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The compactness theorem states 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 follows by using the same model for every finite subset. Conversely, suppose every finite subset of $T$ has a model. If $T$ were inconsistent, a formal derivation of $\perp$ from $T$ would use only finitely many assumptions, say those in $T_0\subseteq T$. Then $T_0$ would be inconsistent. By part (b), $T_0$ would have no model, contrary to the hypothesis. Thus $T$ is consistent, and part (b) gives a model of $T$.

Suppose $T_G$ is inconsistent. By the compactness theorem, some finite subset $\Delta \subseteq T_G$ is inconsistent. Only finitely many added symbols $f_g$ occur in $\Delta$; let $X\subseteq G$ be the finite set of their indices. Every sentence of $\Delta$ belongs to the sublanguage $L_X$, soConsequently $T_G\cap L_X$ is inconsistent.

Compactness theorem says a set of first-order sentences has a model whenever each finite subset has a model.

For a type space $S_n^M(A)$, the basic open set $[\phi ]$ consists of complete types containing the formula $\phi$. These sets are clopen, satisfy $[\phi ]\cap [\psi ]=[\phi \wedge \psi ]$, and distinguish distinct types. The compactness theorem makes this topology compact; it is Hausdorff and has a basis of clopen sets.

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.