121 Theorem 5.1.10 Let Dbe a subset of R. Asubset Kof Dis closed in Dif and only if there exists a closed subset F of Rsuch that K=D∩F. Proof: Suppose Kis a closed set inD. ThenD\Kis open inD. By Theorem 5.1.8, there exists an open set Gsuch that D\K=D∩G. It follows that K=D\(D\K)=D\(D∩G)=D\G=D∩Gc. Let F =Gc. ThenF is a closed subset of RandK=D∩F. Conversely, suppose that there exists a closed subset F of Rsuch 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, Kis closed inDby definition. □ ■ Example 5.1.8 Let D=[0,1) andK=[ 1 2,1). We can write K=D∩[ 1 2,2]. Since [ 1 2,2] is closed in R, we conclude from Theorem 5.1.10 that Kis closed in D. Notice that Kitself is not a closed subset of R. Corollary 5.1.11 Let Dbe a subset of R. Asubset Kof Dis closed in Dif and only if for every sequence {xn} in Kthat converges to a point x0 ∈Dit follows that x0 ∈K. Proof: Let Dbe a subset of R. Suppose Kis closed in D. By Theorem 5.1.10, there exists a closed subset F of Rsuch that K=D∩F. Let {xn}be a sequence in Kthat converges to a point x0 ∈D. Since {xn}is also a sequence inF and F is a closed subset of R, x0 ∈F. Thus, x0 ∈D∩F =K. Let us prove the converse. Suppose by contradiction that Kis not closed in Dor D\Kis not open inD. Then there exists x0 ∈D\Ksuch that for everyδ >0, one has B(x0;δ)∩D⊈D\K. In particular, for everyn∈N, B x0; 1 n ∩ D⊈D\K. For each n∈N, choose xn ∈B(x0; 1 n)∩Dsuch that xn /∈D\K. Then {xn} is a sequence in Kand, moreover, {xn} converges to x0 ∈D. Then x0 ∈K. This is a contradiction. We conclude that Kis closed inD. □ The following theorem is a direct consequence of Theorems 5.1.10 and 5.1.2.
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