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Let $a_i$ be a cardinal number for every $i \in\ I$.

Let $\{A_i\}$ and $\{A'_i\}$ be families of sets and let $A_i$, $A'_i$ and $a_i$ be equipotent for every $i \in I$.

Then show that $\prod_i\ A_i $ is equipotent with $\prod_i\ A'_i $.

This seems obviously true but I don't know how to actually show the bijection between them..

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    Note that this is basically a reformulation of [this question](http://math.stackexchange.com/questions/140467/one-to-one-correspondence/)2012-05-15

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For each $i\in I$ you have a bijection $\varphi_i:A_i\to A'_i$. Define $\varphi:\prod_{i\in I}A_i\to\prod_{i\in I}A'_i:\langle a_i:i\in I\rangle\mapsto\Big\langle\varphi_i(a_i):i\in I\Big\rangle$ and prove that it’s a bijection.

This is what I call a follow-your-nose proof: there really is only one reasonable thing to try. All you’re given is the equipotence of $A_i$ and $A'_i$ for $i\in I$. All that gives you is the existence of the bijections $\varphi_i$, so either the result is very hard (unlikely) or somehow it must be possible to use those bijections to get the one that you want. Since a typical element of $\prod_iA_i$ is just a function $\langle a_i:i\in I\rangle$ from $I$ to $\bigcup_iA_i$, about the only thing to try is to apply the bijections $\varphi_i$ to the components of $\langle a_i:i\in I\rangle$.