## § Using compactness to argue about covers

I've always seen compactness be used by starting with a possibly infinite coverm and then filtering it into a finite subcover. This finite subcover is then used for finiteness properties (like summing, min, max, etc.). I recently ran across a use of compactness when one starts with the set of all possible subcovers , and then argues about why a cover cannot be built from these subcovers if the set is compact. I found it to be a very cool use of compactness, which I'll record below:

#### § Theorem:

If a family of compact, countably infinite sets S_a have all finite intersections non-empty, then the intersection of the family S_a is non-empty.

#### § Proof:

Let S = intersection of S_a. We know that S must be compact since all the S_a are compact, and the intersection of a countably infinite number of compact sets is compact. Now, let S be empty. Therefore, this means there must be a point p ∈ P such that p !∈ S_i for some arbitrary i.

#### § Cool use of theorem:

We can see that the cantor set is non-empty, since it contains a family of closed and bounded sets S1, S2, S3, ... such that S1 ⊇ S2 ⊇ S3 ... where each S_i is one step of the cantor-ification. We can now see that the cantor set is non-empty, since:
1. Each finite intersection is non-empty, and will be equal to the set that has the highest index in the finite intersection.
2. Each of the sets Si are compact since they are closed and bounded subsets of R
3. Invoke theorem.