# Why is the intersection of countably many homogeneously Suslin subsets of $\,^{\omega} \omega$ homogeneously Suslin?

I found this assertion in these notes: The derived model theorem (Steel) right in the beginning on page 3, together with the remark that this is 'not too hard to show'. Unfortunately, I'm stuck at the moment. Could somebody point me to where I can find a proof? In the books I checked I can find similar results only on the weakly homogeneously Suslin sets. (edit: I did find the result in Woodin's "The axiom of determinacy", but again without proof.)

edit: The only thing I can come up with right now is: If $T_n$ is a tree on $\omega\times Z_n$ with $A_n = p[T_n]$ for $n<\omega$, then $T$ on $\omega\times \Pi_{n<\omega} Z_n$ could be defined by $(s,(t_n\, |\, n<\omega))\in T$ iff $(s,t_n) \in T_n$ for all $n$, indeed yielding $p[T]= \bigcap_{n<\omega} A_n$; then for each $s\in\,^{<\omega} \omega$, I need a measure on $\Pi_{n<\omega}T_{n,s}$ with measures on each individual $T_{n,s} = \{ t\in\,^{<\omega} Z\, |\, (s,t)\in T_n\}$ given. But I know of no general way to define a measure on the (entired powerset of the) product.

• Are you concerned with the statement towards the bottom on p.3 which states, »It is not too hard to show $\operatorname{Hom}_\kappa^Y$ is closed under countable intersections.«? This doesn't appear to deal with Suslin homogeneity, which is defined on the following page. – user642796 Oct 13 '15 at 20:05
• You are right, that doesn't seem to be the same. I don't know if the proof would be different. Still, in "The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal", Lemma 2.11. on p. 27, the statement appears for homogeneously Suslin sets. – user35359 Oct 14 '15 at 5:57