# Existence of a winning strategy in the $\boldsymbol{\Delta}_2^0$ game

Consider the following infinite perfect information game with two players (the name I gave in the title of the post it totally made up): at each round $$i \in \omega$$, player $$\mathrm{I}$$ picks a natural number $$x_i$$ and a $$\boldsymbol{\Delta}_2^0$$ subset of the Baire space $$\omega^\omega$$ denoted by $$A_i$$ such that $$A_{i-1}\subseteq A_i$$ for all $$i$$; player $$\mathrm{II}$$ then picks a boolean value $$y_i \in \{0,1\}$$.

So at the end of the game player $$\mathrm{I}$$ will have built an infinite sequence $$x = (x_n)_{n \in \omega}$$ and an increasing chain (wrt set inclusion) of $$\boldsymbol{\Delta}_2^0(\omega^\omega)$$ sets $$(A_n)_{n \in \omega}$$ whilst player $$\mathrm{II}$$ will have built a boolean sequence $$y = (y_n)_{n \in \omega}$$ (an element of the Cantor space). Player $$\mathrm{II}$$ wins the play if at least one of these conditions is satisfied:

• $$\bigcup_{n \in \omega} A_n \not\in \boldsymbol{\Delta}_2^0(\omega^\omega)$$
• The sequence $$y$$ is eventually constant, i.e. $$\exists k \ \forall n \ge k \ y_n = y_k$$. Moreover The sequence $$y$$ is eventually equal to $$1$$ if and only if $$x \in \bigcup_{n \in \omega} A_n$$. Intuitively we require player $$\mathrm{II}$$ to "guess" whether the real played by $$\mathrm{I}$$ will or won't belong to the set $$\mathrm{I}$$ is building.

Now I'm wondering whether player $$\mathrm{II}$$ has a winning strategy in this game, or if the game is determined, and, eventually, under which hypotheses.

In a simpler game, in which player $$\mathrm{I}$$ does not keep changing the sets $$A_i$$, player $$\mathrm{II}$$ has a winning strategy (see, for this result and a wider discussion, this paper by Raphael Carroy Playing in the first Baire class, specifically Proposition 3.10).

Any idea? Thanks

Player I has a winning strategy: First play a singleton $$A_0=A_1=\ldots=\{z_0\}$$, for some real $$z_0$$, and the $$x_n$$'s consistent with $$z_0$$, until player II plays their first 1, if they ever do. After they play a 1 at stage $$n$$, continue with $$A_{n+1}=A_{n+2}=\{z_0\}$$, but make $$x_{n+1}$$ inconsistent with $$z_0$$, and proceed in this way until player II plays their first 0 after this stage, say at stage $$m$$. Then play $$A_{m+1}=\{z_0,z_1\}$$ with some $$z_1$$ where $$(x_0,x_1,\ldots,x_m)\subseteq z_1$$, and keep playing $$A_{m+2}=A_{m+3}=\ldots=\{z_0,z_1\}$$ and the $$x_i$$'s consistent with $$z_1$$ until player I again plays a 1. Then continue playing $$\{z_0,z_1\}$$ but play the $$x_i$$'s inconsistent with $$z_1$$, until player II next plays a $$0$$ at stage $$m_1$$, etc. The $$A_n$$'s and $$\bigcup_{n<\omega}A_n$$ are easily boldface-$$\Delta^0_2$$, and it's easy to see that player I wins.
(In the first edit I said something suggesting that every countable set is boldface-$$\Delta^0_2$$, but that's false, as e.g. the rationals are not boldface-$$\Pi^0_2$$, as they are dense but not comeager. If $$\bigcup_{n<\omega}A_n$$ is infinite above, note that its complement is the union of an open set with a singleton (the singleton contains the limit of the $$z_n$$'s), which is therefore boldface-$$\Sigma^0_2$$.)