# Compactness of $\mathcal K$ in the Hausdorff distance [duplicate]

Possible Duplicate:
The Class of Non-empty Compact Subsets of a Compact Metric Space is Compact

Let $(M,d)$ be a metric space and let $\mathcal K(M)$ denote the set of all non-empty compact subsets of $M$. This collection is a metric space when equipped with the Hausdorff distance $h$.

I want to prove$$(M,d)\mbox{ is compact}\implies(\mathcal K,h)\mbox{ is compact}.$$ The statement is true according to the book [V. I. Istratescu, Fixed Point Theory: An Introduction], but the proof is omitted. I have already shown that $M$ is complete implies that $\mathcal K$ is complete.

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## marked as duplicate by t.b., Norbert, Nate Eldredge, J. M., Davide GiraudoAug 11 '12 at 12:07

Since $M$ is totally bounded, for each $\epsilon > 0$ there is a finite set $\{x_1, \ldots, x_n\} \subseteq M$ such that $\min_{i=1}^n d(x,x_i) < \epsilon$ for all $x \in M$. For any nonempty compact $C \subseteq M$, if $S = \{x_i: d(C, x_i) < \epsilon\}$ we have $h(S,C) < \epsilon$. Thus every $C \in {\cal K}(M)$ is within Hausdorff distance $\epsilon$ of one of the finitely many nonempty subsets of $\{x_1,\ldots,x_n\}$. This shows that ${\cal K}(M)$ is totally bounded. Since you already know that ${\cal K}(M)$ is complete, it is compact.