Compactness of First-Order Logic

Theorem
Let $$\Gamma$$ be any countable set of first-order formulas, and suppose every finite subset of $$\Gamma$$ is satisfiable.

Then $$\Gamma$$ is satisfiable.

Proof
Suppose $$\Gamma$$ is unsatisfiable.

Since consistency implies satisfiability, $$\Gamma$$ is inconsistent, i.e., it proves a contradiction.

But first-order proofs are by definition finite, so there is some subset $$\Gamma_{0}$$ of $$\Gamma$$ such that $$\Gamma_{0}$$ is inconsistent.

Now since satisfiability implies consistency, $$\Gamma_{0}$$ is unsatisfiable.

Hence if all finite subsets of $$\Gamma$$ are satisfiable, so is $$\Gamma$$.

Consequences

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