Summation of families over arbitrary index sets

Remarks

Equivalently, the net $(s_J)$, indexed by the finite subsets of $I$ ordered by inclusion, converges to $s$. The sum is unique when it exists. The same definition works in any Hausdorff abelian topological group, with neighbourhoods of $s$ in place of balls; see summation in topological groups. If all $x_i$ are nonnegative reals, the family is summable if and only if the finite sums $s_J$ are bounded, and then $s = \sup_J s_J$. Example: the family $(2^{-m} 3^{-n})_{(m, n) \in \N \times \N}$ is summable with sum $3$. Its finite sums are bounded by $\sum_{m \le M} 2^{-m} \sum_{n \le M} 3^{-n} \le 2 \cdot \tfrac32$, and the squares $\set{0, \ldots, M}^2$ give sums tending to $3$. Non-example: the family $((-1)^n / (n+1))_{n \in \N}$ is not summable, although the series $\sum (-1)^n/(n+1)$ converges. Any finite set $J_0$ can be enlarged by many even indices $n$, which adds more than $1$ to $s_J$, because $\sum 1/(2k+1)$ diverges. In a Banach space, the family is summable if and only if for every $\varepsilon > 0$ there is a finite $J_0$ with $\norm{s_K} < \varepsilon$ for every finite $K$ disjoint from $J_0$. A summable family in a normed space has at most countably many nonzero terms.