Proposition 6

— Let $E$ be a fully lattice-ordered space, $M$ a subset of $E$, and $B$ the band generated by $M$. Let $M_1$ be the set of elements $\geqslant 0$ of $E$ each of which is $\leqslant$ some element of the form $\sum_i |x_i|$, where $x_i \in M$; let $M_2$ be the set of suprema of those subsets of $M_1$ that are bounded above; then the set $M_2$ is identical with the set of elements $\geqslant 0$ of $B$.

Clearly $M_2 \subset B$ by the definition of a band; on the other hand, if $B'$ is the band of elements that are alien to every element of $M_1$, Theorem 1 shows that $E$ is the ordered direct sum of $B$ and $B'$. But Proposition 5 shows that every element $\geqslant 0$ of $E$ is the sum of an element of $M_2$ and an element of $B'$, whence the proposition.