# How often do incomplete types meet the hypotheses of the omitting types theorem?

I find the following formulation of the hypothesis (namely, non-isolation) for the omitting types theorem. A type $p$ over $T$ is "isolated" iff there is a formula $\phi(\vec{x})$ such that $\exists \vec{x} (\phi(\vec{x}))$ is consistent with $T$ and $T \models \forall \vec{x} (\phi(\vec{x}) \rightarrow \delta(\vec{x}))$ for all $\delta$ in $p$. (Source) I also find the (more common?) formulation that a complete type $p$ is isolated iff $\{p\}$ is open in the Stone space.

I gather that the omitting types theorem is often formulated as applying just to complete types, but it can also be applied to other types (using, I presume, the other definition of "isolated" which I gave, which makes no reference to the Stone space). But I also get the general impression that it may be quite "hard" to meet the hypotheses for being non-isolated if you are not a complete type. I'm wondering how correct that impression is.

So my question (and it's a fairly soft one) is: how "often" is an incomplete type non-isolated? Is this something that can happen fairly easily, or is it a rare thing? And in particular: is a finite type ever non-isolated? Thank you!

-

Relative to most theories, there are many non-isolated incomplete types! All you need a set of formulas which is realized in some model of the theory, but not every model. A characteristic example is $\{x > 1, x > 2 , x > 3, \dots\}$, relative to the theory of arithmetic.
You're right, however, that all finite types are isolated, since the set of formulas $\{\phi_1,\dots,\phi_n\}$ is equivalent to the conjunction $\land_{i = 1}^n \phi_i$.