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In this question, it is shown that all Archimedean ordered groups are isomorphic to an ordered subgroup of $\mathbb R$. Additionally, it is shown that if such a group is complete, then it is isomorphic to the trivial group, $\mathbb Z$, or $\mathbb R$.

I'm curious if a similar result holds it we loosen the group condition to being a monoid. In particular,

  • Are all Archimedean linearly ordered monoids isomorphic to ordered submonoids of $\mathbb R$?
  • If the order is complete, must it be isomorphic to $\mathbb R^{\ge 0}$, $\mathbb R$, or some ordered submonoid of $\mathbb Z$?

If not, are there additional properties like cancellation or commutativity that could be assumed to make these true?

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  • $\begingroup$ You definitely need some cancellativity conditions. For example, consider the two-element commutative monoid, which is linearly ordered. $\endgroup$ – Emil Jeřábek May 10 '19 at 17:14
  • $\begingroup$ For another kind of counterexample, consider the complete submonoid $\mathbb N+\alpha\mathbb N\subseteq\mathbb R$, where $\alpha>0$ is irrational. $\endgroup$ – Emil Jeřábek May 10 '19 at 17:17
  • $\begingroup$ When changing a definition from groups to monoids, it's useful to say explicitly the definition (here, being "linearly ordered"), since some equivalent ways to formulate it might no longer be equivalent once the invertibility axiom is erased. $\endgroup$ – YCor May 10 '19 at 18:05
  • $\begingroup$ As regards your second question, a closed submonoid of $\mathbf{R}_{\ge 0}$ is $\mathbf{R}_{\ge 4}\cup\{0,2,3\}$. $\endgroup$ – YCor May 10 '19 at 18:07
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By a classic result of Alimov (1950), a fully ordered semigroup $S$ is order isomorphic to a subsemigroup of the additive group of real numbers if and only if it is cancellative and contains no anomalous pairs. Archimedean totally ordered monoids need not satisfy either of these conditions, and so the answer to your first question is no. For a thorough discussion of your questions, see Positively Ordered Semigroups by M. Satyanarayana, Marcel Dekker, 1979. Moreover, for an excellent discussion of when the answer is positive, see pp. 162-169 of Partially Ordered Algebraic Systems by L. Fuchs where the classical results of Hölder, Clifford, Alimov, Huntington and Fuchs are discussed.

Edit: Two distinct elements of $a,b$ of $S$ are said to be anomalous if for all $n>0$ $$na<(n+1)b \;{\rm and} \;nb<(n+1)a$$ or $$na>(n+1)b \;{\rm and} \;nb>(n+1)a.$$ The first alternative may occur if $a,b>0$ and the second if $a,b<0$.

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  • $\begingroup$ I see. I never even thought of the idea of anomalous pairs. Does requiring the existence of differences (that is, for all $a, b$, there exists $c$ with $a+c = b$ or $b + c = a$) affect anything? $\endgroup$ – eyeballfrog May 11 '19 at 21:45
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    $\begingroup$ @eyeballfrog. Somewhat. The result for fully ordered monoids can be obtained using the following classic result (due to Hölder (1901) and Clifford (1954). Let $S$ be an Archimedean naturally fully ordered semigroup. Then $S$ is order isomorphic to a subsemigroup of one of the following fully ordered semigroups: (i) the additive semigroup of nonnegative reals; (ii) the real numbers in the interval $[0,1]$ with the usual ordering and $ab=min(a+b, 1)$; (iii) the real numbers in the interval $[0,1]$ with the usual ordering and $ab=a+b$ if $a+b \le 1$ and $ab=\infty$ if $a+b > 1$. $\endgroup$ – Philip Ehrlich May 13 '19 at 13:45
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    $\begingroup$ @eyeballfrog. Another relevant result (in light of Alimov's result) is: Let $S$ be a cancellative fully ordered semigroup. If $S$ is Archimedean and naturally ordered, then it contains no anomalous pairs. $\endgroup$ – Philip Ehrlich May 13 '19 at 15:17

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