118 5.1 Topology of the Real Line Proof: Let A be a nonempty closed set that is bounded above. Then supA exists. 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 no a ∈ A with m− δ < a ≤ m. This contradicts the fact that m is the least upper bound of A (see Proposition 1.5.1). Therefore, m ∈ A and hence maxA exists. □ Defnition 5.1.3 A subset A of R is called compact if for every sequence {an} in A, there exists a 1 subsequence {ank } that converges to a point a ∈ A. ■ 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 in A. Since a ≤ an ≤ b for 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 ≤ b for all k, it follows from Theorem 2.1.2, that a ≤ s ≤ b and, hence, s ∈ A as desired. We conclude that A is compact. Theorem 5.1.5 — Heine-Borel. A subset A of R is compact if and only if it is closed and bounded. Proof: Suppose A is a compact subset of R. Let us frst show that A is bounded. Suppose, by contradiction, that A is not bounded. Then for every n ∈ N, there exists an ∈ A such that |an|≥ n. Since A is 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 A is bounded. Let us now show that A is closed. Let {an} be a sequence in A that converges to a point a ∈ R. By the defnition of compactness, {an} has a subsequence {ank } that converges to b ∈ A. On the other hand, every subsequence must converge to a. Therefore a = b ∈ A and, hence, A is closed by Theorem 5.1.3. For the converse, suppose A is 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 A is 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. Defnition 5.1.4 Let A be a subset of R. A point a ∈ R (not necessarily in A) is called a limit point of A if for every δ > 0, the open ball B(a;δ ) contains a point p of A, p ̸ = a. 1This defnition of compactness is more commonly referred to as sequential compactness.
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