For types, is being definable is strictly stronger than being isolated?

Question

Let $A$ be an $L$-structure and let $X\subset A$ and $a\in A$ be a tuple, and consider the type $p=\mbox{tp}_A(a/X)$. Let $T$ be the theory of $A$.

Definition 1: $p$ is said to be isolated if for some formula $\phi$ we get that $T\vdash \phi \to \psi$ for any $\psi\in p$.

Definition 2 (Hodges, Model Theory, p. 153): $p$ is said to be definable over $X$ if for any formula $\phi(x,y)$ we may find a formula $d\phi(y,z)$ and a tuple $b$ in $X$ such that: for every tuple $a\in X$ we get that $\phi(x,a)\in p$ iff $A\models d\phi(a,b)$.

It is clear from these definition that if $p$ is isolated by an $L_X$ formula then it is definable over $X$. Specifically, this corresponds to the special case where there are a specific formula $\psi$ and tuple $b\in X$ which would work for any $L$-formula.

The question is, why isn't this condition sufficient? I am fairly positive that it isn't, but I was not able to convince myself of that, or to find a counterexample.

The only lead I had was trying to come up with a type of $\mathbb{C}$ which is definable over $\mathbb{Q}$ but could not be isolated in $L_{\mathbb{Q}}$. The only types I actually have the tools to tackle are types of algebraic elements, but this are not useful because they are easily seen to be isolated by a polynomial equation.

Is there maybe a more rudimentary approach to this?

Thanks in advance.

Answer 1

This is strictly stronger. Let $T$ be a theory asserting that $c_n$ are infinitely many distinct constants. The type of an element which is distinct then all these constants is not isolated (it is omitted in the atomic model where all elements are constant). It is definable, however. By quantifier elimination it is sufficient to define $d \phi$ for $\phi$ atomic or its negation (since every type is generated by its quantifier free part, and we can put $d(\phi \vee \psi) = d\phi \vee d\psi$). So just put $d(x = c_i) = (x \ne x)$ and $d(x \ne c_i) = (x = x)$

In fact a similar argument shows tha the 1-type of a transcendental element over the field of algebraic numbers is definable but not isolated

Answer 2

Every type over a model in a stable theory is definable, but not isolated in general. For a specific example let $\bar {\mathbb Q}$ be the set of complex algebraic numbers. Let $a \in \mathbb C \setminus \bar {\mathbb Q}$. Then $\mathrm {tp}(a/\bar {\mathbb Q})$ is definable (hint: $\phi(x, \bar b) \in \mathrm {tp}(a/\bar {\mathbb Q})$ if and only if it defines an infinite set), but not isolated (it is omitted in $\bar {\mathbb Q}$).