If $\{a_i:i\in I\}$ is a set of nonnegative real numbers, then the unordered sum $\sum_{i\in I}a_i$ is defined as $\sup \Bigl\{ \sum_{i\in A}a_i\,\big| A \text{ finite, } A \subset I\Bigr\}$. And $\sum_{i\in I}a_i$ can only be finite if all but countably many of the $a_i$’s is zero.

My question is, can we make the same definition for a set of nonnegative hyperreal numbers? And in that case is it no longer true that $\sum_{i\in I}a_i$ can only be finite if all but countably many of the $a_i$’s is zero?

Intuitively the answer would seem to be yes, but I just want to make sure.

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    $\begingroup$ I don't think this will behave well at all. For example, taking any set $A$ - countable or not - of infinitesimals, any finite sum of elements of $A$ is still infinitesimal. And there's a whole separate issue around forming suprema in the hyperreals, too. $\endgroup$ – Noah Schweber Mar 31 at 16:33
  • $\begingroup$ What do you mean by "the same" definition? Literally the same? Or some kind of tranferred version where your sets are internal and you allow hyperfinite sums? $\endgroup$ – Eric Wofsey Mar 31 at 21:26
  • $\begingroup$ @EricWofsey I don’t really care about the definition being the same, I just want the notion being defined in some way or the other. $\endgroup$ – Keshav Srinivasan Mar 31 at 21:47
  • $\begingroup$ Well, what do you want to do with the definition? Different definitions will give totally different behavior. Your question taken literally is pointless, since of course you can make the same definition; you can make whatever definition you want. The question is whether the definition is useful for anything. $\endgroup$ – Eric Wofsey Mar 31 at 21:52
  • $\begingroup$ @EricWofsey Well it would be nice to have a definition that gave them the same results as the real number definition when the hyperreal numbers you’re adding happen to be real numbers, and which allows an uncountable sum of infinitesimals to be finite. $\endgroup$ – Keshav Srinivasan Apr 1 at 4:40

[Here I assume that by "hyperreals" you mean a countable ultrapower of the reals; if you use a different framework similar comments apply but some technical details may be different.]

The only well-behaved definition of this sort you can make is just the transferred version of the definition for real numbers. Namely, given an internal (indexed) set of nonnegative hyperreals, you can define its sum as the supremum of all sums of hyperfinite subsets. In terms of ultrapowers, this just means that the set you are taking the sum of is the ultraproduct of some sequence of sets of reals, and the sum of the ultraproduct is defined as the hyperreal obtained by taking the ordinary real sum of each set in the sequence.

By transfer, this definition will have all the same properties as sums of real numbers have when stated in the appropriate language. In particular, the sum can only exist if the set of nonzero terms is internally countable (i.e., there exists and internal injection to the hypernatural numbers; note that an infinite internal set is never actually countable).

It should be stressed that this notion of sum is very much not a means of summing arbitrary infinite sets of hyperreals but is instead a construction in the internal language of the hyperreals. For instance, this notion of summation cannot be applied to an ordinary countable sum of real numbers like $\sum_{n\in\mathbb{N}}1/2^n$; instead it can only be applied to the hyperreal analogue $\sum_{n\in{}^*\mathbb{N}}1/2^n$ (which will be equal to $2$ just like the real sum).

Alternatively, you could of course define the sum of a set of nonnegative hyperreals literally as the supremum of its finite partial sums, if this supremum exists. However, this definition is totally useless because it makes no use of the richer structure of the hyperreals beyond just being an ordered abelian group. Indeed, it turns out that the supremum will actually never exist if infinitely many terms are nonzero. (Proof sketch: if there is a countable cofinal set of partial sums, the supremum cannot exist by countable saturation. The only way a countable cofinal set of partial sums can fail to exist is if after removing finitely many terms, for all $i$ there exists $j$ such that $a_j>na_i$ for all $n\in\mathbb{N}$. But then the supremum cannot exist either, since if $b$ is any upper bound then so is $b/2$.)


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