The Axiom of Completeness can be formulated as:

There exists a set $R$ such that:

  1. $R$ is an ordered Archimedean field
  2. Any nonempty subset of $R$ with an upper bound has a least upper bound.

Recently, I read something that suggested this is logically equivalent to the following:

There exists a set $R$ such that:

  1. $R$ is an ordered Archimedean field
  2. No proper superset of $R$ is an ordered Archimedean field.

Is the second version logically equivalent to the first? (That would explain why we call it the axiom of completeness and not of least upper bound.) If so, how do we show that? If not, what would be a counter example?

The second version also suggests a dual: There exists a set $Q$ such that $Q$ is an ordered Archmidean field, and no proper subset of $Q$ is an ordered Archimedean field.


See https://hsm.stackexchange.com/questions/17261/when-and-why-was-the-concept-of-having-a-least-upper-bound-dubbed-completenes/17263 which seems to show that indeed, the Axiom of Completeness was first introduced by Hilbert in 1899, and it defined completeness not in terms of least upper bounds, but that a set is complete if it is impossible to add an element to it without violating the axioms:

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1 Answer 1


I've never heard anyone state the axiom of completeness as "there exists a set $R$ such that...". Phrasing it this way makes it sound like an axiom of set theory. Usually the axiom of completeness is an axiom about an ordered Archimedean field, i.e., it just says "any bounded nonempty subset of $R$ has a least upper bound". And then it's a theorem that an ordered Archimedian field satisfying this axiom exists.

Also, I have to nitpick about the way you asked the question: both of the statements you write are provable (say in ZF set theory), so yes, they are equivalent over ZF set theory. On the other hand, they are certainly not logically equivalent (i.e., equivalent on the basis of no axioms).

I think the proper way to frame the question is as follows: Let $R$ be an ordered Archimedean field. Is it true that $R$ is complete (any bounded subset of $R$ has a least upper bound) if and only if $R$ is maximal ($R$ is not a proper subfield of an ordered Archimedean field). The answer is yes. I'll sketch the proof.

In one direction, suppose $R$ is complete. Let $R\subseteq R'$, where $R'$ is an ordered Archimedian field. Let $x\in R'$, and let $L_x = \{r\in R\mid r<x\}$. Since $R'$ is Archimedian, $L_x$ is non-empty and bounded, so it has a least upper bound $y\in R$. If $x\neq y$, then $y-x$ is a positive infinitesimal, contradicting the assumption that $R'$ is Archimedean. Thus $x = y\in R$, so $R' = R$. Thus $R$ is maximal.

Conversely, suppose $R$ is maximal. Construct the Dedekind completion $R'$ of $R$. By maximality, $R' = R$, so $R$ is complete.

On the "dual version": Yes, $\mathbb{Q}$ is the minimal ordered Archimedean field in the sense that no proper subfield of $\mathbb{Q}$ is an ordered Archimedean field. But writing "ordered Archimedean" twice is overkill here. $\mathbb{Q}$ is a minimal field in the sense that has no proper subfields at all. You can just as well characterize $\mathbb{Q}$ as the unique minimal field of characteristic $0$ up to isomorphism.

  • $\begingroup$ Great answer. "Construct the Dedekind completion $R′$ of $R$. By maximality, $R′=R$." Doesn't that assume the Dedekind completion of an ordered Archimedean field is an ordered Archimedean field? How have we proven that? $\endgroup$ Feb 25 at 19:58
  • $\begingroup$ Well yes, that's something you need to check. It's tedious, but entirely standard - basically, it's the same as the proof that $\mathbb{R}$ (defined as the Dedekind completion of $\mathbb{Q}$) is an Archimedean ordered field. $\endgroup$ Feb 25 at 20:23
  • $\begingroup$ @SRobertJames Sure, I've looked at it... do you have a question about it? $\endgroup$ Feb 27 at 0:22
  • $\begingroup$ Doesn't it seem to indicate that Axiom of Completeness, at least as initially phrased, was about maximality, and not about least upper bounds? True that Hilbert was talking about points, not numbers, but it seems that he defined Axiom of Completeness as "a certain set (or 'system') is maximal" and from there deduced least upper bound property. Do you agree? $\endgroup$ Feb 27 at 1:50
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    $\begingroup$ @SRobertJames Well, I'm not a historian, and I don't know whether your quote is the earliest occurrence of the term "axiom of completeness". But I agree that your quote indicates that Hilbert, in 1899, preferred to phrase completeness in terms of maximality. $\endgroup$ Feb 27 at 2:30

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