Suppose you have infinitely many different singleton sets. If you took the infinite cartesian product of all your singleton sets, would the result be one giant ordered pair? Would it matter whether you had countably or uncountably infinite many singletons?
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1The answer depends on how you define "infinite cartesian product". – Ethan Bolker Jul 11 '18 at 21:40
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@Holo The OP asks "cartesian product of all *your singleton sets" [emphasis mine]. It's not a cartesian product of the entire class of singleton sets, it's the cartesian product of a particular set of singleton sets. – Acccumulation Jul 11 '18 at 21:47
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@Acccumulation , I see, noted, but this link is still relevant, OP, you should read that definition – ℋolo Jul 11 '18 at 21:51
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@EthanBolker are there different ways to define an infinite cartesian product? – hydrangea Jul 11 '18 at 22:01
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@J.P.Escarcega: If you know one definition, why not reveal what it is instead of playing games? – hmakholm left over Monica Jul 11 '18 at 22:22
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@HenningMakholm: It may be genuine surprise, rather than game-playing. – Cameron Buie Jul 11 '18 at 22:41
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@CameronBuie: Granted, I can think of only one too. But my suspicion is that the OP knows (or knew) no definition of infinite cartesian product. At least, if they know one where speaking about "one giant ordered pair" makes sense, this definition would be news to me and I'd be interesting in hearing it ... – hmakholm left over Monica Jul 11 '18 at 22:52
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@HenningMakholm That's why I commented as I did. The single answer (so far) uses the standard definition. – Ethan Bolker Jul 12 '18 at 01:23
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The result would be a set of one element: a fixed function, essentially. If the singletons are $\{a_i\}$ for $i \in I$, $I$ being some index set, then:
$$\prod_{i \in I} \{a_i\} = \{f\}$$ where $f: I \to \{a_i : i \in I\}$ is defined by $f(i) = a_i$.
So one element regardless of the size of $I$ (countable, finite, uncountable..)
Henno Brandsma
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