Proposition 2

Let $(u_\lambda)_\lambda$ be a summable family of elements of $\mathbf{A}[[\mathbf{I}]]$. (i) The family $(1 + u_\lambda)_\lambda$ is multipliable. (ii) Let $\mathfrak{T}$ be the set of all finite subsets of $L$. For any $M \in \mathscr{S}$ put $u_M = \prod_{\lambda \in M} u_\lambda$. Then the family $(u_M)_{M \in \mathfrak{T}}$ is summable and we have

Let us define the ideals $\mathfrak{a}_\beta$ as at the beginning of this No., and let $\beta \in \mathbf{N}^{(I)}$. There exists a finite subset $L_0$ of $L$ such that $u_\lambda \in \mathfrak{a}_\beta$ for $\lambda \notin L_0$. Then for every $M \in \mathscr{S}$ such that $M \not\subset L_0$ we have $u_M \in \mathfrak{a}_\beta$. It follows that the family $(u_M)_{M \in \mathfrak{T}}$ is summable. On the other hand, for any finite subset $M_0$ of $L$ we have

Taken along the filtered ordered set $\mathfrak{T}$, the left-hand side has as limit $\sum_{M \in \mathfrak{T}} u_M$. Hence the right-hand side has as limit $\sum_{M \in \mathfrak{T}} u_M$, which proves (i) and (ii) at the same time.