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Definitions:

  • KC (P100): each compact subset of the space is closed.
  • $k_2$-Hausdorff (P171): for every compact $T_2$ space and continuous map $f:K\to X$ and points $k,l\in K$ with $f(k)\not=f(l)$, there exist open neighborhoods $U,V$ of $k,l$ with $f[U],f[V]$ disjoint.
  • Alexandrov extension: the space $Y=X\cup\{\infty\}$ such that neighborhoods of $\infty$ are the complements of compact-and-closed subsets of $X$.
  • Compactification: a space $Y\supseteq X$ such that $Y$ is compact and $X$ is dense.

It's known that the Alexandrov extension is the unique Hausdorff one-point compactification of a space whenever the Alexandrov extension is Hausdorff. It's also known that the Alexandrov extension of any $KC$ space is $k_2$-Hausdorff.

Is the Alexandrov extension the unique $k_2$-Hausdorff one-point compactification of a KC space?

PatrickR
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Steven Clontz
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  • Also, from the proof in the linked answer, you could replace "Hausdorff" with $KC$, ie, the proof there shows that if the Alexandrov extension of a space is $KC$, then it is the unique $KC$ one-point compactification. – M W Jan 05 '24 at 23:32

1 Answers1

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This is not the case, even if the original uncompactified space is locally compact Hausdorff. For a counterexample, consider $X=\mathbb R$ with the Euclidean topology, and let $Y=X\cup \{\infty\}$ with the neighborhoods of $\infty$ consisting of sets $U\cup\{\infty\}$ where $U\subseteq \mathbb R$ is dense, open, and cocompact.

A very similar space ($[0,1]\cup\{\infty\}$) is constructed in this question and is shown there to be $k_2$-Hausdorff. The exact same argument shows our space $Y$ is $k_2$-Hausdorff as well.

Moreover, $Y$ is certainly compact, since any open cover of $Y$ includes a neighborhood of the form $U\cup\{\infty\}$, with $U\subseteq \mathbb R$ cocompact. And $X$ is dense in $Y$, since $\infty$ is not isolated.

Thus even $\mathbb R$ has another $k_2$-Hausdorff one-point compactification.

Remark.

$Y$ is not $KC$, as compact intervals in $\mathbb R$ are not closed in $Y$. This is to be expected, since as observed in the comments, the arguments in this answer show that a space can have at most one $KC$ one-point compactification (everything in the linked answer is stated for Hausdorff spaces, but only $KC$ is used.)

M W
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