Ehrenfeucht-Mostowski Model

The Ehrenfeucht-Mostowski construction is a construction which produces models in which few types are realized.

'''Theorem. ''' Let $$T$$ be a complete theory with infinite models. Then for any $$\kappa \ge |T|$$, there is a model $$M \models T$$ of cardinality $$\kappa$$ such that if $$A \subset M$$, then at most $$|A| + |T|$$ types over $$A$$ are realized.

Proof. First suppose that $$T$$ has definable Skolem functions. Let $$\{a_\lambda\}_{\lambda < \kappa}$$ be a non-constant indiscernible sequence of length $$\kappa$$ in some model $$M_0$$. We can find such an indiscernible sequence by extracting an indiscernible sequence from a non-constant sequence in an infinite model. Let $$M$$ be $$\operatorname{dcl}(\{a_\lambda : \lambda < \kappa\})$$. Because $$T$$ has definable Skolem functions, $$M \preceq M_0$$, so $$M \models T$$ and $$\{a_\lambda\}_{\lambda < \kappa}$$ is still indiscernible within $$M$$. The set $$\{a_\lambda\}$$ is called the spine. Let $$A$$ be a subset of $$M$$. Each element of $$A$$ is in the definable closure of some finite subset of the spine, so we can find some $$A'$$ contained in the spine, with $$A \subset \operatorname{dcl}(A')$$, and $$|A'| \le |A| + |T|$$. An element’s type over $$A$$ is determined by its type over $$A'$$, because $$A \subset \operatorname{dcl}(A')$$. So it suffices to show that at most $$|A'| + |T| = |A| + |T|$$ types over $$A'$$ are realized.

First we check that at most $$|A'|$$ types are realized by tuples from the spine. We can write $$A'$$ as $$\{a_\lambda : \lambda \in S\}$$ for some $$S \subset \kappa$$ with $$|S| = |A'|$$. The indiscernibility of the spine implies that $$\operatorname{tp}(a_{\lambda_1}a_{\lambda_2}\cdots a_{\lambda_n}/A')$$ is entirely determined by how the $$\lambda_i$$ relate to each other and how they relate to elements of $$S$$. That is, $$\operatorname{tp}(a_{\lambda_1} \cdots a_{\lambda_n}/A')$$ is entirely determined by the following pieces of data:


 * Whether $$\lambda_i \le \lambda_j$$ for each $$i, j$$.
 * The set $$\{x \in S : x \le \lambda_i\}$$ for each $$i$$.
 * The set $$\{x \in S : x < \lambda_i\}$$ for each $$i$$.

Because $$S$$ is well-ordered, there are only about $$|S| + \aleph_0$$ choices for the second and third bullet points. All told, there are therefore only $$|S| + \aleph_0 \le |A'| + |T|$$ possibilities for $$\operatorname{tp}(a_{\lambda_1}\cdots a_{\lambda_n}/A')$$.

Now if $$f(x_1,\ldots,x_n)$$ is a 0-definable function, then $$\operatorname{tp}(f(a_{\lambda_1},\ldots,a_{\lambda_n})/A')$$ depends only on $$\operatorname{tp}(a_{\lambda_1},\ldots,a_{\lambda_n}/A')$$, so there are at most $$|A'| + |T|$$ types over $$A'$$ realized by elements of the form $$f(a_{\lambda_1},\ldots,a_{\lambda_n})$$. But all of $$M$$ is in the definable closure of the spine, so every element of $$M$$ is of this form. Since there are only $$|T|$$-many 0-definable functions, the total number of types over $$A'$$ realized in $$M$$ is at most $$(|A'| + |T|) \times |T| = |A'| + |T|.$$ So at most $$|A'| + |T|$$ types over $$A'$$ are realized, completing the proof (in the case where we had definable Skolem functions).

Now suppose $$T$$ is arbitary. We can find a theory $$T'$$ expanding $$T$$, which does have Skolem functions. This can easily be done in such a way that $$|T'| = |T|$$. By the above argument one gets a model $$M'$$ of $$T'$$ of size $$\kappa$$ with the property that for every subset $$A$$ of $$M'$$, at most $$|A| + |T|$$ types over $$A$$ are realized in $$M'$$. Let $$M$$ be the reduct of $$M'$$ to the original language. Then $$M \models T$$. If $$A \subset M$$, and $$a$$ and $$b$$ have the same $$T'$$-type over $$A$$, then they certainly have the same $$T$$-type over $$A$$ within $$M$$, because $$T$$ has fewer definable sets and relations to work with than $$T'$$. So there are at most as many $$T$$-types over $$A$$ as there are $$T'$$-types over $$A$$, which is at most $$|A| + |T|$$. QED

An important consequence of this result is the following, which is the first step of the proof of Morley’s Theorem.

'''Corollary. ''' Let $$T$$ be a complete countable theory which is $$\kappa$$-categorical for some $$\kappa \ge \aleph_1$$. Then $$T$$ is $$\aleph_0$$-stable (hence totally transcendental).

Proof. Let $$\mathbb{U}$$ be the monster model of $$T$$. Suppose $$T$$ is not $$\aleph_0$$-stable. Then we can find a countable set $$A$$ over which there are uncountably many types. Realize $$\aleph_1$$ of these types and let $$B$$ be the set of these realizations. Then $$|A \cup B| \le \aleph_1 \le \kappa$$, so by Löwenheim-Skolem we can find a model $$M$$ of cardinality $$\kappa$$ containing $$A \cup B$$. By the Ehrenfeucht-Mostowski construction, we can find a model $$M'$$ of cardinality $$\kappa$$ in which at most countably many types are realized over countable sets. By $$\kappa$$-categoricity, $$M \cong M'$$. So $$M$$ also has the property that over countable sets, countably many types are realized. But over the countable set $$A \subset M$$, uncountably many types are realized in $$B \subset M$$, a contradiction.

(It is a general fact that $$\aleph_0$$-stable theories are totally transcendental. The proof goes as follows: if $$T$$ failed to be totally transcendental, then $$RM(D) = \infty$$ for some set $$D$$. Then one inductively builds a tree $$D, D_0, D_1, D_{00}, D_{01}, D_{10}, \ldots$$ of non-empty $$\mathbb{U}$$-definable sets such that $$D_w$$ is the disjoint union of $$D_{w0}$$ and $$D_{w1}$$ for every $$w \in \{0,1\}^{< \omega}$$, and such that each $$D_w$$ has Morley rank $$\infty$$. This is the same construction used to prove that perfect sets in Polish spaces have cardinality $$2^{\aleph_0}$$. At any rate, there are countably many $$D_w$$’s, so the $$D_w$$’s are all definable over some countable set $$A$$. Now each path through the tree yields a different type over $$A$$, so that there are uncountably many types over $$A$$, contradicting $$\omega$$-stability.) QED