If I have a compact metric space $(X,d)$, I can define the Hausdorff metric on the set $K(X)$ of all non-empty compact (equivalently, closed) subsets of $X$ as $$d_H(A,B) = \max ( \sup_{x \in A} \inf_{y \in B} d(x,y), \sup_{y \in A} \inf_{x \in B} d(x,y))$$

Now, I'm told, "the topology on $K(X)$ depends only on the topology of $X$, as any two metrics on $X$ are equivalent". Should I interpret the any two metrics are equivalent as "they generate the same topology"? Then, the implication of the Hausdorff metrics being equivalent is not clear at all.

On the other hand, if I interpret two metrics being equivalent as $c d_1<d_2 < Cd_1$, then the Hausdorff metrics producing the same topology seems clear, but the fact that any two metrics on $X$ are equivalent is suspicious. Does the compactness of $X$ force this?

Even if I know one of the interpretation to be correct, I can then try and prove the statement. In fact, I would prefer that, over a complete proof of the statement.

As an aside, is there a standard notation for my $K(X)$? The book I'm reading uses $2^X$, although I suspect due to typographical limitations.

  • $\begingroup$ $2^X$ generally denotes the power set of a set $X$, so your book's usage is highly non-standard in that regard; $K(X)$, I think, is pretty much standard for what you're actually discussing. $\endgroup$ – Branimir Ćaćić May 27 '13 at 10:04
  • $\begingroup$ @BranimirĆaćić: It's actually quite standard in this context to denote the set of (nonempty) closed sets by $2^X$ while the nonempty compact subsets are denoted by $K(X)$ (here it doesn't matter). This goes back at least to E. Michael's work on hyperspaces. ronno: try to show that $K(X)$ is compact. $\endgroup$ – Martin May 27 '13 at 10:31
  • $\begingroup$ @Martin Well, I know that $K(X)$ is compact. Why can't the topologies still be different? I also see that it's enough for the topologies to be comparable (in the finer, coarser sense). $\endgroup$ – ronno May 27 '13 at 10:33
  • 1
    $\begingroup$ A continuous map $f \colon X \to Y$ induces a continuous map $K(X) \to K(Y)$. // Another way to proceed is to show that the collection of sets formed by $U^+ = \{C \mid C \cap U \neq \emptyset\}$ and $U^- = \{C \mid C \subseteq U\}$ where $U$ runs through the open sets of $X$ forms a subbasis for the topology induced by the Hausdorff metric. $\endgroup$ – Martin May 27 '13 at 10:35
  • $\begingroup$ @Martin so the first interpretation is the correct one then? I think I get both your arguments, and will work out the details. The topological characterization of the Hausdorff distance seems nice. Would you mind re-posting your comment(s) as an answer? $\endgroup$ – ronno May 27 '13 at 10:39

Your first interpretation is the correct one.

The topology on $K(X)$ induced by the Hausdorff metric has a description as a hit-and-miss topology. That is, for a nonempty open set $U \subseteq X$ put $$ U^+ = \{C \in K(X) \mid C \cap U \neq \emptyset\} $$ and $$ U^- = \{C \in K(X) \mid C \subseteq U\}. $$ The set $U^+$ contains the compact sets that meet $U$ and the set $U^-$ contains the compact sets that miss $X \setminus U$.

The sets $U^+$ and $U^-$ where $U$ runs through the non-empty open sets of $X$ form a subbasis for the so-called Vietoris topology on $K(X)$. The sets $[U; V_1,\dots,V_n] := U^- \cap V_{1}^+ \cap \cdots \cap V_{n}^+$ form a convenient basis for the Vietoris topology.

The point is that one can prove that the Hausdorff metric induces the Vietoris topology on $K(X)$.

It may be helpful to prove at some point that for a countable dense subset $D$ of $X$ the collection of non-empty finite subsets of $D$ forms a countable dense subset of $K(X)$.

A nice and detailed exposition of these ideas can be found e.g. in Srivastava, A course on Borel sets, section 2.6, see Spaces of Compact sets, pages 66ff.

Added: The second interpretation is too strong. It is not very hard to show that a fat Cantor set and the usual ternary Cantor set are homeomorphic, but not bi-Lipschitz homeomorphic.

  • $\begingroup$ The book "Hyperspaces, Fundamentals and Recent Advances" (Ilanes,Nadler) is pretty nice, but expensive. It focuses on continua mostly. There is also a book on Selection theorems (selecting a point from each member in a nice way), though I don't recall a title. In general the non-empty closed sets on a space $X$ with the Vietoris topology is called the Hyperspace ($H(X)$) and other topologies on the set are also studied. But $H(X)$ metric implies $X$ compact metric and then the Hausdorff metric comes in play... $\endgroup$ – Henno Brandsma May 27 '13 at 16:57
  • $\begingroup$ Thanks for these addenda. You probably mean the nice recent book by Jayne and Rogers, Selectors. I would add G. Beer, Topologies on Closed and Closed Convex Sets as an accessible introduction to these more advanced topics. $\endgroup$ – Martin May 27 '13 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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