Introduction to Mathematical Analysis I - Second Edition

60 2.6 OPEN SETS, CLOSED SETS, COMPACT SETS, AND LIMIT POINTS Example 2.6.8 Let D = [ 0 , 1 ) and K = [ 1 2 , 1 ) . We can write K = D ∩ [ 1 2 , 2 ] . Since [ 1 2 , 2 ] is closed in R , we conclude from Theorem 2.6.9 that K is closed in D . Notice that K itself is not a closed subset of R . Corollary 2.6.10 Let D be a subset of R . A subset K of D is closed in D if and only if for every sequence { x k } in K that converges to a point ¯ x ∈ D it follows that ¯ x ∈ K . Proof: Let D be a subset of R . Suppose K is closed in D . By Theorem 2.6.9 , there exists a closed subset F of R such that K = D ∩ F . Let { x k } be a sequence in K that converges to a point ¯ x ∈ D . Since { x k } is also a sequence in F and F is a closed subset of R , ¯ x ∈ F . Thus, ¯ x ∈ D ∩ F = K . Let us prove the converse. Suppose by contradiction that K is not closed in D or D \ K is not open in D . Then there exists ¯ x ∈ D \ K such that for every δ > 0, one has B ( ¯ x ; δ ) ∩ D * D \ K . In particular, for every k ∈ N , B ¯ x ; 1 k ∩ D * D \ K . For each k ∈ N , choose x k ∈ B ( ¯ x ; 1 k ) ∩ D such that x k / ∈ D \ K . Then { x k } is a sequence in K and, moreover, { x k } converges to ¯ x ∈ D . Then ¯ x ∈ K . This is a contradiction. We conclude that K is closed in D . The following theorem is a direct consequence of Theorems 2.6.9 and 2.6.2 . Theorem 2.6.11 Let D be a subset of R . The following hold: (a) The subsets /0 and D are closed in D . (b) The intersection of any collection of closed sets in D is closed in D . (c) The union of a finite number of closed sets in D is closed in D . Example 2.6.9 Consider the set D = [ 0 , 1 ) and the subset A = [ 1 2 , 1 ) . Clearly, A is bounded. We showed in Example 2.6.8 that A is closed in D . However, A is not compact. We show this by finding a sequence { a n } in A for which no subsequence converges to a point in A . Indeed, consider the sequence a n = 1 − 1 2 n for n ∈ N . Then a n ∈ A for all n . Moreover, { a n } converges to 1 and, hence, every subsequence also converges to 1. Since 1 6∈ A , it follows that A is not compact. Exercises 2.6.1 Prove that a subset A of R is open if and only if for any x ∈ A , there exists n ∈ N such that ( x − 1 / n , x + 1 / n ) ⊂ A . 2.6.2 Prove that the interval [ 0 , 1 ) is neither open nor closed. 2.6.3 I Prove that if A and B are compact subsets of R , then A ∪ B is a compact set.