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
 A: 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-$\Sigma^0_2$.)
