# Set of winning strategies for union of winning sets

Suppose that $G( \omega, A, X)$ denotes a sequential game of perfect information in which player I and player II play an element in $A$ in each turn with a total number of $\omega$ moves. $X$, the winning set, is a subset of $A ^{\omega}$. If the outcome is in $X$, then player I wins. Otherwise, player II wins.

$G(\omega, A, X_1)$ and $G(\omega, A, X_2)$ are two games in which player I has a winning strategy. Clearly in the game $G(\omega, A, X_1\cup X_2)$, player I also has a winning strategy. Let $W_{X_1}$, $W_{X_2}$ and $W_{X_1 \cup X_2}$ be the sets of player I's winning strategies respectively. I'm interested to know their relationship given their non-emptiness in general.

In particular, is it true that $W_{X_1} \cap W_{X_2} =\varnothing$, then $W_{X_1 \cup X_2} = W_{X_1} \cup W_{X_2}$?

• for $X$ closed or open in $A^\omega$, for the product topology with $A$ discrete, the game $G(A,X)$ is determined.
• with the axiom of choice, there is a subset $X$ of $\omega^\omega$ for which the game $G(\omega,X)$ is not determined.
I am not aware of any result saying that if $X_1$ and $X_2$ are determined, then so is their union or intersection. I once asked an expert in the field, who says this might well be wrong in ZFC.
For your particular question, the answer is no. Take $A=2$, $$X_1=\{01x\mid x\in 2^\omega\}\cup\{10x\mid x\in 2^\omega\}$$ and $$X_2=2^\omega \setminus X_1=\{11x\mid x\in 2^\omega\}\cup \{00x\mid x\in 2^\omega\}$$ then every strategy for player I is winning in $X_1\cup X_2=2^\omega$, while there are strategies for player I which are not winning neither in $G(2,X_1)$ nor in $G(2,X_2)$, take for example the strategy which consists in playing $1$ all the time.