One of the central theorems in real analysis states that any complete ordered field $X$ is isomorphic to $\mathbb R$. Here, I mean "complete" in the sense of the least upper bound property, but I am open to the possibility that by considering a different form of the completeness axiom (e.g. Cauchy-completeness), my question might have a more interesting answer.
What I am wondering about is whether we can relax the hypotheses placed on $X$, since most of the obvious ways don't work:
- If we drop the assumption that $X$ is complete, then the conclusion
no longer follows: $\mathbb Q$ is an ordered field, but it is
certainly not isomorphic to $\mathbb R$. - If we drop the assumption that $X$ is a field, and replace it with
the assumption that $X$ is an ordered set, then again the conclusion
no longer follows: the empty set is a complete ordered set! (And there are other non-trivial examples, such as the long line.) - The assumption that $X$ is ordered cannot, by itself, be dropped, since we need $X$ to be ordered to make sense of any of the forms of the completeness axiom.
There is a positive result: up to an isomorphism, the only complete ordered abelian groups are the trivial group, the integers, and the real numbers. But note that in a way this makes our job harder: this means that there is a complete ordered commutative ring which is not isomorphic to $\mathbb R$, namely the integers.
Question: Is there a non-trivial way of relaxing the hypotheses on $X$ in such a manner that it still follows that $X$ is isomorphic to $\mathbb R$?
It should be noted that to change the hypotheses, we might also have to change the type of isomorphism being considered.
Best Answer
Turning my comment into an answer:
$(\mathbb{R};+)$ is up to isomorphism the unique nontrivial divisible completely orderable abelian group.
$(\mathbb{R};<)$ is up to isomorphism the unique nonempty complete separable linear order without endpoints.
Interestingly, a slight tweak of that second definition takes us into serious set theory: $\mathsf{ZFC}$ alone cannot decide whether every nonempty dense complete linear order without endpoints with the countable chain condition is isomorphic to $(\mathbb{R};<)$. A linear order satisfies the c.c.c. iff it has no uncountable family of pairwise-disjoint nontrivial open intervals. This question is Suslin's problem, and its solution(?) required the introduction of iterated forcing by Solovay and Tennenbaum.