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closed set in a subspace∗ yark† 2013-03-21 19:47:48 In the following, let X be a topological space. Theorem 1. Suppose Y ⊆ X is equipped with the subspace topology, and A ⊆ Y . Then A is closed in Y if and only if A = Y ∩ J for some closed set J ⊆ X. Proof. If A is closed in Y , then Y \ A is open in Y , and by the definition of the subspace topology, Y \ A = Y ∩ U for some open U ⊆ X. Using properties of the set difference, we obtain A = Y \ (Y \ A) = Y \ (Y ∩ U ) = Y \U = Y ∩ U {. On the other hand, if A = Y ∩ J for some closed J ⊆ X, then Y \ A = Y \ (Y ∩ J) = Y ∩ J { , and so Y \ A is open in Y , and therefore A is closed in Y. Theorem 2. Suppose X is a topological space, C ⊆ X is a closed set equipped with the subspace topology, and A ⊆ C is closed in C. Then A is closed in X. Proof. This follows from the previous theorem: since A is closed in C, we have A = C ∩ J for some closed set J ⊆ X, and A is closed in X. ∗ hClosedSetInASubspacei created: h2013-03-21i by: hyarki version: h37460i Privacy setting: h1i hTheoremi h54B05i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1