116 5.1 Topology of the Real Line (b) The sets A =(−∞,c) and B =(c,∞) are open, but the set C =[c,∞) is not open. The reader can easily verify that A and B are open. Let us show that C is not open. Assume by contradiction that C is 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 m ≠ 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/ and R are open. (ii) The union of any collection of open subsets of R is open. (iii) The intersection of a fnite number of open subsets of R is 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 every α ∈ I. Let us show that the set [ G = Gα α∈I 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, G is open. (iii) Suppose Gi,i = 1,...,n, are open subsets of R. Let us show that the set n\ G = Gi i=1 is also open. Take any a ∈ G. Then a ∈ Gi for i = 1,...,n. Since each Gi 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, G is open. □ Defnition 5.1.2 A subset S of R is called closed 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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