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.
Best Answer
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.