So, I have set $A$ and family of sets $\{A_i: i\in I \}$, such that $A_i\cap A_j=\emptyset$ for $i\neq j$. Given that $A\sim A_i$ for all $i\in I$, and $I\sim B$. I want to show that $A\times B\sim \bigcup_{i\in I}A_i$.
My work so far:
$A\sim A_i$ for all $ i\in I$. Given that $I\sim B$, so $A\times B\sim A\times I$ or $A\times B\sim A_i\times I$ for all $ i\in I$, so we left to prove: $A_i\times I\sim \bigcup_{i\in I}A_i$
I thought defining a bijection, $f:A_i\times I\to \bigcup_{i\in I}A_i$ by $f(
Is that correct so far? And if so, what next?