I read here:



The Archimedean axiom guarantees the existence of a metric of line segments m(U) as a function defined on a set of all line segments with the specific characteristics properties (it is positive, additive, monotonic).

Is this true (and would it be true more generally, e.g. if we were not dealing with line segments but rather real numbers)? Is there a more reliable reference for this statement (in a book or published article)?


To begin with, if you want to define arithmetic on line segments, you do not initially need the Archimedean axiom; you do need the parallel postulate to be able to multiply line segments. (To construct a line segment of length $ab$ from line segments of lengths $1$, $a$, and $b$, we use similar triangles, and triangle similarity theorems run on the parallel postulate.) For a complete development of this idea, see for example Chapter 4 (Segment Arithmetic) of Hartshorne's Geometry: Euclid and Beyond.

Without further assumptions, segment lengths (that is, equivalence classes of congruent line segments) are the positive elements of a uniquely-determined ordered field $F$. So for some notion of "metric", we do not need the Archimedean axiom to have a positive, additive, monotone metric. (But usually we want distances to be real numbers, not just elements of a field!)

The Archimedean axiom in our geometry translates to the Archimedean property of $F$. That is, for any $x,y \in F$ with $0 < x < y$, there is a natural number $n$ such that $nx > y$. (The natural numbers are contained in $F$: they are the elements $0, 1, 1+1, 1+1+1, \dots$. Because these all have inverses, the rational numbers are also contained in $F$.)

In particular, Archimedean ordered fields are all subfields of the real numbers. See here for a complete proof. The idea is that an ordered field $F$ contains the rational numbers as a subfield, and to any $x \in F$ we may assign a set $\rho(x) = \{q \in \mathbb Q : q < x\}$. If $F$ is Archimedean, then $\rho(x)$ is a Dedekind cut, so $\rho$ is a map $F \to \mathbb R$. We can then check that it's a (necesarily injective) field homomorphism, making $F$ isomorphic to its image under $\rho$, a subfield of $\mathbb R$.

So when your source says "the Archimedean axiom guarantees the existence of a metric $m(U)$", what that means is that it guarantees the existence of a positive real number length assigned to any line segment. Without the Archimedean axiom, the field $F$ could be some really weird object.

Finally, the Cantor axiom guarantees that $F$ is not just a subfield of the real numbers but includes all of $\mathbb R$.

  • $\begingroup$ Do you really mean 'the natural numbers are all elements of F'? Perhaps you mean they are the elements $0_F ,e, e+e, e+e+e ....$ such that $e$ is mapped to 1 by the metric, and $0_F$ is mapped to 0 by the metric. $\endgroup$ – IIM Mar 15 '18 at 15:57
  • $\begingroup$ No, the metric doesn't map $F$ to anywhere: the metric maps line segments to elements of $F$. $\endgroup$ – Misha Lavrov Mar 15 '18 at 16:00
  • $\begingroup$ Oh do you mean all the natural numbers are elements of F? Because the way you put it I thought you meant that F consists purely of natural numbers. $\endgroup$ – IIM Mar 15 '18 at 16:04
  • 1
    $\begingroup$ The metric is on line segments, but before the Archimedean axiom is used, it is only a "metric": it is not real-valued but $F$-valued. Only after we prove that $F$ is a subfield of $\mathbb R$ do we realize that we've actually assigned real numbers to every line segment. (And yes, I just meant that $F$ contains $\mathbb N$, not that $F = \mathbb N$; I'll correct that.) $\endgroup$ – Misha Lavrov Mar 15 '18 at 16:12

You can find a construction of a measure function based on Archimedes axiom in neutral geometry (without the parallel postulate) on page 228 of Franz Rothe book here: http://math2.uncc.edu/~frothe/3181all.pdf.

If one assume the parallel postulate and not the Archimedes axiom, it is possible to define an pythagorean ordered field from the geometric axioms, then one can define the length of a segment using the usual formula based on coordinates: $length(AB) = \sqrt{(x_B-x_A)^2+(y_B-y_A)^2} $.

  • $\begingroup$ In your second paragraph, how are you defining the coordinates of a point? $\endgroup$ – Misha Lavrov Mar 18 '18 at 11:20
  • $\begingroup$ @MishaLavrov First your build an isosceles right triangle IOJ, and project the point P on OI and OJ to get two points X and Y coordinates. Then you copy the distance OX and OY on a given line which is used to perform the arithmetic operations using geometric constructions. $\endgroup$ – Julien Narboux Mar 19 '18 at 12:34

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