# a compact set in a Hausdorff space is closed

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.ÃÂ
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 ACompactSetInAHausdorffSpaceIsClosed 2013-03-22 13:34:31 2013-03-22 13:34:31 mathcam (2727) mathcam (2727) 6 mathcam (2727) Theorem msc 54D10 msc 54D30 ClosedSubsetsOfACompactSetAreCompact