# Does uncountable summation, with a finite sum, ever occur in mathematics?

Obviously, “most” of the terms must cancel out with opposite algebraic sign.

You can contrive examples such as the sum of the members of R being 0, but does an uncountable sum, with a finite sum, ever occur naturally as part of a larger scenario (eg, proof of a theorem, or preparation for a definition)?

• There is no natural order in which to sum such a series, so algebraic cancellation can't really help. In other words, for a conditionally convergent (countable) series, you can rearrange the order of summation to get any value you want, sp that for an uncountable summation, the sum wouldn't be well-defined if it only converged in some kind of conditional sense. Robjohn's answer below shows that it can never absolutely converge. – Cheerful Parsnip Oct 5 '11 at 22:18
• I just noticed (looking at the related questions) that this is pretty close to Transfinite series: Uncountable sums. – robjohn Oct 5 '11 at 22:52
• @Jim: Good point. I tried to address the cancellation issue when I reread the question and noticed the part about cancellation. When I mentioned needing an order for conditional convergence, I was thinking about the rearrangement of conditionally convergent series, but it would have been better if I had mentioned it explicitly. – robjohn Oct 5 '11 at 22:56
• Here is another related question: Use of $\sum$ for uncountable indexing set – Martin Sleziak Jul 2 '15 at 8:38
• Possible duplicate of use of $\sum$ for uncountable indexing set – Santropedro Mar 15 '17 at 2:19

If $I$ is a set and $f:I\to\mathbb R$ is a function, there is a standard definition of what it means that $f$ be summable.

One compact way of giving it is as follows. Let $\mathscr P_{\mathrm{fin}}(I)$ be the set of all finite subsets of $I$, and let $\bar f:\mathscr P_{\mathrm{fin}}(I)\to\mathbb R$ be function such that $$\bar f(A)=\sum_{a\in A}f(a)$$ for each $A\in\mathscr P_{\mathrm{fin}}(I)$. The set $\mathscr P_{\mathrm{fin}}(I)$ is partially ordered by inclusion, and with this order it is directed, so we can view $\bar f:\mathscr P_{\mathrm{fin}}(I)\to\mathbb R$ as a net in $\mathbb R$.

We say that the function $f$ is summable if the net $\bar f$ converges, and the limit of the net is by definition the sum of $f$. This notion extends absolute summability of series; I am not aware of any sensible meaning of conditional summability of such a family -- so I don't think one deals at all with "cancellation".

Now, since $\mathbb R$ is a topological space with a denumerable basis, one can easily check that

if a function $f:I\to\mathbb R$ is summable according to this definition, then the set $\{i\in I:f(i)\neq0\}$ is countable.

Only if all but a countable number of terms are $0$. Consider the number of terms bigger than $1$, there can only be a finite number. Then consider the number of terms bigger than $1/2$, there can only be a finite number. Continuing, there can only be a finite number of terms bigger than $1/k$ for each k. Thus, since any term greater than $0$ must be greater than $1/k$ for some $k$, there can only be a countable number (a countable union of finite sets) of terms greater than $0$.

More: I forgot to mention about cancellation. You can only have cancellation, as in conditional convergence, if you can order the terms, which you can't do with an uncountable number of terms. Thus, it only makes sense to work with absoluely convergent, uncountable sums.

• And an absolutely convergent (possibly uncountable) sum is a Lebesgue integral over the counting measure. With this identification, in concrete situations it may make sense to use something like Cauchy principal values to assign values to conditionally convergent sums. – hmakholm left over Monica Oct 5 '11 at 22:25
• @Henning: do you have an idea of how to define something like Cauchy principal value on a discrete, uncountable topology? – robjohn Oct 5 '11 at 22:48
• This answer seems to assume the terms are real numbers or something quite similar. – Michael Hardy Oct 5 '11 at 23:26
• @Michael: The OP did specify R, which I assumed was $\mathbb{R}$. Perhaps I was misreading that part. Is there an example using another field where uncountable sums of non-zero terms make more sense? – robjohn Oct 5 '11 at 23:45
• @robjohn: I'm imagining a situation where the summands are indexed by a set with some extra structure that can be used here. For example if, the index set is $\mathbb R$, such that we have a function $f:\mathbb R\to\mathbb R$ with countable support, it is not per se meaningless to consider whether $\lim_{n\to\infty} \sum\{f(x)\mid -n<x<n\}$ exists. (Of course this limit will be irrelevant unless the structure of the index set has some meaningful relation to the process that generated the summands). – hmakholm left over Monica Oct 6 '11 at 0:17

One application of uncountable sums (or, to be more precise, sums along arbitrary index set) I am aware of is the definition of the Hilbert space $\ell_2(A)$.

A very basic example of a Hilbert space is the space $\ell_2=\ell_2(\mathbb N)$. The elements of this space are sequences such that $\sum\limits_{i\in\mathbb N} x(i)^2<\infty$. It is endowed with the inner product is given by $\langle x,y \rangle =\sum\limits_{i\in\mathbb N} x(i)y(i)$.

If we allow summation over arbitrary sets, then we can define $\ell_2(A)$ using almost the same construction; in this case, we take all functions $x\colon A\to\mathbb R$ such that $$\sum_{i\in A} x(i)^2 < \infty$$ and the inner product will be $$\langle x,y \rangle = \sum\limits_{i\in A} x(i)y(i).$$

It can be shown that this is indeed a Hilbert space and that every Hilbert space $X$ is isomorphic to $\ell_2(A)$ for some set $A$. Cardinality of $A$ is precisely the "Hilbert dimension", i.e. cardinality of orthonormal basis for $X$. This result is, in some sense, a classification of all Hilbert spaces.

These results can be found, for example, in:

• Chapter 13 Roman's Advanced Linear Algebra;
• Chapter IX of Dixmier's General Topology;
• Chapter II of Retherford's Hilbert space;
• Corollary 1.4.19 in Tao's Epsilon of Room...

And there are probably many other places to look. Just google for some reasonable phrases, for example:

• the problem is that the expression $\sum_{i\in A} x(i)^2 < \infty$ is still meaningless, m mean, how you knows if such a sum is less than infinity? – Masacroso May 24 '19 at 14:32
• @Masacroso Using the usual definitions - which can be found in the linked posts. However, for non-negative numbers the situation becomes easier - so $\sum\limits_{i\in A} x(i)^2<+\infty$ is equivalent to $\sup\{\sum\limits_{i\in F} x(i)^2; F\subseteq A, F\text{ is finite ]\}<+\infty$. (In fact, some texts take this as a definition - typically in cases where they don't need to deal with negative values.) – Martin Sleziak May 24 '19 at 14:52
• The formula in the previous comment which is not rendered (due to my mistake) is supposed to say that it is equivalent to $\sup\{\sum\limits_{i\in F} x(i)^2; F\subseteq A, F\text{ is finite}\}<+\infty$. – Martin Sleziak May 28 '19 at 13:17