Saturated Models of same Cardinality are Isomorphic

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Theorem

Let $T$ be an $\LL$-theory.

Let $\kappa$ be an infinite cardinal.

If $\MM$ and $\NN$ are saturated models of $T$ and the cardinality of the universes of $\MM$ and $\NN$ are both $\kappa$, then $\MM$ and $\NN$ are isomorphic.

Proof

Outline

The idea of the proof is that since the models are saturated, we can define an isomorphism $f: \NN \to \MM$ by picking the image $\map f x$ of an element $x \in \NN$ as something which realizes the type that $\map f x$ would need to satisfy in $\MM$.

This is done using a (transfinite) recursive construction which extends an elementary map at each step.

The proof is very similar to that of Saturated Implies Universal; in both proofs we attempt to construct a formula-preserving map.

In this case, however, the map must be bijective between the universes of the models.

Body of Proof

First, since $\MM$ and $\NN$ have cardinality $\kappa$, we can denote their elements by $m_\alpha$ and $n_\alpha$ respectively for ordinals $\alpha < \kappa$.

For each $\alpha < \kappa$, let $A_\alpha$ denote the domain of the $f_\alpha$ we define below. Note that we do not know in advance what the domains will be.

Base case $\alpha = 0$:

Define $f_0 = \O$.

Note that $f_0$ is trivially an elementary embedding from $A_0 = \O$ into $\MM$.

$\Box$

Limit ordinals $\alpha$, assuming $f_\beta: A_\beta \to \MM$ is defined and elementary, and that $\card {A_\beta} < \kappa$ for all $\beta < \alpha$:

Let $\ds f_\alpha = \bigcup_{\beta < \alpha} f_\beta$.

If $\phi$ is an $\LL$-sentence with parameters from $A_\alpha$, then since it involves only finitely many such parameters, they must all be contained in some $A_\beta$ for $\beta < \alpha$.

But $f_\alpha \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well.

Note that $\card {A_\alpha} < \kappa$ by Cardinality of Infinite Union of Infinite Sets.

$\Box$

Successor ordinals $\alpha = \beta + 1$, assuming $f_\beta: A_\beta \to \MM$ is defined and elementary, and that $\card {A_\beta} < \kappa$:

We need to extend $f_\beta$ to some $f_\alpha$ so that the domain includes $n_\beta$, the image includes $m_\beta$, and so that $\LL$-formulas with parameters from $A_\alpha$ are preserved by $f_\alpha$.

First we add $n_\beta$ to the domain:

Consider the subset:

$p = \set {\map \phi {v, \map {f_\beta} {\bar a} }: \bar a \text { is a tuple from } A_\beta \text { and } \NN \models \map \phi {n_\beta, \bar a} }$

of the set of $\LL$-formulas with one free variable and parameters from the image $\map {f_\beta} {A_\beta}$ of $A_\beta$ under $f_\beta$.

The set $p$ is a $1$-type over the image $\map {f_\beta} {A_\beta}$ in $\MM$.

Since $\card {A_\beta} < \kappa$ by the inductive hypothesis and since by assumption $\MM$ is $\kappa$-saturated, this means that $p$ is realized in $\MM$ by some element $b$.

Thus $f'_\alpha = f_\beta \cup \set {\tuple {n_\beta, b} }$ is an elementary embedding $A_\beta \cup \set {n_\beta} \to \MM$.

Now we add $m_\beta$ to the image:

This is done similarly to the above.

Consider the subset:

$q = \set {\map \phi {v, \bar a}: \bar a \text { is a tuple from } A_\beta \cup \set {n_\beta} \text { and } \MM \models \map \phi {m_\beta, \map {f'_\alpha} {\bar a} } }$

of the set of $\LL$-formulas with one free variable and parameters from $A_\beta \cup \{n_\beta\}$.

The set $q$ is a $1$-type over $A_\beta \cup \{n_\beta\}$ in $\NN$.

Since $\card {A_\beta} < \kappa$ by the inductive hypothesis and hence $\card {A_\beta \cup \set {n_\beta} } < \kappa$ as well, and since by assumption $\NN$ is $\kappa$-saturated, this means that $q$ is realized in $\NN$ by some element $c$.

Thus $f_\alpha = f'_\alpha \cup \set {\tuple {c, m_\beta} } = f_\beta \cup \set {\tuple {n_\beta, b}, \tuple {c, m_\beta} }$ is an elementary embedding $A_\beta \cup \set {n_\beta} \to \MM$ which includes $m_\beta$ in its range.

Finally define $\ds f = \bigcup_{\alpha < \kappa} f_\alpha$.

The map $f$ is an elementary embedding $\NN \to \MM$ since $\ds \bigcup_{\alpha < \kappa} A_\alpha = \NN$, any finite set of parameters from $\NN$ must belong to some single $A_\alpha$, and $f\restriction A_\alpha$ is elementary.

$f$ is onto $\MM$ since we have constructed it so that $m_\alpha$ is in the image of $f_{\alpha + 1}$.

So, $f$ is a bijection.

Since elementary embeddings are by definition $\LL$-embeddings, this means that $f$ is an isomorphism by definition.

$\blacksquare$