Proposition 1

For an ordered vector space E to be fully lattice-ordered, it is necessary and sufficient that E be a Riesz space and that it satisfy one of the following two conditions:

a) every nonempty subset A, consisting of elements $\geq 0$ of E, that is bounded above and directed for the relation $\leq$, has a supremum in E;

b) every nonempty subset $A$, consisting of elements $\geqslant 0$ of $E$ and directed for the relation $\geqslant$, has an infimum in $E$.

The conditions are obviously necessary. Conversely, suppose that $E$ is a Riesz space satisfying the condition a). Let $B$ be a nonempty subset of $E$ that is bounded above; the set $C$ consisting of the suprema of the finite subsets of $B$ is directed for the relation $\leqslant$; let $a$ be one its elements and $C_a$ the set of $x \in C$ that are $\geqslant a$; if we prove that $C_a$ has a supremum then it will also be the supremum of $B$. Now, $C_a - a$ is a set of elements $\geqslant 0$, bounded above and directed for the relation $\leqslant$; it therefore has a supremum $b$, consequently $a + b$ is the supremum of $C_a$.

On the other hand, the condition b) implies a): for, if $F$ is a nonempty set of elements $\geqslant 0$ of $E$, bounded above and directed for $\leqslant$, and if $c$ is an upper bound for $F$, then $c - F$ is a set of elements $\geqslant 0$ that is directed for $\geqslant$; if it has an infimum $m$, then $c - m$ is the supremum of $F$.