If $A$ is $\kappa\text{-Suslin}$ then so is $pA$.

Question

So there is this proposition in "The Higher Infinite", that I have a problem with, and I need some help.

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The prop.

Proposition. Suppose that $A \subseteq {^k(^\omega\omega)}$ and that $\kappa \gt 1$. If $A$ is $\kappa\text{-Suslin}$, then so is $pA$.
Proof. If a tree $T$ on ${^k\omega} \times \kappa$ is such that $A = p[T]$, then that tree recast as $T'$ on ${^{k-1}\omega} \times (\omega \times \kappa)$ is such that $pA = p[T']$. The result follows by applying a bijection:$\omega \times \kappa \rightarrow \kappa$.

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So I am totally fine with the proof except for one small detail: what happens when $\kappa$ is finite? We can't have the bijection in the proof. So I naturally tried using a surjection but that has counter-examples to it. So my question is, is this true for finite $\kappa$? If so, can you give me a hint towards the solution please?

Answer 1

I don't recall any uses of $\kappa$-Suslin for finite $\kappa$. But Asaf's intuition is correct: $A\subseteq {^k(^\omega\omega)}$ is $n$-Suslin if and only if $A$ is closed.

The direct implication holds because if $A$ is $Y$-Suslin for $Y$, then $A$ is the projection of a closed subset of $^k(^\omega\omega)\times {}^\omega Y$, and if $Y$ is finite then $^\omega Y$ is compact. It is generally true that that the projection of a closed subset of $B\times C$ onto $B$ (where both $B,C$ are separable metrizable and $C$ is compact) must be closed. The reverse implication is trivial.

In any case, the correct statement of the quoted proposition is

Suppose that $A \subseteq {^k(^\omega\omega)}$ and that $\mathbf{k}>1$. If $A$ is $\boldsymbol\kappa$-Suslin, …

So it is just a typographical confusion.