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Every infinite game of perfect information on a set-sized move alphabet with a Borel payoff in its discrete product topology has a winning strategy in an infinite game for one player. A proof strengthens the induction hypothesis to existence of arbitrary-depth-preserving game coverings unravelling the payoff. The closed-set announcement construction starts the induction. For countable unions, successive coverings stabilize each finite level and their inverse limit makes every constituent payoff clopen; their union is open and can be unravelled once more.

With the axiom of choice, enumerate both players' binary-move strategies in an infinite game in length $\mathfrak{c}$. Each strategy admits $\mathfrak{c}$ compatible plays. At stage $\alpha$, reserve a fresh play following I's strategy as a loss for I, and a fresh play following II's strategy as a win for I. Avoid all previously reserved plays, which number less than $\mathfrak{c}$. The collection of the second reservations is a payoff set against which neither player has a winning strategy in an infinite game. This uses no regularity assumption on $\mathfrak{c}$.

An infinite game of perfect information is determined when one of its two players has a winning strategy in an infinite game.

An infinite game of perfect information with an open set of winning plays for I in the product topology on a discrete move set has a winning strategy in an infinite game for one player. Construct a winning-position attractor in an infinite game by transfinite recursion; ranks give I a terminating descent strategy inside it, and its complement gives II a strategy avoiding every winning prefix. The move set need not be countable, in the usual set theory with axiom of choice.

In an infinite game of perfect information, the players alternately choose elements of a nonempty set $A$, producing $x\in A^{\omega }$. Player I wins when $x$ belongs to the payoff set $B$. A strategy in an infinite game chooses a move at each position where its player moves, and a winning strategy in an infinite game wins against every possible opposing play. The Gale-Stewart theorem states that such a game is determined when $B$ is open in the product topology of discrete $A$. The same holds for closed payoff sets and, more strongly, for every $G_{\delta }$ payoff set.

For open determinacy, let $P=A^{<\omega }$ be the positions and $D=\{s:[s]\subseteq B\}$ the prefixes certifying a win for I, where $[s]$ is the extension cylinder set. Form the winning-position attractor in an infinite game by transfinite recursion: start at $W_0=D$, at a successor stage add I-positions with some successor already in $W_{\alpha }$ and II-positions with all successors in $W_{\alpha }$, and take unions at limits. This increasing sequence stabilizes, since its positions form a set. Call the stable set $W$.

Each position in $W\setminus D$ has a least entry ordinal, necessarily a successor. I chooses a successor with smaller entry ordinal; every II successor has smaller entry ordinal. An infinite strictly decreasing sequence of ordinals is impossible, so play reaches $D$ in finitely many moves. This gives I a winning strategy in an infinite game from every position in $W$. Outside $W$, all I moves stay outside, while II can choose a move outside. The axiom of choice selects those moves simultaneously. Such a play never reaches $D$, so it is outside the open payoff set. Thus II wins from the complement of $W$, proving determinacy. Interchanging the players and complementing the payoff gives closed determinacy.

We prove the $G_{\delta }$ strengthening via Buchi games, including their determinacy argument. On a set of positions $S$ with nonempty successor sets, putFor a specified set $F$ of marked positions, I's objective is to visit $F$ infinitely often. For each $X\subseteq S$, let $H(X)$ be the least fixed point obtained by starting $Y_0=\varnothing$ and iteratingwith unions at limits. Its inner entry ranks let I force a visit to $F\cap \mathrm{Pre}(X)$. The operation $H$ is monotone. Starting at $X_0=S$, iterate $X_{\alpha +1}=H(X_{\alpha })$ with intersections at limits. This descending sequence stabilizes at $W=H(W)$.

From $W$, the inner ranks force a marked position with a next move still in $W$. I chooses such a move, and the construction repeats. At II-positions all necessary moves stay in the indicated inner attractor, or in $W$ at a marked position. Every round takes finitely many moves by decreasing inner ranks. I therefore forces infinitely many marked visits.

