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 $B\_y=\emptyset$?



Convergence and winning strategies
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 $f(0)$.


