Remark

Let E be a totally ordered set. The set of open intervals of E (bounded or not) is stable under finite intersection. It is a basis of a topology $\mathscr{T}_0(E)$ on E (TG, I, p. 91, exerc. 5). The topology $\mathscr{T}_0(\mathbf{I})$ is identical with the topology induced on $\mathbf{I}$ by that of $\mathbf{R}$. It follows that, in prop. 16, every isomorphism of ordered sets of $\mathbf{I}$ onto E is a homeomorphism of $\mathbf{I}$ onto the topological space obtained by endowing E with the topology $\mathscr{T}_0(E)$.