Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 158 / 2

Let $T\subseteq (X\times \omega {)}^{<\omega }$ be a Suslin representation of $A$, and let $f:X\to Y$ be injective. Apply $f$ coordinatewise to the first coordinate of every node and putThe injectivity of $f$ ensures that a sequence is a branch of $T_f$ exactly when its first coordinate decodes to a branch of $T$. Hence $p[T_f]=p[T]=A$, proving that every X-Suslin set is $Y$-Suslin whenever $X$ injects into $Y$.

Use one label for each member of $A$. More explicitly, setAn infinite branch through this tree has a constant first coordinate $a\in A$ and second coordinate $a$, so $p[T_A]=A$. Thus $A$ is $A$-Suslin. Since $A\subseteq {\omega }^{\omega }$ injects into a set of cardinality $2^{{\aleph }_0}$, part i makes $A$ a $2^{{\aleph }_0}$-Suslin set. This proves that Every set of reals is continuum-Suslin.

Let $A=p[T]$ for a tree $T\subseteq ({\kappa }^{+}\times \omega {)}^{<\omega }$. For each $\alpha <{\kappa }^{+}$, restrict the first-coordinate labels to $\alpha$ and writeEvery countable subset of the successor cardinal ${\kappa }^{+}$ is bounded in ${\kappa }^{+}$, so the first coordinates of any branch through $T$ all lie below some $\alpha <{\kappa }^{+}$. ConsequentlyEvery $\alpha <{\kappa }^{+}$ has cardinality at most $\kappa$, hence injects into $\kappa$. Part i shows that each $A_{\alpha }$ is $\kappa$-Suslin. This is the Successor-Suslin decomposition.

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