Theorem 1 (F. Riesz)

— Let $A$ be a subset of a fully lattice-ordered space $E$. The set $A'$ of elements that are alien to every element of $A$ is a band; the band $A''$ of elements that are alien to every element of $A'$ is identical to the band generated by $A$, and $E$ is the ordered direct sum of the bands $A'$ and $A''$.

The properties of alien elements, reviewed in No. 1, and the definition of a band, show at once that $A'$ is a band, hence so is $A''$. By Proposition 5 and the definition of a band, every element $x \geqslant 0$ of $E$ may be written $x = y + z$, with $y \in A'$ and $z \in A''$, $y$ and $z$ being $\geqslant 0$; since every element of $E$ is the difference of two elements $\geqslant 0$, we have $E = A' + A''$; on the other hand, since $0$ is the only element alien to itself, we have $A' \cap A'' = \{0\}$, which proves that $E$ is the direct sum of $A'$ and $A''$; finally, since the components in $A'$ and $A''$ of an element $\geqslant 0$ of $E$ are $\geqslant 0$, $E$ is the ordered direct sum of $A'$ and $A''$ (No. 4, Prop. 4).

It remains to show that $A''$ is identical to the band $B$ generated by $A$. Now, $E$ is the direct sum of $B$ and the band $B'$ formed by the elements alien to all the elements of $B$; since $A \subset B$, we have $B' \subset A'$; on the other hand $B \subset A''$ and $E$ is also the direct sum of $A'$ and $A''$; therefore necessarily $B = A''$, $B' = A'$.

Theorem 1 and Proposition 5 make it possible to give another definition of the band generated by a set of elements of $E$: