Introduction to Mathematical Analysis I - Second Edition

59 Then G is a union of open subsets of R , so G is open. Moreover, V ⊂ G ∩ D = ∪ a ∈ V [ B ( a ; δ a ) ∩ D ] ⊂ V . Therefore, V = G ∩ D . Let us now prove the converse. Suppose V = G ∩ D , where G is an open set. For any a ∈ V , we have a ∈ G , so there exists δ > 0 such that B ( a ; δ ) ⊂ G . It follows that B ( a ; δ ) ∩ D ⊂ G ∩ D = V . The proof is now complete. Example 2.6.7 Let D = [ 0 , 1 ) and V = [ 0 , 1 2 ) . We can write V = D ∩ ( − 1 , 1 2 ) . Since ( − 1 , 1 2 ) is open in R , we conclude from Theorem 2.6.7 that V is open in D . Notice that V itself is not an open subset of R . The following theorem is now a direct consequence of Theorems 2.6.7 and 2.6.1 . Theorem 2.6.8 Let D be a subset of R . The following hold: (a) The subsets /0 and D are open in D . (b) The union of any collection of open sets in D is open in D . (c) The intersection of a finite number of open sets in D is open in D . Definition 2.6.6 Let D be a subset of R . We say that a subset A of D is closed in D if D \ A is open in D . Theorem 2.6.9 Let D be a subset of R . A subset K of D is closed in D if and only if there exists a closed subset F of R such that K = D ∩ F . Proof: Suppose K is a closed set in D . Then D \ K is open in D . By Theorem 2.6.7 , there exists an open set G such that D \ K = D ∩ G . It follows that K = D \ ( D \ K ) = D \ ( D ∩ G ) = D \ G = D ∩ G c . Let F = G c . Then F is a closed subset of R and K = D ∩ F . Conversely, suppose that there exists a closed subset F of R such that K = D ∩ F . Then D \ K = D \ ( D ∩ F ) = D \ F = D ∩ F c . Since F c is an open subset of R , applying Theorem 2.6.7 again, one has that D \ K is open in D . Therefore, K is closed in D by definition.