# Finite sets are dense with respect to Hausdorff distance

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

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