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

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.