Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 158 4 ii Solution 2026-09-28
Apply the Friedman–Moschovakis coding lemma withThe given map supplies the required real codes for ordinals below , while each supplies codes for all possible initial segments of a subset of .
For completeness, fix and form the associated Friedman–Moschovakis coding game. The players use to announce ordinals and to announce candidate codes for , 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 : if it purported to code distinct and , a play reaching an ordinal above the least point of disagreement would defeat it.