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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?2012-03-23
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    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.2012-03-23
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    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.2012-03-23
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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 $\langle X,d\rangle$.2012-03-24
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    @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?2012-03-24
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    Yes, it is. Since $F\subseteq K$, it’s clear that $F\subseteq K_\epsilon$ for any $\epsilon$, so everything depends on getting $K\subseteq F_\epsilon$, which is exactly what we’ve done.2012-03-24
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    Correct, you need local total boundedness to show Fin is dense in BC...2012-03-24

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