Exercise 14.6

It is very important to note that, in order to apply the definition to prove that a set K is compact, we must examine an arbitrary collection of open sets whose union contains K, and show that K is contained in the union of some finite number of sets in the given collection. That is, it must be shown that any open cover of K has a finite subcover. On the other hand, to prove that a set H is not compact, it is sufficient to exhibit one specific collection G of open sets whose union contains H, but such that the union of any finite number of sets in G fails to contain H. That is, H is not compact if there exists some open cover of H that has no finite subcover.

Proof. We have shown in Theorem 11.2.4 that a compact set in R must be closed and bounded. To establish the converse, suppose that K is closed and bounded, and let G ¼ f g Ga be an open cover of K. We wish to show that K must be contained in the union of some finite subcollection from G. The proof will be by contradiction. We assume that: