# Maxed-out Hausdorff metric

Let $(Y,d)$ be a non-degenerate compact metric space, and let $d_H$ be the Hausdorff metric (https://en.wikipedia.org/wiki/Hausdorff_distance) on $K(Y)$ generated by $d$.

Here $K(Y)$ is the set of non-empty compact subsets of $Y$.

Let $M$ be the maximum value of $d_H$.

If $A\in K(Y)$, and $d_H(A,Y)=M$, then is $A$ necessarily nowhere dense in $Y$?.

• No. $Y=\{ y\}$ is a counterexample. – Christian Remling Jul 25 '18 at 3:22
• I assume >1 point (edited) – Forever Mozart Jul 25 '18 at 5:02
• Then another counterexample is $Y=\{ x,y \}$. – Christian Remling Jul 25 '18 at 6:53
• I think the statement in the question is true if $d:Y\times Y \to [0, s[$ is surjective where $s = \sup\{d(x,y): x,y\in Y\}$. I am trying to prove this. – Dominic van der Zypen Jul 25 '18 at 9:22
• @DominicvanderZypen: A counterexample -- take two copies of (the usual metric on) $[0,1]$, and declare two points in different copies to be distance $1$ apart. Taking $A$ to be one of the two copies maximizes $d_H(A,Y)$. – Will Brian Jul 25 '18 at 10:02

If start with a maximizing set $A$ in $Y$, and then consider $(\{0\} \times A) \cup (\{1\} \times Y)$ in $\mathbf{2} \times Y$, we see that it is maximizing again. Here we need to use a reasonable metric $d'$ on $\mathbf{2} \times Y$, eg $d'((i,x),(i,y)) = d(x,y)$ and $d'((i,x),(j,y)) = M$ for $i \neq j$.