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$(i) X\cap(\bigcup\limits_{n=1}^\infty A_n) = \bigcup\limits_{n=1}^\infty (X\cap A_n)$

$(ii) X\cup(\bigcap\limits_{n=1}^\infty A_n) = \bigcap\limits_{n=1}^\infty (X\cup A_n)$

Any suggestions on how to approach this?

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6  
Use the same basic ‘element-chasing’ approach that I used in this answer to your similar question about DeMorgan’s laws. –  Brian M. Scott Oct 18 '12 at 16:31

3 Answers 3

This reduces from set theory to logic, where you can show $$x\in X\land \exists n\in\mathbb N\colon x\in A_n\iff \exists n\in\mathbb N\colon (x\in X\land x\in A_n)$$ and $$x\in X\lor \forall n\in\mathbb N\colon x\in A_n\iff \forall n\in\mathbb N\colon (x\in X\lor x\in A_n)$$ Once again, there is no reason to restrict the task to merely countable index sets..

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An equation $A=B$ between sets is usually proved by showing the 2 inclusions: $$A\subseteq B \ \text{ and }\ B\subseteq A$$ For such kind of equations, containing $\bigcup$ and $\bigcap$, one direction usually comes easily: for example in (ii), the left hand side is contained in each member on the rhs, so $\subseteq$ follows. For the otheer direction, use the element chasing method, as commented by Brian: Consider an element of the rhs. in (ii) then show it is in the lhs.

Similarly for (i).

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Hint

For i), On the left side:

Let $a \in X \cap \bigcup _{n=1}^\infty A_n$
Then $ a \in X$ and $a \in $ every $A_n$. So $a \in X$ and $A_1, a \in X $ and $A_2$,... Since $a$ is in X and every $A_n$, $a$ will be in the union of intersections.

For ii) you can apply the same type of reasoning. Hope it helps.

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