Theorem 2 (Associativity of the sum)

Let $(x_i)_{i \in I}$ be a summable family in a complete group G, and let $(I_\lambda)_{\lambda \in L}$ be any partition of I. If $s_\lambda$ denotes $\sum_{i \in I_\lambda} x_i$, then the family $(s_\lambda)_{\lambda \in L}$ is summable and has the same sum as the family $(x_i)_{i \in I}$.

Thus if we have a summable family in a complete group, we can associate its terms arbitrarily into subfamilies and form the sum of each subfamily thus obtained; the family of these partial sums is again summable and its sum is equal to that of the given family.

Let $s = \sum_{i \in I} x_i$, and let $V$ be any closed neighbourhood of 0 in $G$. Then there is a finite subset $J_0$ of $I$ such that, for each finite subset $J$ of $I$ which contains $J_0$, we have $\sum_{i \in J} x_i \in s + V$. Let $K_0$ be the subset of $L$ consisting of indices $\lambda$ such that $J_\lambda = I_\lambda \cap J_0$ is not empty: $K_0$ is clearly finite. Let $K$ be any finite subset of $L$ which contains $K_0$; we shall show that $\sum_{\lambda \in K} s_\lambda \in s + V$, which will establish the theorem.

Now $s_\lambda$ is very close to a finite partial sum of $(x_i)$, whose indices all belong to $I_\lambda$; more precisely, given any symmetric neighbourhood $W$ of 0, there exists for each $\lambda \in K$ a finite subset $H_\lambda$ of $I_\lambda$, containing $J_\lambda$ and such that $s_\lambda - \sum_{i \in H_\lambda} x_i \in W$. Let $J = \sum_{\lambda \in K} H_\lambda$; then $J$ is a finite subset of $I$ containing $J_0$, and we have

by the associativity of finite sums. Hence by reason of the choice of $J_0$ and the $H_\lambda$, we have

where $n$ is the number of elements in $K$; this relation holds for each $W$, hence also $\sum_{\lambda \in K} s_\lambda \in s + V$, since $V$ (being closed) is the intersection of the neighbourhoods $V + nW$ [§ 3, no. 1, formula (1)].

We can thus write the associativity formula for sums:

(I)

which is valid whenever the family $(I_\lambda)$ is a partition of its union and the right-hand side is defined. In particular, if the index set is a product $I = L \times M$, and if the "double" family $(x_{\lambda \mu})_{(\lambda, \mu) \in L \times M}$ is summable, we have the formula of change of order of summation

(2)

It should be remarked that the left-hand side of (1) can have a meaning, without the right-hand side being defined. Consider for example the case where $I = LX \{1, 2\}$ and $L$ is infinite, and $I_\lambda$ consists of the two elements $(\lambda, 1)$ and $(\lambda, 2)$; if we take $x_{\lambda, 1} = a,\ x_{\lambda, 2} = -a$, where $a$ is any non-zero element of $G$, then all the partial sums corresponding to the $I_\lambda$ are zero, and therefore the left-hand side of (1) is defined and equal to 0, whereas the right-hand side of (1) has no meaning, as Proposition 1 shows.

Likewise, it can happen that the left-hand side of (2) is not defined but that each of the two "double sums" in (2) has a meaning; and the elements of $G$ which they represent need not be equal (see Chapter IV, § 7, Exercise 17).

Thus although it is always possible to "associate" the terms of a sum, it is not possible, on the other hand, to "dissociate" into their elements those terms of a sum which appear themselves as sums. Nevertheless, this operation is legitimate whenever the number of these "dissociable" terms is finite.