Proposition 1

Let E be a linearly ordered set with a smallest element ω; let I be a subset of E such that ω ∈ I and such that the relations x ∈ I, y ≤ x imply y ∈ I; let (x, y) → xy be a law of composition on E, defined for x ∈ I and y ∈ I. Then, if the axioms (GR_I), (GR_{II}), (GR_{III}) and (GR_{IV}) are satisfied, there exists a strictly increasing mapping f of I into the set R_+ of real numbers ≥ 0, such that

whenever x ∈ I, y ∈ I and xy ∈ I; moreover, the intersection of f(I) with every interval [0, f(b)] of R is dense in this interval, where b denotes any element of I.

Given any two elements x, y of I such that y ≠ ω, let us denote by (x : y) the largest integer n ≥ 0 such that y^n is defined and ≤ x (); this integer exists by (GR{IV}); if (x : y) = p, then y^{p+1} is defined and > x. If x ∈ I, y ∈ I and xy ∈ I, we have

(I)

For let (x : z) = p, (y : z) = q; then we have z^p ≤ x, z^q ≤ y; since xy ∈ I, z^p z^q is defined and belongs to I, therefore z^{p+q} is defined and z^{p+q} = z^p z^q ≤ xy; moreover, if z^{p+q+2} is defined, we have z^{p+q+2} > xy because z^{p+1} > x and z^{q+1} > y.

Next we establish the inequalities

(2)

Let (x : y) = p and (y : z) = q; then y^p ≤ x and z^q ≤ y, so that (z^q)^p is defined and ≤ x; it belongs therefore to I; consequently z^{pq} is defined and we have z^{pq} = (z^q)^p ≤ x, from which the first inequality follows. On the other hand, if z^{(p+1)(q+1)} is defined, we have z^{(p+1)(q+1)} > x, because y^{p+1} > x and z^{q+1} > y; hence the second inequality.

Let $\tilde{\mathscr{X}}$ denote the filter of sections of the ordered set of elements > ω in I, with respect to the relation ≥; the intervals ]ω, z[, where z runs through the set of all elements > ω, form a base of $\tilde{\mathscr{X}}$. Given two elements a and x of I such that a > ω, we shall show that the ratio $\frac{(x : z)}{(a : z)}$, which is defined for z ≤ a and is a rational number > 0,

(*) When E = I is the set of natural integers, the law of composition being addition, (x : y) is the integral part of x/y.

is a function of $z$ which has a limit with respect to $\mathfrak{F}$. This is obvious if $x = \omega$, for then $(x : z) = 0$ for all $z$. If $x > \omega$, we shall show that the image $\mathscr{G}$ of $\mathfrak{F}$ under the mapping $z \to \frac{(x : z)}{(a : z)}$ (restricted to the set of those $z > \omega$ which are $\leq x$ and $\leq a$) is a Cauchy filter base for the uniform structure of the multiplicative group $\mathbf{R}_+^*$, and therefore converges to a real number $> 0$. Note first that, $u > \omega$ being given, $(u : z)$ has limit $+\infty$ with respect to $\mathfrak{F}$: for there exists $z > \omega$ such that $z^{2^n} \leq u$, so that $(u : z) \geq 2^n > n$. Now take a number $\varepsilon > 0$ arbitrarily; there exists $t > \omega$ such that $(x : t) \geq 1 / \varepsilon$ and $(a : t) \geq 1 / \varepsilon$. Consider the double inequality

which follows immediately from the inequalities (2). There exists $z_0 > \omega$ such that $z \leq z_0$ implies $(t : z) \geq 1 / \varepsilon$, so that

which shows that $\mathscr{G}$ is a Cauchy filter base for the multiplicative uniformity.

Fix once and for all the element $a > \omega$ (the "unit of measure") and for each $x \in I$ put

From what has already been proved, we have $f(\omega) = 0$, $f(x) > 0$ for $x > \omega$, and $f(a) = 1$. If we divide the inequality (1) throughout by $(a : z)$ and pass to the limit with respect to $\mathfrak{F}$, we see that

whenever $x \in I, y \in I$ and $xy \in I$. Likewise, the relation $x \leq y$ implies $(x : z) \leq (y : z)$, whence by dividing by $(a : z)$ and passing to the limit we have $f(x) \leq f(y)$, so that $f$ is increasing on $I$. We deduce that $f$ is strictly increasing on $I$; for if $x < y$, there exists $z > \omega$ such that $xz \leq y$, whence $f(xz) \leq f(y)$; and since $xz \in I$,

but $f(z) > 0$, so that indeed $f(x) < f(y)$.

Finally, if $b \in I$, the intersection of $f(I)$ and the interval $[0, f(b)]$ of $\mathbf{R}$ is dense in this interval. For if $n$ is any integer $> 0$, there exists $x > \omega$ such that $f(x) \leq 2^{-n}$ (take $x$ such that $x^{2^n} \leq a$); if $p$ is the smallest integer such that $x^{p+1} > b$, we have $(p+1)f(x) > f(b)$ and $qf(x) \leq f(b)$ for $1 \leq q \leq p$; therefore every interval contained in $[0, f(b)]$ and of length $> 2^{-n}$ contains at least one point of the form $qf(x) = f(x^q) \in f(I)$. The proof of Proposition 1 is therefore complete.