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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 $Y$?.

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    $\begingroup$ No. $Y=\{ y\}$ is a counterexample. $\endgroup$
    – Christian Remling
    Commented Jul 25, 2018 at 3:22
  • $\begingroup$ I assume >1 point (edited) $\endgroup$
    – Forever Mozart
    Commented Jul 25, 2018 at 5:02
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    $\begingroup$ Then another counterexample is $Y=\{ x,y \}$. $\endgroup$
    – Christian Remling
    Commented Jul 25, 2018 at 6:53
  • $\begingroup$ 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. $\endgroup$
    – Dominic van der Zypen
    Commented Jul 25, 2018 at 9:22
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    $\begingroup$ @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 $d_H(A,Y)$. $\endgroup$
    – Will Brian
    Commented Jul 25, 2018 at 10:02

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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 $i \neq j$.

Thus, being maximal is a local property (for a sufficiently large notion of local), and cannot imply a global property such as nowhere dense.

[While writing this, Will Brian posted essentially the same idea in the comments.]

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