I got this from Thomson et al.'s freely available "Elementary Real Analysis" p.356.
They introduce Baire's category theorem through a game where, given two players (A) and (B)
Player (A) is given a subset $A$ of $\mathbb{R}$, and player (B) is given the complementary set $B = \mathbb{R} \backslash A$. Player (A) first selects a closed interval $I_1 \subset \mathbb{R}$; then player (B) chooses a closed interval $I_2 \subset I_1$. The players alternate moves, a move consisting of selecting a closed interval inside the previously chosen interval.
The play of the game thus determines a descending sequence of closed intervals \begin{align} I_1 \supset I_2 \supset \ldots I_n \supset \ldots\end{align} where player (A) chooses those with odd index and player (B) those with even index. If \begin{align} A \: \bigcap_n^{\infty} I_n \neq \emptyset \end{align} then player (A) wins; otherwise player (B) wins.
Then they argue that if player (A) is dealt the irrational set and (B) is dealt the rational set, (A) always has a strategy to win.
I'm confused in several ways by this argument. One confusion is about the term "closed interval". Does, for example, the closed interval [1,1] count as an interval? Because if that's the case, can't whoever has the first turn just end the game then and there without regard to whether he has the rationals or irrationals? Say, if (A) received the rationals, he can just pick [0.5,0.5]. Game over. (But I'm guessing probably not, because the game wouldn't happen)
If that is not the case, and an interval has to be defined $[a,b]$ s.t. $a < b$, isn't it always true that for any $I_{2n+1} = [a_{2n+1}, b_{2n+1}]$ for the odd numbered turns of (A), and where (A) has the irrationals, there exists some $q \in [a_{2n+1},b_{2n+1}]$ s.t. $q$ is rational? Because $\mathbb{Q}$ is also dense on the real line. Or is this argument relying on the fact that a countable intersection of closed sets is always closed? And that it can converge to a single point. But if it converges to a single point, it might be a closed set, but is it still a closed interval? And does that still count as winning? And can't (A) still play this same game even if he were dealt the rationals if all that is needed is that his choice of intervals converge to a single point?
I've been thinking about this for a few days now and none of the ways I've approached it convince me that their discussion of the game is true (though I do trust that it is true, since the authors are mathematicians and I'm not), so any help would be appreciated.