= Solution
The proposed implication is not symmetric. In $X=\mathbb R$, take
$$
A=\{0,2\},\qquad B=\{0\},\qquad r=1.
$$
Then $B\subseteq E_1(A)$, but $2\notin E_1(B)$, so $A\nsubseteq E_1(B)$.
For a point and a nonempty set, write $d(x,B)=\inf_{b\in B}d(x,b)$. The given <Hausdorff distance> is equivalently
$$
d_H(A,B)=\max\left\{
\sup_{a\in A}d(a,B),
\sup_{b\in B}d(b,A)
\right\}.
$$
It is nonnegative and symmetric. If $d_H(A,B)=0$, then every $a\in A$ has $d(a,B)=0$, so $a\in\overline B=B$ because $B$ is closed; hence $A\subseteq B$, and symmetry gives $A=B$.
For the triangle inequality, the ordinary triangle inequality gives
$$
d(a,C)\leq d(a,B)+\sup_{b\in B}d(b,C).
$$
Taking suprema over $a\in A$, and then repeating with $A$ and $C$ interchanged, yields
$$
d_H(A,C)\leq d_H(A,B)+d_H(B,C).
$$
Thus $d_H$ is a metric on $H(X)$. Closedness is essential: if it is omitted, then the distinct bounded sets $(0,1)$ and $[0,1]$ have Hausdorff distance zero.
For singleton sets,
$$
d_H(\{x\},\{y\})=d(x,y),
$$
so $\theta(x)=\{x\}$ is an isometry. Its image is closed. Indeed, if $\{x_n\}$ converges in Hausdorff distance to $A\in H(X)$, then for every $\varepsilon>0$ and all sufficiently large $n$,
$$
A\subseteq E_\varepsilon(\{x_n\}).
$$
Hence any $a,b\in A$ satisfy $d(a,b)\leq d(a,x_n)+d(x_n,b)\leq2\varepsilon$. Letting $\varepsilon\downarrow0$ gives $a=b$, so the nonempty set $A$ is a singleton. This is the <closed singleton embedding in a Hausdorff hyperspace>.
If $H(X)$ is complete, its closed subspace $\theta(X)$ is complete. Since $\theta$ is an isometry, $X$ is complete. Hence <completeness is reflected by the Hausdorff hyperspace>.
Back to article page
