Let $T$ be a countable, complete theory, $M\models T$ and $A\subset M$. Now, Theorem 5.12.16 in Srivastava's book "A Course on Mathematical Logic, 2nd Edition" says that $S_{n}^M(A)$ must be countable if the isolated types are dense in $S_{n}^M(A)$.
But $A=M$ is a counterexample of this statement (if $M$ is uncountable), isn't it? Now we look what happens if $A$ should be countable.
Let $T=DLO$, $M=\mathbb{R}$ (with the natural ordering) and $A=\mathbb{Q}$. Then we have the following:
- Every two distinct tuples of reals have different types over $\mathbb{Q}$ since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus $S_{n}^{\mathbb{R}}(\mathbb{Q})$ is uncountable.
- If $\varphi(x_1,\ldots, x_n)$ is a formula of $L_A$ ($L=L_<$) such that $\{\varphi\}\cup Th(\mathbb{R}_{\mathbb{Q}})$ is satisfiable, then $\varphi$ is satisfiable by a tuple of rational numbers since $\mathbb{Q}$ is an elementary substructure of $\mathbb{R}$. Thus the isolated types are dense in $S_{n}^\mathbb{R}(\mathbb{Q})$.
This is contradictory to the theorem mentioned above. Where is the mistake?
I don't have the book to check, but you are right. If the isolated types in $S_n(A)$ are dense, it does not imply that $S_n(A)$ is countable.
What is true, is the converse: if $S_n(A)$ is countable, then the isolated types in $S_n(A)$ are dense. Curiously this is a purely topological fact. The space $S_n(A)$ is a Bair space, as a compact Hausdorff space. Let $Y$ by the set of isolated points and assume that $S_n(A) \setminus \bar Y$ is countable and nonempty. Then $S_n(A) \setminus \bar Y$ is itself Bair as an open subset of a Bair space. But then singletons are nowhere dense and it is a countable union of nowhere dense subsets. A contradiction.