117 (b) Since [a,b]c =(−∞,a) ∪ (b,∞), [a,b]c is open by Theorem 5.1.1. Also, single element sets are closed since, say, {b}c =(−∞,b) ∪ (b,∞) which is open. (c) The set Z is closed as follows from Example 5.1.1(c) Theorem 5.1.2 The following hold: (i) The sets 0/ and R are closed. (ii) The intersection of any collection of closed subsets of R is closed. (iii) The union of a fnite number of closed subsets of R is closed. Proof: The proofs for these are simple using the De Morgan’s law. Let us prove, for instance, (ii). Let {Sα : α ∈ I} be a collection of closed sets. We will prove that the set \ S = Sα α∈I is also closed. We have ! c \ [ Sc Sc = Sα = α . α∈I α∈I Thus, Sc is open because it is a union of opens sets in R (Theorem 5.1.1(ii)). Therefore, S is closed. □ ■ Example 5.1.3 It follows from part (iii) and Example 5.1.2 that any fnite set is closed. Theorem 5.1.3 Let A be a subset of R. The following are equivalent: (i) A is closed. (ii) for any sequence {an} in A that converges to a point a ∈ R, it follows that a ∈ A. Proof: Assume frst that (i) holds. Let {an} is a sequence in A that converges to a. Suppose by contradiction that a ̸ ∈ A. Since A is closed, Ac is open. Hence, there exists ε > 0 such that B(a;ε)=(a − ε,a + ε) ⊂ Ac . Since {an} converges to a, there exists N ∈ N such that a− ε < aN < a + ε. This implies aN ∈ Ac, a contradiction. Let us now prove the converse. Assume (ii) holds. Suppose by contradiction that A is not closed. Then Ac is not open. Since Ac is not open, there exists a ∈ Ac such that for any ε > 0, one has B(a;ε) ∩ A ̸ = 0/. In particular, for such an a and for each n ∈ N, there exists an ∈ B(a; 1 ) ∩ A. It is n 1 clear that the sequence {an} is in A and it is convergent to a (because |an − a| < , for all n ∈ N). n This contradicts (ii) since a ̸ ∈ A. Therefore, A is closed. □ The following result relates closed sets to the existence of maximum and minimum elements (see Remark 1.5.1). Theorem 5.1.4 If A is a nonempty subset of R that is closed and bounded above, then maxA exists. Similarly, if A is a nonempty subset of R that is closed and bounded below, then minA exists
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