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)?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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:

• "every hilbert space" isomorphic "l2" Google Books, Google, Google Scholar
• "every hilbert space" isomorphic cardinality Google Books, Google, Google Scholar
• "every hilbert space" "linearly isometric" Google Books, Google, Google Scholar

See also: Is every Hilbert space an $L^2$ space?