Proof. Showing that it is a topology is left as an exercise.

Hausdorff: Fix $p,q\in S_n^M(A)$ distinct. Then there is a $\phi (\overline{x})$ $L_A$-formula such that $\phi (\overline{x})\in p$ and $\neg \phi (\overline{x})\in q$. Then $p\in [\phi (\overline{x})]$ and $q\in [\neg \phi (\overline{x})]$ – these are disjoint.

Compactness: Sufficient to consider open covers consisting of basic open sets. SUppose we have $L_A$-formulas ${({\phi }_i(\overline{x}))}_{i\in I}$ such that $S_n^M(A)={\bigcup }_{i\in I}[{\phi }_i(\overline{x})]$. Let $\mathrm{\Sigma }=\{\neg {\phi }_i(\overline{x}):i\in I\}$.

Claim: $\mathrm{\Sigma }\cup {Th}_A(M)$ is inconsistent.

Proof of claim: Otherwise $N\models {Th}_A(M)$, $\overline{a}\in N^n$ such that $\overline{a}\models \mathrm{\Sigma }$. Let $p={tp}^N(\overline{a}∕A)\in S_n^M(A)$. But $p\notin [{\phi }_i(\overline{x})] \forall i\in I$, contradiction.

So by compactness we have finite $I_0\subseteq I$ with

inconsistent.

Claim: $S_n^M(A)={\bigcup }_{i\in I_0}[{\phi }_i(\overline{x})]$.

Proof: Fix $p\in S_n^M(A)$. We can realise $p$ in some $N\models {Th}_A(M)$, i.e. we have $\overline{a}\in N^n$ with $\overline{a}\models p$.

By ($\ast$), we have $N\models {\phi }_i(\overline{a})$ for some $i\in I_0$. So ${\phi }_i(\overline{x})\in p$, so $p\in [{\phi }_i(\overline{x})]$.

Totally disconnected: In a compact Hausdorff space, we have totally disconnected if and only if two points are separated by a clopen set. All basic sets are clopen, so totally disconnected. □