIRC the game in question is this: there is a set $B \subseteq I_0$, where $I_0$ is a closed interval (all intervals here are required to have positive length). The first player $(A)$ chooses a closed interval $I_1 \subseteq I_0$, and then they alternate forever, choosing a closed subinterval $I_{k+1}$ of the previous interval $I_k$. Player (A) wins if the intersection $\bigcap_k I_k$ is disjoint from $B$, otherwise player (B) wins.

In answer to your question: $I_1$ is a closed subinterval of $I_0$ such that $I_1 \cap B$ is a set of first category. How we find it will depend on how $B$ is defined. If $B$ has the property of Baire, it differs from some open set $U$ by a set of first category. Then $I_1$ exists if and only if $U$ is not dense in $I_0$; if so, $I_1$ can be a small interval around any point not in the closure of $U$.