I'm dreaming of a way to define an uncountable product of real numbers. Of course any sensible definition should only converge for a sequence with only finitely many terms outside $[0, 1]$. It should also be the case that a product of unaccountably many 1's is again 1, and for any $0\leq r<1$ the product of infinitely many $r$ should also be zero.
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Eric Wofsey
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Kristaps John Balodis
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1What does "converge" mean when we have an uncountable product? – vadim123 Mar 14 '18 at 04:28
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1Near-duplicate: https://math.stackexchange.com/questions/70194/does-uncountable-summation-with-a-finite-sum-ever-occur-in-mathematics – Eric Wofsey Mar 14 '18 at 05:50
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1Also https://math.stackexchange.com/questions/106102/use-of-sum-for-uncountable-indexing-set – Eric Wofsey Mar 14 '18 at 05:52
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I do not know of a good way to do this, although it would certainly be interesting if there was such a notion.
One thing that makes it tricky is that for countable products, you could construct a sequence of finite products giving a sequence. Then you can take the limit of the sequence. It seems difficult to generalize this to what you are trying to do here.
I can see where it would probably be the uncountable product of the values from a totally ordered set to the reals (like with countable products).
Also, perhaps to get a start, begin by considering the set of all countable products. Say for example, $f:\mathbb{R} \to \mathbb{R}$ and you are trying to take talk about $\displaystyle\prod_\mathbb{R} f(x)$, then you might want to consider the set $\{\displaystyle\prod_{i=1}^\infty f(x_n):\{x_n\} \text{strictly increasing}\}$, and work with something like that to figure out a definition.
Also, this artical on product integrals may be of interest to you as they seem to related to this topic.
Jonathan Dunay
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