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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? :)

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  • $\begingroup$ 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? $\endgroup$ – Alex Kruckman Jan 9 at 19:46
  • $\begingroup$ Yeah. I don't get, why he assumes that $y_i\in U_i^{(n)}$. $\endgroup$ – MLS Jan 9 at 20:18
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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 $y_1\in A_p$.

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  • $\begingroup$ That really helped me - thank you so much! $\endgroup$ – MLS Jan 9 at 20:34
  • $\begingroup$ @MLS You're welcome! Instead of thanking me in a comment, you can upvote my answer (if you think it's a good answer) and/or accept it (if it settles the question). $\endgroup$ – Alex Kruckman Jan 9 at 20:40
  • $\begingroup$ I'm afraid I can't upvote, I'm too new here. $\endgroup$ – MLS Jan 9 at 20:42
  • $\begingroup$ @MLS Ah, that's right, I forgot about the reputation restriction on voting. $\endgroup$ – Alex Kruckman Jan 9 at 20:43

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