Overflow Theorem

Theorem
Let $$A$$ be a set of first-order formulas which has finite models of arbitrarily large size.

Then $$A$$ has an infinite model.

Corollary
The class of finite models is not $$\Delta$$-Elementary.

That is, there is no set of formulas $$F$$ such that $$F$$ is satisfied by a model $$\mathcal{M}$$ iff $$\mathcal{M}$$ is finite.

Proof
For each $$n$$, let $$A_{n}$$ be the formula:


 * $$\exists x_{1} \exists x_{2} \ldots \exists x_{i}: (x_{1} \ne x_{2} \and x_{1} \ne x_{3} \and \ldots \and x_{n} \ne x_{n-1})$$

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, from Compactness of First-Order Logic.

So $$\Gamma$$ is satisfiable in some model $$\mathcal{M}$$.

But since $$\mathcal{M} \models A_{i}$$ for each $$i$$, $$\mathcal{M}$$ must be infinite.

So $$A$$ has an infinite model.

Proof of Corollary
Follows directly from the Overflow Theorem above.