Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 6 / d

For cut types over the named rational order, choose an irrational real cut, for example $r=\sqrt{2}$, and formEvery finite subset is realized by a rational between its finite lower and upper bounds, so compactness makes it a consistent partial type; quantifier elimination for dense linear orders without endpoints makes these comparisons determine a complete type in the language naming all rationals. No rational realizes the cut, so $Q^{+}$ omits it.

Add a new constant realizing this type and apply compactness, followed by downward Löwenheim-Skolem in the countable expanded language. This gives a countable model of $T$ realizing it. An isomorphism in $L_{\mathbb{Q}}$ fixes all named rationals and preserves realization of the cut, so it cannot be isomorphic to $Q^{+}$. Thus $T$ is not countably categorical. Omission by itself would need this additional realization-and-countable-model argument to establish noncategoricity.

No infinite cardinal is categorical for this theory, as described by no categorical cardinal for the complete named rational diagram. For an arbitrary infinite $\kappa$, choose a dense linear order $I_{\kappa }$ without endpoints of size $\kappa$; for example the lexicographic order on $\kappa \times \mathbb{Q}$ works, since every rational block is dense and has no endpoints.

Construct one order by inserting $I_{\kappa }$ into a fixed irrational cut of the named rational order. Every element of this order is bounded above by some named rational. Construct a second order by putting an $I_{\kappa }$ tail above all named rationals, as the ordered sum $\mathbb{Q}+I_{\kappa }$. Both are dense linear orders without endpoints, both have cardinality $\kappa$, and both contain the named rational copy as an elementary substructure by quantifier elimination. Hence both satisfy its complete diagram $T$.

They are not isomorphic over the named constants: the second has an element greater than every named rational, and the first has none. This property need not be expressible by a single sentence in the infinite language to be an isomorphism invariant. Finite cardinalities have no models at all because the constants name infinitely many distinct Elements, so the existence clause excludes categoricity there too.

Yes: $Q^{+}$ is prime. In any model of its complete diagram, map each rational to the interpretation of its named constant. The diagram ensures this is an embedding, and quantifier elimination for dense linear orders without endpoints makes it elementary. Alternatively every finite tuple of $Q^{+}$ has its complete type isolated by the conjunction equating its coordinates to their named rational constants. Thus it is a countable atomic model, and the earlier prime-model equivalence applies. A prime model may omit many types; primeness is not saturation.

There are $2^{{\aleph }_0}$ distinct irrational cuts of the named rational order. Each defines a consistent complete one-type as above, already over the empty additional parameter set, since the rational constants are part of the language. Different cuts cannot be realized by the same element: a named rational between two cuts gives contradictory required inequalities.

An ${\aleph }_0$-saturated model must realize every one of these types, so it has at least $2^{{\aleph }_0}$ Elements. Consequently there is no countable ${\aleph }_0$-saturated model of $T$. This does not contradict existence of a saturated model from the preceding construction, whose size was unrestricted, or existence of the countable prime model, whose requirement is different. No continuum-hypothesis assumption is involved.