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Recall that the Fell topology $\tau_F$ is a topology on the hyperspace $F(X)$ of closed subsets of a Hausdorff space (maybe you can define it in a more general context, but I am interested in $\mathbb{R}^n$). A prebase for the Fell topology is given by the sets of the form

$$\{ F\in F(X) : F\cap K =\emptyset \} $$ $$\{ F\in F(X) : F\cap U \neq \emptyset \} $$

for some $K\subset X$ compact and some non-empty $U\subset X$ open.

I read in [1, Thm. 5.1.5] that $(F(\mathbb{R}^n),\tau_F)$ is compact metrizable, and becomes Polish if we restrict it to $(F(\mathbb{R}^n)\backslash \{\emptyset\},\tau_F)$. I assume that it is implicitly saying that $(F(\mathbb{R}^n),\tau_F)$ is not complete.

How can this be proved explicitly? I would guess that, since $(F(\mathbb{R}^n),\tau_F)$ is metric, there should be a Cauchy non-convergent sequence, but it is not clear to me what is the metric on $(F(\mathbb{R}^n),\tau_F)$.

[1]: Beer, Gerald, Topologies on closed and closed convex sets, Mathematics and its Applications (Dordrecht). 268. Dordrecht: Kluwer Academic Publishers. xi, 340 p. (1993). ZBL0792.54008.

Manlio
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  • Since every compact metric space is complete, Beer's Theorem 5.1.5 says that $F(\mathbb{R}^n)$ ($=2^{\mathbb{R}^n}$) is completely metrisable. – Tyrone Nov 15 '20 at 14:53
  • Oh good point, you're right! I guess I should have thought about that... But does this mean that $F(\mathbb{R}^n)$ is Polish then? I wonder why Beer didn't say it explicitly, while he did for $F(\mathbb{R}^n\backslash \emptyset)$ – Manlio Nov 15 '20 at 15:39
  • He basically does if you read the proof of the implication $(2)\Rightarrow(3)$. – Tyrone Nov 15 '20 at 15:45
  • Well, thanks, I guess this answers the question. If you post it as an answer I will accept and upvote it! – Manlio Nov 15 '20 at 15:50

1 Answers1

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It's true that every compact metric space is complete. Since Beer proves in Theorem 5.1.5 that $F(\mathbb{R}^n)$ is compact and metrisable, it must be completely metrisable. I agree that Beer does not openly record this, but he does use it implicitly in the proof of the theorem.

btw, +1 for the well written question.

Tyrone
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