121 Theorem 5.1.10 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 5.1.8, 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 ∩ Gc . Let F = Gc . 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∩ Fc . Since Fc is an open subset of R, applying Theorem 5.1.8 again, one has that D \ K is open in D. Therefore, K is closed in D by defnition. □ 1 1 ■ Example 5.1.8 Let D =[0,1) and K =[2 ,1). We can write K = D ∩ [ 1 2 ,2]. Since [2 ,2] is closed in R, we conclude from Theorem 5.1.10 that K is closed in D. Notice that K itself is not a closed subset of R. Corollary 5.1.11 Let D be a subset of R. A subset K of D is closed in D if and only if for every sequence {xn} in K that converges to a point x0 ∈ D it follows that x0 ∈ K. Proof: Let D be a subset of R. Suppose K is closed in D. By Theorem 5.1.10, there exists a closed subset F of R such that K = D ∩ F. Let {xn} be a sequence in K that converges to a point x0 ∈ D. Since {xn} is also a sequence in F and F is a closed subset of R, x0 ∈ F. Thus, x0 ∈ D ∩ F = K. Let us prove the converse. Suppose by contradiction that K is not closed in D or D \ K is not open in D. Then there exists x0 ∈ D \ K such that for every δ > 0, one has B(x0;δ ) ∩ D ⊈ D\ K. In particular, for every n ∈ N, 1 B x0; ∩ D ⊈ D \ K. n For each n ∈ N, choose xn ∈ B(x0; 1 ) ∩ D such that x n ∈/ D \ K. Then {xn} is a sequence in K and, n moreover, {xn} converges to x0 ∈ D. Then x0 ∈ K. This is a contradiction. We conclude that K is closed in D. □ The following theorem is a direct consequence of Theorems 5.1.10 and 5.1.2.
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