# 6.3 Saturated Models

Saturated models provide a way to control which types are realized inside a structure, and they serve as canonical large models in which all consistent descriptions over small parameter sets are already realized, so that no further extension is needed to witness those types.

Throughout this section, let $L$ be a first order language, let $T$ be an $L$-theory, and let $\mathcal M \models T$ be an $L$-structure with domain $M$. We study how types over subsets of $M$ behave inside $\mathcal M$.

### Definition 6.52 (Realization of Types Over a Set)

Let $A \subseteq M$, and let $p(x)$ be a partial type over $A$. We say that $p(x)$ is realized in $\mathcal M$ if there exists $a \in M^n$ such that:
$$
\mathcal M \models \varphi(a)
$$
for every formula $\varphi(x) \in p(x)$.

This definition specializes the earlier notion of realization to the case where the ambient structure is fixed, and it emphasizes that realization depends on both the type and the structure.

### Definition 6.53 (Saturated Model)

Let $\kappa$ be an infinite cardinal. The structure $\mathcal M$ is said to be $\kappa$-saturated if for every subset $A \subseteq M$ with:
$$
|A| < \kappa,
$$
every type $p(x)$ over $A$ that is consistent with $\operatorname{Th}(\mathcal M)$ is realized in $\mathcal M$.

Thus $\kappa$-saturation means that $\mathcal M$ realizes all types over parameter sets of size strictly less than $\kappa$, provided that those types are consistent.

When $\kappa = |M|$, the model is simply called saturated.

### Interpretation

A saturated model contains as many elements as possible with respect to the theory it satisfies, in the sense that any consistent description over a small parameter set already has a witness inside the model, and therefore no extension is needed to realize that description.

### Example 6.54

Let:
$$
\mathcal N = (\mathbb N; <).
$$

Consider the type over $\mathbb N$:
$$
p(x) = \{x > n : n \in \mathbb N\}.
$$

This type is consistent with the theory of $(\mathbb N; <)$ and is finitely satisfiable, but it is not realized in $\mathbb N$ itself, since no natural number is greater than all natural numbers.

Therefore $(\mathbb N; <)$ is not $\aleph_0$-saturated.

### Example 6.55

Let $\mathcal M$ be a nonstandard model of arithmetic. Then $\mathcal M$ contains elements that are larger than every standard natural number, and such elements realize the type:
$$
\{x > n : n \in \mathbb N\}.
$$

Thus nonstandard models are richer with respect to realization of types, and sufficiently large ones can be saturated.

### Lemma 6.56 (Saturation Implies Realization of Finitely Satisfiable Types)

Let $\mathcal M$ be $\kappa$-saturated, let $A \subseteq M$ with $|A| < \kappa$, and let $p(x)$ be a type over $A$ that is finitely satisfiable in $\mathcal M$. Then $p(x)$ is realized in $\mathcal M$.

Proof

Since $p(x)$ is finitely satisfiable in $\mathcal M$, every finite subset of $p(x)$ is realized in $\mathcal M$. By compactness, it follows that $p(x)$ is consistent with $\operatorname{Th}(\mathcal M)$.

Since $\mathcal M$ is $\kappa$-saturated and $|A| < \kappa$, every type over $A$ that is consistent with $\operatorname{Th}(\mathcal M)$ is realized in $\mathcal M$. Therefore $p(x)$ is realized in $\mathcal M$.

### Definition 6.57 (Countably Saturated Model)

A model $\mathcal M$ is countably saturated if it is $\aleph_1$-saturated, that is, every type over a finite or countable parameter set is realized in $\mathcal M$.

This notion is particularly important in analysis and algebra, where many constructions involve countable parameter sets.

### Lemma 6.58 (Extension to Saturated Models)

Let $\mathcal M$ be a model of a theory $T$, and let $\kappa$ be an infinite cardinal. Then there exists an elementary extension $\mathcal N \succ \mathcal M$ such that $\mathcal N$ is $\kappa$-saturated.

Proof

We construct $\mathcal N$ by a transfinite process, adding realizations of types step by step.

Enumerate all types over subsets of $M$ of size less than $\kappa$, up to logical equivalence, as:
$$
(p_\alpha : \alpha < \lambda),
$$
where $\lambda$ is a sufficiently large ordinal.

We build a chain of elementary extensions:
$$
\mathcal M = \mathcal M_0 \prec \mathcal M_1 \prec \mathcal M_2 \prec \cdots
$$

At stage $\alpha$, if the type $p_\alpha$ is not yet realized in $\mathcal M_\alpha$, then extend $\mathcal M_\alpha$ to a model $\mathcal M_{\alpha+1}$ in which $p_\alpha$ is realized. This is possible because $p_\alpha$ is consistent with the theory.

At limit stages, take unions:
$$
\mathcal M_\beta = \bigcup_{\alpha < \beta} \mathcal M_\alpha.
$$

The union of an elementary chain is again an elementary extension of each stage.

At the end of the construction, define:
$$
\mathcal N = \bigcup_{\alpha < \lambda} \mathcal M_\alpha.
$$

By construction, every type over a subset of size less than $\kappa$ is realized in $\mathcal N$, so $\mathcal N$ is $\kappa$-saturated.

### Lemma 6.59 (Back and Forth Method)

Let $\mathcal M$ and $\mathcal N$ be two $\kappa$-saturated models of the same complete theory $T$, with:
$$
|M| = |N| = \kappa.
$$

Then $\mathcal M$ and $\mathcal N$ are isomorphic.

Proof

We construct an isomorphism between $\mathcal M$ and $\mathcal N$ by a back and forth argument.

Enumerate:
$$
M = \{a_\alpha : \alpha < \kappa\},
\quad
N = \{b_\alpha : \alpha < \kappa\}.
$$

We build a sequence of partial isomorphisms:
$$
f_\alpha : M_\alpha \to N_\alpha,
$$
where $M_\alpha \subseteq M$ and $N_\alpha \subseteq N$ are finite or small sets.

At each step, we extend the map to include the next element from one structure, using saturation to find a matching element in the other structure that satisfies the same type over the already matched elements.

More precisely, suppose $f_\alpha$ is defined. To extend it to include $a_\alpha$, consider the type of $a_\alpha$ over $M_\alpha$, and transport this type via $f_\alpha$ to a type over $N_\alpha$. Since $\mathcal N$ is $\kappa$-saturated, this type is realized in $\mathcal N$, so we can choose $b$ in $N$ matching $a_\alpha$.

Similarly, we extend in the other direction to ensure surjectivity.

At limit stages, take unions of the partial maps.

At the end, we obtain a bijection:
$$
f : M \to N
$$
that preserves all relations and functions, and hence is an isomorphism.

### Saturation and Definability

Saturation interacts strongly with definability. In a saturated model, every consistent description over a small parameter set corresponds to an actual element of the model, so types can be treated as points.

In particular, definable sets can be understood in terms of types: a definable set corresponds to a collection of types that all contain a certain formula, and in a saturated model every such type is realized by an element.

This perspective allows one to move freely between syntactic descriptions, given by formulas and types, and semantic objects, given by elements and subsets of the model.

### Remark 6.60

Saturated models are often used as canonical representatives of a theory, because they eliminate accidental limitations caused by small size, and they ensure that all logically possible configurations over small parameter sets are already present inside the model.

In practice, many arguments in model theory are simplified by working inside a sufficiently saturated model, where types can be realized without leaving the structure.