Hi there, I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e. for all $x,y,z$ we have that if $y\leq z$ then $x\oplus y \leq x\oplus z$, and I am wondering if this already suffices to show that the group is archimedean. (i.e., that for all $a,b$ with $a \leq b$ there exists an $n$ such that $\underbrace{a\oplus\ldots\oplus a}_n \geq b$).

I strongly suppose it is, as I couldn't find any counterexample, but neither can I find an convincing proof. (My further goal is to show that the group is isomorphic to $(\mathbb{R};\leq,+,0,-)$ which should be immediate once I can proof archimedeanness)

thanks in advance, Chris

  • 2
    $\begingroup$ What does "over the reals" mean here? $\endgroup$ – Qiaochu Yuan May 28 '12 at 20:08
  • $\begingroup$ I agree that this question needs some work to be coherent enough to answer. But two unexplained downvotes don't seem to be helping. I am canceling one out. Chros, please hit edit then proofread your question (totally ordered group? etc.) and define terms which you are not sure are standard. $\endgroup$ – Noah Stein May 28 '12 at 21:39
  • $\begingroup$ Let me try and guess that "over the reals" means that the carrier of the structure is the reals. On another hand, I don't understand why the OP speaks of an ordered group but specifies what looks like the signature of an ordered monoid. $\endgroup$ – Salvo Tringali May 28 '12 at 22:03

I guess that the notation $(\mathbb R;\leq,\oplus,0)$ is intended to say that your ordered group is the set of reals with its standard ordering $\leq$ but with a possibly strange group operation $\oplus$ (whose identity element is nevertheless the standard 0). In this case, archimedianness can be proved as follows. Consider some positive $a$ and its multiples $na=\underbrace{a\oplus\ldots\oplus a}_n$; you want these to be cofinal in $\mathbb R$. If they were not cofinal, they would have a least upper bound $b$ (because the ordering is standard). Then $b-a$, being strictly smaller than $b$ (because $a$ is positive), is $<na$ for some $n$; but then $b<(n+1)a$, a contradiction.

(If my guess is wrong and you intended to allow a non-standard interpretation of $\leq$, then archimedianness does not follow, simply because there are non-archimedian totally ordered groups of the cardinality of the continuum. Furthermore, archimedian examples would not have to be isomorphic to the reals, since $\mathbb R$ has proper subgroups of the cardinality of the continuum.)


There is no first order property of a totally ordered group $G$ which

(a) implies that $G$ is archimedean

(b) is satisfied by the real numbers (with the usual order and usual addition).

EDIT: In view of Andreas Blass' interpretation and answer, this may be irrelevant now, but here are two proof sketches:

  1. "logical proof": Take the first order theory of the reals, add constants $c,d$ to the language, and add the axioms $0\lt c\lt d$, $c+c\lt d$, $c+c+c\lt d$, etc. The resulting theory is consistent (by compactness) and hence has a model - the desired non-archimedean counterexample.

  2. "Algebraic proof": Let $U$ be a non-principal ultrafilter on the natural numbers $\mathbb N$. Let $M$ be the ultrapower $\mathbb R^{\mathbb N}/U$. Compare the class of the identity function and any constant function (say: 1) to see that $M$ is not archimedean. By Łoś' theorem, $M$ satisfies the same first order theory as the real numbers.

  • $\begingroup$ Can you give any references for your claims? $\endgroup$ – Igor Rivin May 29 '12 at 1:14
  • $\begingroup$ Could not think of references, so I added proof sketches. They could be exercises in any model theory book. $\endgroup$ – Goldstern May 29 '12 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.