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 Zis closed as follows from Example 5.1.1(c) Theorem 5.1.2 The following hold: (i) The sets 0/ andRare closed. (ii) The intersection of any collection of closed subsets of Ris closed. (iii) The union of a finite number of closed subsets of Ris 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=\ α∈I Sα is also closed. We have Sc = \ α∈I Sα! c =[ α∈I Sc α. Thus, Sc is open because it is a union of opens sets inR(Theorem 5.1.1(ii)). Therefore, Sis closed. □ ■ Example 5.1.3 It follows from part (iii) and Example 5.1.2 that any finite set is closed. Theorem 5.1.3 Let Abe a subset of R. The following are equivalent: (i) Ais closed. (ii) for any sequence {an}inAthat converges to a point a∈R, it follows that a∈A. Proof: Assume first that (i) holds. Let {an} is a sequence in A that converges to a. Suppose by contradiction that a̸ ∈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∈Nsuch 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 Ais 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)∩A̸=0/ . In particular, for such an a and for each n∈N, there exists an ∈B(a; 1 n)∩A. It is clear that the sequence {an} is in Aand it is convergent to a (because |an −a| < 1 n, for all n∈N). This contradicts (ii) since a̸∈A. Therefore, Ais 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 Ais a nonempty subset of Rthat is closed and bounded above, then maxAexists. Similarly, if Ais a nonempty subset of Rthat is closed and bounded below, then minAexists
RkJQdWJsaXNoZXIy NTc4NTAz