Is \bigcap(F \cup G) equal to (\cap F)\bigcap(\cap G)?

[Klungo]

Homework Statement

(Title is wrong)

I was able to prove the similar \bigcup(F \cup G)=(\cup F)\bigcup(\cup G) but I'm not to sure how to go about this one.

Let F and G be nonempty families of sets. Prove
\bigcap(F \cup G)=(\cap F)\bigcap(\cap G)

2. The attempt at a solution
To prove \bigcap(F \cup G) \subset (\cap F)\bigcap(\cap G),
let x \subset \bigcap(F \cup G) be arbitrary. Clearly, x \in S for some set S \subset \bigcap F \cup G containing all common elements in F \cup G. We have S \in F \cup G.

Do I go by..

We have two cases: (only one needs to be proven really.)
Case 1: If S \in F, clearly S \subset \cup F...
(I'm stuck here)

or Do I go by...

Suppose S \in F and S \in G. Clearly S \subset \cup F and S \subset \cup G. (I'm stuck here)

I'll try proving the converse and restart the proof since I misread the problem.

 

[voko]

By definition of the operation \bigcap, any x in \bigcap(F\cup G) must be contained in EVERY SET of F\cup G. Continue from here.