use of $\sum $ for uncountable indexing set

Solution 1:

It makes sense when it is given a precise definition that makes sense. If $I$ indexes a set of elements of a Hausdorff topological abelian group, then $\sum\limits_{i\in I}a_i$ can be used to denote the limit of the net of finite sums, with the finite subsets of $I$ directed by inclusion, when this limit exists.

In particular, if each $a_i$ is a nonnegative real number, then $\sum\limits_{i\in I}a_i$ exists as an element of $\mathbb R$ if and only if $\sup\left\{\sum\limits_{i\in F}a_i:F\text{ is a finite subset of }A\right\}<\infty$, and in that case $\sum\limits_{i\in I}a_i$ equals that supremum. Finiteness of this sum implies that $\{i\in I:a_i\neq 0\}$ is countable.

Solution 2:

You can define the sum of the elements of an infinite set $S \subseteq \mathbb{R}_{>0}$ by

$$ \sum_{s \in S} s = \sup \left \{ \sum_{s \in F} s\, :\, F \subseteq S \text{ finite} \right \} $$

However, the sum of uncountably many strictly positive reals is always infinite, so this often isn't too useful. However the definition still coincides with the countable case.

When negative numbers are included it becomes impossible to define what you mean by the sum. With sums of countably many reals the idea is that you add one element at a time in a particular order, so that sums like $\displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$ are well-defined despite the fact that by taking the elements of the sequence $-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, \dots$ in a different order you can obtain any real number. For an uncountable set this is harder to make sense of.