Compactness Theorem

Compactness Theorem

A set of sentences $\Sigma$ has a model if and only if every finite subset has a model.

Proof. If $\Sigma$ has a model, then this model is also a model of every finite subset of $\Sigma$. Conversely, if every finite subset of $\Sigma$ has a model, then by the Soundness Theorem (version in Corollary 3.6) every finite subset of $\Sigma$ is consistent. Hence $\Sigma$ is consistent, and so $\Sigma$ has a model by the Completeness Theorem. $\square$

Theorem

If a set of sentences $\Sigma$ has finite models of arbitrarily large finite cardinality, then $\Sigma$ has an infinite model.

Proof. Let

${\Sigma }^{\ast }=\Sigma \cup \left\{\exists x_1\dots \exists x_n{\bigwedge }_{1\le i<j\le n}{x}_i\ne x_j|n\ge 2\right\}.$

Every finite subset of ${\Sigma }^{\ast }$ has a model and so ${\Sigma }^{\ast }$ has a model by compactness. But this model is infinite and is a model of $\Sigma$. $\square$