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Consider sets $\mathcal{B} = \{1,\ldots,B\}$ and $\mathcal{A}_b = \{1,\ldots,A_b\}$ for each $b\in\mathcal{B}$.

Now consider pairs of the form $(b,a)\in\mathcal{B}\times\mathcal{A}_b$. I want to write the set of all such pairs (i.e. the powerset). I am stuck as to how to write this. Any ideas?

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I'm not aware of any better method than $$\{(b,a): b \in B, a \in \mathcal{A}_b\}$$ although if your audience is type-theoretically inclined, they may recognise the dependent sum type $$\sum_{b : B} A_b$$

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$\cup_b$ B×$A_b,$
the collection of all such pairs, is not a power set.

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  • $\begingroup$ Doesn't this cause the dependency to be lost? Let $\mathcal{B} = \{1,2\}$ and $\mathcal{A}_1=\{1,2,3\}$, $\mathcal{A}_2=\{1,2\}$, wouldn't the above definition allow for pairs of the form $(b,a)=(2,3)$? Notice this shouldn't happen because $3$ is not in $\mathcal{A}_2$. $\endgroup$ – jonem May 27 '18 at 21:58
  • $\begingroup$ The notation in the Question is somewhat infelicitous. $\mathcal B$ is described as $\{1,\ldots,B\}$, yet there is a family of sets $\mathcal A_b$ indexed by $b\in \mathcal B$. Worse, the elements of $\mathcal A_b$ are described as $\{1,\ldots,A_b\}$. In short the mix of fonts and multiple uses of $b\in \mathcal B$ as an index/subscript leave us somewhat unclear what exactly the dependence of $\mathcal A_b$ or $A_b$ on $b$ is supposed to mean. $\endgroup$ – hardmath May 28 '18 at 3:24
  • $\begingroup$ Amen @hardmath. $\endgroup$ – William Elliot May 28 '18 at 8:30

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