closed set in a compact space is compact
Proof. Let be a closed set in a compact space . To show that is compact, we show that an arbitrary open cover has a finite subcover. For this purpose, suppose be an arbitrary open cover for . Since is closed, the complement of , which we denote by , is open. Hence and together form an open cover for . Since is compact, this cover has a finite subcover that covers . Let be this subcover. Either is part of or is not. In any case, is a finite open cover for , and is a subcover of . The claim follows.
|Title||closed set in a compact space is compact|
|Date of creation||2013-03-22 13:34:02|
|Last modified on||2013-03-22 13:34:02|
|Last modified by||mathcam (2727)|