Proposition 2

If a linearly ordered set $E$ and a subset $I$ of $E$ satisfy the axioms (GRI), (GR_{II}), (GR_{IIIa}) and (GR_{IVa}), there exists a strictly increasing mapping $f$ of $I$ onto an interval of $\mathbf{R}$, with $0$ as its left-hand end-point, such that $f(\omega) = 0$ and $f(xy) = f(x) + f(y)$ whenever $x, y$ and $xy$ belong to $I$.*

Let us first show that axiom (GR_{IV}) is satisfied. We argue by contradiction: suppose that there exist $x, y \in I$ such that $x > \omega$, $x^n$ is defined and $x^n \leq y$ for all integers $n > 0$. The increasing sequence $(x^n)$ has a least upper bound $b \in I$ by (GR$_{\text{IV}_a}$). Since $x < b$, there exists $c \in I$ such that $xc = b$ by (GR$_{\text{III}_a}$), and we have $c < b$ since $x > \omega$. Now, for every $n$, we have $x^{n+1} \leq b = xc$, whence $x^n \leq c$ by (GR$_{\text{II}}$) : the upper bound $b$ of the $x^n$ is therefore $\leq c$, which is a contradiction.

We are therefore in a position to apply Proposition 1. It remains to show that, if $\gamma = f(c)$ ($c > \omega$) is any element of $f(I)$, and if $\beta$ is any real number such that $0 < \beta < \gamma$, there exists $b \in I$ such that $f(b) = \beta$ (Chapter IV, § 2, no. 4, Proposition 1). Since the intersection of $f(I)$ and $[0, \gamma]$ is dense in $[0, \gamma]$, there exists an increasing sequence $(x_n)$ of elements of $I$ such that $f(x_n)$ has $\beta$ as limit. Let $b$ be the least upper bound of the sequence $(x_n)$ in $I$; we have $f(b) \geq f(x_n)$ for all $n$, hence $f(b) \geq \beta$; but $f(b) > \beta$ is impossible, otherwise there would exist $y \in I$ such that $\beta < f(y) < f(b)$, and since $\beta$ is the least upper bound of the sequence $(f(x_n))$, we should have $f(x_n) < f(y) < f(b)$ for all $n$, whence $x_n < y < b$ for all $n$, which is absurd. Hence $f(b) = \beta$, and Proposition 2 is therefore proved.