Theorem (Compactness): A set $\Gamma$ of first-order sentences has a model if and only if every finite subset of $\Gamma$ has a model.

We prove compactness using Gödel's completeness theorem.

Direction ($\Rightarrow$): If $\Gamma$ has a model $\mathcal{M}$, then $\mathcal{M}$ is also a model of every subset of $\Gamma$, including all finite subsets. This direction is trivial.

Direction ($\Leftarrow$): Assume every finite subset of $\Gamma$ has a model. We must show $\Gamma$ has a model.

Step 1: Prove $\Gamma$ is consistent

Suppose for contradiction that $\Gamma$ is inconsistent, meaning $\Gamma \vdash \bot$ (proves a contradiction). By the definition of formal proof, any derivation of $\bot$ uses only finitely many axioms. Therefore, there exists a finite subset $\Gamma_0 \subseteq \Gamma$ such that $\Gamma_0 \vdash \bot$.

By soundness of first-order logic, if $\Gamma_0 \vdash \bot$, then $\Gamma_0 \models \bot$, which means $\Gamma_0$ has no model. This contradicts our assumption that every finite subset of $\Gamma$ has a model.

Therefore, $\Gamma$ must be consistent.

Step 2: Apply completeness to obtain a model

By Gödel's completeness theorem, every consistent set of sentences has a model. Since we've shown $\Gamma$ is consistent, it follows that $\Gamma$ has a model.

Conclusion: We've shown that if every finite subset of $\Gamma$ has a model, then $\Gamma$ itself has a model, completing the proof.

Alternative Proof via Ultraproducts (sketch):

For readers familiar with ultraproducts, here's a model-theoretic proof that doesn't rely on completeness:

1. Let $\mathcal{F}$ be the set of all finite subsets of $\Gamma$
2. For each $\Delta \in \mathcal{F}$, let $\mathcal{M}_\Delta$ be a model of $\Delta$ (exists by assumption)
3. Let $\mathcal{U}$ be an ultrafilter on $\mathcal{F}$ containing all sets of the form $\{\Delta \in \mathcal{F} : \phi \in \Delta\}$ for $\phi \in \Gamma$
4. Form the ultraproduct $\mathcal{M} = \prod_{\Delta \in \mathcal{F}} \mathcal{M}_\Delta / \mathcal{U}$
5. By Łoś's theorem, for any sentence $\phi$: $\mathcal{M} \models \phi \iff \{\Delta \in \mathcal{F} : \mathcal{M}_\Delta \models \phi\} \in \mathcal{U}$
6. For any $\phi \in \Gamma$, the set $\{\Delta \in \mathcal{F} : \phi \in \Delta\} \subseteq \{\Delta \in \mathcal{F} : \mathcal{M}_\Delta \models \phi\}$ is in $\mathcal{U}$
7. Therefore $\mathcal{M} \models \phi$ for all $\phi \in \Gamma$, so $\mathcal{M}$ is a model of $\Gamma$

Historical Note: The compactness theorem was first proved by Gödel in 1930 as a consequence of his completeness theorem. The ultraproduct proof came later with the development of model theory in the 1950s. Compactness is now recognized as one of the most fundamental properties of first-order logic, with applications spanning all areas of mathematics.