Intersection distributing over an infinite union, union over an infinite intersection I'm not really sure how to type this in notation...but I hope this makes sense. I help in proving that:
$$F \cap \bigcup\limits_{i=1}^\infty E_i = \bigcup\limits_{i=1}^\infty (F\cap E_i)$$
And
$$F \cup \bigcap\limits_{i=1}^\infty E_i = \bigcap\limits_{i=1}^\infty(F\cup E_i)$$
I really don't know where to begin with this one. 

For the first one:
We prove that the left hand side is contained in the right hand side, and conversely.
Suppose that $x\in F\cap\bigcup\limits_{i=1}^{\infty}$. That means that $x\in F$, and $x\in \bigcup\limits_{i=1}^{\infty}E_i$; therefore, $x\in F$ and there exists $n$ such that $x\in E_n$. Therefore, there exists $n$ such that $x\in F$ and $x\in E_n$, so $x\in F\cap E_n$. Thus, there exists at least one $n$ such that $x\in F\cap E_n$, so $x\in \bigcup\limits_{i=1}^{\infty}(F\cap E_i)$, as desired.
For the converse inclusion, suppose that $y\in \bigcup\limits_{i=1}^{\infty}(F\cap E_i)$. That means that there exists an $n$ such that $y\in F\cap E_n$. Therefore, $y\in F$, and there is an $n$ such that $y\in E_n$; the latter condition implies that $y\in\bigcup\limits_{i=1}^{\infty}E_i$. Since we also have $y\in F$, then $y$ lies in the intersection $F\cap\bigcup\limits_{i=1}^{\infty}E_i$.
Thus, we have shown that
$$F\cap\bigcup_{i=1}^{\infty}E_i \subseteq \bigcup_{i=1}^{\infty}(F\cap E_i)\text{ and } \bigcup_{i=1}^{\infty}(F\cap E_i) \subseteq F\cap\bigcup_{i=1}^{\infty}E_i.$$
Hence, the two sets are equal.
A similar argument holds for the second equality, remembering that lying in the intersection means lying in every one of the sets.
