0
$\begingroup$

Let $(X,d)$ be a compact metric space. Let $(K,h)$ be the space of non-empty compact subsets of $X$ with the Hausdorff metric. Show that $K$ is compact.

First of all, I've found 2 related questions on stackexchange, but the answers only hint that the limit of a Cauchy sequence $\{A_n\}_{n\geq1}$ in $K$ is $A=\{x\in X:$ there exists a sequence $\{a_n\}_{n\geq 1}$ with $a_n\in A_n$ such that it converges to $x\}$. I've also found lecture slides from Harvard about the same problem but all of the sources think it's easy to show $\{A_n\}$ converges to $A$, which is not obviously to me at all. How can we bound the Hausdorff distance between $A_n$ and $A$ based on our definition of $A$?

$\endgroup$
4
  • $\begingroup$ Another way is to show that the Hausdorff metric induces the Vietoris topology on the hyperspace of non-empty compact subsets. In that case, compactness of $K$ is a simple consequence of Alexander's subbase lemma. $\endgroup$ Jan 17, 2019 at 19:04
  • $\begingroup$ @Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :( $\endgroup$ Jan 17, 2019 at 19:05
  • $\begingroup$ Duplicate here: math.stackexchange.com/questions/181158/… $\endgroup$
    – T. Fo
    Jan 17, 2019 at 19:40
  • $\begingroup$ math.stackexchange.com/questions/2493757/… $\endgroup$
    – Guy Fsone
    Jul 14, 2022 at 15:24

1 Answer 1

0
$\begingroup$

You have to show that $K_n$ converges in the Hausdorff sense to $A$, and also that $A$ is compact.

  • To show that $A $ is compact, it is sufficent to show that $A $ is closed. By sequential caracterisation, it not too hard to make it.

  • To show the other point, i will give some notation.

Let for $B$ included in $X$, $B_\epsilon$ the union of balls of radius $\epsilon $ centered in a point of $B $ (i.e. the points at à distance $\leq \epsilon $ of $B$).

Then the Hausdorff distance between 2 sets $B$ and $C$ is the infinimum of the epsilon such that $B_\epsilon$ contains $C$ and $C_\epsilon$ contains $B$. You can show that the distance between $B$ and $C$ is the same as the distance between the closure of $B$ and the closure of $C$

$(K_n)$ is Cauchy. So for $n$ and $m$ greater than a $n_0$, $K_n$ and $K_m$ are at a Hausdorff distance $\leq \epsilon$. So it is the same for $K_n$ and the closure of the union of $K_{n'}$ for n' greater than $n_0$. This last set contains $A $, so you can include $A $ in ${K_n}_\epsilon $.

The other inclusion, let $x$ in $K_n$ and build à sequence $(x_{n'}) $ of elements in $K_{n'}$ at à distance $\leq \epsilon$ of $x$. Extract a convergent subsequence, then the limit will be in $A$ and at a distance $\leq \epsilon$ of $x$. This concludes for the inclusion $K_n$ in $A_\epsilon$.

Then, to conclude, you need to show that a sequence of compacts has Cauchy subsequence. Hint to show it :

  • first show it if $X$ is a finite set with the discrete metric.

  • then, cover $X$ by à finite number of balls of radius < 1 ; by using this and the preceding point (look at the indices of the balls which intersects with $K_n$), you can extract a subsequence of $(K_n)$ on which the terms are espaced of a hausdorff distance <1. Then continue with $1/2^n$ and do a diagonal argument.

$\endgroup$
3
  • $\begingroup$ For the second inclusion. How do you extract a convergent subsequence? Generally it's not true that a closed all in a metric space is compact. $\endgroup$ Jan 17, 2019 at 22:58
  • $\begingroup$ You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'? $\endgroup$ Jan 17, 2019 at 23:05
  • $\begingroup$ But it is true that a closed space included in a compact space is compact. (That is why I said it is sufficent to show A closed to show A compact here). Kn and the (closure of) the union of Kn' are still at a Hausdorff distance $\leq \epsilon $. $\endgroup$
    – Dlem
    Jan 17, 2019 at 23:20

You must log in to answer this question.

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