118 5.1 Topology of the Real Line Proof: Let Abe a nonempty closed set that is bounded above. Then supAexists. Let m=supA. To complete the proof, we will show that m∈A(and therefore m=maxA). Assume by contradiction that m/ ∈A. Then m∈Ac, which is an open set. So there exists δ >0 such that (m−δ,m+δ) ⊂Ac. This means there exists noa∈Awith m−δ <a≤m. This contradicts the fact that mis the least upper bound of A(see Proposition 1.5.1). Therefore, m∈Aand hence maxAexists. □ Definition 5.1.3 A subset Aof Ris calledcompact if for every sequence {an}inA, there exists a subsequence {ank}that converges to a point a∈A. 1 ■ Example 5.1.4 Let a,b∈R, a≤b. We show that the set A= [a,b] is compact. Let {an} be a sequence inA. Since a≤an ≤bfor all n, then the sequence is bounded. By the Bolzano-Weierstrass theorem (Theorem 2.4.1), we can obtain a convergent subsequence {ank}. Say, limk→∞ank =s. We now must show that s ∈A. Since a≤ank ≤bfor all k, it follows from Theorem 2.1.2, that a≤s ≤b and, hence, s ∈Aas desired. We conclude that Ais compact. Theorem 5.1.5 — Heine-Borel. A subset Aof Ris compact if and only if it is closed and bounded. Proof: Suppose A is a compact subset of R. Let us first show that A is bounded. Suppose, by contradiction, that Ais not bounded. Then for every n∈N, there exists an ∈Asuch that |an| ≥n. Since Ais compact, there exists a subsequence {ank}that converges to some a∈A. Then |ank| ≥nk ≥k for all k. Therefore, limk→∞|ank| =∞. This is a contradiction because {|ank|} converges to |a|. Thus Ais bounded. Let us now show that Ais closed. Let {an}be a sequence inAthat converges to a point a∈R. By the definition of compactness, {an} has a subsequence {ank} that converges to b∈A. On the other hand, every subsequence must converge toa. Therefore a=b∈Aand, hence, Ais closed by Theorem 5.1.3. For the converse, suppose Ais closed and bounded and let {an} be a sequence in A. Since A is bounded, the sequence is bounded and, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it has a convergent subsequence, {ank}. Say, limk→∞ank =a. It now follows from Theorem 5.1.3 that a∈A. This shows that Ais compact as desired. □ We now revisit the notion of limit point with new terminology and illustrate it with more general examples of subsets of R. Definition 5.1.4 Let Abe a subset of R. A point a∈R(not necessarily in A) is called a limit point of Aif for every δ >0, the open ball B(a;δ) contains a point pof A, p̸=a. 1This definition of compactness is more commonly referred to as sequential compactness.
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