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An analytic subset of a Polish space is a continuous image of a Borel set, equivalently a projection of a closed subset of its product with the Baire space of sequences.

Descriptive set theory studies definable subsets of spaces such as the Baire space of sequences, with particular emphasis on hierarchies of complexity, regularity properties and infinite games.

For $A\subseteq \lambda$, the Friedman–Moschovakis coding game asks the players to present coherent local codes for initial segments $A\cap \xi$ while challenging each other at larger ordinals. The diagonal and boundedness argument rules out a winning strategy for Player I; a winning strategy for Player II determines at most one set $A$. Coding strategies by reals therefore gives a surjection from the Baire space of sequences onto $\mathcal{P}(\lambda )$.

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.

By part ii every subset of the Baire space of sequences is $2^{{\aleph }_0}$-Suslin. If $2^{{\aleph }_0}\le {\aleph }_2$, part i would make every such set ${\aleph }_2$-Suslin. Therefore

Now suppose $2^{{\aleph }_0}>{\aleph }_2$. The axiom of choice gives a set $B\subseteq {\omega }^{\omega }$ of cardinality exactly ${\aleph }_2$. If $B$ were ${\aleph }_1$-Suslin, the Aleph-one-Suslin decomposition into analytic sets would write it as a union of ${\aleph }_1$ analytic sets. If all those analytic sets were countable, their union would have cardinality at most ${\aleph }_1$, so one of them is uncountable. The perfect set property for analytic sets then makes that member, and hence $B$, have cardinality $2^{{\aleph }_0}$, contradictingThus $B$ is not ${\aleph }_1$-Suslin, and