Proposition 16

Let E be a totally ordered set not reduced to one element. Suppose that every subset of E has a supremum and that there exists a countable subset of E which meets every open interval $]x, y[$, where $x$ and $y$ are elements of E such that $x < y$. There then exists an isomorphism of ordered sets of $\mathbf{I}$ onto E.

Since $\emptyset$ and E each have a supremum in E, E has a smallest element $a$ and a greatest element $b$. These are distinct since E is not reduced to one element. Let $D'$ be a countable subset of E which meets every open interval of E of the form $]x, y[$, with $x < y$. Put $D = D'\cup \{a, b\}$. The set D is totally ordered and without gaps. By lemma 6, there exists an isomorphism of ordered sets $f$ of $\mathbf{I}\cap \mathbf{Q}$ onto D.

For every $t\in \mathbf{I}$, let $g(t)$ be the supremum of $f([0, t]\cap \mathbf{Q})$ in E. For every $x\in E$, let $h(x)$ be the supremum of $f^{-1}([a, x]\cap D)$ in $\mathbf{I}$. The mappings $g:\mathbf{I}\rightarrow E$ and $h: E\rightarrow \mathbf{I}$ thus defined are increasing, $g$ coincides with $f$ in $\mathbf{I}\cap \mathbf{Q}$ and $h$ coincides with $f^{-1}$ in D.

We therefore have $g(h(y)) =y$ for every $y\in D$. Let $x\in E$. If one had $g(h(x))> x$, the interval $]x, g(h(x))[$ would contain a point $y$ of D and the relations $g(h(y)) =y < g(h(x))$ would contradict the fact that $g\circ h$ is increasing. Analogously, one does not have $g(h(x))< x$. We therefore have $g(h(x)) =x$, which proves that $g\circ h$ is the identity mapping of E. One proves analogously that $h\circ g$ is the identity mapping of $\mathbf{I}$. Thus, $g:\mathbf{I}\rightarrow E$ and $h: E\rightarrow \mathbf{I}$ are reciprocal isomorphisms of ordered sets.