# For the existence of one-point compactification, do we need locally compactness?

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)

• If you remove a point from a compact Hausdorff space, you get a locally compact space. – Lord Shark the Unknown Dec 26 '18 at 8:53
• 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 – Alessandro Codenotti Dec 26 '18 at 9:02
• For every space there is the one-point-Alexandrov compactification. – drhab Dec 26 '18 at 9:03
• "we can choose a compact set in $X$..." last paragraph – Alessandro Codenotti Dec 26 '18 at 9:07
• In topology a property P is called hereditary when ($X\subset Y$ and $Y$ has property P$)\implies (X$ has property P). – DanielWainfleet Dec 27 '18 at 17:27

## 1 Answer

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.