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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?

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  • $\begingroup$ 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). $\endgroup$ – Barry Cipra Mar 3 at 23:16
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No. A simple winning strategy for the empty-intersection player:

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.

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