I’ve written up a proof of the compactness theorem before, but I recently looked it over and think that the proof can be expressed more, heh heh, compactly, than before.

So, what is the compactness theorem? It is the statement that if a set of statements is finitely satisfiable (each of its finite subsets has a model), then it’s satisfiable. The converse of this (a satisfiable set of statements is also finitely satisfiable) is trivial: a model of a set of sentences is also a model of every subset of that set. The following proof will be for propositional logic, but can be easily extended to first order logic.

The short version

Suppose that a set of sentences A is finitely satisfiable. We’ll extend A to a larger set B by giving A an “opinion” on every sentence. We can build up this extension in a series of stages: B_{0} is just A. Now, take any sentence a. If B_{0} ⋃ {a} is finitely satisfiable, define B1 to be B_{0} ⋃ {a}. Otherwise, define B_{1} to be B_{0} ⋃ {¬a}. Either way, B_{1} will be finitely satisfiable, because B_{0} cannot be inconsistent with both a and ¬a. When we’ve gone through every sentence, we take the union of all these extensions to form B. B is finitely satisfiable, since every finite subset of B is also a finite subset of one of the extensions that make it up. Now, we define a truth assignment V that assigns True to each propositional variable p if and only if p is in B. V satisfies B (which you can show by induction on sentences), and since A is a subset of B, V also satisfies A. So A is satisfiable.

The long(er) version

Suppose that a set of sentences A is finitely satisfiable. If a set of sentences A’ is finitely satisfiable, then for any sentence a, at least one of A’ ⋃ {a} and A’ ⋃ {¬a} is finitely satisfiable. If neither were, then we’d have two finite sets, one that entails ¬a and the other that entails a, and the union of these would be an unsatisfiable finite subset of A’.) So given any well-ordering of the sentences of the language, we can extend A one sentence at a time, maintaining finite satisfiability at each step. The union of all these extensions, call it B, is still finitely satisfiable, because any finite subset of B is also a finite subset of one of the extensions. B is also complete: for any sentence b either b ∈ B or ¬b ∈ B.

Now, define the truth assignment V as follows: for any atomic sentence b, V(b) = True iff b ∈ B. V satisfies B, which we prove by induction on sentences:

- If b is an atomic sentence, then V(b) = True iff b ∈ B, by construction.
- If b = ¬c for some c (for which V(c) = True iff c ∈ B), then V(b) = True iff V(c) = False iff c ∉ B iff b ∈ B.
- If b = c ∧ d for some c and d that satisfy the induction hypothesis, then V(b) = V(c ∧ d) = True iff V(c) = V(d) = True iff c ∈ B and d ∈ B. If both c and d are in B, then ¬(c ∧ d) can’t be in B by finite satisfiability, so c ∧ d ∈ B by completeness of B. And if c ∧ d ∈ B, then neither ¬c nor ¬d can be in B by finite satisfiability. So c and d are both in B by completeness of B.

This proof by induction covers the connectives ¬ and ∧, with which you can express all other connectives, so this shows that V satisfies B. And since B is a superset of A, V satisfies A as well. This shows that any finitely satisfiable set of sentences is also satisfiable. Note that we used the axiom of choice implicitly with the instruction to well-order the language. As the language can be any size, this is equivalent to the well-ordering principle, which is equivalent in ZF to the axiom of choice. The same sort of assumption arises in the proof of the completeness theorems for first-order and propositional logics. If you reject choice, then you should also be skeptical that propositional logic and first-order logics are complete!