# Compactness

Theorem 1. (Compactness Theorem)

1. If $$\Gamma \models \varphi$$, then for some finite $$\Gamma_0 \subseteq \Gamma$$ we have $$\Gamma_0 \models \varphi$$.
2. If every finite subset $$\Gamma_0$$ of $$\Gamma$$ is satisfiable, then $$\Gamma$$ is satisfiable.

Proof.

1. \begin{align*} \Gamma \models \varphi &\implies \Gamma \vdash \varphi \\ &\implies \Gamma_0 \vdash \varphi \text{ for some finite } \Gamma_0 \subseteq \Gamma \text{, deductions being finite} \\ &\implies \Gamma_0 \models \varphi \end{align*}
2. If every finite subset $$\Gamma_0$$ of $$\Gamma$$ is satisfiable, then by soundness $$\Gamma_0$$ is consistent. Thus $$\Gamma$$ is consistent (since deductions are finite). By completeness, $$\Gamma$$ is also satisfiable.