## FANDOM

78 Pages

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