Past exam of the logic and set theory course of the University of Oxford / 2010 / A / 3 / d

Use the axioms for a strict dense linear order without endpoints:The first three give strict linear order; the next gives density; the final two exclude endpoints. Domains are nonempty under the adopted first-order semantics, and these axioms force them to be infinite.

Yes. The usual order on the rational numbers is strict and linear. Between $a<b$ lies the rational midpoint $(a+b)/2$, and $a-1,a+1$ witness absence of endpoints. Thus it is a dense linear order without endpoints.

Yes. The usual order on the real numbers is strict and linear, midpoints witness density, and translation by $\pm 1$ supplies points on either side. It therefore satisfies the same dense linear order without endpoints axioms.

No. With its induced usual order, the closed interval has a least element and a greatest element. At zero there is no smaller domain element, and at one there is no larger domain element. It fails the no-endpoint axioms, although it is densely ordered.

Yes. The induced order is strict and linear and remains dense. For $0<x<1$, the points $x/2$ and $(x+1)/2$ lie in the domain and are strictly below and above $x$. Hence the open interval is a dense linear order without endpoints.

Yes. Removing zero leaves the induced order strict and linear, with no endpoints. For any $a<b$ there are infinitely many points in the real interval between them, so one can choose an intermediate point different from zero. More explicitly, the midpoint works unless it is zero, in which case $(2a+b)/3$ is nonzero and lies strictly between the endpoints. Thus density also survives.

The missing point is a cut, not a pair of adjacent elements; the axioms do not require all cuts to be filled.

Enumerate both countable models. Build an increasing sequence of finite order-preserving partial bijections. At an even stage, take the first unmatched element of the first model. Its position relative to the finitely many already matched points determines an open interval or exterior ray in the second model. Density and absence of endpoints supply an unused point there, so the map extends.

At an odd stage perform the symmetric operation starting from the second model, extending the inverse map. Every required interval has infinitely many points, so avoiding the finitely many used values is possible. The union is a total bijection, by the alternating enumeration requirements, and it preserves and reflects order because every finite stage does. Hence it is an isomorphism of structures.

This is the back-and-forth method, proving countable categoricity of the dense-order theory.

The theory is consistent because the rational order is a model. If some sentence $\theta$ were not decided by the theory, both extensions by $\theta$ and by $\neg \theta$ would be consistent. By Gödel's completeness theorem and the countable-model consequence of Löwenheim-Skolem theorem, each would have a countable model. The two models would disagree on $\theta$, so could not be isomorphic. This contradicts part (iii).

Therefore the theory decides every sentence and is complete. Under the standard convention that a theory includes all its logical consequences, this makes it maximal consistent.

If one takes only the displayed finite list of axioms as a raw set, that list is not literally maximal consistent: one can add further theorems without inconsistency. The conclusion applies to its deductive closure, which is the complete theory of dense linear order without endpoints.