a compact set in a Hausdorff space is closed
Proof. Let be a compact set in a Hausdorff space . The case when is empty is trivial, so let us assume that is non-empty. Using this theorem (http://planetmath.org/APointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods), it follows that each point in has a neighborhood , which is disjoint to . (Here, we denote the complement of by .) We can therefore write
Since an arbitrary union of open sets is open, it follows that is closed.
- 1 J.L. Kelley, General Topology, D. van Nostrand Company, Inc., 1955.
- 2 I.M. Singer, J.A.Thorpe, Lecture Notes on Elementary Topology and Geometry, Springer-Verlag, 1967.
|Title||a compact set in a Hausdorff space is closed|
|Date of creation||2013-03-22 13:34:31|
|Last modified on||2013-03-22 13:34:31|
|Last modified by||mathcam (2727)|