Introduction to Mathematical Analysis I - Second Edition

57 Let us now prove the converse. Suppose by contradiction that A is not closed. Then A c is not open. Since A c is not open, there exists a ∈ A c such that for any ε > 0, one has B ( a ; ε ) ∩ A 6 = /0. In particular, for such an a and for each n ∈ N , there exists a n ∈ B ( a ; 1 n ) ∩ A . It is clear that the sequence { a n } is in A and it is convergent to a (because | a n − a | < 1 n , for all n ∈ N ). This is a contradiction since a / ∈ A . Therefore, A is closed. Theorem 2.6.4 If A is a nonempty subset of R that is closed and bounded above, then max A exists. Similarly, if A is a nonempty subset of R that is closed and bounded below, then min A exists Proof: Let A be a nonempty closed set that is bounded above. Then sup A exists. Let m = sup A . To complete the proof, we will show that m ∈ A . Assume by contradiction that m / ∈ A . Then m ∈ A c , which is an open set. So there exists δ > 0 such that ( m − δ , m + δ ) ⊂ A c . 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, max A exists. Definition 2.6.3 A subset A of R is called compact if for every sequence { a n } in A , there exists a subsequence { a n k } that converges to a point a ∈ A . 1 Example 2.6.4 Let a , b ∈ R , a ≤ b . We show that the set A = [ a , b ] is compact. Let { a n } be a sequence in A . Since a ≤ a n ≤ b for all n , then the sequence is bounded. By the Bolzano-Weierstrass theorem (Theorem 2.4.1 ) , we can obtain a convergent subsequence { a n k } . Say, lim k → ∞ a n k = s . We now must show that s ∈ A . Since a ≤ a n k ≤ b for all k , it follows from Theorem 2.1.5 , that a ≤ s ≤ b and, hence, s ∈ A as desired. We conclude that A is compact. Theorem 2.6.5 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 first show that A is bounded. Suppose, by contradiction, that A is not bounded. Then for every n ∈ N , there exists a n ∈ A such that | a n | ≥ n . Since A is compact, there exists a subsequence { a n k } that converges to some a ∈ A . Then | a n k | ≥ n k ≥ k for all k . Therefore, lim k → ∞ | a n k | = ∞ . This is a contradiction because {| a n k |} converges to | a | . Thus A is bounded. Let us now show that A is closed. Let { a n } be a sequence in A that converges to a point a ∈ R . By the definition of compactness, { a n } has a subsequence { a n k } that converges to b ∈ A . Then a = b ∈ A and, hence, A is closed by Theorem 2.6.3 . For the converse, suppose A is closed and bounded and let { a n } be a sequence in A . Since A is bounded, the sequence is bounded and, by the Bolzano-Weierstrass theorem (Theorem 2.4.1 ) , it 1 This definition of compactness is more commonly referred to as sequential compactness .