Exercise 11, § 1

Let E be a fully lattice-ordered space, $(x_\iota)_{\iota \in I}$ a family of elements of E. For every finite subset H of I, set $s_H = \sum_{\iota \in H} x_\iota$; the family $(x_\iota)_{\iota \in I}$ is said to be summable for the order structure of E if the mapping $H \mapsto s_H$ has a limit for this order structure, with respect to the directed ordered set $\mathfrak{F}(I)$ of finite subsets of I; this limit s is then called the sum of the family $(x_\iota)_{\iota \in I}$ and is denoted $\sum_{\iota \in I} x_\iota$.

a) For a family $(x_\iota)_{\iota \in I}$ to be summable, it is necessary and sufficient that: 1° for every finite subset H of I, the set of $|s_K|$, where K runs over the set of finite subsets of I that do not intersect H, has a supremum $r_H$ in E; 2° $\inf_H r_H = 0$ as H runs over $\mathfrak{F}(I)$ (make use of Exer. 9 a)).

b) Generalize to summable families, for the order structure of E, the properties of summable families in commutative topological groups (GT, III, §5, Props. 2, 3 and Th. 2).

c) Let $(x_\iota)_{\iota \in I}$ be a family of elements $\geq 0$ of E; for the family to be summable, it is necessary and sufficient that the set of finite partial sums $s_H$ be bounded above (make use of Exer. 9 b)). If $(x_\iota)_{\iota \in I}$ is summable and if $(y_\iota)_{\iota \in I}$ is a family of elements such that $0 \leq y_\iota \leq x_\iota$ for all $\iota$, show that the family $(y_\iota)$ is summable and that $\sum_{\iota \in I} y_\iota \leq \sum_{\iota \in I} x_\iota$, with equality holding only if $x_\iota = y_\iota$ for all $\iota \in I$.

d) Let $(x_\iota)$ be a family of elements of E; show that if the family $(|x_\iota|)$ is summable for the order structure, then so is the family $(x_\iota)$.