TITLE: Intersection distributing over an infinite union, union over an infinite intersection
QUESTION [13 upvotes]: 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.

REPLY [12 votes]: Probably the easiest way to prove these statements is by what I call element-chasing: show that every element of the left-hand side is also an element of the right-hand side and vice versa. Let’s look at the first one. Suppose that $x \in F \cap \bigcup\limits_{i=1}^\infty E_i$. The definition of intersection tells you that $x \in F$ and $x \in \bigcup\limits_{i=1}^\infty E_i$. Since $x \in \bigcup\limits_{i=1}^\infty E_i$, the definition of union tells you that $x$ belongs to at least one of the sets $E_1,E_2,E_3,\dots$, so there is some positive integer $k$ such that $x \in E_k$. (There might be more than one such integer, but all we need is one.) Now we know that $x \in F$ and $x \in E_k$, so $x \in F \cap E_k$. But $x \in F \cap E_k$ is one of the sets that goes into the union $\bigcup\limits_{i=1}^\infty (F\cap E_i)$ on the right-hand side, so automatically $F \cap E_k \subseteq \bigcup\limits_{i=1}^\infty (F\cap E_i)$, and therefore $x \in \bigcup\limits_{i=1}^\infty (F\cap E_i)$. We’ve now shown that every member of $F \cap \bigcup\limits_{i=1}^\infty E_i$ is also a member of $\bigcup\limits_{i=1}^\infty (F\cap E_i)$, or in other words that $$F \cap \bigcup\limits_{i=1}^\infty E_i \subseteq \bigcup\limits_{i=1}^\infty (F\cap E_i).$$ To finish the argument, you have to show that $$\bigcup\limits_{i=1}^\infty (F\cap E_i)\subseteq F \cap \bigcup\limits_{i=1}^\infty E_i.$$ You do this the same way, by starting with an arbitrary $x \in \bigcup\limits_{i=1}^\infty (F\cap E_i)$ and showing that it must belong to $F \cap \bigcup\limits_{i=1}^\infty E_i$ as well.
You can handle the second statement the same way: first show that every member of $F \cup \bigcap\limits_{i=1}^\infty E_i$ is a member of $\bigcap\limits_{i=1}^\infty(F\cup E_i)$, and then show the opposite inclusion. I’ll get you started. Suppose that $x \in F \cup \bigcap\limits_{i=1}^\infty E_i$; then $x \in F$ or $x \in \bigcap\limits_{i=1}^\infty E_i$ (or both), and you’ll have to distinguish two cases.
(1) If $x \in F$, then $x \in F \cup E_i$ for every positive integer $i$, so $x \in \bigcap\limits_{i=1}^\infty(F\cup E_i)$ by the definition of intersection.
(2) If $x \in \bigcap\limits_{i=1}^\infty E_i$, then ... what? I’ll leave this case to you, as well as the opposite inclusion.