For $s\notin W$, let $\alpha$ be its removal rank: $s\in X_{\alpha }\setminus X_{\alpha +1}$. Since $s\notin H(X_{\alpha })$, at I's turn all successors lie outside $H(X_{\alpha })$, and at II's turn there is a successor outside it. II chooses such successors. If also $s\in F$, then $s\notin \mathrm{Pre}(X_{\alpha })$; II can choose outside $X_{\alpha }$, or every I move lies outside $X_{\alpha }$. Along this strategy removal ranks never increase and strictly decrease after each marked visit. Infinitely many visits would give an infinite ordinal descent, so II wins. This proves Buchi game determinacy on a set-sized arena without any countability restriction.

Now write the original payoff as $B={\bigcap }_{n<\omega }{U}_n$, with each $U_n$ open. Add a deterministic monitor to the finite position, initially $n=0$. At each new prefix, if its cylinder is contained in $U_n$, increment the monitor and mark that step. For a play $x$, the monitor increments infinitely often exactly when $x\in B$. In one direction every increment certifies the next required open set. In the other, after the monitor reaches $n$, openness of $U_n$ supplies a later certifying prefix, so it eventually reaches $n+1$. The expanded game is a Buchi game; its monitor is determined by the observed finite play, so its winning strategy transfers directly to the original game. ConsequentlyComplementing and interchanging players also gives determinacy for $F_{\sigma }$ payoff sets. These arguments work on pruned game subtrees as well, and assigned terminal winners can be incorporated by replacing each terminal position by an absorbing position with the corresponding outcome.

Represent the arena by a pruned set-theoretic tree $T$ of legal finite positions. Players I and II alternate; every position has a legal continuation. The usual terminal-position convention can also be handled by adding terminal wins and losses at the initial stage. Give the branches the product topology generated by finite-position cylinders, and let $A$ be I's winning open set.

Define the winning-position attractor in an infinite game by transfinite recursion. Start withAt a successor stage retain $W_{\alpha }$ and add I-positions with at least one child in $W_{\alpha }$, and II-positions with every child in $W_{\alpha }$. At a limit stage take the union. The increasing sequence stabilizes because the positions form a set: strict increases at arbitrarily many stages would select too many distinct positions. Write $W$ for the stable set. The rank of a position in $W$ is its first entry stage; any positive such rank is a successor ordinal.

If the initial position belongs to $W$, at every I-position of positive rank choose a child that was present at the preceding stage. Every move by II also leads to a child of smaller rank, by the rule for II-positions. There is no infinite strictly decreasing sequence of ordinals, so every play following this strategy in an infinite game reaches $W_0$ after finitely many moves. Its complete branch then belongs to $A$. Hence this is a winning strategy in an infinite game for I.

If the initial position is outside $W$, every move of I stays outside $W$. At a II-position outside $W$ at least one child is outside $W$: otherwise the entry ranks of its set of children have a common ordinal bound, and that position would enter at the next stage. Choose such a child. This gives II a strategy in an infinite game that never enters $W_0$. A resulting branch cannot belong to $A$, because $A$ is an open set and would therefore have a finite prefix whose whole cylinder lies in $A$, namely a member of $W_0$. Thus II wins. This proves the Gale-Stewart theorem for an arbitrary set-sized arena, in the ordinary metatheory with choice. Countability of the move set was never used.

The restriction on the payoff is essential. Under the axiom of choice, already for binary moves there is an undetermined game. There are $\mathfrak{c}=2^{{\aleph }_0}$ strategies for each player, since a strategy is a binary-valued map on a countable set of positions. List I's strategies ${\sigma }_{\alpha }$ and II's strategies ${\tau }_{\alpha }$ for $\alpha <\mathfrak{c}$. Each strategy is compatible with $\mathfrak{c}$ different plays, obtained by varying all the opponent's binary moves.

