6
$\begingroup$

For each perfect Polish space $X$, let $H[X]$ be the set of all compact non-empty subsets of $X$. If $x ∈ X$ and $A ∈ H[X]$, put $$d(x,A) = \inf \{d(x, y) : y ∈ A\}$$ where on the right $d$ is the distance function on $X$. The Hausdorff distance between two compact sets is defined by $$d_H(A,B) = \max \{ \sup \{d(x,B) : x ∈ A\}, \sup\{d(y,A) : y ∈ B\}\}$$ Prove that $d_H$ is a metric on $H[X]$.

This is an exercise on page 13, Descriptive Set Theory, Yiannis N. moschovakis(2009). I got stuck on how to show $d_H(A,B)+d_H(B,C) \ge d_H(A,C)$.

$\endgroup$
1
  • 2
    $\begingroup$ How about distinguishing two distance functions (default distancing function and Hausdorff distance on H(X)) by different notations? $\endgroup$
    – Hamilton
    Sep 23, 2021 at 13:17

1 Answer 1

14
$\begingroup$

Let $d\colon X\times X\to E^1$ be the distance in $X$ and $d_H\colon H[X]\times H[X]\to E^1$ be the Hausdorff metric on $H[X]$.

We have $$d(a,C)\le d(a,b)+d(b,C)\le d(a,b)+d_H(B,C) $$ for all $b\in B$, so, taking $\inf_{b\in B}$, we have $d(a,C)\le d(a,B)+d_H(B,C)$ which is $\le d_H(A,B)+d_H(B,C)$ then take $\sup_{a\in A}$, and similarly, we can show that $d(A,c)\le d_H(A,B)+d_H(B,C)$ as well.

$\endgroup$
3
  • $\begingroup$ (For the first inequality: $d(a,C)\le d(a,c)\le d(a,b)+d(b,c)$, and take $\inf_{c\in C}$ on the right hand side.) $\endgroup$
    – Berci
    Mar 28, 2013 at 22:27
  • 2
    $\begingroup$ How about distinguishing two distance functions (default distancing function and Hausdorff distance on H(X)) by differnt notations? $\endgroup$
    – Hamilton
    Sep 23, 2021 at 0:50
  • $\begingroup$ I followed the notation in the original question. $\endgroup$
    – Berci
    Sep 23, 2021 at 5:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .