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 $$\Gamma.$$

Overflow Theorem
Suppose A is a set of first-order formulas which has finite models of arbitrarily large size. Then A has an infinite model.

Proof: For each n, let $$A_{n}$$ be the formula $$\exists x_{1} \exists x_{2} \ldots \exists x_{i}(x_{1} \neq x_{2} \wedge x_{1} \neq x_{3} \wedge \ldots \wedge x_{n} \neq x_{n-1}).$$ Then $$A_{i}$$ is true in an interpretation I iff I has at least i elements. Now, take $$\Gamma \equiv A \cup \bigcup_{i=1}^{\infty}A_{i}.$$ Since A has models of arbitrarily large size, every finite subset of $$\Gamma$$ is satisfiable, so $$\Gamma$$ is satisfiable in some model M. But since $$M\models A_{i}$$ for each i, M must be infinite, so A has an infinite model.

The Class of Finite Models is not $$\Delta$$-Elementary
There is no set of formulas F such that F is satisfied by a model M iff M is finite.

Proof: By the Overflow Theorem above.

Existence of Non-Standard Models of Arithmetic
Let $$P\ $$ be the set of axioms of Peano arithmetic and let $$Q = P \cup \{\neg x = 0, \neg x = s0, \neg x = ss0, ... \}$$ where $$x\ $$ is a variable of the language. Then each finite subset of $$Q$$ is satisfied by the standard model of arithmetic, hence $$Q$$ is satisfiable by the Compactness theorem. But any model satisfying $$Q$$ must assign $$x$$ to an element which cannot be obtained by iterating the successor operator on zero a finite number of times.