= Solution
The <compactness theorem> for first-order logic says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. The forward implication is immediate. Conversely, if every finite subset is satisfiable, then $T$ is syntactically consistent: any proof of a contradiction would use only finitely many premises. The <Godel completeness theorem> then gives a model of $T$.
The <Upward Lowenheim-Skolem theorem> says that a first-order theory with an infinite model has arbitrarily large models. Given an infinite model of $T$ and any cardinal $\kappa$, enlarge the language by constants $c_\alpha$ for $\alpha<\kappa$ and add
$$
c_\alpha\ne c_\beta\qquad(\alpha\ne\beta).
$$
Every finite subset can interpret its finitely many constants as distinct elements of the original infinite model. By the <compactness theorem>, the enlarged theory has a model containing at least $\kappa$ distinct elements; its reduct is a model of $T$.
Finally, suppose a set $S$ and a finite set $T$ axiomatize the same class. Let $\tau$ be the conjunction of the sentences in $T$. Since $S\models\tau$, the set $S\cup\{\neg\tau\}$ is unsatisfiable. Compactness gives a finite $S_0\subseteq S$ such that $S_0\models\tau$. Because $\tau\models S$, we also have $S_0\models S$, while $S\models S_0$ is immediate. Therefore \b[yes: the finite subset $S_0$ axiomatizes the same theory.] This is <finite subaxiomatization from an equivalent finite theory>.
Back to article page
