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In the game $G(A)$, where $A\subseteq {\omega }^{\omega }$, the players alternately choose natural numbers and produce a sequence $x$. Player I wins exactly when $x\in A$. A strategy in an infinite game assigns a move to every finite position where its player is to move; it is winning if every play following it is won by that player. A game is determined if one player has a winning strategy. Both cannot have winning strategies, since their joint play would have to be won by both. The axiom of determinacy asserts

Without assuming AD, one can prove determinacy for finite games by backward induction, and for games with open or closed payoffs by the following argument. Give ${\omega }^{\omega }$ its product topology of discrete move coordinates. If $A$ is open, let $W_0$ be the finite positions whose entire cylinder of extensions lies in $A$. Form a winning-position attractor in an infinite game by transfinite recursion: add an I-position if some child has already been added, and add an II-position if all its children have already been added. Take unions at limits. The process stabilizes because there are only set-many positions and each genuine successor change adds a new one.

An added nonterminal position receives its least entry ordinal. At an I-position in the stable attractor, choose a child of smaller rank; at an II-position every child has smaller rank. Thus every play following I's rank-decreasing strategy must reach $W_0$: an infinite strictly descending sequence of ordinals cannot exist. The resulting play is in $A$, so I wins from any position in the attractor.

Outside the attractor, every child of an I-position remains outside, and an II-position has some child outside. Otherwise the ranks of all its children would have a set-sized supremum and the parent would have been added at a later stage. II chooses such a child and keeps the play outside $W_0$ forever. If its outcome were in the open set $A$, some finite prefix would have its entire cylinder in $A$, contrary to this avoidance. Therefore II wins from outside. With natural-number moves the least suitable child may be chosen at each position, so no choice axiom is needed for these strategies. This proves open determinacy. For a closed payoff, apply the same argument to its open complement with the players' winning roles interchanged. Clopen payoffs are included in both results.

Now assume AC and construct an undetermined game. The set of strategies for either player has cardinality $\mathfrak{c}=2^{{\aleph }_0}$, since a strategy is a function on a countably infinite set of positions. Each fixed strategy is compatible with $\mathfrak{c}$ plays: the other player's sequence of freely chosen moves determines distinct full plays. Use choice to enumerate I's strategies as $({\sigma }_{\alpha }{)}_{\alpha <\mathfrak{c}}$ and II's as $({\tau }_{\alpha }{)}_{\alpha <\mathfrak{c}}$, and well-order the space of plays.

Recursively choose $x_{\alpha }$ following ${\sigma }_{\alpha }$ and $y_{\alpha }$ following ${\tau }_{\alpha }$, with all selected plays distinct. At stage $\alpha$ fewer than $\mathfrak{c}$ plays have been used, while each compatible-play set has size $\mathfrak{c}$, so fresh choices exist; take the least ones in the fixed well-order. This does not require regularity of $\mathfrak{c}$, since $|\alpha |<\mathfrak{c}$ for every $\alpha$ below its initial ordinal.

Put $A=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. Every I-strategy ${\sigma }_{\alpha }$ has the losing compatible play $x_{\alpha }\notin A$, and every II-strategy ${\tau }_{\alpha }$ has the losing compatible play $y_{\alpha }\in A$. Therefore neither player has a winning strategy. This choice diagonalization of an undetermined game provesThe elementary determinacy proofs above concern specified definable payoff classes; the diagonal construction explains why extending them to every arbitrary subset of the play space is incompatible with choice.

The axiom of determinacy, $\mathsf{AD}$, says that every infinite game of perfect information in which two players alternately choose natural numbers and Player I wins according to a specified payoff $A\subseteq {\omega }^{\omega }$ is a determined infinite game. A strategy specifies a move at every finite position where its player moves; it is winning if every compatible infinite play is won by that player.

Finite games are determined by backward induction. For infinite games, open determinacy has a direct proof. Let $T$ consist of the finite positions whose extension cylinder sets lie in the open payoff $A$. Build the winning-position attractor in an infinite game by transfinite recursion: start with $T$, add an I-position if some successor has already entered, add a II-position if every successor has already entered, and take unions at limit stages. The process stabilizes because there are only countably many finite positions.

