a compact set in a Hausdorff space is closedACompactSetInAHausdorffSpaceIsClosed2003-04-17 14:04:032003-05-03 12:57:05Theoremcompact% this is the default PlanetMath preamble. as your knowledge
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{\bf Theorem.} A compact set in a Hausdorff space is closed.
\emph{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 \PMlinkname{this theorem}{APointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods},
it follows that each point
$y$ in $A^{\comp}$ has a neighborhood $U_y$, which
is disjoint to $A$. (Here, we denote the complement of $A$
by $A^{\comp}$.)
We can therefore write
\begin{eqnarray*}
A^{\comp} &=& \bigcup_{y\in A^{\comp}} U_y.
\end{eqnarray*}
Since an arbitrary union of open sets is open, it follows that $A$ is
closed. $\Box$
{\bf Note.} \\
The above theorem can, for instance, be found in \cite{kelley} (page 141),
or \cite{singer} (Section 2.1, Theorem 2).
\begin{thebibliography}{9}
\bibitem{kelley}
J.L. Kelley,
\emph{General Topology},
D. van Nostrand Company, Inc., 1955.
\bibitem{singer}
I.M. Singer, J.A.Thorpe,
\emph{Lecture Notes on Elementary Topology and Geometry},
Springer-Verlag, 1967.
\end{thebibliography}