Proposition 1

In a well-ordered set E, every segment of E other than E itself is an interval $]\leftarrow, a[$, where $a \in \mathrm{E}$.

Let S be a segment of E such that $\mathrm{S} \neq \mathrm{E}$. Since $\mathrm{E} - \mathrm{S}$ is not empty, it has a least element $a$. By virtue of Definition 2, the relation $x \geqslant a$ implies $x \notin \mathrm{S}$; otherwise we would have $a \in \mathrm{S}$, which is absurd. Hence $\mathrm{E} - \mathrm{S}$ is the interval $[a, \rightarrow[$, and S is the interval $]\leftarrow, a[$.

¶ For every element $x$ in a totally ordered set E the segment $]\leftarrow, x[$ is called the segment with endpoint x, and is denoted by $\mathrm{S}_x$.

Note that if E is well-ordered and not empty, $\mathrm{S}_x$ has a least element $\alpha$ and consequently is also the interval $[\alpha, x[$.

Let E be a totally ordered set. The union A of the $\mathrm{S}_x$, as $x$ runs through E, is E if E has no greatest element; and if E has a greatest element $b$, we have $\mathrm{A} = \mathrm{E} - \{b\}$.