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Y is compact if and only if every open cover of Y has a finite subcover∗ mathcam† 2013-03-21 15:53:29 Theorem. Let X be a topological space and Y a subset of X. Then the following statements are equivalent. 1. Y is compact as a subset of X. 2. Every open cover of Y (with open sets in X) has a finite subcover. Proof. Suppose Y is compact, and {Ui }i∈I is an arbitrary open cover of Y , where Ui are open sets in X. Then {Ui ∩ Y }i∈I is a collection of open sets in Y with union Y . Since Y is compact, there is a finite subset J ⊂ I such that Y = ∪i∈J (Ui ∩ Y ). Now Y = (∪i∈J Ui ) ∩ Y ⊂ ∪i∈J Ui , so {Ui }i∈J is finite open cover of Y . Conversely, suppose every open cover of Y has a finite subcover, and {Ui }i∈I is an arbitrary collection of open sets (in Y ) with union Y . By the definition of the subspace topology, each Ui is of the form Ui = Vi ∩ Y for some open set Vi in X. Now Ui ⊂ Vi , so {Vi }i∈I is a cover of Y by open sets in X. By assumption, it has a finite subcover {Vi }i∈J . It follows that {Ui }i∈J covers Y , and Y is compact. The above proof follows the proof given in [?]. References [1] B.Ikenaga, Notes on Topology, August 16, 2000, available online http://www.millersv.edu/ bikenaga/topology/topnote.html. ∗ hYIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcoveri created: h2013-0321i by: hmathcami version: h34179i Privacy setting: h1i hTheoremi h54D30i † 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