Explicitly constructing a non determined game on $\mathcal{P}(2^{\mathbb{N}})$ I am trying to solve Ex 20.10 on Kechris descriptive set theory book. Throughout this post, a "game on $A$" refers to the infinite game $G(T,X)$ where $T$ is a pruned tree $T\subset A^{<\omega}$ and $X\subset [T]$.
Ex 20.10 states:
Define explicitly a game with moves in $A=\mathcal{P}(2^{\mathbb{N}})$ which is not determined.
Kechris then gives this remark: It is easy to define such a game explicitly and then show that it is not determined using AC.
I am familiar with the construction of a non determined game using AC, but in this exercise, I think what Kechris wants is a construction that doesn't depend on AC, where one can argue using AC that the constructed game is not determined.
I have thought about this for quite a long time but couldn't get closer.
 A: Very belatedly, let me turn my comment into an answer:
The first two moves of the game $\mathfrak{G}$ consist of having player $1$ open with a set $A\subseteq 2^\mathbb{N}$ and player $2$ responding with "Stay" or "Switch." If player $2$ says "Stay," the remainder of the game is played by having players $1$ and $2$ alternate elements of $\{0,1\}$, with player $1$ going first and winning iff the sequence built is in $A$. If player $2$ says "Switch," the game continues in the same way but with player $2$ going first and winning if the sequence built is in $A$.
I've phrased this a bit informally; I'll leave it as an exercise to formalize $\mathfrak{G}$ as a game literally on $\mathcal{P}(2^\mathbb{N})$.

Basically, player $1$ picks a game on $2$ and player $2$ decides whether to go first or second in that game.

Now working in $\mathsf{ZF}$ alone, we argue as follows:

*

*Player $1$ does not have a winning strategy in $\mathfrak{G}$.

Proof: Supposing they did, let $A$ be their opening move according to such a strategy. Now player $2$ has two choices - "Stay" or "Switch" - and player $1$ must be able to beat them no matter which they pick. Being able to beat player $2$ if they pick "Stay" amounts to $A$ being a win for the first player, and being able to beat player $2$ if they pick "Switch" amounts to $A$ being a win for the second player. Both can't be true at the same time!


*Player $2$ does not have a winning strategy in $\mathfrak{G}$.

Proof: If they did, this would amount to uniform determinacy - the existence of a function sending every game on $2$ to a winning strategy in that game. On the one hand, uniform determinacy clearly implies determinacy in the usual sense. On the other hand, uniform determinacy would also give a choice function $c$ for $2^\mathbb{N}\setminus\{0\}$ (given a nonempty set of reals $A$, consider the game with payoff set $\{(a_i)_{i\in\mathbb{N}}: (a_{2i})_{i\in\mathbb{N}}\in A\}$). Iterating such a choice function gives a well-ordering of the reals by setting $r_\alpha=c(\{r_\beta: \beta<\alpha\})$, and if the reals are well-ordered then there is an undetermined game, contradicting uniform determinacy.
