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Let $(X,d)$ be a metric space and define for $B\subset X$ bounded, i.e.

$$\operatorname{diam}(B)= \sup \{ d(x,y) \colon x,y\in B \} < \infty,$$

the measure

$$\beta(B) = \inf\{r > 0\colon\text{there exist finitely many balls of radius r which cover } B\},$$

or equivalently,

$$\beta(B)=\inf\big\lbrace r > 0|\exists N=N(r)\in{\bf N} \text{ and } x_1,\ldots x_N\in X\colon B\subset\bigcup_{k=1}^N B(x_k,r)\big\rbrace,$$

where $B(x,r)=\{y\in X\colon d(x,y)\}$ denotes the open ball of radius $r$ centered at $x\in X$. Let ${\bf K}(X)$ denote the collection of (non-empty) compact subsets in $X$.

I would like to prove

$$\beta(B)=d_H\big(B,{\bf K}(X)\big),$$ where $d_H$ is the Hausdorff distance.

I proved $d_H\big(B,{\bf K}(X)\big)\le\beta(B)$. Is there someone that knows how to prove the other inequality?

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1 Answer 1

Let $d_0=d_H(B,K(X))$. So for $d>d_0$ we can find compact $K$ with $d_H(B,K)<d$. In particular, $B \subseteq \cup_{k\in K} B(k,d)$.

As $K$ is compact, for any $\epsilon>0$ we can find $k_1,\cdots,k_n\in K$ with $K\subseteq \cup_i B(k_i,\epsilon)$. For $b\in B$, we can find $k\in K$ with $d(b,k)<d$. Then we can find $i$ with $d(k,k_i)<\epsilon$, and so $d(b,k_i)<d+\epsilon$. Thus $\beta(B)\leq d+\epsilon$.

As $\epsilon>0$ and $d>d_0$ were arbitrary, we find that $\beta(B) \leq d_0$ as required.

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