What is the winning strategy for this deterministic game? I was trying to solve one of the questions in Herrlich's book "The Axiom of Choice" and got stuck on the second part of this problem. The game in question as described by Herrlich is:
Let $X$ be a non-empty set and $A$ a subset of $X^\mathbb{N}$. The game $G(X,A)$ is played as follows:
Two players choose alternate elements $x_0,x_1,x_2, \dots$ in $X$, such that each player knows, (besides $X$ and $A$), whenever it is his term, the tuple of previously chosen elements. The first player (who chooses $x_0,x_2,x_4,\dots$) wins if the resulting sequence $(x_n)$ belongs to $A$. Otherwise the second player (who chooses $x_1,x_3,x_4,\dots$) wins. The game is determined if one of the players has a winning strategy.
Consider $A=[{(x_n) \in 2^{\mathbb{N}} : \forall_n x_{2n+1}=x_{2n}}]$ and the constant sequence $(0)$.
1)The second player has a winning strategy for $G(2,A)$.
2)The first player has a winning strategy for $G(2,A \cup \left\{ (0) \right\})$.
My solution for (1) is as follows:
Player 2 can easily win this game by ensuring for whatever $x_0$ player 1 first plays, they play $x_1 \in 2^{\mathbb{N}}$, ensuring $x_1 \neq x_0$, resulting in the sequence necessarily not being in $A$.
However I don't see how the first player can have a winning strategy for part (2) such that $(x_n) \in A$. Am I miss understanding what the constant sequence $(0)$ means?
Any help is appreciated!
 A: This is indeed a typo; in particular, $(0)$ is already in $A$ as defined! Unfortunately, given that the exercise is so short, it's hard to figure out what the author meant.
Here's one possible interpretation. Clearly the author intends $(0)$ to be pivotal in some way; basically, we can assume that "both players $1$ and $2$ wait for the other to play $1$" is a plausible outcome of playing the original game $G(2,A)$ (of course player $2$ should ultimately win here, but player $1$ shouldn't have an incentive to play a $1$ early). This suggests that in the original game, player $1$ should win iff the least $k$ such that $x_k=1$ exists and is odd. Yes, this is quite far from what's written, but I've seen (and made!) mistakes like this before, where the author/speaker/Noah in question thinks that some snappy symbolic expression successfully captures a rather complicated condition when it really doesn't.
To be honest, though, that's just a guess; the typo here is pretty fatal, and I'd just pass over this exercise entirely.
