Type Space is Compact

Theorem
Let $\MM$ be an $\LL$-structure.

Let $A$ be a subset of the universe of $\MM$.

The type space $\map {S_n^\MM} A$ of $n$-types over $A$ is compact.

Proof
It will suffice to show that every open cover of $\map {S_n^\MM} A$ by the basic open sets $[\phi]$ of the topology has a finite subcover.

Let $\UU = \set { [\phi_i] : i \in I}$ be a cover of $\map {S_n^\MM} A$ by basic open sets.

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

We will find a finite subcover of $\UU$.

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

Then $\map {\operatorname{Th}_\AA} M \cup \Gamma$ cannot be satisfied, since if $\NN \models \map {\operatorname{Th}_\AA} M \cup \map \Gamma {\bar b}$, then the type $\map {\operatorname{tp}_\NN} {\bar b / A}$ is a complete $n$-type in $\map {S_n^\MM} A$ which does not contain any $\phi_i$.

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

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

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

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

Since $\FF$ is clearly a subset of $\UU$, we need only show that every $p \in \map {S_n^\MM} A$ is contained in some $[\phi_i] \in \FF$.

That is, we must show that each $p \in \map {S_n^\MM} A$ contains one of these $\phi_i$.

Let $p \in \map {S_n^\MM} A$.

By definition, this means that $p \cup \map {\operatorname{Th}_\AA} M$ is satisfiable by some $\NN$.

But, as mentioned above, since $\NN \models \map {\operatorname{Th}_\AA} M$, we have that $\NN \not\models \neg \phi_i$ 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 $\NN \models \neg \phi_i$, which contradicts $\NN \not\models \neg \phi_i$.

Thus, $p$ contains $\phi_i$.

This demonstrates that $\FF$ is a finite subcover for $\UU$.

Thus, $\map {S_n^\MM} A$ is compact.