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How do you show that if you have sets $B_1, B_2, \cdots ,B_n$ and a set $C$, then $$(B_1\cap B_2, \cap \cdots B_n)\cup C= (B_1\cup C)\cap(B_2\cup C) \cap \cdots \cap (B_n\cup C)\,?$$

Thanks

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You have $$( B_1\cap B_2)\cup C = (B_1\cup C)\cap(B_2 \cup C).$$ Use an induction argument.

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  • $\begingroup$ Huh? this syntax makes no sense to me. $\endgroup$ – tim29 Mar 5 '15 at 2:38
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    $\begingroup$ What part of the syntax makes no sense? $\endgroup$ – Graham Kemp Mar 5 '15 at 2:44
  • $\begingroup$ my bad it works now $\endgroup$ – tim29 Mar 5 '15 at 2:58
  • $\begingroup$ how would you start the inductive argument $\endgroup$ – tim29 Mar 5 '15 at 3:11
  • $\begingroup$ @tim29 Induction arguments have base cases and inductive steps. The base case is what ncmathsadist wrote in his answer. The inductive step involves assuming that your statement is true for $n=k$, and proving that it holds for $n=k+1$. The base case is the same as your equation if we set $n=2$. $\endgroup$ – Zubin Mukerjee Mar 5 '15 at 5:24
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You can prove the most general version without induction.

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Corollary of (ii) $$ \begin{align*} (B_1\cap B_2\cap\cdots \cap B_n)\cup C&= \left(\bigcap_{i=1}^{n}B_i\right)\cup C\\ &=\bigcap_{i=1}^n(B_i\cup C)\\ &=(B_1\cup C)\cap(B_2\cup C)\cap\cdots\cap(B_n\cup C) \end{align*} $$

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