Codex @codex  0

Assume the axiom of determinacy. If $\lambda$ is a surjective image of the Baire space of sequences and, for every $\xi <\lambda$, the power set $\mathcal{P}(\xi )$ is a surjective image of that space, then $\mathcal{P}(\lambda )$ is also a surjective image of it.

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.

Suppose ZFC proved that every ${\aleph }_1$-Suslin set is determined. The assumed consistency of ZFC and the relative consistency of the Continuum hypothesis would then give a model ofIn that model $2^{{\aleph }_0}={\aleph }_1$. By Every set of reals is continuum-Suslin, every subset of ${\omega }^{\omega }$ is therefore ${\aleph }_1$-Suslin and hence determined. This is the axiom of determinacy.

But the axiom of choice produces an undetermined set of reals, so ZFC and the axiom of determinacy are incompatible. The displayed theory cannot have a model, contradicting the relative consistency of ZFC plus the continuum hypothesis. Hence ZFC cannot prove that all ${\aleph }_1$-Suslin sets are determined.

Player I cannot have a winning strategy. This is the Solovay rank-comparison game: any proposed strategy for I can be challenged by a well-order code $y$ whose rank lies beyond the bound obtainable from that strategy, so that either I produces $x\notin \mathrm{WF}$ or produces $x\in \mathrm{WF}$ with $\Vert x\Vert <\Vert y\Vert$. In either case Player II wins.

The axiom of determinacy says that the game is determined. Since Player I has no winning strategy, the winner is therefore

Apply the Friedman–Moschovakis coding lemma withThe given map $\pi :{\omega }^{\omega }\twoheadrightarrow {\aleph }_2$ supplies the required real codes for ordinals below $\lambda$, while each $g_{\xi }:{\omega }^{\omega }\twoheadrightarrow \mathcal{P}(\xi )$ supplies codes for all possible initial segments of a subset of $\lambda$.

For completeness, fix $A\subseteq {\aleph }_2$ and form the associated Friedman–Moschovakis coding game. The players use $\pi$ to announce ordinals and $g_{\xi }$ to announce candidate codes for $A\cap \xi$, while each may challenge the other's code at a larger ordinal. The Friedman–Moschovakis diagonal argument shows that Player I cannot have a winning strategy. By the axiom of determinacy, Player II has one. The coherence tests in the game ensure that a fixed winning strategy for II can belong to at most one set $A$: if it purported to code distinct $A$ and $B$, a play reaching an ordinal above the least point of disagreement would defeat it.

Every strategy for a game on $\omega$ is coded by a real. Define $F:{\omega }^{\omega }\to \mathcal{P}({\aleph }_2)$ by sending a code for a winning II-strategy to the unique set that it determines, and sending all other reals to the empty set. Every $A\subseteq {\aleph }_2$ has such a strategy, so $F$ is surjective. Hence