If there is a hyperimaginary $a_{E}$ and sets: $A,B\subset M$, such that for every automorphism F which fix A , F fix $a_{E}$, and for every automorphism F which fix B , F fix $a_{E}$. Does it true that for every automorphism F which fix $A\cap B$ , F fix $a_{E}$?
1 Answer
The answer is not necessarily.
For a counterexample, consider the first-order structure $M$ consisting of a naked set with exactly two points $M=\{a,b\}$, in the first order language having just equality. Let $E$ be the equivalence relation of equality, which is type definable and indeed outright definable in this language, and consider the equivalence class $a_E=\{a\}$ of the first point $a$. This is a hyperimaginary (the equivalence class of a type definable equivalence relation in a model). Meanwhile, let $A=\{a\}$ and $B=\{b\}$. Note that there are exactly two automorphisms of $M$, namely, the identity and the automorphism that swaps $a$ and $b$. But the only automorphism of $M$ fixing $A$ is the identity, which therefore fixes $a_E$ as well, and similarly for automorphisms fixing $B$. But the automorphism swapping $a$ and $b$ fixes $A\cap B$, since this set is empty, but takes $a_E$ to $b_E$.
This same idea as in this example can be made to work in much more elaborate settings.