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Given a topological space $X$, under what conditions is $X$ the one-point compactification of some other space $Y$? Obviously $X$ must be compact but it seems unlikely to me that every compact space arises from one-point compactification.

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    Certainly finite sets under the discrete topology are (counter)examples. Would it not suffice for $X$ to contain a limit point $x$, so that $Y=X-{x}$ works? – Greg Martin Nov 15 '19 at 18:30
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    I'd disagree: I would say a discrete space with $n+1$ points is the one-point compactification of a discrete space with $n$ points @GregMartin – Angina Seng Nov 15 '19 at 18:43
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    @LordSharktheUnknown: It is the Alexandroff extension, but not the one-point compactification, because it is not a compactification, because the discrete space with $n$ points is not dense in the discrete space with $n+1$ points. The Alexandroff extension is a compactification exactly if the original space is not compact. – celtschk Nov 15 '19 at 21:08
  • @GregMartin: You also need that ${x}$ is closed. – celtschk Nov 15 '19 at 21:19
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    @GregMartin: And even that is not sufficient. Counterexample: Be $Y=\mathbb R$ with the usual topology, and $X=\mathbb R\cup{\infty}$ with the topology where a set is open if it is either an open subset of $\mathbb R$ or the full space. This clearly is a compact space, $\infty$ is a limit point (so it is indeed a compactification), and ${\infty}$ is closed (the condition I added above). Yet $X$ is not the one-point compactification of $Y=\mathbb R$ because e.g. ${0}$ is not closed in $X$, while in the one-point compactification of $\mathbb R$ it is. – celtschk Nov 15 '19 at 21:35
  • Cool, thanks for articulating your expertise! So let's see, we'd need something like: the intersection of all neighborhoods of $x$ is ${x}$...? – Greg Martin Nov 15 '19 at 21:39
  • @GregMartin: No, because it might be that this is not the case in $Y$ (there was no restriction to T1 spaces), and then it is also not the case in the one-point compactification of $Y$. What is needed is that all closed compact sets of $Y$ need still be closed in $X$. However at the moment I don't see an easy criterion that is simpler than “$X$ is the one-point compactification of $Y$ up to homeomorphism,” – celtschk Nov 15 '19 at 21:52

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Any compact Hausdorff space $X$ with a non-isolated point $p\in X$ (so $\{p\}$ is not open) is homeomorphic to the one-point compactification of $Y:=X\setminus \{p\}$ by standard theorems.

A finite discrete space shows why we need the condition for a non-isolated point.

Henno Brandsma
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