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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 addEvery 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 yes: the finite subset $S_0$ axiomatizes the same theory. This is finite subaxiomatization from an equivalent finite theory.

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One implication follows by taking the same model. Conversely, if $T$ had no model, then by the Godel completeness theorem it would prove a contradiction. A formal proof uses only finitely many assumptions, so some finite subset of $T$ would already have no model. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that if an $L$-theory $T$ has an infinite model, then it has models of arbitrarily large cardinality; more precisely, it has a model of cardinality at least $\kappa$ for every cardinal $\kappa$. Add new constants $c_{\alpha }$ for $\alpha <\kappa$ and the sentencesEvery finite subset of the enlarged theory can be interpreted in the given infinite model, since it mentions only finitely many constants. Compactness supplies a model of the whole enlarged theory, in which the $c_{\alpha }$ are pairwise distinct. Its reduct to $L$ is a model of $T$ having at least $\kappa$ elements. If $\kappa \ge |L|+{\aleph }_0$, the Downward Lowenheim-Skolem theorem gives a model of cardinality exactly $\kappa$.

This class is not axiomatisable. It contains an infinite model, for example the total order $\mathbb{Q}$. If a theory $T$ in any fixed expanded language axiomatized it, choose a cardinalThe Upward Lowenheim-Skolem theorem would give a model of $T$ of cardinality at least $\kappa$. Its underlying order cannot be isomorphic to a subset of $\mathbb{R}$, because every such subset has cardinality at most $|\mathbb{R}|$. This contradiction proves the claim.

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One direction is immediate: a model of $T$ models every subset. Conversely, suppose every finite subset is satisfiable. If $T$ had no model, then $T\models \perp$. The Godel completeness theorem would give a formal proof $T⊢\perp$. Every proof is finite, so it would use only a finite subset $T_0\subseteq T$, making $T_0$ inconsistent and hence unsatisfiable. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that a first-order theory with an infinite model has models of arbitrarily large cardinality. Fix a cardinal $\kappa$ and expand the language by constants $c_{\alpha }$ for $\alpha <\kappa$. Add the sentencesto $T$. Any finite subset mentions only finitely many new constants, which can be interpreted as distinct elements of the given infinite model. Compactness therefore gives a model of the expanded theory in which the $c_{\alpha }$ are all distinct. Its reduct is a model of $T$ of cardinality at least $\kappa$. Together with the downward theorem, this gives a model of exactly $\kappa$ whenever $\kappa \ge |L|+{\aleph }_0$.

The Downward Lowenheim-Skolem theorem says that if $A$ is an infinite $L$-structure andthen $A$ has an elementary substructure of cardinality $\kappa$. To see why, first choose $\kappa$ elements. For every existential formula, add a Skolem function selecting a witness whenever one exists, and repeatedly close the chosen set under all language operations and these witness functions. There are at most $\kappa$ functions of finite arity, so the closure still has size $\kappa$. Every existential statement true in $A$ with parameters from the closure has a witness in the closure; the Tarski--Vaught test therefore says that the resulting substructure is elementary.