# Who has a winning strategy in Choquet game on rational numbers?

I know that in a real-numbers-variant of Choquet game, the player aiming for non-empty intersection has a winning strategy. Is the same true for rational numbers?

• The answer can be found in the linked-to Wikipedia article, in the paragraph with a link to "Baire space" ($\mathbb{Q}$ is not a Baire space, so Player 1 has a winning strategy). – Barry Cipra Mar 3 at 23:16

Enumerate the rationals. On the $$n$$th move, choose an interval that excludes the $$n$$th rational $$r_n$$.
One way to do so: If $$r_n$$ is not an element of $$V_{n-1}$$, we set $$U_n=V_{n-1}$$. If $$r_n$$ is an element of $$V_{n-1}$$, we set $$U_n=V_{n-1}\cap (r_n,\infty)$$. This isn't empty, because $$U_n$$ contains some interval $$(r_n-\epsilon,r_n+\epsilon)$$, and there are rational numbers in $$(r_n,r_n+\epsilon)$$.
Then, for each $$n$$, $$r_n$$ is not in the intersection $$\bigcap_i U_i$$. Every rational is some $$r_n$$, and the intersection is empty.