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 ${A}^{\mathrm{\complement}}$ has a neighborhood^{} ${U}_{y}$, which is disjoint to $A$. (Here, we denote the complement of $A$ by ${A}^{\mathrm{\complement}}$.) We can therefore write
${A}^{\mathrm{\complement}}$ | $=$ | $\bigcup _{y\in {A}^{\mathrm{\complement}}}}{U}_{y}.$ |
Since an arbitrary union of open sets is open, it follows that $A$ is closed. $\mathrm{\square}$
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 |
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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 |