At stage $\alpha$, choose a fresh play $x_{\alpha }$ following ${\sigma }_{\alpha }$ and then a different fresh play $y_{\alpha }$ following ${\tau }_{\alpha }$, avoiding every previously chosen play. Fewer than $\mathfrak{c}$ plays have been used at that stage, even when $\mathfrak{c}$ is singular, so these choices are possible. Let $B=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. No I-strategy wins $B$, because its selected $x_{\alpha }$ is outside $B$; no II-strategy wins, because its selected $y_{\alpha }$ is inside $B$. Future choices avoid all earlier $x_{\alpha }$, so they never spoil the first assertion. This is the choice diagonalization of an undetermined game. Consequently determinacy for all payoff sets is incompatible with the axiom of choice. The attractor still makes sense for a nonopen payoff, but an avoiding play may then enter the payoff without ever having a winning finite prefix.

For a finite position $s\in A^{<\omega }$, let $[s]$ be the cylinder set of infinite plays extending $s$. Because the factors have the discrete topology, these cylinder sets form a basis for the product topology. PutMembership in $D$ means that I has already secured the outcome, regardless of future moves. Since $G$ is an open set, a play belongs to $G$ exactly when one of its prefixes belongs to $D$.

Build the winning-position attractor in an infinite game by transfinite recursion. Start with $W_0=D$, use unions at limit ordinals, and setThis sequence is increasing: the defining operation is order-preserving, and $W_0\subseteq W_1$. It stabilizes at some ordinal. Indeed, if it strictly increased at every successor stage below the successor cardinal number of $|A^{<\omega }|$, choosing one newly added position per stage would inject that larger cardinal into $A^{<\omega }$. Denote the stable set by $W$. This constructs the needed fixed point directly, without an appeal to an unstated fixed-point theorem.

Every position in $W$ has a least entry ordinal, its rank. Rank zero means membership in $D$. A positive rank is a successor $\alpha +1$: at an I-position some extension belongs to $W_{\alpha }$, while at a II-position every extension does. If the empty word belongs to $W$, I always chooses an extension of smaller rank until $D$ is reached. II's choices also decrease rank before that happens. There is no infinite strictly decreasing sequence of ordinals, so some finite prefix lies in $D$. This is a winning strategy in an infinite game for I.

If the empty word is outside $W$, the fixed point equation says that every extension of an I-position outside $W$ is still outside $W$, and that a II-position outside $W$ has at least one extension outside $W$. II chooses such an extension. Every resulting prefix avoids $D$, hence the whole play avoids $G$ by openness. This is a winning strategy in an infinite game for II.

The choices can be made simultaneously by fixing a well-order of $A$, as allowed by the usual axiom of choice. The argument does not assume that $A$ is countable. Every such open game is determined:

TakeThus ${\Phi }_M$ is equivalent in ZF to ${\mathsf{AC}}_{M^{<\omega }}(M)$. Indeed, this choice principle chooses one move from every nonempty value of a winning quasistrategy, turning it into a winning strategy in an infinite game. The converse encodes an arbitrary $M^{<\omega }$-indexed family of nonempty subsets of $M$ into a quasidetermined game. This is the choice characterization of quasideterminacy.

Consider the following game on the real numbers. On their first moves, Player I plays $y\in \mathbb{R}$ and Player II replies with $x\in \mathbb{R}$; later moves are ignored. Declare Player II the winner whenPlayer I cannot have a winning strategy in an infinite game: its first move is some fixed $y$, and if $y\notin pA$ then II wins automatically, while if $y\in pA$ then II can reply with an $x$ satisfying $(x,y)\in A$.

The axiom of determinacy for games on $\mathbb{R}$ therefore gives Player II a winning strategy $\tau$. For every $y\in pA$, define $f(y)$ to be II's first response to the move $y$. The winning condition forcesso $f:pA\to \mathbb{R}$ is the required uniformization of a binary relation. This is the direct game proof of uniformization from determinacy.

An infinite game of perfect information is quasidetermined when one of the players has a winning quasistrategy. Choosing one allowed move at every relevant position refines a winning quasistrategy to a winning strategy in an infinite game whenever the required restricted choice principle holds.