# Player II has a winning strategy in the *-game, G*(A), iff A is countable

I have a question concerning the proof of this theorem in Kechris' "Classical Descriptive Set Theory." The rules of the game i as follows: Let $$X$$ be a nonempty perfect Polish space with compatible complete metric $$d$$. Fix a basis $$\{V_n\}$$ of nonempty open sets for $$X$$. Given $$A\subseteq X$$, consider the game $$G^*(A)$$, where Player 1 (I) plays pairs of basic open sets (with certain requirements, but they are not necessary for my question), while Player 2 (II) moves by choosing one of those open sets, I plays two open sets from the set that II chose, and so on... (p. 149, Kechris).

Now, in the proof of $$\Rightarrow$$, Kechris does as follows: Assume $$\sigma$$ is a winning strategy for II. Given $$x\in A$$, call a position \begin{align*} p=((U_0^{(0)},U_1^{(0)}), i_0,...,(U_0^{(n-1)},U_1^{(n-1)}),i_{n-1}), \end{align*} $$i_j\in \{0,1\}$$, "good" for $$x$$, if it has been played according to $$\sigma$$ and $$x\in U_{i_{n-1}}^{(n-1)}$$. He then concludes that for each $$x\in A$$ there is a maximal good $$p$$ for $$x$$. Now, if $$p$$ is maximal good for $$x$$, then \begin{align*} x\in A_p =\{y\in U_{i_{n-1}}^{(n-1)}:\forall \,\, \text{legal} \,\, (U_0^{(n)},U_1^{(n)}), \,\, \text{if i is what}\,\, \sigma\,\, \text{requires II to play next, then}\,\, y\not \in U_i^{(n)}\}. \end{align*} He then claims that $$A_p$$ contains at most one point, and this is what I cannot convince myself of! Can anyone help me? :)

• Kechris explains why $A_p$ contains at most one point in the proof ("Now notice that $A_p$ contains at most one point, since..."). What part of this explanation did you not understand? – Alex Kruckman Jan 9 at 19:46
• Yeah. I don't get, why he assumes that $y_i\in U_i^{(n)}$. – MLS Jan 9 at 20:18

Suppose for contradiction that $$A_p$$ contains more than one point. So we have distinct points $$y_0\neq y_1$$ in $$A_p\subseteq U^{(n-1)}_{i_{n-1}}$$. Now because we're in a Polish space, we can find basic open neighborhoods $$y_0\in U^{(n)}_0$$ and $$y_1\in U^{(n)}_1$$, each with diameter less than $$2^{-n}$$, such that $$\overline{U^{(n)}_0}\cap \overline{U^{(n)}_1} = \emptyset$$ and $$\overline{U^{(n)}_0\cup U^{(n)}_1} \subseteq U^{(n-1)}_{i_{n-1}}$$.
So it is legal for Player I to play $$(U^{(n)}_0, U^{(n)}_1)$$, and the strategy $$\sigma$$ requires Player II to respond with $$i\in \{0,1\}$$. If $$i = 0$$, then $$y_0\in U^{(n)}_i$$, contradicting $$y_0\in A_p$$. And if $$i = 1$$, then $$y_1\in U^{(n)}_i$$, contradicting $$y_1\in A_p$$.