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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!

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up vote 2 down vote accepted

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.

I'm not how to answer your question about how "often" incomplete types are isolated, but it seems to me that for types, the properties of being complete and being isolated are completely orthogonal.

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 $\land_{i = 1}^n \phi_i$.

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