Type Space is Compact

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Theorem

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


The type 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 the type $\operatorname{tp}_\mathcal N (\bar{b}/A)$ 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.


$\blacksquare$