Proposition 5

— Let $E$ be a fully lattice-ordered space, $A$ a non-empty subset of $E$ consisting of elements $\geqslant 0$, such that: 1) $A + A \subset A$, and 2) the relations $x \in A,\ 0 \leqslant y \leqslant x$ imply $y \in A$. Let $M$ be the set of suprema in $E$ of those subsets of $A$ that are bounded above in $E$. Under these conditions, every element $x \geqslant 0$ of $E$ may be written in the form $y + z$, where $y \in M$ is the supremum of the elements $v \in A$ such that $v \leqslant x$, and where $z$ is an element $\geqslant 0$ that is alien to every element of $M$.

At any rate $y \leqslant x$, so it all comes down to showing that $z = x - y$ is alien to every element $t \in A$ (No. 1), in other words that $u = \inf(z, t)$ is zero. By hypothesis, $u \in A$ and $u \leqslant x - y$, thus $u + y \leqslant x$; for every $v \in A$ such that $v \leqslant x$, by definition $v \leqslant y$, thus $u + v \leqslant u + y \leqslant x$; since $u + v \in A$ by hypothesis, also $u + v \leqslant y$ by the definition of $y$; finally, since $u + y$ is the supremum in $E$ of the elements $u + v$ such that $v \in A$ and $v \leqslant x$, we have $u + y \leqslant y$, whence $u \leqslant 0$, which completes the proof.