# Compactness of Hausdorff metric

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 $$A$$?

• 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. Jan 17, 2019 at 19:04
• @Herno Brandsma I just learnt about compactness. I know nothing about the things you mentioned :( Jan 17, 2019 at 19:05
• Duplicate here: math.stackexchange.com/questions/181158/… Jan 17, 2019 at 19:40
• math.stackexchange.com/questions/2493757/… Jul 14, 2022 at 15:24

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.

• 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. Jan 17, 2019 at 22:58
• You lost me when you introduced Kn' for proving the first inclusion. What is the same for Kn and union of Kn'? Jan 17, 2019 at 23:05
• 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 $\leq \epsilon$.
– Dlem
Jan 17, 2019 at 23:20