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 |
---|---|
Canonical name | ClosedSetInACompactSpaceIsCompact |
Date of creation | 2013-03-22 13:34:02 |
Last modified on | 2013-03-22 13:34:02 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 9 |
Author | mathcam (2727) |
Entry type | Proof |
Classification | msc 54D30 |
Related topic | ClosedSubsetsOfACompactSetAreCompact |