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I am currently looking for the simplest way to characterize the real numbers. Usually they are described as the complete archimedean field, showing that all such fields are isomorphic.

Archimedean means that for all $x$ in the field, there exists a natural number $n$ such as $1+\dots+1$, summed $n$ times, is greater than $x$.

Now in the usual definition, the order is understood to be total, ie for all $x, y$, either $x\leq y$ or $y\leq x$. Is this assumption redundant with the rest of the definition of a complete archimedean field? I wonder if the integer bounds given by the archimedean property could be used to compare all real numbers, because the order on the rational numbers is total.

The rest of the assumptions remain the same. For multiplication it means, for all $x, y$ in the field, if $0\leq x$ and $0\leq y$ then $0\leq xy$. We do assume that if $0<x$, then $0<\frac{1}{x}$.

For the completeness, a sequence $x_n$ of the field is said to converge towards a limit $l$ in the field iif, $$\forall \varepsilon>0,\exists n\in\mathbb{N}, \forall p\geq n,\; l-\varepsilon < x_p < l+\varepsilon$$ Cauchy sequences are defined similarly and the field is said to be complete if all Cauchy sequences converge.

V. Semeria
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    So there is a partial order? And how it is related to the field operations? And what is the archimedean property? – GEdgar Mar 29 '19 at 13:02
  • Yes, it is not at all clear how your partial order would act with respect to multiplication. For a non-total ordering, the ordering of rationals is not unique with the standard definition. For example, I think you could have the odd integers and even integers incomparable. – tomasz Mar 29 '19 at 13:10
  • How do you define complete? – tomasz Mar 29 '19 at 14:00
  • @V.Semeria: if you are interested in the different characterizations of $\mathbb{R}$ (then you can decide whether there is a "simplest"), I recommend you to check 3.6 and 3.11 of https://arxiv.org/pdf/1101.5652.pdf, all the necessary definitions are included in the pdf. – Chilote Mar 29 '19 at 18:57
  • @tomasz completeness added to the question. But I think we do not need it for the relations between the order and the archimedean property. – V. Semeria Mar 29 '19 at 22:45
  • @chilote thanks. This paper assumes total orderings right from the start. Is there a way to prove that this totality is redundant and can be proven by the rest of the assumptions? – V. Semeria Mar 29 '19 at 22:47

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Not an answer

Here is some terminology from
Birkhoff, Garrett, Lattice theory, New York: American Mathematical Society (AMS). 3rd Edition (1967).

Def. p. 290
A p.o. group $G$ (written additively) is called Archimedean iff for all $a,b \in G$: $$ n a \le b\;\text{ for all } n \in \mathbb Z \quad\Longrightarrow\quad a=0 $$

Def. p. 397
A p.o. ring is a ring which is also a partially ordered set under a relation $\ge$ which satisfies $$ x \ge y \qquad \Longrightarrow \qquad a+x \ge a+y \tag{1}$$

$$ x \ge 0 \text{ and } y \ge 0 \qquad\Longrightarrow\qquad xy \ge 0 \tag{2}$$

Def. p. 398
A p.o. ring if called Archimedean when it is Archimedian with respect to its operation of addition.

GEdgar
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