Introduction to Mathematical Analysis I - Second Edition

56 2.6 OPEN SETS, CLOSED SETS, COMPACT SETS, AND LIMIT POINTS (c) Suppose G i , i = 1 , . . . , n , are open subsets of R . Let us show that the set G = n \ i = 1 G i is also open. Take any a ∈ G . Then a ∈ G i for i = 1 , . . . , n . Since each G i is open, there exists δ i > 0 such that B ( a ; δ i ) ⊂ G i . Let δ = min { δ i : i = 1 , . . . , n } . Then δ > 0 and B ( a ; δ ) ⊂ G . Thus, G is open. Definition 2.6.2 A subset S of R is called closed if its complement, S c = R \ S , is open. Example 2.6.2 The sets [ a , b ] , ( − ∞ , a ] , and [ a , ∞ ) are closed. Indeed, ( − ∞ , a ] c = ( a , ∞ ) and [ a , ∞ ) c = ( − ∞ , a ) which are open by Example 2.6.1 . Since [ a , b ] c = ( − ∞ , a ) ∪ ( b , ∞ ) , [ a , b ] c is open by Theorem 2.6.1 . Also, single element sets are closed since, say, { b } c = ( − ∞ , b ) ∪ ( b , ∞ ) . Theorem 2.6.2 The following hold: (a) The sets /0 and R are closed. (b) The intersection of any collection of closed subsets of R is closed. (c) The union of a finite 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, (b) . Let { S α : α ∈ I } be a collection of closed sets. We will prove that the set S = \ α ∈ I S α is also closed. We have S c = \ α ∈ I S α ! c = [ α ∈ I S c α . Thus, S c is open because it is a union of opens sets in R (Theorem 2.6.1 ( b)). Therefore, S is closed. Example 2.6.3 It follows from part (c) and Example 2.6.2 that any finite set is closed. Theorem 2.6.3 A subset A of R is closed if and only if for any sequence { a n } in A that converges to a point a ∈ R , it follows that a ∈ A . Proof: Suppose A is a closed subset of R and { a n } is a sequence in A that converges to a . Suppose by contradiction that a 6∈ A . Since A is closed, there exists ε > 0 such that B ( a ; ε ) = ( a − ε , a + ε ) ⊂ A c . Since { a n } converges to a , there exists N ∈ N such that a − ε < a N < a + ε . This implies a N ∈ A c , a contradiction.