In Stevo Todorcevic's "A dichotomy for P-ideals of countable sets" (link, page 261 at the bottom [page 11 in the pdf]), the following confusing situation comes up:
(Context: $\mathcal I$ is a P-Ideal (of countable sets) on an uncountable ordinal $\theta$ and $Y_0$ is a subset of $[\mathcal I]^\omega$ such that for every $A\in Y_0$ there exists some finite $F\subset\theta$ witnessing a certian property; $M$ is just a countable elementary submodel of some $H_\kappa$.)
I quote:
It follows that $Y_0=[\mathcal I]^\omega$. By the Pressing Down Lemma, applied in $M$, there is a finite set $F_0$ in $M$ and a stationary subset $S$ of $[\mathcal I]^\omega$ in $M$ such that $F_0$ witnesses $a\in Y_0$ for all $a\in S$.
My question is: Can anybody give me a reference for the variant of the Pressing Down Lemma he's using here? I.e. it seems it would have to work for functions $f:[\mathcal I]^\omega\to[\theta]^{<\omega}$...
Asaf Karagila was so nice as to suggest (link), but it doesn't seem to be applicable here, or at least I don't see how to quite make it work in this situation...
If I’m reading it correctly, for each $a\in[\mathcal{I}]^\omega$ you have a finite $F_a\subseteq A_a$. $A_a\subseteq\bigcup a$, so $F_a$ is contained in the union of a finite $f_a\subseteq a$. Thus, we actually have a map
$$\varphi:[\mathcal{I}]^\omega\to[\mathcal{I}]^{<\omega}:a\mapsto f_a$$
with the property that $f_a\subseteq a$ for each $a\in[\mathcal{I}]^\omega$. Replacing $\mathcal{I}$ with $[\mathcal{I}]^{<\omega}$ turns out to require only minor modifications to the usual arguments.
Suppose that $\varphi$ is not constant on any stationary subset of $[\mathcal{I}]^\omega$. For each $f\in[\mathcal{I}]^{<\omega}$ let $C_f$ be a club set in $[\mathcal{I}]^\omega$ disjoint from $\{a\in[\mathcal{I}]^\omega:f_a=f\}$.
Let $C=\{a\in[\mathcal{I}]^\omega:a\in C_f\text{ for each }f\in[a]^{<\omega}\}$. Since $\left|[\mathcal{I}]^{<\omega}\right|=\big|[\mathcal{I}]\big|$, the usual proof that diagonal intersections of club sets are club goes through with very minor modification to show that $C$ is a club set. Now fix $a\in C$; then on the one hand $a\in C_{f_a}$, since $f_a\in[a]^{<\omega}$, but on the other hand $a\notin C_{f_a}$ by the definition of $C_{f_a}$. This contradiction shows that $\varphi$ is constant on some stationary subset of $[\mathcal{I}]^\omega$.
Added: (Oops; I forgot to finish the main argument.) Say that $\varphi$ is constant with value $f$ on a stationary $S\subseteq[\mathcal{I}]^\omega$. For each $F\in\left[\bigcup f\right]^{<\omega}$ let $S_F=\{a\in S:F_a=F\}$; then $S$ is the union of the countable family of sets $S_F$ for $F\in\left[\bigcup f\right]^{<\omega}$. But the union of countably many non-stationary subsets of $[\mathcal{I}]^\omega$ is non-stationary, so one of the $S_F$ must be stationary, as desired.