Compactness

The Compactness Theorem states that if T is a collection of first-order statements and every finite subset of T is consistent, then T is itself consistent. A set of statements is consistent if it has a model.

By abuse of terminology, the following related fact is also frequently referred to as "compactness." Let M be a $$\kappa$$-saturated model, $$D \subset M^n$$ be a definable set, and $$\Sigma$$ be a collection of definable subsets of $$M^n$$, with $$\Sigma$$ of size less than $$\kappa$$. If $$\Sigma$$ covers D, then some finite subset of $$\Sigma$$ covers D. This fact follows from the definition of $$\kappa$$-saturation. Compactness is used to prove the existence of $$\kappa$$-saturated models, however.

Common Corollaries and Uses of Compactness
The compactness theorem has a number of commonly-used corollaries. These corollaries are often used implicitly in proofs, and explained only as "compactness."

Lemma: Let M be a model, and $$\Sigma(x)$$ be a consistent partial type over M. Then there is an elementary extension $$M' \succeq M$$ in which $$\Sigma(x)$$ is realized.

Proof: Apply the Compactness Theorem to the union of the elementary diagram of M and the statements $$\Sigma(c)$$, where $$c$$ is a new constant symbol. QED

Lemma: Let T be a theory. Let $$\phi(x)$$ be a formula, and let $$\{\psi_i(x) : i \in I\}$$ be a collection of formulas. Suppose that in every model M of T, we have
 * $$ \phi(M) \subseteq \bigcup_{i \in I} \psi_i(M) $$.

Then there is a finite subset $$I_0 \subset I$$ such that in every model M of T,
 * $$ \phi(M) \subseteq \bigcup_{i \in I_0} \psi_i(M) $$.

Proof: Apply compactness to the union of T and the statements
 * $$ \{ \phi(c) \} \cup \{ \neg \psi_i(c) : i \in I\} $$

where $$c$$ is a new constant symbol. By assumption, this collection of statements is inconsistent, so by compactness, some finite subset is inconsistent. This yields $$I_0$$. QED

Lemma: Let T be a theory, and $$\phi(x)$$ be a formula. Suppose that in every model M of T, the set $$\phi(M)$$ is finite. Then there is a number n such that $$|\phi(M)| < n$$ for every model M.

Proof: Apply compactness to the union of T and the set of sentences $$\psi_n$$ asserting for each n that at least n elements of the model satisfy $$\phi$$. By assumption, this is inconsistent. Consequently, there is some n such that $$\psi_n$$ is inconsistent with T. This means exactly that $$|\phi(M)| < n$$ in every model of T. QED

The analogous statement in saturated models is the following: Lemma: Let M be a $$\aleph_1$$-saturated model. Suppose that $$\phi(x;y)$$ is a formula such that for every $$b \in M$$, the set $$\phi(M;b)$$ is finite. Then there is a uniform bound n on the size of $$\phi(M;b)$$.

Other important and prototypical applications of compactness include the following:
 * The Upwards Löwenheim-Skolem Theorem
 * The existence of indiscernible sequences
 * The existence of $$\kappa$$-saturated models
 * The equivalence between the existence of Skolem functions and the property that every substructure is an elementary substructure.
 * The equivalence between elimination of imaginaries and the interdefinability of every imaginary element with a real element.
 * The statement that if every substructure of a model of T is a model of T, then T is equivalent to a set of universal statements.

Interpretation as Compactness of Stone Space
Let S denote the space of all complete theories (in some fixed first-order language). For each sentence $$\phi$$, let
 * $$[\phi] = \{ T \in S | \phi \in T \}$$

There is a natural topology on S in which the sets of the form $$[\phi]$$ are the basic open (and basic closed) subsets. The compactness theorem says exactly that S is compact with this topology.

Proofs of Compactness
Compactness follows easily from some forms of Gödel's Completeness Theorem. Specifically, a theory $$T$$ is inconsistent if and only if $$\exists x : x \ne x$$ holds in all models of $$T$$. By the Completeness Theorem, this holds if and only if $$\exists x : x \neq x $$ can be proven from $$T$$. But proofs are finitary, so any proof must take only finitely many steps, and must use only a finite subset of $$T$$. In particular, if $$T$$ proves $$\exists x : x \neq x $$, then so does some finite subset $$T_0 \subset T$$. So if $$T$$ is inconsistent, so is a finite subset.

Compactness can alternatively be proven from Łoś's Theorem.

Proof: Let T be a collection of statements, with every finite subset of T being consistent. Let X be the set of all finite of T. For each finite subset $$S \subset T$$, let
 * $$X_S = \{ T' \in X | S \subset T'\}$$

Note that $$X_S \cap X_T = X_{S \cup T}$$ and that $$X_S \ne \emptyset$$ for any S. It follows that any finite intersection of $$X_S$$'s is non-empty. Therefore we can find an ultrafilter $$\mathcal{U}$$ on X such that $$X_S \in \mathcal{U}$$ for every S. In other words, for each S, $$\mathcal{U}$$ thinks that "most" elements of X contain S.

For each finite subset S of T, we can find a model $$M_S$$ of S, by assumption. Consider the ultraproduct
 * $$M = \prod_{S \in X} M_S / \mathcal{U}$$

We claim that M is the desired model of T. Let $$\phi$$ be a formula in T. Then
 * $$ M_S \models \phi $$

for every $$ S \ni \phi$$, or equivalently, for every $$ S \in X_{\{\phi\}}$$. But
 * $$X_{\{\phi\}} \in \mathcal{U}$$.

Consequently, the set of S such that $$M_S \models \phi$$ is "large" with respect to the ultrafilter $$\mathcal{U}$$. By Łoś's Theorem, $$\phi$$ holds in the ultraproduct M. QED