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 $\kappa \geq |T|$ , 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 $|A'|\leq |A|+|T|$ . 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 $\lambda _{i}\leq \lambda _{j}$ for each $i,j$ .
The set $\{x\in S:x\leq \lambda _{i}\}$ 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 $|S|+\aleph _{0}\leq |A'|+|T|$ 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 $\kappa \geq \aleph _{1}$ . 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 $|A\cup B|\leq \aleph _{1}\leq \kappa$ , 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