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Let $X$ be a Hausdorff, locally compact but non-compact topological space. If the (Alexandroff) one-point compactification is connected, can $X$ have compact connected components?

Stefan Hamcke
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iago
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2 Answers2

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I think I proved the following

Lemma

Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$.

in this answer.

For the present problem this implies a negative answer, since a compact set in a locally compact Hausdorff space has a compact neighbourhood. Let $C$ be a compact component of $X$, and let $K$ be some compact neighbourhood of $C$ in $X$. Applying the lemma one way then shows that $C$ is a component of $K$, and then applying it the other way shows that $C$ is a component of the compactification, since $K$ remains a compact neighbourhood of $C$.

Community
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Niels J. Diepeveen
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  • +1. That Lemma seems correct (after going through it) and it is quite powerful :) Thanks again for pointing out my mistake in my answer below. – Stefan Hamcke Feb 08 '14 at 23:16
  • @StefanHamcke: Thanks for confirming that. I wasn't too confident since hardly anyone else seemed to even have looked at it. – Niels J. Diepeveen Feb 09 '14 at 01:13
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Well, no. Assume $X=Y\sqcup C$, where $C$ is compact. Then the 1PC is $X\cup\{\infty\}$ with the topology: $U$ is open iff $U\subset X$ is open in the previous topology or $\infty\in U$ and $U^c$ is compact. Then $C = (U\cup\{\infty\})^c$ is compact, and thus $U\cup\{\infty\}$ is open. $C$ is also open (it was open in $X$). Thus the 1PC of $X$ is not connected.

Daniel Robert-Nicoud
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