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Let $(X,d)$ be a metric space. Let $F(X)$ denote the collection of finite subsets of $X$ and let $K(X)$ be the collection of all non-empty compact subsets of $X$. I want to show that $K(X)$ is equal to the closure $F(X)$ w.r.t. the Hausdorff distance $d_H$ (see http://en.wikipedia.org/wiki/Hausdorff_distance).

I have done the following: By definition, $$d_H(F,K)=\inf\lbrace\varepsilon>0\colon F\subset K_\varepsilon\text{ and } K\subset F_\varepsilon\rbrace,$$where $$F_\varepsilon=\bigcup_{x\in F}\lbrace y\in X\colon d(y,x)<\varepsilon\rbrace\quad\text{and}\quad K_\varepsilon=\bigcup_{x\in K}\lbrace y\in X\colon d(y,x)<\varepsilon\rbrace.$$

It can be shown that that completeness of $S$ implies that $K(X)$ is complete w.r.t. $d_H$.

Let $K\subset X$ be compact. Then, by definition, for any $\varepsilon>0$ the open cover $\lbrace B(x,\varepsilon)\colon x\in K\rbrace$ has a finite subcover ($B(x,\varepsilon)=\lbrace y\in X\colon d(x,y)<\varepsilon\rbrace)$. Therefore, there are $x_1,\ldots x_n\in X$ and put $F=\lbrace x_1,\ldots, x_n\rbrace$ such that $$K\subset\bigcup_{k=1}^nB(x_k,\varepsilon).$$ Then, $$F_{\varepsilon}=\bigcup_{x\in F}\lbrace y\in X\colon d(y,x)<\varepsilon\rbrace=\bigcup_{k=1}^nB(x_k,\varepsilon).$$ So $K\subset F_{\varepsilon}.$

Is there some that knows how to finish the proof?

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Your argument provides $x_1,\dots,x_k$ that are not only in $X$ but in $K$. So $F\subseteq K\subseteq K_\varepsilon$. –  Andreas Blass Mar 29 '12 at 22:28
1  
Yes, that's very simple: $F\subset K$, which makes trivial the other inclusion $F \subset K_\epsilon$. –  Pietro Majer Mar 29 '12 at 22:29
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