Completeness of the Hausdorff hyperspace

For any nonempty complete metric space $X$, the family of its nonempty closed bounded subsets is a complete metric space under the Hausdorff distance. Boundedness of the entire ambient space is unnecessary in this general form.

For a Cauchy sequence $(F_n)$ of these sets, define $A_n=\overline{{\bigcup }_{m\ge n}{F}_m}$ and ${\delta }_n={\sup }_{p,q\ge n}{d}_H(F_p,F_q)$. A Cauchy sequence in any metric space is bounded, so its tail sets are contained in a common finite-radius neighbourhood of one bounded $F_N$; hence every $A_n$ is bounded. The distance to a set is one-Lipschitz, so estimates pass to the closure, giving $d_H(F_n,A_n)\le {\delta }_n$ and $d_H(A_n,A_m)\le {\delta }_n$ for $m\ge n$. Choose $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. The Summably nested Hausdorff intersection theorem gives $A_{n_j}\to G\ne \varnothing$. The triangle inequality now gives $F_n\to G$, proving completeness.