Suppose $A = \bigcup_{i=1}^{N} E_i$ and $B = \bigcup_{j=1}^{N} F_j $. Does it follow that

$$ A \cap B = \bigcup_{i=1}^{N} \bigcup_{j=1}^{N} (E_i \cap F_j) = \bigcup_{j=1}^{N} \bigcup_{i=1}^{N} (E_i \cap F_j)$$


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    $\begingroup$ Yes. !!!!!$\!\!$ $\endgroup$ – Trevor Wilson Nov 19 '13 at 6:33

Do you know the distributive property? Namely, that $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$? If, so all you need to do is expand the expression with some careful algebra. I will do it for collections of $3$ sets and maybe you can try and do it for $N$ sets. You could either do this directly, or using induction (induction may give you less of a headache).

So say $A = E_1 \cup E_2 \cup E_3$ and $B = F_1 \cup F_2 \cup F_3$ then $$A \cap B = (E_1 \cup E_2 \cup E_3) \cap (F_1 \cup F_2 \cup F_3)$$ $$=((E_1 \cup E_2 \cup E_3) \cap F_1) \cup ((E_1 \cup E_2 \cup E_3) \cap F_2) \cup ((E_1 \cup E_2 \cup E_3) \cap F_3) $$ $$=((E_1 \cap F_1) \cup (E_2 \cap F_1) \cup (E_3 \cap F_1)) \cup ((E_1 \cap F_2) \cup (E_2 \cap F_2) \cup (E_3 \cap F_2)) \cup ((E_1 \cap F_3) \cup (E_2 \cap F_3) \cup (E_3 \cap F_3))$$ $$=\bigcup_{i=1}^3 (E_i \cap F_1) \cup \bigcup_{i=1}^3 (E_i \cap F_2) \cup \bigcup_{i=1}^3 (E_i \cap F_3)$$

And then we get: $$ A \cap B = \bigcup_{j=1}^3 \bigcup_{i=1}^3 (E_i \cap F_j)$$

Similarly, by distributing the other way you get $$A \cap B = \bigcup_{i=1}^3 \bigcup_{j=1}^3 (E_i \cap F_j)$$

  • $\begingroup$ I think there is a mistake here. Let $E_1 = E_2 = E_3 = V$. Then the union of the $F_i$'s somehow turns into the intersection of the $F_i$'s. $\endgroup$ – Trevor Wilson Nov 19 '13 at 7:16
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    $\begingroup$ Apparently, I don't know the distributive property! Let me fix that... $\endgroup$ – Ryan Sullivant Nov 19 '13 at 7:21

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