Proposition 8

— *Let P be the set of positive elements of an ordered group G. For G to be lattice ordered, it is necessary and sufficient that $G = P - P$, and that in addition P, equipped with the induced ordering, satisfies one or other of the following conditions:

a) Each pair of elements of P has a supremum in P. b) Each pair of elements of P has an infimum in P.

The necessity of these conditions is obvious : indeed the relation $G = P - P$ says that G is filtered (VI, p. 4, Prop. 4); on the other hand the supremum and infimum in G of two elements of P are positive, so are also their supremum and infimum in P.

Conversely, note first that under hypothesis a) (resp. b)), every pair of elements x, y of P has a supremum (resp. infimum) in G equal to its supremum a (resp. infimum b) in P. This is obvious for a, since any upper bound for x and y is positive; for b, let $z \in G$ be a lower bound for x and y; then there exists $u \in P$ such that $z + u \in P$, since $G = P - P$; now $\inf_P(x + u, y + u)$ is greater than $b + u$, and so is of the form $b + c + u$ ($c \geq 0$); since $b + c$ is less than x and less than y, we have $c = 0$; hence $\inf_P(x + u, y + u) = b + u$, which implies $z + u \leq b + u$, thus $z \leq b$ and b is certainly the infimum of x and y in G. Now if x and y are arbitrary elements of G we translate them into P: let $v \in P$ be such that $x + v$ and $y + v$ are positive (VI, p. 5, Prop. 5); under hypothesis a) (resp. b)) there exists a supremum (resp. infimum) for $x + v$ and $y + v$ in $P$, and hence also in $G$ by what we have just seen; by translation $x$ and $y$ have a supremum (resp. infimum) in $G$; the existence of one of these implies the other, by Prop. 7, and this shows that the conditions are sufficient.