So a lurking thought in the back of my mind for a month or so has been the notion of a sum over an uncountable set, say

$$\sum_{a \in I} a$$

for some $I$ whose cardinality is greater than $\aleph_0$. This came up as a sort of "crossing of wires" because, in my coursework, I've often seen sums over countable sets (think of geometric series, power series, Riemann sums, Laurent series, Fourier series, and so on and so forth), and in my abstract algebra class this semester we introduced the notion of a direct (internal and external) product - which is instead called the direct sum if the groups involved are abelian - that generalizes the Cartesian product. These products/sums can be defined over sets which may not be finite, possibly even uncountable.

So in a twisted sense, there's one way to sum over uncountable sets, even if it's not a sum in the arithmetic sense. But it made me wonder - could we do sums over uncountable sense in normal arithmetic? This question actually popped up a few times on MSE (here and here, just as a couple of examples) and those answers were to my satisfaction (this being my favorite).

But then a further question is begged...

Countable and Uncountable Products:

We have the notion of products over countable sets; for example, one formulation of the Riemann zeta function:

$$\zeta(s) = \prod_{p \; \text{prime}} \frac{1}{1-p^{-s}}$$

But can we define a product over an uncountable set, some $I$ with cardinality greater than $\aleph_0$? What would that definition even be?

I wasn't able to find much on MSE to this effect, nor elsewhere via Google (usually just stuff pertaining to the algebraic structure stuff I noted earlier). A previous question did come up regarding the product analogue of integration (the product integral), but that only takes care of sets with cardinality the continuum, to my understanding - I'm trying to get something even more general than that.

On thinking on the matter, there are some immediate observations, a few I found interestingly similar to the algebraic structure version. Let $I$ be a set - it may be finite or infinite, countable or uncountable, just any set. We consider the product:

$$\prod_{a\in I} a$$

We can consider a few cases, as so...

  • Suppose $I$ is finite. Then this is the usual iterated product we're all familiar with. Nothing special there.

  • Suppose there exists $a\in I$ such that $a=0$. Then $\prod \limits_{a\in I} = 0$.

  • Suppose $I$ consists only of $1$'s, except for finitely many $a \in I$. Define the finite set of $a\neq 1$ by $I_0$. Then I believe $$\prod_{a \in I} a = \prod_{a \in I_0}a$$ which basically returns us to the case of $I$ finite.

  • Suppose, for all $a \in I$, $0<|a|<1$. Then it feels like, logically, the product would be zero again.

  • If $I$ is infinite, no $a\in I$ is $0$, and there are an uncountable number of elements $a$ with $|a|>1$, then the product diverges.

The first I felt interestingly reminiscent of how the weak direct product of groups is the regular product when the index set is finite, and the third felt reminiscent of the definition of weak product. Likely coincidental but it amused me.

That said, I can't really think of much else that would be particularly noteworthy. I don't really feel like I have the mathematical background, either, to justify the above observations - they're just something I intuitively feel is correct. They also don't constitute a particularly concrete definition for a product.

Any ideas? Are the above true, and if not, why? How precisely do we define this uncountable product?

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    $\begingroup$ There’s nothing particularly interesting to say. Too many zeros or negative signs and the product is either zero or infinity or doesn’t obviously make sense; otherwise, take the logarithm, and then you’re left with a sum again. $\endgroup$ – Qiaochu Yuan Mar 10 '19 at 2:54
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    $\begingroup$ Also, note that there are infinite products (will, at least countable) with elements $0<|a|<1$ which converge to a positive number. $\endgroup$ – enedil Mar 10 '19 at 2:59

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