Let $\mathcal{K} = \{A \subset \mathbb{R}^N| A \neq \emptyset, A \text{ closed and bounded with respect to the euclidean metric} \}$

Let us define $A_\epsilon = \bigcup_{x \in A}U_\epsilon(x)$, where $U_\epsilon(x)$ is the open ball around $x$ with radius $\epsilon$ with respect to the euclidean metric on $\mathbb{R}^N$.

Now we can also define a map $d_H:\mathcal{K}^2 \to \mathbb{R}$ with $$d_H(A,B) = \inf\{\epsilon > 0 | A \subset B_\epsilon \land B \subset A_\epsilon\}$$

I managed to show that $d_H$ defines a metric on $\mathcal{K}$, so that $(\mathcal{K}, d_H)$ forms a metric space.

Does someone have an idea or even just a hint for showing that $(\mathcal{K}, d_H)$ is complete?


Yes, $\mathcal K$ is complete with respect to Hausdorff metric:

Let $(A_n)_{n\in\mathbb N}$ be a Cauchy sequence. For each $n$, let $h_n=\sup\{\,d_H(A_k,A_n)\mid k>n\,\}$ and $B_n=\overline {(A_n)_{h_n}}$ (where the outer index denotes the dilation by $\epsilon=h_n$). Then $B_n$ is compact and contains all $A_k$, $k\ge n$. Also, $h_n\to 0$ because the sequence is Cauchy.

Let $A=\bigcap B_n$. Then $A$ is closed and bounded - but could it be empty? As each $A_n$ is nonempty, pick $a_n\in A_n$. Then this defines a sequence in $B_1$, hence has a limit point $a\in B_1$. In fact, the sequence $(a_n)$ is eventually in $B_k$ for any $k$, hence $a\in B_k$ for all $k$, i.e., $a\in A$. So we have found a candidate $A\in\mathcal K$ for the limit.

Remains to show that indeed $d_H(A_n,A)\to 0$.


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