I've been studying the axiomatic definition of the real numbers, and there's one thing I'm not entirely sure about.

I think I've understood that the Archimedean axiom is added in order to discard ordered complete fields containing infinitesimals like the hyperreal numbers. Additionally, this property clearly cannot be derived solely from the axioms of ordered field and completeness, since $^*\mathbb{R}$ and $\mathbb{R}$ are two complete ordered fields, two models of the axioms, one of them Archimedean and the other non-Archimedean. Are these ideas correct?


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  • By the way, to my shame I have not yet really learned non-standard analysis. Thus it is not clear to me whether the hyperreals are even complete in the weaker sense of my answer -- i.e., Cauchy complete. I suppose I will guess that they are *not*, but it would be nice to hear from someone who knows. – Pete L. Clark Mar 22 '11 at 02:35
  • @Pete: Dear Pete, I think that the hyperreals are in fact Cauchy complete. The reason is that there are very few Cauchy sequences! For instance, given any sequence of positive hyperreal numbers, there is a hyperreal number smaller than all of them. (Proof: this is clear if we use the ultraproduct construction of the hyperreals by diagonalization. Probably it can be deduced axiomatically, though I don't see it at the moment.) So I think any Cauchy sequence of hyperreals is eventually constant... – Akhil Mathew Mar 22 '11 at 02:43
  • Well, after seeing Akhil's comment I am ready to change my guess. :) – Pete L. Clark Mar 22 '11 at 02:46
  • @Akhil: ah, but perhaps not every Cauchy *net* in the hyperreals converges? – Pete L. Clark Mar 22 '11 at 02:48
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    @Pete: I don't know about *nets*; I've asked this at http://math.stackexchange.com/questions/28427/does-every-cauchy-net-of-hyperreals-converge – Akhil Mathew Mar 22 '11 at 04:00
  • @Akhil: " ...given any sequence of positive hyperreal numbers, there is a hyperreal number smaller than all of them." This follows from a so-called "saturation property" that holds in some, but not all, nonstandard models of $\mathbb R$. In case it fails, then we can, also, find a Cauchy sequence that fails to converge. – GEdgar May 12 '11 at 16:08

2 Answers2


The answer to your question depends critically on what you mean by a "complete ordered field" $(F,<)$. Here are two rival definitions:

1) [added: sequentially] Cauchy complete: every Cauchy sequence in $F$ converges.

2) Dedekind complete: every nonempty subset $S \subset F$ which is bounded above has a least upper bound.

(There are in fact many other axioms equivalent to 2): that every bounded monotone sequence converges, that $F$ is connected in the order topology, the principle of ordered induction holds, and so forth.)

It turns out that there is a unique Dedekind complete ordered field up to (unique!) isomorphism, namely the real numbers $\mathbb{R}$. Famously $\mathbb{R}$ is also Cauchy complete -- or, if you like, Dedekind complete ordered fields satisfy the Bolzano-Weierstass theorem, which is enough to make Cauchy sequences converge -- so that Dedekind completeness implies Cauchy completeness.

The converse is true with an additional hypothesis: an Archimedean Cauchy-complete field is Dedekind complete. I show this in $\S 12.7$ of these notes using somewhat more sophisticated methods (namely Cauchy nets). For a more elementary proof, see e.g. Theorem 3.11 of this nice undergraduate thesis.

On the other hand, just as one can take the "Cauchy" completion of any metric space (or normed field) and get a complete metric space (or complete normed field), one can take the Cauchy completion of a non-Archimedean ordered field and get an ordered field which is Cauchy complete but not Dedekind complete. The easiest example of such a field is probably the rational function field $\mathbb{R}(t)$ with the unique ordering that makes $t$ positive and infinitely large.

For some reason these subtleties seem to be hard to find in standard analysis texts. I myself didn't learn about them until rather recently (so, several years after my PhD). I actually wrote up some of this material as supplemental notes for a sophomore-junior level course I am currently teaching on sequences and series...but I have not as yet been able to make myself inflict these notes on my students. I talked about ordered fields in several lectures and it seemed to be one level of abstraction beyond what they could even meaningfully grapple with (so it started to seem a bit pointless).

Pete L. Clark
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    The fact that Dedekind complete ordered fields are all isomorphic to the reals (without any need to assume the field is Archimedean) is much less well known than it deserves to be. – Carl Mummert Mar 22 '11 at 02:21
  • @Carl: I agree. Not that this is at all hard to show: if an ordered field were Dedekind complete but non-Archimedean, the subset $\mathbb{Z}$ would have to have a least upper bound $M$, i.e., for all $n \in \mathbb{Z}$, $n \leq M$, but for some $n \in \mathbb{Z}$, $n > M-1$: ridiculous! – Pete L. Clark Mar 22 '11 at 02:27
  • Thanks for your very interesting answer. It seems that I was under the assumption that Dedekind and Cauchy completeness were equivalent. In any way, there's a lot I should look into, but now I think that the different constructions of the real numbers are not as easy to compare as I believed, so I'm not entirely certain as to what properties of the real field I can assume. I should probably just go with the "Archimedean Cauchy-complete ordered field" definition for the time being and study others further ahead. One last thing: does Dedekind-completeness imply Archimedeanness? ... – Abel Mar 22 '11 at 04:15
  • ... I'm guessing this is true because of the essential uniqueness of the real field. – Abel Mar 22 '11 at 04:17
  • @Abel: yes, Dedekind completeness implies Archimedeanness -- that is exactly the proof I gave in my comment above! (And yes, it certainly follows from the fact that the only Dedekind complete field is $\mathbb{R}$, which is Archimedean...but actually this fact is used in the proof, as in the notes I linked to in my answer.) – Pete L. Clark Mar 22 '11 at 05:48
  • Great. I'll take a look at your notes. Thanks. – Abel Mar 22 '11 at 09:19
  • If you are going to talk about non-archimedean ordered fields with your stuents, use the example $\mathbb R(t)$, not ${}^*\mathbb R$. – GEdgar May 12 '11 at 14:22
  • @GEdgar: well, I'm quite sure that I'm not -- not with students in that class, anyway -- but my own interest in the subject has only increased. I am just about at the point in my notes where I want to give two different examples of sequentially complete non-Archimedean fields: one is $\mathbb{R}((t))$, where there are plenty of convergent sequences, and one is a field in which the only convergent sequences are eventually constant. – Pete L. Clark May 12 '11 at 18:27

There are non-archemedian completions of the rationals, called p-adic completions. The book Gouvêa, Fernando Q. (2000). p-adic Numbers : An Introduction (2nd ed.). Springer is an excellent introduction to these.

Scott Carter
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    Dear Scott, I think the notion of non-archimedeanness the OP has in mind may be different from the one you describe above (I think (s)he wishes to consider non-archimedeanness as a property of an ordering, not of an absolute value). – Akhil Mathew Mar 22 '11 at 02:02
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    Agreed. But a comment: there are closer connections between these two concepts than I had realized. Indeed, the field $\mathbb{R}((t))$ in its unique ordering which makes $t$ positive and infinitesimal is sequentially complete, and the arrgument is very similar to the one telling you that this field equipped with the $t$-adic valuation is complete! – Pete L. Clark May 12 '11 at 18:29