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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 a 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.)

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    $\begingroup$ What does "over the reals" mean here? $\endgroup$
    – Qiaochu Yuan
    Commented May 28, 2012 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
    Commented May 28, 2012 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
    Commented May 28, 2012 at 22:03
  • $\begingroup$ The right way to formulate the question would be: Let $(A,+,\le)$ be a totally ordered abelian group such that $(A,\le)$ is isomorphic to $(\mathbf{R},\le)$ as ordered set. Is $(A,+,\le)$ Archimedean? (If groups are allowed to be non-abelian, it should be specified what invariance is required: left-invariance or bi-invariance.) $\endgroup$
    – YCor
    Commented Feb 15, 2020 at 17:51

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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.)

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  • $\begingroup$ I guess the question was intended with abelian groups. Anyway, let me provide the argument for arbitrary left-ordered groups $G$. Namely, the claim is that if the topology on the order on $G$ is connected, then $G$ is archimedean. By contradiction, suppose $G$ non-archimedean: there exists $x>1$ such that $X=\{x^n:n\ge 0\}$ is bounded above, hence by connectedness it has a supremum $y$. Hence $y$ is also the sup of $xX$, which by left-invariance implies $y=xy$, so $x=1$, contradiction. (In general this shows that in a left-ordered group, $\{x^n:n\ge 0\}$ has no upper bound for every $x>1$.) $\endgroup$
    – YCor
    Commented Feb 15, 2020 at 18:10
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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.

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  • $\begingroup$ Can you give any references for your claims? $\endgroup$
    – Igor Rivin
    Commented May 29, 2012 at 1:14
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    $\begingroup$ Could not think of references, so I added proof sketches. They could be exercises in any model theory book. $\endgroup$
    – Goldstern
    Commented May 29, 2012 at 6:09

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