# Algebraic closure and automorphism group (model theory)

Let $$M$$ be a homogeneous $$L$$-structure and $$A$$ be a finite subset of $$M$$. Let $$acl(A)$$ denote the algebraic closure of $$A$$ which means the set of all points $$a$$ such that there is $$L_A$$-formula $$\phi(x)$$ such that $$M\models \phi(a)$$ and $$\phi(x)$$ has finitely many solution in $$M$$.

Question 1. Are the following statements true? why?

1. Let $$B$$ be a finite subset of $$M$$ such that for every automorphism $$f\in Aut(M/A)$$ we have $$f(B)=B$$. Then $$B\subseteq acl(A)$$.

2. Let $$B$$ be a finite subset of $$M$$ such that for every $$b\in B$$ the orbit of $$b$$ under the action of $$Aut(M/A)$$ is finite. Then $$B\subseteq acl(A)$$.

Update: I just realized that $$1.\Rightarrow 2.$$ obviously!

In fact, $$1$$ and $$2$$ are equivalent. To see that $$2$$ implies $$1$$, let $$B$$ be finite such that $$f(B) = B$$ for all $$f \in \operatorname{Aut}(M/A)$$. Then every $$b \in B$$ has finite orbit, because its orbit is contained in $$B$$, which is finite. So indeed $$B \subseteq \operatorname{acl}(A)$$.
Without saturation (a counterexample). Consider the language with countably many unary predicates $$P_n(x)$$ for $$n < \omega$$. Let $$M$$ have as underlying set $$2^\omega$$, and for $$\xi \in 2^\omega$$ we set $$M \models P_n(\xi)$$ iff $$\xi(n) = 1$$. Since no two elements have the same type, there are no non-trivial automorphisms on $$M$$ and also no non-trivial partial automorphisms. So $$M$$ is homogeneous and the orbit of every element is finite (has size $$1$$ even). On the other hand, for any finite set $$A \subseteq M$$, its algebraic closure is just $$A$$ itself. To see this, let $$\phi(x)$$ be a formula with parameters in $$A$$, which has a realisation $$\xi \in M - A$$. Since $$\phi(x)$$ only mentions finitely many predicate symbols $$P_{n_1}, \ldots, P_{n_k}$$, it does not care about any of the remaining symbols. So any extension of $$\xi \upharpoonright_{n_k + 1}$$ will be a realisation of $$\phi(x)$$.
With saturation (statements hold). So let's now assume that $$M$$ is $$\omega$$-saturated (which is all we need). We will prove $$2$$, for this it suffices to prove the following: if the orbit of some element $$b \in M$$ is finite under $$\operatorname{Aut}(M/A)$$, then $$b \in \operatorname{acl}(A)$$. We will prove the contraposition, so suppose that $$b \not \in \operatorname{acl}$$. Then $$p(x) = \operatorname{tp}(b/A)$$ does not contain any algebraic formulas. By compactness the following set of formulas is consistent $$\Sigma((x_i)_{i < \omega}) = \{ p(x_i) : i < \omega \} \cup \{x_i \neq x_j : i < j < \omega \}.$$ Using $$\omega$$-saturation we then find realisations $$(b_i)_{i < \omega}$$ of $$\Sigma$$ in $$M$$. By construction all these realisations have the same type over $$A$$ as $$b$$, so they are all in the orbit of $$b$$ under $$\operatorname{Aut}(M/A)$$. We thus see that $$b$$ has an infinite orbit over $$A$$, as required.