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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.

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: