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 {\displaystyle \kappa }-saturated model, {\displaystyle D\subset M^{n}} be a definable set, and {\displaystyle \Sigma } be a collection of definable subsets of {\displaystyle M^{n}}, with {\displaystyle \Sigma } of size less than {\displaystyle \kappa }. If {\displaystyle \Sigma } covers D, then some finite subset of {\displaystyle \Sigma } covers D. This fact follows from the definition of {\displaystyle \kappa }-saturation. Compactness is used to prove the existence of {\displaystyle \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 {\displaystyle \Sigma (x)} be a consistent partial type over M. Then there is an elementary extension {\displaystyle M'\succeq M} in which {\displaystyle \Sigma (x)} is realized.

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

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

{\displaystyle \phi (M)\subseteq \bigcup _{i\in I}\psi _{i}(M)}.

Then there is a finite subset {\displaystyle I_{0}\subset I} such that in every model M of T,

{\displaystyle \phi (M)\subseteq \bigcup _{i\in I_{0}}\psi _{i}(M)}.

Proof: Apply compactness to the union of T and the statements

{\displaystyle \{\phi (c)\}\cup \{\neg \psi _{i}(c):i\in I\}}

where {\displaystyle c} is a new constant symbol. By assumption, this collection of statements is inconsistent, so by compactness, some finite subset is inconsistent. This yields {\displaystyle I_{0}}. QED

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

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

The analogous statement in saturated models is the following: Lemma: Let M be a {\displaystyle \aleph _{1}}-saturated model. Suppose that {\displaystyle \phi (x;y)} is a formula such that for every {\displaystyle b\in M}, the set {\displaystyle \phi (M;b)} is finite. Then there is a uniform bound n on the size of {\displaystyle \phi (M;b)}.

Other important and prototypical applications of compactness include the following:

Interpretation as Compactness of Stone Space[]

Let S denote the space of all complete theories (in some fixed first-order language). For each sentence {\displaystyle \phi }, let

{\displaystyle [\phi ]=\{T\in S|\phi \in T\}}

There is a natural topology on S in which the sets of the form {\displaystyle [\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 {\displaystyle T} is inconsistent if and only if {\displaystyle \exists x:x\neq x} holds in all models of {\displaystyle T}. By the Completeness Theorem, this holds if and only if {\displaystyle \exists x:x\neq x} can be proven from {\displaystyle T}. But proofs are finitary, so any proof must take only finitely many steps, and must use only a finite subset of {\displaystyle T}. In particular, if {\displaystyle T} proves {\displaystyle \exists x:x\neq x}, then so does some finite subset {\displaystyle T_{0}\subset T}. So if {\displaystyle 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 {\displaystyle S\subset T}, let

{\displaystyle X_{S}=\{T'\in X|S\subset T'\}}

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

For each finite subset S of T, we can find a model {\displaystyle M_{S}} of S, by assumption. Consider the ultraproduct

{\displaystyle M=\prod _{S\in X}M_{S}/{\mathcal {U}}}

We claim that M is the desired model of T. Let {\displaystyle \phi } be a formula in T. Then

{\displaystyle M_{S}\models \phi }

for every {\displaystyle S\ni \phi }, or equivalently, for every {\displaystyle S\in X_{\{\phi \}}}. But

{\displaystyle X_{\{\phi \}}\in {\mathcal {U}}}.

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