Proposition 12

Let E be a totally ordered set and let X be a subset of E. For an element $b$ of E to be the least upper bound of X in E, it is necessary and sufficient that (1) $b$ is an upper bound of X, (2) for every $c \in \mathrm{E}$ such that $c < b$, there exists $x \in \mathrm{X}$ such that $c < x \leqslant b$.

The second condition says that no element $c < b$ is an upper bound of X, i.e., $b$ is a minimal element of the set M of upper bounds of X; but this is the same as saying that $b$ is the least element of M, since M is totally ordered (no. 10, Proposition 10).