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I have got a problem concerning generalized commutative law for generalized cartesian product. Let $b \colon I \mapsto I $ be a bijection (we think of it as a permutation). How to prove that generalized cartesian product is commutative: $\Pi \langle A_i \colon \ i \in I \rangle = \Pi \langle A_{b(i)} \colon i \in I\rangle$?

I argue in the following way:

we are trying to show $\subseteq$:

let $f \in \Pi \langle A_i \colon \ i \in I \rangle$, then $f \colon I \mapsto \bigcup_{i \in I} A_i$ is a function such that $f(i) \in A_i$ for every $i \in I$. How to show that $f(i) \in A_{b(i)}$ and therefore $f \in \Pi \langle A_{b(i)} \colon i \in I\rangle$. Is $A_i = A_{b(i)}$? If so, how to demonstrate this step by step?

I must miss something in the reasoning chain, because I am not able to complete the proof. Can somebody help me? Thanks in advance.

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1 Answer 1

up vote 4 down vote accepted

You can't prove that, because it's not true. Exactly because it might be the case that $A_i\neq A_{b(i)}$.

In general, $A\times B\neq B\times A$. For example $A=\Bbb N$ and $B=\Bbb R$. But it is true that there is a natural bijection between the two products. Which induces isomorphisms of any kind of structure you would usually want on products. So often people abuse the language of mathematics and say that the two products are equal, and they mean "canonically isomorphic".

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