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Given some topological space $\mathbf{X}$, we consider the Fell topology on the set of closed subsets of $\mathbf{X}$. This is generated by sets of the form $I_U = \{A \mid A \cap U \neq \emptyset\}$ with $U$ ranging over open subsets of $\mathbf{X}$ together with sets of the form $D_K = \{A \mid A \cap K = \emptyset\}$ where $K$ ranges over compact subsets of $\mathbf{X}$. Let $\mathcal{F}(\mathbf{X})$ be the topological space constructed as such.

It seems to be known that if $\mathbf{X}$ is locally compact, then $\mathcal{F}(\mathbf{X})$ is compact.

What is known about the inverse implication? I am interested in both the general case, and the restriction to countably-based based spaces.

Bonus question: What exactly is meant by "local compactness" here?

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  • $\begingroup$ It's the first time that I hear that the above topology is called Fell topology (sometimes it was called Hausdorff topology, with the understanding that it was a generalized Hausdorff topology. Could you say more about the history of this topology? $\endgroup$
    – Wlod AA
    Jun 23, 2018 at 21:48
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    $\begingroup$ @WlodAA mathoverflow.net/questions/121775/… $\endgroup$
    – Adam Epstein
    Jun 24, 2018 at 7:58
  • $\begingroup$ @AdamEpstein, thank you. Was Fell already born when they invented the topology under consideration? (My classical association of "fell" and topology is the horrible Hurewicz's fall in Mexico which caused him his life). $\endgroup$
    – Wlod AA
    Jun 24, 2018 at 9:02

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Local compactness means that every point has a compact neighborhood. That is, every point is interior to a compact set. The Fell topology on the nonempty closed sets of a Hausdorff space is always compact, without any further assumption. However, local compactness is equivalent to the Fell topology being Hausdorff.

You can find a lot more material on the topic in the 1993 book "Topologies on closed and closed convex sets" by Gerald Beer.

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  • $\begingroup$ What happens without the Hausdorff assumption? $\endgroup$
    – Arno
    Jun 24, 2018 at 11:44
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    $\begingroup$ I'm not sure. Beer writes that he focuses on the Hausdorff case for simplicity, but many people have worked on the non-Hausdorff case. So the assumption might well not be needed. $\endgroup$
    – Michael Greinecker
    Jun 24, 2018 at 11:46

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