Now that we've shown that k-cells, and in particular, closed intervals on the real number line, are compact, I feel like we have actual useful examples of compact sets. Until now we had sort of an abstract definition, and we were working off that definition to prove properties of compact sets, but they weren't really attached to anything. We also had one example which I've used a couple of times, and will use again in this post, but which feels kind of limited. In discussing the proofs I've used a square on a piece of paper, where the paper represents some metric space and the square represents some compact set. It's something of a relief that we can now say that closed rectangles on the complex plane are in fact compact sets.

There's one more fact about compact sets in general that I want to prove, and then we can focus on k-cells some more. Start with a compact set. Take any infinite subset of that set. That subset must have a limit point in the compact set.

I want to go back to the first example we had of a compact set to discuss this. Our compact set includes all rational numbers of the form 1/n, where n is any natural number, and also includes the number 0. Our proof that this set is compact is dependent on the fact that 0 is a limit point of the set, so any open set which contains 0 also contains an infinite number of other points in the set. Now we can turn that around and say that any infinite subset of this set must have 0 as a limit point. 0 itself doesn't have to be in the subset, but the only way we can get to an infinite number of points in the subset is by letting n go to infinity, which makes 0 a limit point of the subset. (Looking at the example, I'm happy thinking informally. I can see this is true, so I don't feel a need to work through the details formally. For the general proof, though, I should be a little more formal.)

For the formal proof, take any compact set and call it K. Take any infinite subset of K and call it E. Assume for the moment that E has no limit points. Pick a point p in K. Since E has no limit points, there exists a neighborhood of p which contains no points in E, unless p is a member of E. If p is a member of E, there exists a neighborhood of p which contains p and no other points in E. This neighborhood of p is an open set. For every point in K, there is an open set which contains that point, and at most one point in E. The union of all of these open sets is an open cover of K. However, no finite subcollection of these neighborhoods can be a cover of K, since there is an infinite number of points in E, and every point in E is in exactly 1 set in the open cover. Since we know that K is compact and therefore a finite subcover must exist, the conclusion is that E must have a limit point in K.

The limit point of E doesn't have to be a member of E, but must be a member of K. From the example, 0 is always the limit point of any infinite subset, but 0 doesn't have to be a member of the subset. However, 0 is a member of the original compact set. If 0 is not a member of the subset in this example, the subset is an open set. If 0 is a member of the subset, the subset is closed, and is therefore also compact.