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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 $$f(<a_i,i>)=a_i$$

Is that correct so far? And if so, what next?

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When you write $X\sim Y$, do you mean $|X|=|Y|$ ($X$ and $Y$ have the same cardinality, there is a bijection between $X$ and $Y$)? – Brian M. Scott Nov 14 '12 at 23:07
@BrianM.Scott: Yes! – 17SI.34SA Nov 14 '12 at 23:11
up vote 1 down vote accepted

Your $f$ is fine, though you might even write directly (with given bijections $\beta\colon B\to I$, $\alpha_i\colon A\to A_i$) $$f(\langle a,b\rangle) = \alpha_{\beta(b)}(a).$$ You need to show that this is a bijection. Note that $f(\langle a,b\rangle)\in A_{\beta(b)}$. Therefore, if $f(\langle a,b\rangle)=f(\langle a',b'\rangle)$ then $\beta(b)=\beta(b')$, hence $b=b'$. But then $\alpha_{\beta(b)}(a)=\alpha_{\beta(b)}(a')$ also implies $a=a'$. Thus $f$ is injective. But $f$ is also surjective because given $x\in A_i$ for some $i\in I$, we see that $x=f(\langle \alpha_i^{-1}(x),\beta^{-1}(i)\rangle) $.

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For each $i\in I$ you have a bijection $f_i:A\to A_i$ and a bijection $g:B\to I$, and you want a bijection

$$f:A\times B\to\bigcup_{i\in I}A_i\;.$$

Suppose that $\langle a,b\rangle\in A\times B$; the natural idea is to use $b$ to pick the right index $i\in I$, and $a$ to pick the right element of $A_i$. The index picked out by $b$ is clearly $i=g(b)$, so you want $f$ to send $\langle a,b\rangle$ to some element of $A_{g(b)}$. Which one? The one that corresponds to $a$ under the map $f_{g(b)}$. Thus, you want to try

$$f:A\times B\to\bigcup_{i\in I}A_i:\langle a,b\rangle\mapsto f_{g(b)}(a)\;;$$

it’s just a matter of proving that this map really is a bijection.

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Thank you, Mr Scott! – 17SI.34SA Nov 14 '12 at 23:39
@17SI.34SA: You’re welcome. – Brian M. Scott Nov 14 '12 at 23:40

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