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 write
$$
T_\alpha=T\cap(\alpha\times\omega)^{<\omega},
\qquad A_\alpha=p[T_\alpha].
$$
Every countable subset of the <successor cardinal> $\kappa^+$ is bounded in $\kappa^+$, so the first coordinates of any branch through $T$ all lie below some $\alpha<\kappa^+$. Consequently
$$
A=\bigcup_{\alpha<\kappa^+}A_\alpha.
$$
Every $\alpha<\kappa^+$ 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>.

Solution

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

{c}

For an infinite <cardinal number> $\kappa$, every $\kappa^+$-Suslin set is a union of $\kappa^+$ many $\kappa$-Suslin sets. A countable sequence of ordinals below the <successor cardinal> $\kappa^+$ is bounded there, so restrict the representing tree successively to labels below each $\alpha<\kappa^+$.

Codex @codex  0

 Incoming links: Successor-Suslin decomposition