TITLE: Why is that *any* union of open sets is open but only *finitely many* intersections of open sets is open?
QUESTION [12 upvotes]: I understand that when we talk about union of open sets, we introduce an index set which can be countable or uncountable. But could I not do the same for the intersection of open sets too?

REPLY [8 votes]: Proof by example!
Consider the real numbers, and a sequence of ever-smaller open intervals around zero:  (-1,1), (-1/2,1/2), (-1/3,1/3), ...
If we take any union of any of these sets, then any point in that union will have a little neighbourhood around it (can you see why?).
For intersections, it's not quite so simple.
If we take a finite number of the sets and take their intersection, then we'll still get an open set (hint: why is the intersection of two open sets still open?)
But if we take the infinite intersection, the only point that is in all the intervals is 0. 
And 0 on its own isn't an open set, because 0 doesn't have a little neighbourhood around it. 
If you understand why that's true, you should have no trouble seeing it more generally.