Proposition 4

Let E be an ordered set in which every well-ordered subset is bounded above; then E has a maximal element.

We shall say that an element $v \in \mathrm{E}$ is a strict upper bound of a subset X of E if $v$ is an upper bound of X and $v \notin \mathrm{X}$. Let $\mathfrak{S}$ be the set of subsets of E which have a strict upper bound in E, and for each $\mathrm{S} \in \mathfrak{S}$ put $p(\mathrm{S}) = \tau_v(v$ is a strict upper bound of S$)$; then $p(\mathrm{S})$ is a strict upper bound of S. Applying Lemma 3 of no. 3 to $\mathfrak{S}$ and $p$, we see that there exists a subset M of E and a well-ordering $\Gamma$ on M which satisfies the conditions of Lemma 3; in particular, M has no strict upper bound in E. Furthermore, the ordering $\Gamma$ is identical with that induced on M by the ordering on E. For in M the relation “$y\in\Gamma\langle x\rangle$ and $x\ne y$” is equivalent to $x\in S_y$; and since $p(S_y)=y$ is an upper bound of $S_y$ (with respect to the ordering on E), it implies $x<y$ in E. But this means that the injection of M into E is a strictly increasing mapping (M being endowed with the ordering $\Gamma$); and since M is totally ordered, it follows that the relations $y\in\Gamma\langle x\rangle$ and $x\leq y$ are equivalent in M (§ 1, no. 12, Proposition 11). This being so, there exists by hypothesis an upper bound $m$ of M in E; but since M has no strict upper bound, it follows that $m$ is a maximal element of E.