# Convergence and winning strategies

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 $$B_y=\emptyset$$?

• for any such converging $x_n$ (not constant), $B=\{(x_n, f): f(0)=n\}$ whose limits points is empty works. – Jing Zhang Jul 23 '19 at 2:24

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.

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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