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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.

Joe
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  • How about "complete ordered nontrivial divisible abelian group" as a hypothesis? It's also easy to prove that $(\mathbb{R};<)$ is the unique-up-to-isomorphism complete separable nonempty linear order without endpoints. – Noah Schweber Dec 19 '22 at 22:29
  • @NoahSchweber: Those both look like interesting answers, thank you! I didn't expect it to be possible to characterise $\mathbb R$ just with the relation $\le$. – Joe Dec 19 '22 at 22:55

1 Answers1

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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.

Noah Schweber
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