2
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.