Exercise 2, Appendix 2

In a totally ordered set I, a nonempty subset J will be called a segment if, whatever the elements $\alpha, \beta$ of J such that $\alpha < \beta$ may be, one has $(\alpha, \beta) \subset J$.

(a) Let $(x_i)_{i \in I}$ be a multipliable family in a complete group $G$. Now prove that, for every segment $J$ of $I$, the family $(x_i)_{i \in J}$ is multipliable (first consider the particular case where the set of lower bounds of $J$, not belonging to $J$, is empty).

(b) A partition $(J_\lambda)_{\lambda \in L}$ of $I$ is said to be an ordered partition if $L$ is a totally ordered set, and if the relations $\lambda < \mu, \alpha \in J_\lambda, \beta \in J_\mu$ imply $\alpha < \beta$; the $J_\lambda$ are then segments of $I$. If $(x_i)_{i \in I}$ is a multipliable family in a complete group $G$, now prove that, if one sets $p_\lambda = \prod_{i \in J_\lambda} x_i$, the family $(p_\lambda)_{\lambda \in L}$ is multipliable and has the same product as $(x_i)_{i \in I}$.

(c) Conversely, if $(J_\lambda)_{\lambda \in L}$ is a finite ordered partition of $I$, and if $(x_i)_{i \in I}$ is such that each of the families $(x_i)_{i \in J_\lambda}$ is multipliable, the family $(x_i)_{i \in I}$ is multipliable.