Lemma 1

Let $(\mathrm{X}_\alpha)_{\alpha \in \mathrm{A}}$ be a family of ordered sets, directed with respect to the relation $\subset$ (in other words, such that for each pair of indices $\alpha$, $\beta$ there exists an index $\gamma$ such that $\mathrm{X}_\alpha \subset \mathrm{X}_\gamma$ and $\mathrm{X}_\beta \subset \mathrm{X}_\gamma$). Suppose that, for each pair of indices $(\alpha, \beta)$ such that $\mathrm{X}_\alpha \subset \mathrm{X}_\beta$, the ordering induced on $\mathrm{X}_\alpha$ by that of $\mathrm{X}_\beta$ is identical with the given ordering on $\mathrm{X}_\alpha$. Under these conditions there exists a unique ordering on the set $\mathrm{E} = \bigcup_{\alpha \in \mathrm{A}} \mathrm{X}_\alpha$ which induces the given ordering on each $\mathrm{X}_\alpha$.

Let $\mathrm{G}_\alpha$ be the graph of the given ordering on $\mathrm{X}_\alpha$. If G is the graph of an ordering on E which induces on each $\mathrm{X}_\alpha$ the ordering whose graph is $\mathrm{G}_\alpha$, then we must have $\mathrm{G}_\alpha \subset \mathrm{G}$ for each $\alpha \in \mathrm{A}$; hence G contains $\bigcup_{\alpha \in \mathrm{A}} \mathrm{G}_\alpha$. On the other hand, for each pair $(x, y)$ of elements of E there exists by hypothesis an index $\alpha \in \mathrm{A}$ such that $x \in \mathrm{X}_\alpha$ and $y \in \mathrm{X}_\alpha$; if $(x, y) \in \mathrm{G}$, we have $(x, y) \in \mathrm{G}_\alpha$, so that $\mathrm{G} \subset \bigcup_{\alpha \in \mathrm{A}} \mathrm{G}_\alpha$. Hence if the required ordering on E exists, its graph is necessarily $\mathrm{G} = \bigcup_{\alpha \in \mathrm{A}} \mathrm{G}_\alpha$. It remains to be shown that this set satisfies the conditions of the lemma. Since $\mathrm{G}_\beta \cap (\mathrm{X}_\alpha \times \mathrm{X}_\alpha) = \mathrm{G}_\alpha$ if $\mathrm{X}_\alpha \subset \mathrm{X}_\beta$, we have $\mathrm{G} \cap (\mathrm{X}_\alpha \times \mathrm{X}_\alpha) = \mathrm{G}_\alpha$ for all $\alpha \in \mathrm{A}$; on the other hand, it follows from the hypothesis that any three elements $x$, $y$, $z$ of E belong to the same $\mathrm{X}_\alpha$. Hence $(x, y) \in \mathrm{G}$ is an order relation on E, and the lemma is proved.

¶ We now take up the proof of Proposition 3. Let us begin by showing that each $\mathrm{X}_\iota$ is a segment of E. Indeed, if $x \in \mathrm{X}_\iota$, $y \in \mathrm{E}$, and $y \leqslant x$, there exists an index $\varkappa$ such that $\mathrm{X}_\iota \subset \mathrm{X}_\varkappa$ and $y \in \mathrm{X}_\varkappa$; since by hypothesis $\mathrm{X}_\iota$ is a segment of $\mathrm{X}_\varkappa$, we have $y \in \mathrm{X}_\iota$, which proves the assertion. The same reasoning proves that for each $x \in \mathrm{X}_\iota$ the segment with endpoint $x$ in $\mathrm{X}_\iota$ is identical with the interval $]\leftarrow, x[$ in E. Next let us show that E is well-ordered. If H is a non-empty subset of E, there is an index $\iota \in \mathrm{I}$ such that $\mathrm{H} \cap \mathrm{X}_\iota \neq \emptyset$; if $a$ is the least element of $\mathrm{H} \cap \mathrm{X}_\iota$ in $\mathrm{X}_\iota$, then $a$ is also the least element of H in E. That is, if $x \in \mathrm{H}$, there exists an index $\varkappa \in \mathrm{I}$ such that $\mathrm{X}_\iota \subset \mathrm{X}_\varkappa$ and $x \in \mathrm{X}_\varkappa$; we cannot have $x < a$, because the interval $]\leftarrow, a[$ is contained in $\mathrm{X}_\iota$, and therefore we have $x \geqslant a$ since $\mathrm{X}_\varkappa$ is totally ordered.

¶ Finally, we must show that a segment of E, other than E itself, is a segment of one of the $\mathrm{X}_\iota$; this is an immediate consequence of the preceding arguments, since such a segment is of the form $]\leftarrow, x[$ (Proposition 1) and since $x$ belongs to some $\mathrm{X}_\iota$.