116 5.1 Topology of the Real Line (b) The sets A=(−∞,c) andB=(c,∞) are open, but the set C=[c,∞) is not open. The reader can easily verify that Aand Bare open. Let us show that Cis not open. Assume by contradiction that Cis open. Then, for the element c ∈C, there exists δ >0 such that B(c;δ)=(c−δ,c+δ) ⊂C. However, this is a contradiction because for any positive δ we have that c−δ/2 /∈C. (c) The set A=Zc is open. To show this, let a∈Zc, that is, a is not a positive integer, and let m be the closest integer to a(see Theorem 1.6.2). Setting δ =|a−m|, we get δ >0 since m̸=a and then B(x;δ)∩Z=0/ , that is, B(x;δ) ⊂A. Theorem 5.1.1 The following hold: (i) The subsets 0/ andRare open. (ii) The union of any collection of open subsets of Ris open. (iii) The intersection of a finite number of open subsets of Ris open. Proof: The proof of (i) is straightforward. (ii) Suppose {Gα : α∈I}is an arbitrary collection of open subsets of R. That means Gα is open for everyα∈I. Let us show that the set G=[ α∈I Gα is open. Take any a∈G. Then there exists α0 ∈I such that a∈Gα0. Since Gα0 is open, there exists δ >0 such that B(a;δ) ⊂Gα0. This implies B(a;δ) ⊂G, because Gα0 ⊂G. Thus, Gis open. (iii) Suppose Gi,i =1,...,n, are open subsets of R. Let us show that the set G= n\ i=1 Gi is also open. Take anya∈G. Then a∈Gi for i =1,...,n. Since eachGi is open, there exists δi >0, i =1,...,n, such that B(a;δi) ⊂Gi. Let δ =min{δi : i =1,...,n}. Then δ >0 and B(a;δ) ⊂B(a,δi) ⊂Gi, for i =1,...,n. Therefre B(a;δ) ⊂G. Thus, Gis open. □ Definition 5.1.2 Asubset S of Ris calledclosed if its complement, Sc =R\S, is open. ■ Example 5.1.2 The following examples illustrate closed sets. (a) The sets (−∞,a], and[a,∞) are closed. Indeed, (−∞,a]c =(a,∞) and[a,∞)c =(−∞,a) which are open by Example 5.1.1.
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