Consider a general metric space $(S,d)$, with $d$ a $1$-bounded metric, and let $X,Y \subset S$ be two closed subsets of $S$. Notice that $X$ and $Y$ are not compact.
Let $\mathcal{P}(X)$ denote the power set of $X$.
My aim is to prove that
$$
H(d)(X,Y) = H(H(d))(\mathcal{P}(X), \mathcal{P}(Y))
$$
where
$$
H(d)(X,Y) = \max \left\{ \sup_{x \in X}\, \inf_{y \in Y}\, d(x,y),\; \sup_{y \in Y}\, \inf_{x \in X}\, d(x,y) \right\}
$$
and
$$
H(H(d))(\mathcal{P}(X),\mathcal{P}(Y)) = \max \left\{ \sup_{A \subseteq X}\, \inf_{B \subseteq Y}\, H(d)(A,B),\; \sup_{B \subseteq Y}\, \inf_{A \subseteq X}\, H(d)(A,B) \right\}
$$
$$
= \max \left\{ \sup_{A \subseteq X}\, \inf_{B \subseteq Y}\, \max \left\{
\begin{array}{l}
\sup_{a \in A}\, \inf_{b \in B}\, d(a,b)\\
\sup_{b \in B}\, \inf_{a \in A}\, d(a,b)
\end{array}
\right\},\;
\sup_{B \subseteq Y}\, \inf_{A \subseteq X}\, \max \left\{
\begin{array}{l}
\sup_{b \in B}\, \inf_{a \in A}\, d(a,b)\\
\sup_{a \in A}\, \inf_{b \in B}\, d(a,b)
\end{array}
\right\}
\right\}
$$
namely that the Hausdorff distance between the two sets equals the Hausdorff lifting of the Hausdorff distance between their power sets.
This result seems natural to me, but I could not prove it. It is in fact clear that (by properties of suprema and infima)
$$
\sup_{x \in X}\, \inf_{y \in Y}\, d(x,y) = \sup_{A \subseteq X}\, \inf_{B \subseteq Y} \big( \sup_{a \in A}\, \inf_{b \in B}\, d(a,b) \big)
$$
but then I do not know how to introduce the maximum.
Can anybody give an hint or suggestion? Or also a reference, if this result is already known.
Thanks.
2025-07-03 07:27:36.1751527656