Download Y is compact if and only if every open cover of Y has a finite subcover

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
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