Automorphisms of saturated models

Question

This is basically Exercise 10.1.5(c) in Hodges's Model theory. First, a reminder of some definitions:

Let $\lambda$ be a cardinal, and let $\Sigma$ be a finitary first-order signature. A $\lambda$-saturated $\Sigma$-structure is a $\Sigma$-structure $A$ such that the following holds:

• If $X$ is a subset of $A$ and $\left| X \right| < \lambda$, then every complete $1$-type over $X$ with respect to $A$ is realised by an element of $A$.

Let $A$ be a $\Sigma$-structure. A definition for an element $a$ with parameters in a subset $X \subseteq A$ is a first-order formula $\phi (y, \vec{x})$ over $\Sigma$ and a finite sequence $\vec{b}$ of elements in $X$ such that $A \vDash \phi[y, \vec{b}] \leftrightarrow y = a$. A definable element of $A$ over $X$ is an element for which there exists a definition with parameters in $X$.

Question. Suppose $A$ is an $\aleph_0$-saturated $\Sigma$-structure and $a$ is not definable without parameters. Why should there be an automorphism of $A$ that moves $a$?

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Obviously, if there is such an automorphism, then there would have to be an element $a'$ in $A$ such that $a \ne a'$ but $a$ and $a'$ have the same type with respect to $A$, since automorphisms preserve types. This is no problem because we can use a compactness argument to build an elementary extension of $A$ where there is such an $a'$, and then use $\aleph_0$-saturation to realise that $a'$ in $A$. Then it suffices to find an automorphism of $A$ that moves $a$ to $a'$... but it is not clear how I am supposed to do this.

Theorem 10.1.8(a) gives a solution in the case where $A$ itself is countable, because then we can just enumerate all the elements of $A$ and do a kind of back-and-forth argument. If $A$ is $\aleph_0$-big then we could apply Exercise 10.1.4(a). What about the general case?

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Addendum. Here is the full text of exercise 10.1.5:

Let $\lambda$ be an infinite cardinal, $A$ a $\lambda$-saturated structure and $X$ a set of fewer than $\lambda$ elements of $A$. Show (a) if an element $a$ of $A$ is not algebraic over $X$, then infinitely many elements of $A$ realise $\textrm{tp}_A(a/X)$, (b) if an element $a$ of $A$ is not definable over $X$, then at least two elements of $A$ realise $\textrm{tp}_A(a/X)$, (c) $\textrm{ACL}(X) = \textrm{acl}(X)$ and $\textrm{DCL}(X) = \textrm{dcl}(X)$.

Parts (a) and (b) are quite easy exercises in using the compactness theorem.

Answer 1

This is not true in general. As you point out in the question, there are many hypotheses you can add to make the statement true. In fact, the equivalences "definable iff fixed by all automorphisms" and "algebraic iff finite orbit under all automorphisms" are two of the reasons why working in a big, or saturated, or sufficiently saturated and sufficiently strongly homogeneous monster model is so conceptually nice.

Anyway, here's a counterexample:

$\Sigma = \{R\}$, a single binary relation.

$A = \{a,a'\}\cup B \cup B'$, where $B$ is a countable set and $B'$ is an uncountable set.

$R = \{(a,b)\,|\,b\in B\}\cup \{(a',b')\,|\,b'\in B'\}$.

The picture is two graphs, with $a$ and $a'$ in their centers, and the elements of $B$ and $B'$ radiating out.

Now $A$ is $\aleph_0$-saturated, and $a$ is not definable over the empty set ($a'$ satisfies all the same formulas without parameters). But $a$ cannot be moved by an automorphism of $A$ - clearly it cannot be moved to any element of $B$ or $B'$, and it cannot be moved to $a'$, since the cardinalities of $B$ and $B'$ differ.

If you're comfortable with $A^{eq}$ (which you can read about in Hodges in the section Imaginary Elements), there's a more obvious counterexample. Just take the theory of two infinite equivalence classes, and let $A$ be a model where the classes have different cardinalities (and check that this is $\aleph_0$-saturated). In $A^{eq}$, there is an extra sort containing two elements, which are generic representatives of the equivalence classes. Neither of these elements is definable, but they cannot be swapped by any automorphism, because doing so would swap the classes.

Answer 2

It is a general fact that for infinite $\kappa$, a model $M$ is $\kappa$-saturated iff it is $\kappa$-homegeneous and $\kappa$-universal.

$\kappa$-homogeneity of $M$ means that for any subset $A$ of $M$ of cardinality less than $\kappa$, and any elementary function $f:A\to M$, and any element $a\in M$, $f$ extends to an elementary function $\overline f :A\cup \{a\}\to M$.

A different, more powerful notion of homogeneity is strong $\kappa$-homogeneity, it means that any elementary function defined on a subset of the model of cardinality less than $\kappa$ can be extended to an automorphism of a model.

As shown by Alex Kruckman's examples, $\kappa$-saturation does not imply strong $\kappa$-homogeneity, however, it is not hard to see that if a model is homogeneous (that is, it is homogeneous in its own cardinality), it is also strongly homogeneous. In particular, saturated models (that is, saturated in their cardinality) are strongly homogeneous, and it is easy to see that for a strongly $\kappa$-homogeneous models, definable small sets (in particular, singletons) are exactly the small sets fixed by automorphisms (and I believe it's actually equivalent condition to strong $\kappa$-homogeneity).

In model theory, we often assume that all considered structures are substructures of some very large structure, often called the monster model, denoted by . Ideally, the "big" structure should be saturated and larger than all other considered models, but existence of large saturated structures in general requires set-theoretic assumptions (for example, the existence of suitably large strongly inaccessible cardinal, because any theory has a saturated model of such a cardinality).

One way to avoid it is to instead construct a model which is saturated and strongly homogeneous in a large enough cardinality (larger than all the models considered, and usually left unspecified), which may nonetheless be smaller than its own, which can be done in ZFC alone, and is for practical purposes just as effective.