Proposition 2

The set E\* of segments of a well-ordered set E is well-ordered by inclusion. The mapping $x \rightarrow \mathrm{S}_x$ is an isomorphism of the well-ordered set E onto the set of segments of E other than E itself.

It is clear that if $x \in \mathrm{E}$ and $y \in \mathrm{E}$, the relation $x \leqslant y$ implies $\mathrm{S}_x \subset \mathrm{S}_y$, and that $x < y$ implies $\mathrm{S}_x \neq \mathrm{S}_y$; the mapping $x \rightarrow \mathrm{S}_x$ is therefore an isomorphism of E onto the set S(E) of segments of E distinct from E itself (§ 1, no. 12, Proposition 11), and consequently S(E) is well-ordered. Moreover, E\* is isomorphic to the well-ordered set obtained from S(E) by adjoining a greatest element.