1
$\begingroup$

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 $Y$?.

$\endgroup$
5
  • 2
    $\begingroup$ No. $Y=\{ y\}$ is a counterexample. $\endgroup$ Jul 25, 2018 at 3:22
  • $\begingroup$ I assume >1 point (edited) $\endgroup$ Jul 25, 2018 at 5:02
  • 3
    $\begingroup$ Then another counterexample is $Y=\{ x,y \}$. $\endgroup$ 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. $\endgroup$ Jul 25, 2018 at 9:22
  • 1
    $\begingroup$ @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
    Jul 25, 2018 at 10:02

1 Answer 1

1
$\begingroup$

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.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.