Okay, enough distractions. Let's get back to compact sets. We know that a compact set is compact in any metric space which contains it. We also know that compact sets are closed. By implication, compact sets are closed in any metric space which contains them, which is not true of closed sets in general. This gives a relationship between a compact set and the world around it. Let's go the other direction and look at the interior of compact sets, by considering subsets of compact sets.

Since compact sets are closed, let's look at a closed subset of a compact set. Now it becomes important to clarify what we mean by that. Suppose X is a metric space, and K is a compact set in that metric space. If F is a subset of K, F could potentially be closed relative to K but not be closed relative to X. So let's be clear that F is closed relative to X (and therefore also closed relative to K).

(I don't have an example at hand of a subset of a compact set which is closed relative to the compact set but not closed relative to the space. The discussion earlier this week of closed sets on the rational number line gave examples of sets which are closed relative to the rational number space, but not closed relative to the real number space. On the other hand, the set of rational numbers is not a compact set in the real number space. I'll have to give some thought to whether it's possible to for a set to be closed relative to a compact space but not closed in the surrounding metric space.)

Take an open cover of F, call it {G_{α}}. {G_{α}} may be infinite, in which case we can use the following process to find a finite subcollection. Since F is closed in X, F^{c} is an open set. Add F^{c} to {G_{α}}, and this is clearly an open cover of X. Since K is a subset of X, this is an open cover of K. Since K is compact, there is a finite subcollection of {G_{α}} and F^{c} which is an open cover of K. Since F is a subset of K, this finite subcollection is also an open cover of F. If F^{c} is part of the finite open cover of K, we can remove it from the open cover of F since F^{c} does not intersect F. The resulting finite open cover of F is a subcollection of {G_{α}}. This is true for any possible starting open cover of F, so F is compact.

There's a little added bonus. Suppose F is any closed set in X. It does not have to be a subset of K. If K is a compact set in X, then K is closed in X. Then we know that the intersection of F and K, F∩K, is closed in X. Furthermore F∩K is a subset of K, so since it is a closed subset of a compact set, it is itself compact. If X is a metric space, F is any closed set in X, and K is any compact set in X, then F∩K is also a compact set.