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This question already exists, but its answer was not helpful to me.

If the elements of a set are of the form $((a,b),(c,d))$ (the elements are pairs of pairs), and the like, how does one denote that? A set of $n$-tuples consisting of numbers all ranging across the exact same set $S$ can be denoted as $S^n$. But what to do if it consists of $x$-tuples of $y$-tuples of $z$-tuples of... etc.?

user110391
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  • Well, $S^a$ is the set of all $a$-tuples of the set $S$. And $(S^a)^b$ is the set of all $b$-tuples of the set $S^a$. And so on. – Lee Mosher Jun 12 '23 at 00:36
  • @LeeMosher That's what I thought too (if you remember my recent, now-replaced post), but $(S^a)^b = S^{ab}$? Thus, it is just the set of all $ab$-tuples? – user110391 Jun 12 '23 at 00:44
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    Your $a$'s, $b$'s and $c$'s are all mixed up. You should use your variables more carefully. – Lee Mosher Jun 12 '23 at 00:49
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    And, you should say what set each variable takes values in. – Lee Mosher Jun 12 '23 at 00:49

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Strictly speaking, $S^{ab}$ and $(S^a)^b$ are not the same. There is, however, a canonical bijection between them given by $$(x_1,...,x_{ab})\mapsto ((x_1,...,x_a), ..., (x_{ab-a+1}, x_{ab}))$$ which we often "mod out" by implicitly (note that this is really the same issue as the technically-non-associative nature of the Cartesian product). In situations where the difference is something you actually want to keep track of as opposed to an annoyance you want to avoid, you obviously don't want to do this. In terms of communicating clearly to the reader, all you need to do is make an explicit remark somewhere pointing this out, so that your readers understand why you aren't "simplifying" the relevant expressions.

Noah Schweber
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  • If they are not the same due to the non-associativity of the Cartesian product, then $X^a \ni ((\cdots(x_1,x_2),x_3)x_4), \ \dots \ ),x_n)$? – user110391 Jun 12 '23 at 01:37