Upward Löwenheim-Skolem Theorem

Theorem
Let $T$ be an $\LL$-theory with an infinite model.

Then for each infinite cardinal:
 * $\kappa \ge \card \LL$

there exists a model of $T$ with cardinality $\kappa$.

Proof
The idea is:
 * to extend the language by adding $\kappa$ many new constants

and:
 * to extend the theory by adding sentences asserting that these constants are distinct.

It is shown that this new theory is finitely satisfiable using an infinite model of $T$.

Compactness then implies that the new theory has a model.

Some care needs to be taken to ensure that we construct a model of exactly size $\kappa$.

Let $\LL^*$ be the language formed by adding new constants:
 * $\set {c_\alpha: \alpha < \kappa}$ to $\LL$.

Let $T^*$ be the $\LL^*$-theory formed by adding the sentences:
 * $\set {c_\alpha \ne c_\beta: \alpha, \beta < \kappa, \ \alpha \ne \beta}$

to $T$.

We show that $T^*$ is finitely satisfiable:

Let $\Delta$ be a finite subset of $T^*$.

Then $\Delta$ contains:
 * finitely many sentences from $T$

along with:
 * finitely many sentences of the form:
 * $c_\alpha \ne c_\beta$

for the new constant symbols.

Since $T$ has an infinite model, it must have a model $\MM$ of cardinality at most:
 * $\card \LL + \aleph_0$

This model already satisfies everything in $T$.

So, since we can find arbitrarily many distinct elements in it, it can also be used as a model of $\Delta$ by interpreting the finitely many new constant symbols in $\Delta$ as distinct elements of $\MM$.

Since $T^*$ is finitely satisfiable, it follows by the Compactness Theorem that $T^*$ itself is satisfiable.

Since $T^*$ ensures the existence of $\kappa$ many distinct elements, this means it has models of size at least $\kappa$.

It can be proved separately or observed from the ultraproduct proof of the compactness theorem that $T^*$ then has a model $\MM^*$ of exactly size $\kappa$.

Since $T^*$ contains $T$, $\MM^*$ is a model of $T$ of size $\kappa$.

Also see

 * Overflow Theorem