 Solution

The proposed implication is not symmetric. In $X=\mathbb{R}$, takeThen $B\subseteq E_1(A)$, but $2\notin E_1(B)$, so $A⊈E_1(B)$.

For a point and a nonempty set, write $d(x,B)={\inf }_{b\in B}d(x,b)$. The given Hausdorff distance is equivalentlyIt 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 givesTaking suprema over $a\in A$, and then repeating with $A$ and $C$ interchanged, yieldsThus $d_H$ 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,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 $\epsilon >0$ and all sufficiently large $n$,Hence any $a,b\in A$ satisfy $d(a,b)\le d(a,x_n)+d(x_n,b)\le 2\epsilon$. Letting $\epsilon \downarrow 0$ 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.