In the book Topology by Munkres, at page 184, it is given the existence and uniqueness of one point compactification of a locally compact Hausdorff space; however, in the existence part, I can't see where we needed the locally compactness of that space, and this raised the question:

Does one point compactification of a Hausdorff space always exist (even though it is not unique) ?

See the proof in the book;

(sorry for the images; they are just for reference for those that doesn't have the book with them)

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    $\begingroup$ If you remove a point from a compact Hausdorff space, you get a locally compact space. $\endgroup$ – Lord Shark the Unknown Dec 26 '18 at 8:53
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    $\begingroup$ If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way $\endgroup$ – Alessandro Codenotti Dec 26 '18 at 9:02
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    $\begingroup$ For every space there is the one-point-Alexandrov compactification. $\endgroup$ – drhab Dec 26 '18 at 9:03
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    $\begingroup$ "we can choose a compact set in $X$..." last paragraph $\endgroup$ – Alessandro Codenotti Dec 26 '18 at 9:07
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    $\begingroup$ In topology a property P is called hereditary when ($X\subset Y$ and $Y$ has property P$)\implies (X $ has property P). $\endgroup$ – DanielWainfleet Dec 27 '18 at 17:27

For any space $X$ we can construct a space $\alpha(X)$, the Aleksandrov extension of $X$ by defining a space $Y$ as Munkres does with the extra provision that we take all complements of closed compact subsets of $X$ as the extra neighbourhoods for $\infty$. One can easily check that $\alpha(X)$ is then compact.

The "closed" is needed in general because if e.g. $X$ is not Hausdorff it could have some compact subset $K$ which is not closed, and then (if we were to omit the closed condition) $(X\setminus K) \cup \{\infty\}$ would be open while its intersection with $X$ would be $X\setminus K$, which was not open, so if we left out the closed condition $X$ would not have the same topology as a subspace of $\alpha(X)$ as originally, going against the idea of an extension/compactification: we want to embed $X$ in a larger space with better properties, so in the larger space it should be a subspace with the same topology that it had originally.

If we want $Y = \alpha(X)$ to be Hausdorff, (so in particular $X$ should then be Hausdorff, as a subspace of $Y$) we need to be able to separate $\infty$ from every point $x$ in $X$. As a neighbourhood of $\infty$ is of the form $\{\infty\} \cup X \setminus C$, with $C$ compact and closed, every point $x$ should then have a neighbourhood that sits inside a compact closed set, i.e. $X$ must be locally compact.

So $\alpha(X)$ can always be defined such that $\alpha(X)\setminus X$ is a point and $X$ is a subspace of $\alpha(X)$ and it is always compact (regardless of $X$) but $\alpha(X)$ is Hausdorff iff $X$ is locally compact and Hausdorff. A special case is when $X$ is already Hausdorff and compact, in which case we add an isolated point $\infty$ (as $X$ can be taken as $C$, a compact closed subset) and we get that $X$ is not dense in $\alpha(X)$.

Normally we only consider Hausdorff compactifications and in that case the local compactness is needed for the Hausdorffness of the construction $\alpha(X)$. And also because then $X$ is an open subset of a compact Hausdorff space and thus locally compact for that reason.


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