Show that if K is compact and nonempty then supK and infK bo

Show that if K is compact and nonempty, then sup(K) and inf(K) both exist and are elements of K.

Solution

Suppose K is compact. Then K is bounded (by Heine-Borel), so there exists a real number M > 0 with x < M for all x œ K. Since x § x < M, we see that M is an upper bound for K. By the axiom of completeness there exists a least upper bound sup K. For each n œ N, there exists a member xn œ A such that s - 1 n < xn . Now one or more of these xn may be equal to s, but if that is the case then s œ A Õ A and we are done, so suppose that xn ¹ s for all n. Let e > 0, and choose n so large that 1 n < e. By costruction we have xn œ V1ênHsL › A, and since V1ênHsL Õ VeHsL, it follows that VeHsL contains a member xn of A other than s. Hence s is a limit point of A , and so s œ A sup K œ K. But K is closed (by Heine-Borel again), so K = K and sup K œ K.