Prove the Distributive Law of indexed sets $(i) X\cap(\bigcup\limits_{n=1}^\infty A_n) = \bigcup\limits_{n=1}^\infty (X\cap A_n)$
$(ii) X\cup(\bigcap\limits_{n=1}^\infty A_n) = \bigcap\limits_{n=1}^\infty (X\cup A_n)$
Any suggestions on how to approach this?
 A: 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..
A: 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).
A: 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.
