is compact if and only if every open cover of has a finite subcover
Theorem.
Let be a topological space and a subset of . Then the following
statements are equivalent
.
-
1.
is compact
as a subset of .
-
2.
Every open cover of (with open sets in ) has a finite subcover.
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].
References
- 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 |
---|---|
Canonical name | YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover |
Date of creation | 2013-03-22 13:34:07 |
Last modified on | 2013-03-22 13:34:07 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 6 |
Author | mathcam (2727) |
Entry type | Theorem |
Classification | msc 54D30 |