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In the answer to this post Ittay Weiss wrote that, "...a sum of infinitely many positive elements can be bounded only if there are countably many elements." Though I asked about a rigorous proof of this fact and a reply was given, as can be easily seen, I still have some doubts which I need to clarify. I also didn't understand how Weiss's comment gives a proof of my question and so I am asking it here again.

  • It seems from the phrasing of the quote that there are other types of sum of positive elements for which the number of elements isn't countably infinite. What are some examples of sums of this type? To be precise, if sum of positive elements for which the number of elements isn't countably infinite then should it be necessarily uncountable?

  • If so, then how do we define a "sum" of uncountable number of positive quantities?

  • What is meant when we say that the "sum" is unbounded?

  • Why is a sum of infinitely many positive elements can be bounded only if there are countably many elements?

Any help will be appreciated.

Community
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    See here http://math.stackexchange.com/questions/724573/why-is-a-summable-family-at-most-countable – Kelenner Sep 09 '15 at 13:48
  • This has been answered before, but maybe I can clear up a few points of confusion. The "sum" of uncountably many positive real numbers is undefined precisely because for any $M > 0$ we can find a finite subset of them whose sum is greater than $M$. Indeed if you wish we can find a countable subset of the uncountable set whose "sum" is infinite in just this sense. That is what it means that the sum is unbounded. I'll find some previous Questions that discuss the technical details, but that should clear up much of the confusion. – hardmath Sep 09 '15 at 13:51
  • See also The sum of an uncountable number of positive numbers. – hardmath Sep 09 '15 at 13:55

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The definition of a sum of uncountable quantities is as follows: If $\{x_i : i \in I\}$ is a subset of a normed linear space, then we write $$ \mathcal{F} := \{F \subset I : F \text{ is finite}\} $$ Then for each $F\in \mathcal{F}$, define the partial sum to be $$ s_F := \sum_{i\in F} x_i $$ We say that $\{s_F: F \in \mathcal{F}\}$ converges to $s$ if for each open set $U$ containing $s, \exists F_0 \in \mathcal{F}$ such that $$ s_F \in U \quad\forall F\in \mathcal{F} \text{ such that } F_0 \subset F $$ If $\{s_F : F \in \mathcal{F}\}$ converges, then (as mentioned in the comment above), the set $$ \{i \in I : x_i \neq 0\} $$ must be countable.

Prahlad Vaidyanathan
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    How would this help a Reader (such as the OP) who does not know how to complete the circle of thoughts in your last line? Doesn't this merely repeat the claim referred to in the Question? Only the barest of hints is given in this Answer of why countably infinite and uncountably infinite sets (of positive "elements") should have different potentials for "sums". – hardmath Sep 09 '15 at 14:05