I was wondering whether it makes sense to use the $\sum $ notation for uncountable indexing sets. For example it seems to me it would not make sense to say

$$ \sum_{a \in A} a \quad \text{where A is some uncountable indexing set e.g. some $A \subset \mathbb{R}$ } $$

Would it be better to avoid the above notation in general for uncountable indexing sets ? Any help in making better sense of this would be very appreciated.

  • 5
    $\begingroup$ Sometimes one takes sums over uncountable indexing sets, assuming that only countably many terms are nonzero (implicitly assuming that the sum of zeros over any set is zero). Also, sometimes one wishes to consider "formal sums". In other words, it depends on the context, as long as you define what you mean by your notation in the given setting. $\endgroup$
    – user2093
    Feb 5, 2012 at 21:24
  • $\begingroup$ See also here: Does uncountable summation, with a finite sum, ever occur in mathematics?. $\endgroup$ Jul 3, 2012 at 16:08

5 Answers 5


It makes sense when it is given a precise definition that makes sense. If $I$ indexes a set of elements of a Hausdorff topological abelian group, then $\sum\limits_{i\in I}a_i$ can be used to denote the limit of the net of finite sums, with the finite subsets of $I$ directed by inclusion, when this limit exists.

In particular, if each $a_i$ is a nonnegative real number, then $\sum\limits_{i\in I}a_i$ exists as an element of $\mathbb R$ if and only if $\sup\left\{\sum\limits_{i\in F}a_i:F\text{ is a finite subset of }A\right\}<\infty$, and in that case $\sum\limits_{i\in I}a_i$ equals that supremum. Finiteness of this sum implies that $\{i\in I:a_i\neq 0\}$ is countable.

  • $\begingroup$ Why would it only make sense for Hausdorff topological group? As far as I know the construction via nets is for any plain index set. $\endgroup$ Nov 2, 2015 at 3:37
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    $\begingroup$ Freeze_S: I didn't say only, but a topology allows limits to be defined, a Hausdorff topology allows limits to be unique, an abelian group (or semigroup) structure allows the finite sums to make sense. Together these don't imply topological group, that was perhaps overly restrictive, but still general enough to get started on such an open ended question. If you want to describe something else, you can post your answer. $\endgroup$ Nov 2, 2015 at 6:50
  • $\begingroup$ Ah my missreading: "If $I$ indexing a set ..." instead of "If $I$ an index set ...". Anyway nice answer!! :) $\endgroup$ Nov 2, 2015 at 10:54
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    $\begingroup$ @Freeze_S: I didn't write either of those things, but I don't know what different meanings you attribute to them. Other than the incorrect grammar, they both seem the same as what I said. $\endgroup$ Nov 2, 2015 at 13:55

You can define the sum of the elements of an infinite set $S \subseteq \mathbb{R}_{>0}$ by

$$ \sum_{s \in S} s = \sup \left \{ \sum_{s \in F} s\, :\, F \subseteq S \text{ finite} \right \} $$

However, the sum of uncountably many strictly positive reals is always infinite, so this often isn't too useful. However the definition still coincides with the countable case.

When negative numbers are included it becomes impossible to define what you mean by the sum. With sums of countably many reals the idea is that you add one element at a time in a particular order, so that sums like $\displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$ are well-defined despite the fact that by taking the elements of the sequence $-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, \dots$ in a different order you can obtain any real number. For an uncountable set this is harder to make sense of.

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    $\begingroup$ You can allow negative numbers too, just not in general. Effectively, you are taking an integral with respect to counting measure. A function can be integrable and have both positive and negative values. $\endgroup$ Feb 5, 2012 at 21:42
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    $\begingroup$ It is not true that in every uncountable well ordered set there is an element between any two elements, so I am not sure what you mean there. When negative numbers are included the sum $\sum_{i\in I}a_i$ can be defined to be the limit of the net $\left\{\sum_{i\in F}a_i\right\}_{F\subset I,F\text{ finite}}$ indexed by the finite subsets of the index set. (Convergence ends up being equivalent to absolute convergence, so your example of the alternating harmonic series wouldn't converge in this sense.) $\endgroup$ Feb 5, 2012 at 21:46
  • $\begingroup$ Thanks, I should have noticed that; post edited accordingly. $\endgroup$ Feb 5, 2012 at 23:54

For sums of real numbers, I have always thought of all sums as integrals with respect to a suitable counting measure (which is somewhat limiting, I know, since I am handling everything in the Lebesgue sense, which messes up conditional convergence). At the very least, that interpretation reduces to something most graduate students have seen a well-developed theory for.

As an aside, it is not uncommon to see sums over possibly uncoutnable index sets with interpretation in the above sense. One of the classical examples (you can find it in Rudin's Real and Complex Analysis) arises in the study of Fourier series over a Hilbert Space and if you would like an example in both the statement and proof of the Parseval identity. Consider $\mathcal{l}^2 (\mathbb{R})$ (that is, the Lebesgue space of $\mathbb{R}$ equipped with the counting measure). Appealing to Zorn's Lemma we know this space has a maximal Hilbert Space basis (uncountable) $e_\alpha$ and that for any $x \in \mathcal{l}^2 (\mathbb{R})$ we have $||x||^2$ = $\sum_\alpha |\langle e_\alpha, x\rangle|^2$


Suppose $I$ is an index set, and let $n_i$ be a real number for all $i$ in $I$. If $n_i$ is almost always zero, i.e., there exists a finite subset $S\subset I$ such that $n_i =0$ for all $I\backslash S$, then I see no harm in using the notation $\sum_{i\in I} n_i$.

In general, it's not well-defined.

  • $\begingroup$ By the way, "almost always" may also be interpreted in a somewhat standard way to mean "outside a measure zero set." Also, why not take $S$ countable? There is a well-defined concept of (conditional) convergence for countable sums. $\endgroup$
    – user2093
    Feb 5, 2012 at 21:27

There's no problem with summing over an uncountable indexing set, but over the reals you don't really get anything interesting by moving from the countable case to the uncountable case.

Suppose we have such an uncountable summation of nonnegative real numbers: $\sum_{r \in \Gamma} x_r$. Then we have a few cases:

(i) One of the $x_r = \infty$. Then our sum, $\sum_{r \in \Gamma} x_r$, is also $\infty$.

(ii) Suppose none of the $x_r =\infty$. Now we utilize dyadic decomposition to write the positive reals as a countable union of sets: $(0, \infty] = \cup_{j \in \mathbb{Z}} (2^j, 2^{j+1}]$. By the pigeonhole principle, either (a) There is a $j$ such that there are uncountably many nonzero $x_r$ in $(2^j, 2^{j+1}]$, in which case our sum is $\infty$ again or (b) there are only countably many nonzero $x_r$ in each interval for all $j$. In the latter case, we're back to our countable sum, as a countable union of countable sets is still countable.

Now, if our indexing set $\Gamma$ is essentially countable in this way, we can motivate our sum as usual by defining $\sum_{r \in \Gamma} x_r = \sup_{E \subset \Gamma} \sum_{r \in E} x_r$, where $E$ is a finite set. So an uncountable sum can be defined in the same way, without really picking up many problems (or interesting cases, sadly).


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