## § 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:
- Each finite intersection is non-empty, and will be equal to the set thathas the highest index in the finite intersection.
- Each of the sets
`Si`

are compact since they are closed and bounded subsets of `R`

- Invoke theorem.