Consider a Gale-Stewart game with a Borel winning set $W$ and a finite action space $A$ at each stage, which we denote as $G(W)$.

For a winning set $C$, $V(G(C))$, the value of $G(C)$, is $1$, if Player $1$ wins, that is the outcome is in $C$. $V(G(C))=0$, if Player $2$ wins.

Does there necessarily exist two winning sets, an open set $U$ and a closed set $F$,and $F \subseteq W \subseteq U$, such that $V(G(F))=V(G(U))$?

Added: I initially asked whether it's necessarily the case there exists an open set $U$ and a closed set $F$,and $U \subseteq W \subseteq F$, $V(G(F))=V(G(U))$ with the order of $F \subseteq W \subseteq U$ reversed. It's not true since there exists a winning set with an empty interior that player $1$ has a winning strategy.


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