Can we add an uncountable number of positive elements, and can this sum be finite?

I always have trouble understanding mathematical operations when dealing with an uncountable number of elements. Any help would be great.

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    $\begingroup$ Somewhat related: Back in Leibniz's time, the concept of an integral was first conceived as a sum of infinitely many infinitesimal quantities. The idea of infinities being countable or uncountable had not yet been invented then, but if it had, an integral would have been described as a sum of uncountably many infinitesimals, one for each possible value of the integration variable. It later turned out that it is difficult to make this intuition rigorous, but I suspect non-standard analysis might have some way to do it. $\endgroup$ – hmakholm left over Monica Aug 29 '15 at 19:06
  • $\begingroup$ You can find several older related posts. For example here (and in the linked questions) or here (an in the linked questions).. $\endgroup$ – Martin Sleziak Sep 2 '15 at 12:30
  • $\begingroup$ In particular, this post shows that an uncountable sum cannot be finite if all summands are non-zero: The sum of an uncountable number of positive numbers $\endgroup$ – Martin Sleziak Sep 2 '15 at 12:33
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    $\begingroup$ @Henning Makholm integral is a sum of COUNTABLE number of infinitelimals. $\endgroup$ – Anixx Jul 17 '17 at 14:32
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    $\begingroup$ @Anixx: No -- the sum of countably many infinitesimals would itself be infinitesimal (each partial sum is less than every $\frac1n$, so their limit would also be). $\endgroup$ – hmakholm left over Monica Jul 17 '17 at 14:47

Suppose $\{s_i : i\in\mathcal I\}$ is a family of positive numbers.$^\dagger$ We can define $$ \sum_{i\in\mathcal I} s_i = \sup\left\{ \sum_{i\in\mathcal I_0} s_i : \mathcal I_0 \subseteq \mathcal I\ \&\ \mathcal I_0 \text{ is finite.} \right\} $$ (If both positive and negative numbers are involved, then we have to talk about a limit rather than about a supremum, and then the definition is more complicated and we have questions of conditional convergence and rearrangements.)

Now consider \begin{align} & \{i\in\mathcal I : s_i \ge 1\} \\[4pt] & \{i\in\mathcal I : 1/2 \le s_i < 1 \} \\[4pt] & \{i\in\mathcal I : 1/3 \le s_i < 1/2 \} \\[4pt] & \{i\in\mathcal I : 1/4 \le s_i < 1/3 \} \\[4pt] & \quad \quad \quad \vdots \end{align} If one of these sets is infinite, then $\sum_{i\in\mathcal I} s_i=\infty$. But if all are finite, then $\mathcal I$ is at most countably infinite.

Thus the sum of uncountably many positive numbers is infinite.

I don't know whether by some arguments about rearrangements one could somehow have some sensible definition of a sum of numbers not all having the same sign that could give us a somehow well defined sum of uncountably many numbers and get a finite number.

$^\dagger$ In the initial edition of this answer, I said "Let $S$ be a set of positive numbers and then went on to say $$ \sum S = \left\{ \sum S_0 : S_0\subseteq S\ \&\ S_0\text{ is finite.} \right\} $$ However, Dustan Levenstein pointed out in comments that "this definition fails to allow for the same number to occur twice in a sum". Rather than "twice", I'd say "more than once", since a number might even occur an uncountably infinite number of times.

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    $\begingroup$ I don't think such definition would be actually useful for anything because already the conditionally convergent countable sums are pathological and absolutely aren't good generalization of finite addition. $\endgroup$ – Blazej Aug 29 '15 at 18:06
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    $\begingroup$ @AyushKhaitan Blazej's comment is referring to the case of including negative numbers. Michael Hardy's definition is perfectly sensible for the case of summing up a system of only positive numbers. $\endgroup$ – Dustan Levenstein Aug 29 '15 at 18:22
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    $\begingroup$ Although, tiny quibble: this definition fails to allow for the same number to occur twice in a sum. $\endgroup$ – Dustan Levenstein Aug 29 '15 at 18:22
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    $\begingroup$ You misread Michael's statement. He said if any of those intersections is infinite, then the sum is automatically infinite. If they're all finite, then there are only countably many (positive) elements of $S$. $\endgroup$ – Dustan Levenstein Aug 29 '15 at 18:27
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    $\begingroup$ I've drastrically edited the answer to allow for some terms to occur more than once. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 29 '15 at 18:39

We have the next proposition

Proposition 1. Let $X$ be an at most countable set, and let $f\colon X\to\mathbf R$ be a function. Then the series $\sum_{x\in X} f(x)$ is absolutely convergent if and only if $$\sup\left\{\sum_{x\in A}|f(x)|:A\subseteq X, A\text{ finite}\right\}<\infty.$$

Inspired by this proposition, we may now define the concept of an absolutely convergent series even when the set $X$ could be uncountable.

Definition 2. Let $X$ be a set (which could be uncountable), and let $f\colon X\to\mathbf R$ be a function. We say that the series $\sum_{x\in X} f(x)$ is absolutely convergent if and only if $$\sup\left\{\sum_{x\in A}|f(x)|:A\subseteq X, A\text{ finite}\right\}<\infty.$$

Note that we have not yet said what the series $\sum_{x\in X} f(x)$ is equal to. This shall be accomplished by the following proposition.

