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Suppose we have a set $B\subseteq 2^\omega\times\omega^\omega$ and a sequence $(x_n)$ in $2^\omega$ such that for each $n$, Player I (the one trying to get into the payoff set) has a winning strategy in the Gale-Stewart game where the payoff set is given by the corresponding slice: $B_{x_n}=\{f\in\omega^\omega:(x_n,f)\in B\}$.

Question: If $(x_n)$ converges to some $y$ in $2^\omega$, what can we say about the game with payoff set $B_y$? In particular, must Player I have a winning strategy in this game?

I am most interested in the cases when $B$ is closed or even clopen, in which case all of its slices are closed or clopen (and thus determined), respectively.

For what it's worth, I suspect the answer to my question is "no" in either case, but haven't been able to see why. Perhaps one could come up with an example where $B_y=\emptyset$?

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    $\begingroup$ for any such converging $x_n$ (not constant), $B=\{(x_n, f): f(0)=n\}$ whose limits points is empty works. $\endgroup$ – Jing Zhang Jul 23 '19 at 2:24
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Let $B(x,y)$ if and only if $x$ is not the constant $0$ sequence.

Let $x_n \in {}^\omega 2$ be such that $$x_n(k) = \begin{cases} 0 & \quad k \neq n \\ 1 & \quad k = n \end{cases}$$ For each $n \in \omega$, $B_{x_n} = {}^\omega\omega$. $y = \lim_{n \rightarrow \infty} x_n$ is the constant $0$ sequence.

Player I has a winning strategy in $B_{x_n} = {}^\omega\omega$. Player I does not have a winning strategy in $B_y = \emptyset$.


You may be interested in notion of the game quantifier $\Game$, scales, and the third periodicity theorem.

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After thinking about Jing and William's answers, as well as something else I already had in mind, here is a clopen counterexample:

Let $C\subseteq 2^\omega\times\mathbb{Z}^\omega$ be the set of all $(x,f)$ such if $f(0)>0$ and $s$ is the initial segment of $x$ having length that of the binary expansion of $f(0)$, then each coordinate of $s$ is $\leq$ to that of the corresponding binary expansion of $f(0)$, listed with the coefficient of $2^n$ in the $n$th place.

For example, if $f(0)=19=2^0+2^1+2^4$, then in order for $(x,f)$ to be in $C$, the initial 5 coordinates of $x$ must be coordinatewise $\leq$ to $(1,1,0,0,1)$.

It is easy to check that $C$ is clopen. Notice that if $y$ is the constant $0$ sequence, then $(y,f)\in C$ for any $f$.

Let $B$ be the complement of $C$, so $B_y=\emptyset$.

Let $x_n(k)=\begin{cases}0&\text{if $k\neq n$}\\1&\text{if $k=n$}\end{cases}$, so $(x_n)$ converges to $y$.

For each $n$, we claim that Player I has a one-move winning strategy into the payoff set $B_{x_n}$: Let I play $f(0)=2^{n+1}$ on their first move. Since the $n$th coordinate of $x_n$ is $1$, which is not $\leq$ to the $n$th binary digit of $2^{n+1}$ (namely, $0$), $(x_n,f)\in B$ for any $f\in\mathbb{Z}^\omega$ extending this $f(0)$.

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