At a position in the attractor but outside $T$, I can choose a successor with strictly smaller entry rank, and every move of II has smaller entry rank. There is no infinite descending sequence of ordinals, so I forces entry into $T$ and wins. Outside the attractor, every I-move stays outside it, while II can choose a successor outside it; that strategy avoids $T$ forever. Since $A$ is open, a play belongs to $A$ exactly when some prefix belongs to $T$. Thus II wins from outside the attractor. This proves determination for open payoffs; swapping the players proves it for closed payoffs too. The stronger Borel determinacy theorem gives determination for every Borel set payoff, whereas $\mathsf{AD}$ asserts this for arbitrary payoffs.

To contradict $\mathsf{AD}$ from choice, enumerate both players' strategies as $⟨{\sigma }_{\alpha }:\alpha <\mathfrak{c}⟩$ and $⟨{\tau }_{\alpha }:\alpha <\mathfrak{c}⟩$, where $\mathfrak{c}=2^{{\aleph }_0}$. There are $\mathfrak{c}$ strategies because the set of finite positions is countable; there are also $\mathfrak{c}$ plays following any fixed strategy because the other player's moves are arbitrary. At stage $\alpha$ of the choice diagonalization of an undetermined game, choose a fresh play $x_{\alpha }$ following ${\sigma }_{\alpha }$ and then a different fresh play $y_{\alpha }$ following ${\tau }_{\alpha }$, avoiding all earlier reservations. This is possible because fewer than $\mathfrak{c}$ plays have been reserved at every $\alpha <\mathfrak{c}$; regularity of $\mathfrak{c}$ is not required.

Set $A=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. Every ${\sigma }_{\alpha }$ loses on $x_{\alpha }\notin A$, and every ${\tau }_{\alpha }$ loses on $y_{\alpha }\in A$. Thus neither player has a winning strategy, and AD and AC are inconsistent together.

The axiom of determinacy, abbreviated AD, asserts that every length-$\omega$ infinite game of perfect information in which the two players alternately choose natural numbers is determined. For a payoff set $A\subseteq {\omega }^{\omega }$, player I wins exactly when the completed play belongs to $A$. A strategy specifies the next move at every finite position where its player is to move; determination means that one player has a winning strategy.

Finite games are determined by backward induction, choosing the least move leading to a guaranteed win when several are available. In ZF one can also prove open determinacy, hence determinacy for closed payoffs. Here is the full open-payoff argument. Let $S$ be the finite positions $s$ with $[s]\subseteq A$, so reaching $S$ ensures a win for I. Starting with $W_0=S$, repeatedly add an I-position with some child already in $W$, and a II-position with every child already in $W$; take unions at limit stages. This increasing process stabilizes because the set of finite positions is a set: continuing strict enlargement through its Hartogs number would choose distinct new positions, contradicting that bound. Assign to each admitted position its least entry stage.

At an I-position in the final $W$, choose the least child of smaller entry stage. At a II-position in $W\setminus S$, every child has smaller entry stage. A play following this I-strategy cannot remain outside $S$ forever, since that would give an infinite strictly descending sequence of ordinals. It reaches $S$ and wins. Outside $W$, every child of an I-position remains outside $W$, and a II-position has at least one child outside $W$. Player II chooses the least such child. The play never reaches $S$, so it cannot belong to the open set $A$. The initial position is either inside or outside $W$, proving determination. For a closed payoff apply the same argument to the open complementary payoff for the other player, with the appropriate player assigned to each level. None of these strategies needs a choice principle because moves are natural numbers and least available moves are definable.

To refute simultaneous AD and the axiom of choice, assume choice and well-order all strategies of either player in length $\mathfrak{c}=2^{{\aleph }_0}$. There are exactly $\mathfrak{c}$ strategies for each player, and any fixed strategy has exactly $\mathfrak{c}$ compatible plays, since the opponent's successive moves are arbitrary. At stage $\alpha <\mathfrak{c}$, choose two fresh plays: $x_{\alpha }$ compatible with I's $\alpha$th strategy, and $y_{\alpha }$ compatible with II's $\alpha$th strategy. Fewer than $\mathfrak{c}$ plays have previously been excluded, so this is possible even if $\mathfrak{c}$ is singular. Put $A=\{y_{\alpha }:\alpha <\mathfrak{c}\}$. No $x_{\alpha }$ ever enters $A$. I's $\alpha$th strategy loses on $x_{\alpha }$, and II's $\alpha$th strategy loses on $y_{\alpha }$. Thus this game has no winning strategy for either player, and

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.