Compactness Theorem

Theorem

Let $T$ be a set of $\mathcal{L}$-sentences where $\mathcal{L}$ is a first-order language. Then $T$ is satisfiable if and only if every finite subset of $T$ is satisfiable.

Note

Since the direction

$T$ satisfiable implies $T$ finitely satisfiable

is trivial, the proofs below all justify the converse

$T$ finitely satisfiable implies $T$ satisfiable.

Proof using Ultraproducts

The idea is to construct an ultraproduct using a purposefully selected ultrafilter and collection of models so that each sentence in $T$ will be realized as a result of Łoś's theorem.

Let $\Sigma$ be the set of finite subsets of $T$, and for each sentence $E\in T$, let $\uparrow (E)$ denote the collection $\{\sigma\in\Sigma\mid E\in\sigma\}$ of finite subsets of $T$ containing $E$.

Since $\{\uparrow(E)\mid E\in T\}$ is a filter, there is an ultrafilter $\mathcal{U}$ on $\Sigma$ which contains it.

Since every finite subset of $T$ is satisfiable and hence has a model, we can associate to each $\sigma\in\Sigma$ a model $\mathcal{M}_\sigma\models\sigma$.

Let $\mathcal{M}$ be the ultraproduct $\displaystyle{\left(\prod_{\sigma\in\Sigma}\mathcal{M}_\sigma\right)/\mathcal{U}}$.

We verify that $\mathcal{M}$ is a model of $T$:

Suppose $E\in T$.

Since $E\in\sigma$ implies that $\mathcal{M}_\sigma \models E$ by choice of $\mathcal{M}_\sigma$, it follows that $\uparrow (E)\subseteq \{\sigma\in\Sigma \mid\mathcal{M}_\sigma \models E\}$.

By choice of the ultrafilter $\mathcal{U}$, this means that $\{\sigma\in\Sigma \mid\mathcal{M}_\sigma \models E\}$ is in $\mathcal{U}$.

Thus, by Łoś's theorem, the ultraproduct $\mathcal{M}$ satisfies $E$.

$\blacksquare$

Proof using Gödel's Completeness Theorem

This proof is by contraposition. The idea is to exploit the finiteness of proofs and the relation between satisfiability and deducibility to show that if $T$ is not satisfiable, then it must have a finite subset which can be used to prove to a contradiction.

Suppose $T$ is not satisfiable.

Since $T$ has no models, it vacuously follows that $T\models \phi\wedge\neg\phi$ for some sentence $\phi$. By Gödel's Completeness Theorem, this implies that $\phi\wedge\neg\phi$ is deducible from $T$. But, any deduction from $T$ involves only finitely many sentences from $T$, so this means that there is a finite subset $\Delta$ of $T$ such that $\phi\wedge\neg\phi$ is deducible from $\Delta$.

By Soundness of First-Order Logic, this means that $\Delta\models\phi\wedge\neg\phi$. Thus, $\Delta$ is not satisfiable.

$\blacksquare$

Proof using Henkin Construction

This proof actually demonstrates a stronger form of the Compactness Theorem by showing

If $T$ is finitely satisfiable and $\kappa$ is an infinite cardinal such that $\kappa > |\mathcal{L}|$, then $T$ is satisfiable by a model whose cardinality is at most $\kappa$.

This is stronger than the original statement, since it provides extra information about the model that is claimed to exist.

The proof is said to use a Henkin Construction because it involves the construction of a model all of whose elements are interpretations of the constant symbols of some language. Such a model was used in Henkin's proof of the Completeness Theorem. This special feature of the constructed model is what allows us to control its cardinality.

First, by this theorem, we can extend $\mathcal{L}$ and $T$ to find a language $\mathcal{L}^*$ of cardinality at most $\kappa$ and an $\mathcal{L}^*$-theory $T^*$ such that all finitely satisfiable $\mathcal{L}^*$-theories containing $T^*$ have the witness property.

Then, since finitely satisfiable theories have maximal finitely satisfiable extensions, we can find a finitely satisfiable $\mathcal{L}^*$-theory $T'$ containing $T^*$ such that $T'$ contains either $\phi$ or $\neg\phi$ for each $\mathcal{L}^*$-sentence $\phi$.

Note that $T'$ has the witness property since it contains $T^*$.

Finally, by Maximal Finitely Satisfiable Theory with Witness Property is Satisfiable, $T'$ has a model. Moreover, since $\mathcal{L}^*$ has cardinality at most $\kappa$ and hence has at most $\kappa$-many constant symbols, this theorem ensures that the model of $T'$ can be taken to have cardinality at most $\kappa$.

$\blacksquare$

Consequences

• Overflow Theorem
• The Class of Finite Models is not $\Delta$-Elementary
• Existence of Non-Standard Models of Arithmetic