Proposition 6

Let E be an ordered set. Then the following statements are equivalent :

(a) Every non-empty subset of E has a maximal element. (b) Every increasing sequence $(x_n)$ of elements of E is stationary.

We show first that (a) implies (b). Let X be the set of elements of the sequence $(x_n)$, and let $x_m$ be a maximal element of X. If $n \geqslant m$, we have by hypothesis $x_n \geqslant x_m$, and therefore $x_n = x_m$ by the maximality of $x_m$. Conversely, suppose that there exists a non-empty subset A of E which has no maximal element. For each $x \in A$, let $T_x$ be the set of all $y \in A$ such that $y > x$. By assumption, $T_x \neq \emptyset$ for all $x \in A$; hence there exists a mapping $f$ of A into A such that $f(x) > x$ for all $x \in A$ (Chapter II, § 5, no. 4, Proposition 6). If $a \in A$, then the sequence $(x_n)_{n \in \mathbf{N}}$ defined inductively by the conditions $x_0 = a$, $x_{n+1} = f(x_n)$ is evidently increasing and not stationary.