Non-existence of prime extensions

Question

Let $\mathcal{R}=\langle\mathbb{R};\ <,Q\rangle$ be the expansion of the ordered set of the reals by a unary predicate, interpreted as the rationals. I want to prove that there is no model of $T=Th(\mathcal{R})$ which is prime over $\mathbb{Q}$. For this, I need to prove that the isolated types in $S_1^{\mathcal{R}}(\mathbb{Q})$ are not dense, i.e. that there exists a formula $\varphi(x)$ which is consistent with $T$ and which does not belong to any isolated type. I have a candidate for $\varphi$, namely $\neg Q(x)$. Since every isolated type is realised, it is enough to understand which real numbers do have an isolated type over $\mathbb{Q}$ in this structure. Clearly all the rationals do. My guess is that no irrational number has an isolated type (hence $\varphi$ does not belong to any isolated type). However, I'm not sure how to prove this formally: in the reduct of $\mathcal{R}$ to the pure order, this follows by quantifier elimination. Should I try to prove QE for $T$? Or is there a more direct and obvious argument?

Answer 1

Unless I'm missing something, there's a straightforward automorphisms argument:

Suppose $p(x)=tp(r)$, with $r$ irrational, is isolated by the formula $\varphi(x)$. There can only be finitely many (constants naming) rationals $q_1, ..., q_n$ appearing in $\varphi$; let $L$ be the reduct of our language to the language of order, the predicate for the rationals, and these finitely many constant symbols.

Now consider an order-automorphism $\mu$ of $\mathbb{R}$ which

• sends rationals to rationals and irrationals to irrationals,

• moves $r$, and

• fixes each rational $q_i$.

We can easily build such a $\mu$ via a back-and-forth argument. But now $\mu(r)$ also satisfies $\varphi$, even though the types of $r$ and $\mu(r)$ are different (since they're separated by a rational).