For example, what is the sum of $[0, 0, 0, 0, …]$ with as many 0's as ordinal numbers?
Is it 0, because I am adding up 0's and nothing else?
Here is my way of adding up unsetly many numbers:
Take the supremum of the sums of all initial segments of the list (for uncountable sums, take the supremum of all countable subsets)
Is that right? If so, is there another way? If not, is adding up unsetly many real numbers even well-defined?
Edit: By 'unsetly' I mean too many to fit in any set, meaning that there are too many numbers to fit in any set (There are $\mathbf{Ord}$ many 0's in said $[0, 0, 0, 0, …]$ list, and $\mathbf{Ord}$ is the proper class of ordinal numbers)
Best Answer
You can do transfinite induction on the proper class of all ordinals (the terminology 'unsetly' should really be replaced by 'proper class'). https://en.wikipedia.org/wiki/Transfinite_induction
Specifically, if you have a map $ORD \to \mathbb{R}$ which assigns $0$ to each ordinal then it follows by transfinite induction that the 'sum' up to every ordinal is $0$ and therefore the only sensible answer to 'what is the sum over all ordinals?' is the supremum of the value at each ordinal, namely $0$.
I suspect you could make sense of this for nonzero values as long as there only countably many ordinals at which the value is nonzero and the sum of the nonzero terms converges in the usual countable sense. If you allow uncountably many nonzero values and assume every value is positive then the only sensible answer is infinite. If you allow negative and positive values for some sort of conditional convergence then I suspect it's not possible to formalize this.