Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2010 / B / 5 / c

Yes. The pure order on the reals is a dense linear order without endpoints and has quantifier elimination for dense linear orders without endpoints. Over a finite parameter set $a_1<\cdots <a_m$, a complete one-variable type either specifies equality to one of these points or specifies one of the open intervals and exterior rays they determine.

Each such interval or ray contains a real number. Hence every complete one-variable type over a finite parameter set is realized. The successive-coordinate characterization gives saturation for all finite tuples. This uses finite parameter sets; it does not say that arbitrary types over a countably infinite set of parameters are realized.

Yes. As an Abelian group, the reals are a nontrivial torsion-free divisible group, equivalently a vector space over the rationals. Rational scalar multiplication is uniquely definable by equations such as $ny=mx$, so adding those operations is a definitional expansion.

In this language, quantifier elimination for torsion-free divisible Abelian groups reduces formulas to Boolean combinations of rational linear equations. Over finitely many parameters, a complete one-variable type either fixes a value in their linear span or excludes every value in that span. The first kind is realized in the span itself; the second is realized by any real number outside it. That span is countable, while the reals are uncountable. Thus every such type is realized and the additive real group is an omega-saturated model.

For finite tuples, the same argument successively supplies any required finite number of new rationally linearly independent elements. The order and multiplication of the real field are absent from this language, so they cannot impose additional type conditions.

No. The parameter-free partial typeis finitely satisfiable in the ordered real field: choose a number larger than the finitely many bounds. By compactness theorem it extends to a complete type over the empty set. But no real number realizes it, by the Archimedean property. Therefore the ordered real field is not an omega-saturated model.

The named unit and addition make every integer bound expressible, even without extra parameters. This is why the obstruction is present in the ordered-field language and absent in the two reducts above.