Open and closed Gale–Stewart games are determined

Let $U=\{p:[T{]}_p\subseteq B\}$, and let $W$ be the positions from which $P$ has a strategy forcing a visit to $U$ at a finite time, allowing time zero. Strategy sets and the set of positions are sets; A1 fixes one such strategy for each $p\in W$ and fixes default legal moves. Every branch in $B$ has a prefix in $U$ by openness and F1.

At $p\mathrm{\notin }U$, if $P$ moves and $p\in W$, the strategy's first chosen child is in $W$ by restriction. Conversely a child in $W$ lets $P$ choose it and follow its selected strategy. If $Q$ moves and $p\in W$, restriction after every possible first opponent move makes all children belong to $W$. Conversely, if all children are in $W$, following the selected continuation for the child the opponent chooses forces a visit. Thus outside $W$, every child of a $P$-node avoids $W$, and at least one child of a $Q$-node avoids $W$.

If the initial position is in $W$, its selected strategy reaches $U$, and every continued branch lies in $B$ because it extends that visited node. If the initial position is outside $W$, use A1 to choose an avoiding child at each $Q$-node outside $W$ and defaults elsewhere. By step 2.1 every consistent branch remains outside $W$ and therefore outside $U$. Step 1.1 shows it is outside $B$. In this case $Q$ wins. This reasoning refers to the actual player at each node, so it also works after either parity of fixed history.

If the specified I payoff is closed, its complement is an open II payoff; apply step 3.1 with $P=II$, retaining the original turn parity. Finally F2 reduces a taboo game to either an already winning reachability case or a pruned residual game; cylinder restriction preserves open and closed sets. Applying the proved pruned result and F2 completes both cases. QED.