Proposition 3. Let $X$ be a set (which could be uncountable), and let $f\colon X\to\mathbf R$ be a function such that the series $\sum_{x\in X} f(x)$ is absolutely convergent. Then the set $\{x\in X:f(x)\ne0\}$ is at most countable.

Because of this, we can define the value of $\sum_{x\in X} f(x)$ for any absolutely convergent series on an uncountable set $X$ by the formula $$\sum_{x\in X} f(x):=\sum_{x\in X:f(x)\ne0} f(x),$$ since we have replaced a sum on an uncountable set $X$ by a sum on the countable set $\{x\in X:f(x)\ne0\}$. (Note that if the former sum is absolutely convergent, then the latter one is also.) Note also that this definition is consistent with the definitions for series on countable sets.

Remark. The definition of series on countable sets that are use is

Definition 4. Let $X$ be a countable set, and let $f\colon X\to\mathbf R$ be a function. We say that the series $\sum_{x\in X}f(x)$ is absolutely convergent iff for some bijection $g\colon\mathbf N\to X$, the sum $\sum_{n=0}^\infty f(g(n))$ is absolutely convergent. We then define the sum of $\sum_{x\in X}f(x)$ by the formula $$\sum_{x\in X}f(x)=\sum_{n=0}^\infty f(g(n)).$$

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Let $H$ be a positive unlimited integer of nonstandard analysis. Then, for example, the sum

$$ \sum_{n=1}^H n = \frac{H(H+1)}{2}$$

is a sum of uncountably many positive numbers... but it's a hyperfinite nonstandard sum, so it exists by the usual methods of nonstandard analysis. The sum is unlimited, though. Other sums can be finite: e.g.

$$ \sum_{n=1}^H \frac{1}{n!}$$

is a finite nonstandard real number that is infinitesimally close to $e$.

That said, IMO, thinking of hyperfinite sums from nonstandard analysis as being sums of uncountably many elements isn't a particularly fruitful line of thought. (also, the sum only works for internal sequences of elements anyways; you can't take an arbitrary uncountable collection)

I bring this up mainly to show that uncountable sums can make sense in some contexts, even if you can't really do much in a standard setting. Each summation operator one might define can have its own sorts of pecularities.

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Here is the more general definition of the sum of any function $f:I\rightarrow E$, where $I=$any non empty set and $E$ is a real or complex topological vector space.

Definition 1 We say that the family $f=(f(i))_{i\in I}$ is summable in $E$ of sum $\omega\in E$ if and only if, for all neighborhood $V$ of $\omega$, there exists a finite subset $A\subset I$ of $I$ such that for all finite parts $K$ of $I$ containing $A$, we have $\sum_{i\in L}f(i)-\omega\in V$. f is said to be absolutely summable if and only if $\|f\|=(\|f(i)\|)_{i\in I}$ is summable in $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$.

Remark 2 This definition does not guarantee the uniqueness of the sum of a given family. The topological vector space $E$ mus be separable for the uniqueness to hold, in which case we write $$\sum_{i\in I}f(i)=\omega.$$

Proposition 3 If $E$ is a normed space, then a family $f=(f(i))_{i\in I}$ is summable of sum $\omega$ if and only if for any $0<\varepsilon\in \mathbb{R}$ there exists a finite part $A$ of $I$ such that for all finite part $K$ of $I$ with $A\subset K$, we have: $$\|\sum_{i\in K}f(i)-\omega\|<\varepsilon.$$

Remark 4 If $f=(f(i))_{i\in I}$ is a family in $E$, we can definie $\mathcal{F}(I)=\{A\subset I:~A~\text{is finite}\}$. Embedding $\mathcal{F}(I)$ with the direction ``$\leq$'' defined by $A\leq B$ in $\mathcal{F}$ iff $A\subset B$, we make $\mathcal{F}(I)$ a directed set (or Riesz space). The if we definite $S:\mathcal{F}(I)\rightarrow E$, by $$S(K)=\sum_{i\in K} f(i),$$ we get a net in $E$. We have the following:

Proposition 5 $f$ is summable if and only if the set $S(K)$, defined above, is convergent.

Notice that here we have defined the sum of any kind of number, countable or uncountable. But it is proved that in case $E=\mathbb{F}$ and $I=\mathbb{N}$, the summability coincide with the absolute summability, which in turns coincide with the commutative convergence of the series $\sum_{n=1}^{+\infty}f(n)$, and the value of the sum of the family is the same as the value of this series.

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  • $\begingroup$ This is a very elaborate Answer but it is hard to spot the true conclusion to be drawn: No, it is not possible for an uncountable set of positive "elements" to have a finite sum. In particular generalizing to complex numbers seems to obscure the facts; complex numbers are not ordered as positive and not positive. $\endgroup$ – hardmath Jun 3 '17 at 15:23
  • $\begingroup$ Yes, you are right. It is not possible. Because for an absolutely summable family in a normed sapce $E$, the $\{i\in I: f(i)\neq 0\}$ is at most countable. $\endgroup$ – Tighana Jan 6 '18 at 7:01

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