is compact if and only if every open cover of has a finite subcover
is compact as a subset of .
Proof. Suppose is compact, and is an arbitrary open cover of , where are open sets in . Then is a collection of open sets in with union . Since is compact, there is a finite subset such that . Now , so is finite open cover of .
Conversely, suppose every open cover of has a finite subcover, and is an arbitrary collection of open sets (in ) with union . By the definition of the subspace topology, each is of the form for some open set in . Now , so is a cover of by open sets in . By assumption, it has a finite subcover . It follows that covers , and is compact.
The above proof follows the proof given in .
- 1 B.Ikenaga, Notes on Topology, August 16, 2000, available online http://www.millersv.edu/ bikenaga/topology/topnote.htmlhttp://www.millersv.edu/ bikenaga/topology/topnote.html.
|Title||is compact if and only if every open cover of has a finite subcover|
|Date of creation||2013-03-22 13:34:07|
|Last modified on||2013-03-22 13:34:07|
|Last modified by||mathcam (2727)|