I specify some definitions that are necessary to understand the exercise I'm working on:
Given a monster model $\mathfrak{U}$ (of inaccessible cardinality) and a set $A\subseteq \mathfrak{U}$, we say that a set $B \subseteq \mathfrak{U}^{|x|}$ (where usually $x$ is a tuple of free variables, maybe infinite) is quasi-invariant over $A$ if, whenever $f_1,\dots,f_n$ is a finite tuple of automorphisms over $\mathfrak{U}$ that pointwise fix $A$, we have that the sets $f_i[B]$ have non-empty intersection.
We say that a type $p(x)\subseteq L(\mathfrak{U})$ is quasi-invariant over $A$ if $\varphi(\mathfrak{U})$ (as a set) is quasi-invariant over $A$ for every conjunction $\varphi(x)$ of formulas in $p(x)$. (Note that here $x$ may be an infinite tuple of free variables)
We say that a set $B$ is strongly quasi-invariant over $A$ if for every definable set $D\subseteq \mathfrak{U}$ we have that at least one of $B\cap D$ and $B \cap \neg D$ is quasi -invariant. The extension of such definition to types is done in the same way as for quasi-invariant types.
Here is the exercise:
Prove that every strongly quasi-invariant type has an extension to a complete strongly quasi-invariant type. Hint: it may help to prove that if $B$ is strongly quasi-invariant then for every definable $D$ either $B\cap D$ or $B \cap \neg D$ is strongly quasi-invariant
I have tried in many ways but I have problems in proving even the claim in the hint. Any help?
EDIT: I managed to prove the result but by strengthening a bit the definition of strongly quasi-invariant sets. The new definition I gave is the following:
We say that a set $B$ is strongly quasi-invariant over $A$ if for every finite equivalence relation $\varepsilon(x;y) \in L(\mathfrak{U})$ there exists a $c \in \mathfrak{U}^{|y|}$ s.t. $B \cap \epsilon(\mathfrak{U};c)$ is quasi-invariant. In other words if for every finite and definable partition of $\mathfrak{U}^{|x|}$, there exists a partition element $D$ s.t. $B\cap D$ is quasi-invariant.
Using this new definition, the claim in the hint follows pretty easily.
I've thought about it and I think that this new definition is strictly stronger than the previous one (i.e. the new strongly quasi-invariant sets are a proper subset of the old strongly quasi-invariant sets ), but I'm not sure about it.
Thanks