Proposition 11

Every strictly monotone mapping $f$ of a totally ordered set E into an ordered set F is injective. If $f$ is strictly increasing, $f$ is an isomorphism of E onto $f(\mathrm{E})$.

For $x \neq y$ implies that either $x < y$ or $x > y$; hence

so that $f(x) \neq f(y)$ in either case. It remains to be shown that if $f$ is strictly increasing, $f(x) \leqslant f(y)$ implies $x \leqslant y$; if not, we should have $x > y$, and therefore $f(x) > f(y)$.