Type Space is Compact

Theorem
Let $\mathcal{M}$ be an $\mathcal{L}$-structure, and let $A$ be a subset of the universe of $\mathcal{M}$.

The space $S_{n}^{\mathcal{M}}(A)$ of $n$-types over $A$ is compact.

Proof
It will suffice to show that every open cover of $S_{n}^{\mathcal{M}}(A)$ by the basic open sets $[\phi]$ of the topology has a finite subcover.

Let $\mathcal{U} = \{ [\phi_i] : i\in I \}$ be a cover of $S_{n}^{\mathcal{M}}(A)$ by basic open sets.

This means that every complete $n$-type over $A$ contains some $\phi_i$.

We will find a finite subcover of $\mathcal{U}$.

Let $\Gamma = \{ \neg\phi_i : i\in I \}$.

Then $\operatorname{Th}_\mathcal{A}(M) \cup \Gamma$ cannot be satisfied, since if $\mathcal{N} \models \operatorname{Th}_\mathcal{A}(M) \cup \Gamma (\bar{b})$, then $\operatorname{tp}_\mathcal{N} (\bar{b}/A)$ is a complete $n$-type in $S_{n}^{\mathcal{M}}(A)$ which does not contain any $\phi_i$.

By the Compactness Theorem, $\operatorname{Th}_\mathcal{A}(M) \cup \Gamma$ must have a finite subset $\Delta$ which is not satisfiable.

Since $\Delta$ is not satisfiable but $\operatorname{Th}_\mathcal{A}(M)$ is, $\Delta$ must contain some of the $\neg\phi_i$ from $\Gamma$.

Furthermore, we must have that every model of $\operatorname{Th}_\mathcal{A}(M)$ fails to satisfy at least one of these finitely many $\neg\phi_i$ in $\Delta$.

We claim that the finite set $\mathcal{F} = \{ [\phi_i] : \neg\phi_i \in \Delta \}$ is a finite subcover of $\mathcal{U}$.

Since $\mathcal{F}$ is clearly a subset of $\mathcal{U}$, we need only show that every $p\in S_{n}^{\mathcal{M}}(A)$ is contained in some $[\phi_i]\in\mathcal{F}$.

That is, we must show that each $p\in S_{n}^{\mathcal{M}}(A)$ contains one of these $\phi_i$.

Let $p\in S_{n}^{\mathcal{M}}(A)$.

By definition, this means that $p\cup \operatorname{Th}_\mathcal{A}(M)$ is satisfiable by some $\mathcal{N}$.

But, as mentioned above, since $\mathcal{N}\models \operatorname{Th}_\mathcal{A}(M)$, we have that $\mathcal{N}\not\models \neg\phi_i$ for some $\neg\phi_i \in \Delta$.

Since $p$ is complete, it must contain either $\phi_i$ or $\neg\phi_i$.

However, if $p$ contained $\neg\phi_i$, then $\mathcal{N}\models \neg\phi_i$, which contradicts $\mathcal{N}\not\models \neg\phi_i$.

Thus, $p$ contains $\phi_i$.

This demonstrates that $\mathcal{F}$ is a finite subcover for $\mathcal{U}$.

Thus, $S_{n}^{\mathcal{M}}(A)$ is compact.