More basic topology, please...?

Question

Thanks for the help with the previous question. I am still new to this topology course and trying to get used to it. My question is, what is the significance of the definition of a topological space T, that the open sets consist of "finite intersections" of elements of T, and "arbitrary unions" of elements of T? I guess "finite intersections" means that there are certain intersections of A and B in T that will also be in T, whereas "arbitrary unions" means that the union of any of elements in T will always be in T. But in my mind, I still imagine them to be the same: the intersection of any two elements in T will be in T, and the union of any two elements in T will be in T.

Does this difference matter in proving that the closure of (the intersection of an arbitrary intersection of A_i's) = the intersection of (the closure of arbitrary A_i's) ? I've been struggling for a while, so any help will be appreciated. Thanks. =)

Answer 1

The idea comes from the analogue on the real line. Consider the following subsets of the real line:

I_k = (-1/k, 1/k)

each I_k is obviously an open interval by definition. Now since each I_k is a subset of I_{k-1}, any finite intersection of these intervals will be I_j for some large j, and still is an open set.

But if you take the infinite intersection over all of the sets I_k, you can as the resulting set the single point {0}. For if you look at x not equal to 0, you can find some k large enough such that |x| > 1/k, so that x does not belong to I_k, and therefore x cannot belong in the infinite intersection. However, a single point on the real line forms a closed, not open, set.

I think what you are not understanding is the meaning of "finite" in finite intersections. It means that the simultaneous intersection of finitely member elements. Compare that with "arbitrary" union, which regards the simultaneous union of any number of sets, even infinite, even uncountably infinite.

As to your second question: not really... recall the definitino of closure: a point x is in the closure of A if every open set containing x has a nonempty intersection with A. So if x is in the closure of (arbitrary intersection of A_i) and N is an open set containing x  then N intersects with every A_i. So in particular x must be in the closure of every A_i. This gives you one direction.

I'll leave the other direction to you.