Proposition 8

Let G be a totally ordered group not reduced to O. The following conditions are equivalent: (a) G is of height 1; (b) for all x > 0 and y ≥ 0 in G, there exists an integer n ≥ 0 such that y ≤ nx; (c) G is isomorphic to a subgroup of the ordered additive group \mathbf{R} which is not reduced to O.

Let x be a positive element of G and let H, be the set of y ∈ G such that there exists an integer n ≥ 0 satisfying |y| ≤ nx. It is easily verified that H, is an isolated subgroup of G and that every isolated subgroup of G containing x contains H,. Condition (a) is therefore equivalent to “H, = G for all x > 0”, that is, to condition (b).

Clearly (c) implies (b). Conversely, suppose condition (b) holds and let Q denote the set of elements >0 of G. Suppose first that Q has a least element x; for all y ∈ Q, let n be the least integer such that y ≤ nx; if y < nx, then also nx − y ≥ x, whence y ≤ (n − 1)x contrary to the choice of n; then y = nx, which shows that G = \mathbf{Z}x is isomorphic to \mathbf{Z} ⊂ \mathbf{R}. Suppose now that Q has no least element; we apply to the ordered set P = Q ∪ (0) Proposition 1 of General Topology, Chapter V, § 2 (which is possible, since condition (b) is just “Archimedes’ axiom”); it is seen that there exists a strictly increasing mapping f of P to \mathbf{R}_+ such that

for x ∈ P and y ∈ P; by linearity f can be extended to an isomorphism of G onto a subgroup of \mathbf{R}, which proves that (b) implies (c).