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Assume $A$ is a countable model of $T$ and its automorphism group is an oligomorphic permutation group. Complete types are constant on automorphism orbits, so only finitely many complete $n$-types are realized in $A$, for each $n$.

These realized types are dense in the Stone topology on a type space $S_n(T)$. Indeed, any nonempty basic clopen set $[\theta ]$ is given by a formula consistent with $T$. Completeness implies $T\models \exists \mathbf{x}\theta$, and thus some tuple in $A$ realizes a type in $[\theta ]$. A finite set of points is closed in a Hausdorff space; if it is also dense, it is the whole space. Therefore $S_n(T)$ equals the finite set of types realized in $A$. This proves (iii) implies (ii), and the previous part then yields (i). We have establishedIn particular, the argument does not assume that an arbitrary countable model is saturated or that its type-orbits coincide before that coincidence has been proved.

The Stone topology on a type space has basic open setswhere $\phi$ ranges over formulas with parameters in $A$ and $n$ free variables. Here points are complete types consistent with $T_A=\mathrm{Th}(\mathcal{M},a{)}_{a\in A}$. Arbitrary open sets are unions of these basic sets.

The identitiesshow that the basic sets are clopen sets and form a basis. Distinct complete types contain opposite choices for some formula, so this topology is Hausdorff. The compactness theorem gives compactness: a family of basic closed sets with the finite intersection property determines a consistent set of formulas, which extends to a complete type. Thus logical formulas provide a basis of clopen tests for distinguishing points of the type space.