Let $X$ be a metrical space. Let $B(X)$ be the set of nonempty, bounded subsets of $X$. Define a semi-metrics on $B(X)$ ($0$ distance can exist) which inducates this convergence: $A_n\rightarrow A$, if for every $\epsilon>0$ there exist $N$ such that for every $n>N$, $A_n\subset A_\epsilon$ and $A\subseteq (A_n)_\epsilon$, where $H_c$ means the open sets with radius $c$ neighbourhood.
Let $K(X)$ be the set of nonempty, compact subsets of $X$. Show that the prevously defined semi-metrics is a metrics on $K(X)$ and if $X$ is complete, then the defined metric space is complete.
I believe the first part of this question is rather easy, still I was not able to figure it out. Any help would be much appreciated!
Given a metric space $(X,d)$, as I understood your notation the required convergence is provided by Hausdorff distance $d_H$, whose restriction on $K(X)$ is a metric and this is proved in Proposition 2-2 of [H]. When $(X, d)$ is complete, then $(K(X), d_H)$ is complete, and this is proved in Theorem 3-3 of [H].
References
[H] Jeff Henrikson, Completeness and total boundedness of the Hausdorff metric, MIT Undergraduate Journal of Mathematics, 1999 69–80.