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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)

3-1-4-One-Five
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    "unsetly" isn't a word. You mean a set (or possibly a sequence) of arbitrarily large cardinality. The only agreed definition I know is for sets of non-negative reals and takes the sum to be the supremum over finite sums. This will invariably be infinite unless the set contains at most countably many non-zero reals. It does give $0$ for a set of all $0$s.. – Rob Arthan Oct 19 '23 at 20:13
  • @RobArthan "unsetly" reminds me of the German word "Unmenge" which means "vast amount". It is composed of the prefix "un" which is also used in English and "Menge" which means "set". – Kritiker der Elche Oct 19 '23 at 22:50
  • To add to someone's answer, you may find the discussion here interesting - basically, there's no need to worry about proper-class-sized lists of real numbers because you can't sensibly go further than uncountable anyway. Your proposed definition seems to assume all the numbers are nonnegative! Is that intended? If so, the linked post shows that your construction always gives $\infty$ when uncountably many entries are nonzero, and otherwise it's just the normal countable sum. – Izaak van Dongen Oct 19 '23 at 23:23

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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.

someone
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