a compact set in a Hausdorff space is closed

Theorem. A compact set in a Hausdorff space is closed.

Proof. Let $A$ be a compact set in a Hausdorff space $X$ . The case when $A$ is empty is trivial, so let us assume that $A$ is non-empty. Using this theorem (http://planetmath.org/APointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods), it follows that each point $y$ in $A^{\complement}$ has a neighborhood $U_{y}$ , which is disjoint to $A$ . (Here, we denote the complement of $A$ by $A^{\complement}$ .) We can therefore write

\displaystyle A^{\complement}

\displaystyle=

\displaystyle\bigcup_{y\in A^{\complement}}U_{y}.

Since an arbitrary union of open sets is open, it follows that $A$ is closed. $\Box$

Note.ÃÂ
The above theorem can, for instance, be found in [1] (page 141), or [2] (Section 2.1, Theorem 2).

References

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
Canonical name	ACompactSetInAHausdorffSpaceIsClosed
Date of creation	2013-03-22 13:34:31
Last modified on	2013-03-22 13:34:31
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	6
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 54D10
Classification	msc 54D30
Related topic	ClosedSubsetsOfACompactSetAreCompact