Compactness theorem

A set of first-order sentences has a model if and only if every finite subset has a model.

If a theory were inconsistent, a proof of contradiction would use a finite inconsistent subset. Therefore consistency of every finite subset implies consistency of the whole theory, and completeness supplies a model.

If a theory claimed that a poset has only finitely many maximal elements, adjoining countably many constants required to be distinct and maximal would remain finitely satisfiable. Compactness would produce a model with infinitely many maximal elements, contradicting the claimed axiomatization.

Suppose a class requires exactly one of two definable sets to be infinite. Add constants witnessing arbitrarily many distinct members of both sets. Every finite fragment can keep one set infinite and make the other large but finite, whereas compactness makes both sets infinite. Therefore such an exclusive disjunction is not first-order axiomatizable.

If a first-order theory had exactly the finite structures in a class containing arbitrarily large members, adjoin constants $c_1,c_2,\dots$ and require them to be pairwise distinct. Every finite fragment has a sufficiently large finite model, while compactness produces an infinite model, a contradiction. In particular, finite groups are not first-order axiomatizable.

If a theory axiomatized the well-orders, adjoin constants $c_0,c_1,\dots$ and the sentences $c_{n+1}<c_n$. Every finite fragment has a finite well-ordered model, but compactness would produce a model with an infinite descending chain.

Suppose theories $S$ and finite $T$ have the same models. Let $\tau$ be the conjunction of $T$. Since $S\models \tau$, compactness supplies a finite $S_0\subseteq S$ with $S_0\models \tau$. Conversely $\tau \models S$, so $S_0$ and $S$ have the same models. Thus an axiomatization equivalent to a finite theory contains a finite equivalent subaxiomatization.