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Let $(X,d)$ be a complete metric space and consider \begin{align*} BC(X)&= \lbrace C\subset X\;|\;C\neq\emptyset\text {, closed and bounded} \rbrace\cr \mathrm{Fin}(X)&= \lbrace F\subset X\;|\;F\text{ is finite} \rbrace\subset BC(X)\cr \mathcal{K}(X)&= \lbrace K\subset X\;|\;K\text{ is compact} \rbrace\cr \end{align*} Consider the metric space $(BC(X), d_H)$ where $d_H$ is the Hausdorff distance (for the definition of $d_H$ see the Wikipedia entry on $d_H$)

I don't know how to prove that $\mathrm{Fin}(X)\subset BC(X)$ is dense in $\mathcal K(X)$ (with respect to $d_H$).

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I assume that when you write that a set $S$ is in $BC(X)$ you treat it as the function $x\mapsto d_H(S,\{x\})$, correct? –  Alex Becker Mar 23 '12 at 23:45
    
For each compact set $C$ consider the open cover $\{B(x,\epsilon):x\in C\}$. Since $C$ is compact, there is a finite set $F\subseteq C$ such that $\{B(x,\epsilon):x\in F\$$ is a cover of $C$. Show that by taking $\epsilon$ small enough, $F$ approximates $C$ arbitrarily well in Hausdorff distance. –  Michael Greinecker Mar 23 '12 at 23:51
    
This is false in general. You will need to assume metric space $X$ is locally compact. (without completeness: locally totally bounded) For an example, consider the unit ball of Hilbert space. –  GEdgar Mar 23 '12 at 23:52
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@GEdgar: You don’t need local compactness to show that $\operatorname{Fin}(X)$ is dense in $\mathcal{K}(X)$. You don’t even need completeness of $\langle X,d\rangle$. –  Brian M. Scott Mar 24 '12 at 0:23
    
@Brian M. Scott: Thank you for the hint. By definition of $d_H$, $F_\epsilon=\bigcup_{j=1}^n B(x_j,\epsilon)$. Since $K\subset F_\epsilon$ must hold, $\epsilon$ must become small for $F\subset K_{\epsilon}$ to hold. I this the right idea? –  mathQ Mar 24 '12 at 0:58

1 Answer 1

Suppose that $K\in\mathcal{K}(X)$ and $\epsilon>0$. Since $K$ is compact, it has a finite cover by $\epsilon$-balls, say $\{B(x_1,\epsilon),\dots,B(x_n,\epsilon)\}$. Let $F=\{x_1,x_2,\dots,x_n\}$. What can you say about $d_H(F,K)$? You may find it easiest to think of $d_H$ in terms of the characterization $$d_H(X,Y)=\inf\{\epsilon>0:X\subseteq Y_\epsilon\text{ and }Y\subseteq X_\epsilon\}$$ given in the Wikipedia article right after the definition.

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