I'm trying to understand a proposition of A. Kechris in chapter 8 of his Classical Descriptive Set Theory, in which given a non empty metrizable separable space $X$ that is dense in a Polish space $Y$, we have to prove that if $X$ is comeager in $Y$ then $X$ is Choquet i.e. the second player has a winning strategy in the Choquet game on $X$. Of course, $X$ contains an intersection of open and dense subsets $W_n$ of $Y$ , that also must be dense because $Y$ is Baire. What is the winning strategy for II? I put $V_n=U_n\cap W_n$ but I'm not sure that is correct.

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    $\begingroup$ Perhaps an indication of how the Choquet game goes might be of use (that is, where $U$s and $V$s come from). $\endgroup$ – Pedro Sánchez Terraf Mar 5 '16 at 21:57

Just a reminder: The Choquet game on $X$ has two players ($I$ and $II$) who alternatively choose open sets in $X$ (resp., $U_n$ and $V_n$ with $n\in\mathbb{N_0}$) subject to the condition $U_n\supseteq V_n\supseteq U_{n+1}$ for all $n\in\mathbb{N_0}$. Player $II$ wins a round if $$ \bigcap_n U_n =\bigcap_n V_n \neq \emptyset. $$

In this context, your proposed strategy won't work as it stands, but almost does. Assume $X= Y = (0,1) =W_n$ for all $n$ and $I$ plays $U_n=(0,\frac{1}{n})$. This example makes it clear that you should use the completeness of $\bigcap_n W_n$ somewhere.

Getting into the argument, $G = \bigcap_n W_n \subseteq X$ is Polish with the relative topology; choose a complete metric $d$ for it. Now the strategy for $II$ is to take $V_n\subseteq U_n\cap W_n$ such that $\overline{V_n}\subseteq U_n$ and $\mathrm{diam}(V_n)<2^{-n}$. Since $W_n$ is open and dense, we can ensure that each such $V_n$ will be nonempty; let $x_n\in V_n$. It is easy to see that $\{x_n\}_n$ is Cauchy in $G$, so it converges to $x\in G$. This $x$ must belong to $$ \bigcap_n \overline{V_n} =\bigcap_n U_n, $$ hence II wins this play.


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