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Let (S, d) be a metric space. Can one always define a metric d# on the power set P(S) of $S$ such that d# ({x} , {y}) = d(x, y) for every x, y ∈ S?

Peter Liu
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    Yes, the Hausdorff distance (https://en.wikipedia.org/wiki/Hausdorff_distance). – Jean Marie Mar 15 '20 at 14:46
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    @JeanMarie then $d$ has to be bounded for $d_H$ to be well-defined. And it's not a metric on all subsets (closedness is minimally needed). E.g. on $[0,2]$ the metric is bounded and $d_H((0,1),[0,1])=0$ while the sets are different. – Henno Brandsma Mar 15 '20 at 14:50
  • @Henno Brandsma You are right. – Jean Marie Mar 15 '20 at 14:52

2 Answers2

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Yes.

Pick $x_0\in S$ and define $$d^\#(A,B)=\begin{cases}d(a,b)&\text{if }A=\{a\}, B=\{b\}\\ 1+d(x_0,b)&\text{if }|A|\ne1, B=\{b\}\\ 1+d(a,x_0)&\text{if }A=\{a\},|B|\ne1\\ 2&\text{if }|A|\ne1, |B|\ne1, A\ne B\\ 0&\text{if }|A|\ne 1, B=A \end{cases} $$

Essentially, we define the distance between singletons as desired, and the distance of all other sets to $\{x_0\}$ as $1$, and everything else from a "path" via $\{x_0\}$.

Hagen von Eitzen
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This question Does every metric on a non empty set have an extension to an arbitrary super set? is related, since you can embed $S$ into $P(S)$ as $S\simeq P_1(S)$

Note there the beautiful solution given by @dtldarek: that basically makes from two metric spaces $X$, $Y$, a metric space on $X\sqcup Y$; intuitively it takes two basic points $x_0\in X$, $y_0\in Y$, and joins $X$, $Y$ by a stick with ends $x_0$, $y_0$, of some length $\delta>0$. The distance $d(x,y)= d_X(x,x_0) + \delta + d_Y(y_0, y)$, the idea being that all traffic between $X$ and $Y$ is channeled through $x_0$ and $y_0$.

orangeskid
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