About the one point compactification

Question

I have a question about the one-point compactification, given an arbitrary topological space $(X,\tau)$, does the one-point compactification always exist?

Thanks!

Answer 1

Yes (provided that the word "compactification" does not involve Hausdorffness, see @Brian M. Scott's answer. For me, a compactification of $X$ is just a compact space containing $X$ as a dense subspace. Under this definition, the one-point compactification of $X$ is a compactification if, and only if, $X$ is not already compact). As a set, it is $X \cup \{\infty\}$ and its open sets are the open sets in $X$ and the complements of closed compact subsets of $X$, where you add $\infty$ to each one of them. In some applications you want the compactification to be Hausdorff, in which case you need to add more assumptions, namely that $X$ itself is Hausdorff and locally compact.

Answer 2

No. A compactification of a space $X$ is an embedding of $X$ as a dense subset of a compact Hausdorff space, so if $X$ isn’t already Hausdorff, it has no compactification. And if $X$ is compact and Hausdorff, it will be closed in any Hausdorff space, so the only compact Hausdorff space in which it is dense is itself. It turns out that $X$ has a one-point compactification if and only if it is a locally compact, non-compact Hausdorff space.

However, one can start with any non-compact space $X$ and form the Alexandroff extension $Y$ by adding a single point in such a way as to get a compact space in which the original space is a dense subspace. (Specifically, the open nbhds of the added point are the complements in $Y$ of closed, compact subsets of $X$.) It is an actual compactification, however, if and only if you start with a locally compact, non-compact Hausdorff space, in which case it’s the one-point compactification of that space. (Note that some people do call it the one-point compactification, not requiring compactifications to be Hausdorff.)