# Compactness Theorem/Proof using Ultraproducts

## Theorem

Let $\LL$ be the language of predicate logic.

Let $T$ be a set of $\LL$-sentences.

Then $T$ is satisfiable if and only if $T$ is finitely satisfiable.

## Proof

By definition, $T$ is finitely satisfiable means that every finite subset of $T$ is satisfiable.

Because the direction:

$T$ satisfiable implies $T$ finitely satisfiable

is trivial, the proof below justifies the converse:

$T$ finitely satisfiable implies $T$ satisfiable.

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 all finite subsets of $T$.

For every $\Sigma_0 \in \Sigma$, define:

$F_{\Sigma_0} = \set {\Delta \in \Sigma: \Sigma_0 \subseteq \Delta}$

That is, $F_{\Sigma_0}$ is the collection of all finite subsets of $T$ including $\Sigma_0$.

Clearly $F_{\Sigma_0}$ is a subset of $\Sigma$.

Now define:

$F = \set {F_{\Sigma_0}: \Sigma_0 \in \Sigma}$

a subset of $\powerset \Sigma$.

We claim this has the finite intersection property.

Take any finite collection $\ds \set {F_{\Sigma_k} }_{k = 1, 2, \dots, n}$ of elements on $F$, and observe that the set $\ds S = \bigcup_{k \mathop = 1}^n \Sigma_k$ is a finite subset of $T$.

That is, $S \in \Sigma$.

$S \in F_{\Sigma_k}$ for each $k$ by construction, and hence is in their intersection.

In fact, their intersection is $F_S$.

So the intersection of any finite collection of elements of $F$ is nonempty as claimed.

Since $F$, which is a set of subsets of $\Sigma$, has the finite intersection property, there is an ultrafilter $U$ on $\Sigma$ including it (by the corollary to the Ultrafilter Lemma).

If $\Sigma_0 \in \Sigma$, then by assumption it has a model $\AA_{\Sigma_0}$.

Define:

$\ds \AA = \paren {\prod_{\Sigma_k \mathop \in \Sigma} \AA_{\Sigma_k} } / U$

We verify $\AA \models T$.

Take any $\phi \in T$.

Observe that $F_{\set \phi} \in U$ by definition of $U$.

Furthermore, if we define $\Phi = \set {\Sigma_0 \in \Sigma: \AA_{\Sigma_0} \models \phi}$, we have $F_{\set \phi} \subseteq \Phi$.

Hence, since $U$ is a filter and $F_{\set \phi} \in U$, $\Phi \in U$.

Then Łoś's Theorem implies that $\AA \models \phi$.

$\blacksquare$

## Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Łoś's Theorem.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.