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given some $n \in \mathbb{N}$, prove that $A \cap ( B_1 \cup \cdots \cup B_n) = (A \cap B_1) \cup \cdots\cup (A \cap B_n)$

Using unary union and intersection and basic set theory.

My approach :

Let $B = \{ B_i \mid 1\leq i \leq n \}$

And let $C = \{ A \cap B_i \mid 1\leq i \leq n \}$, so now i need to prove that

$A \cap \bigcup B = \bigcup C$ but i don't know how to proceed, i know i have to let $x$ be anything and assume that $x \in \bigcup C$ and prove it must be in $x \in A \cap \bigcup B$, but this formal step i don't know how to do it, please help.

Update : one should use unary union

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    $\begingroup$ Hint: Define your unary union operator, for any counting interval $I\subseteq \Bbb N^+$, as $\bigcup_{k\in I}\{S_k\} = \{x: \exists k\in I~(x\in S_k)\}$ and so demonstrate $A\cap\bigcup_{k\in I}\{B_k\} = \ldots = \bigcup_{k\in I}\{A\cap B_k\}$ . $\endgroup$ – Graham Kemp Nov 1 '17 at 3:49
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I find pointwise proofs of set theoretic concepts the most intuitive way to do it. If $x \in A \cap ( B_1 \cup \cdots B_n)$, then $x$ is in $A$, and it is also in some $B_i$. Therefore $x \in A \cap B_i$, and it is certainly in the union of all such $A \cap B_i$. Is it a requirement to use unary union?

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  • $\begingroup$ it seems like that, the lecturer said because the sets are not finite. $\endgroup$ – Ahmad Oct 31 '17 at 6:57

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