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!


1 Answer 1


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 $B \subseteq \operatorname{acl}(A)$.

These are very reasonable statements, but for them to be true you need an important property: saturation. For example, this holds in a monster model (but we do not need the full strength of a monster here). Without saturation, these statements will not hold.

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.


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