Let $T$ be a theory in a language $L$ such that a model of $T$ defines a real closed field. Can we apply the omitting types theorem in order to show the existence of atomic models of $T$? i.e. does the definability of an order impede the countability of the number of non-isolated types over the empty set?
2026-03-28 02:09:19.1774663759
Omitting types and real closed fields
303 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
2
Existence of a total ordering (through a fairly standard argument involving Dedekind cuts and the Cantor tree) implies that over any infinite dense (in itself) set, there are at least $\mathfrak c$ complete types.
In any model of real fields (possibly with some additional structure), rationals are a subset of $\operatorname{dcl}(\emptyset)$ and a dense linear order, so there will always be at least $\mathfrak c$ types over $\emptyset$ (or exactly $\mathfrak c$ if the language is countable; the types correspond to types of real numbers over the rationals in the language of dense linear orderings), so omitting types tells you nothing about existence of atomic models.
As Levon suggested, any o-minimal (with respect to standard ordering) extension of real closed fields will always have an atomic model. O-minimality is not required, though: it's easy to see that real closed fields with added predicate for integers still have (the very same) atomic model consisting of algebraic reals. (In general, it's not hard to see that if a theory has a model $M$ such that every element of $M$ is definable, then in any extension of the theory in which $M$ is still a model, it still has this property, and in particular, it is atomic.)
I can't think of a natural extension of real closed fields which would have no atomic model, but you can probably just add some completely unrelated (to the field) structure to reals (or maybe an extension with infinite elements) to get a theory without an atomic model.