So I've learned in my real analysis class that you can't have uncountably many positive summands "adding" to a finite sum. My question is if it's reasonable to try defining uncountably many summands (which are allowed to be negative) to get a finite sum.
In the spirit of doing things with uncountable sets, a first thought is to use $I = [0, 1)$ as an indexing set, letting $x_i = -1$ for $i \in [0, \frac{1}{2})$ and $x_i = 1$ for $i \in [\frac{1}{2}, 1)$. Is it reasonable to then say that $\sum_{i \in I} x_i = 0$?
More generally, is there a way to do this without running into all the problems normally encountered when one is too greedy with their measures (like the Banach-Tarski paradox)?
Best Answer
Well first we need a definition. If $X$ is any set, "the" definition of $\sum_{x\in X}a_x$ is this:
With that definition it's clear (and easy to prove) that if $\sum_{x\in X}a_x$ exists then $\{x:|a_x|>\epsilon\}$ is finite for every $\epsilon>0$, hence $\{x:a_x\ne0\}$ is countable. (And hence no, the example you give is not a convergent sum.)
Note With that definition, if $X$ is countable a convergent sum must be absolutely convergent. In fact it's easy to see that a sum converges in this sense if and only if the function $x\mapsto a_x$ is integrable with respect to counting measure...