Proposition 5.2.3: Compact means Closed and Bounded
Proof:First, suppose S is closed and bounded. Take a sequence in S. Because S is bounded, the sequence is bounded also, and by the Bolzano-Weierstrass theorem we can extract a convergent subsequence from . Using the theorem about closed sets, accumulation points and sequences, we know that the limit of the subsequence is again in S. Hence, S is compact.
Now assume that S is compact. We have to show that S is bounded and closed.
Suppose S was not bounded. Then for every n there exists a number a n S with | a n | > n. But then no subsequence of the sequence will converge to a finite number, hence S can not be compact. Thus, S must be bounded.
Suppose S was not closed. Then there exists an accumulation point c for the set S that is not contained in S. Since c is an accumulation point, there exists a sequence of elements in S that converges to c. But then every subsequence of that sequence will also converge to c, and since c is not in S, that contradicts compactness. Hence, S must be closed.