Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 158 / 4 / ii

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