9
$\begingroup$

Recall that if we have a metric space $X$ then we can consider the set of its nonempty compact subsets and equip this with a metric called the Hausdorff distance. Denote the resulting metric space $\mathcal{H}(X)$. We of course get an inclusion $X \hookrightarrow \mathcal{H}(X)$. Are any of the homotopy-theoretic properties of this map known? For instance, if $X$ is connected, is $\mathcal{H}(X)$ connected? (I think the answer is yes). Note if $X$ consists of $n$ discrete points, then $\mathcal{H}(X)$ consists of $2^n - 1$ discrete points, so certainly the map is not a weak homotopy equivalence (perhaps it would be fruitful to consider only connected or path-connected compact subsets). Furthermore, I wonder if there are any insights to be had about more categorical structure of this map - $\mathcal{H}$ is clearly a functor; is it a monad? If so, is there something meaningful to be said about it?

$\endgroup$
1

1 Answer 1

8
$\begingroup$

In J. Andres, M. Väth, Calculation of Lefschetz and Nielsen Numbers in Hyperspaces for Fractals and Dynamical Systems, Proc. Amer. Math. Soc. 135 (2007), 479-487, it was shown (esssentially, the result was already implicitly shown in D.W. Curtis, Hyperspaces of noncompact metric spaces, Compositio Math 40 (1980) (2), 139-152, without explicitly noting it):

  1. ${\mathcal H}(X)$ is a union of disjoint open ARs (being its connected components) if and only if $X$ is locally continuum-connected.

  2. ${\mathcal H}(X)$ is an AR if and only if $X$ is locally continuum-connected and connected.

(Locally continuum-connected is a property between locally path-connected and locally connected.)

In particular, ${\mathcal H}(X)$ is contractible in all cases of 2 while $X$, in general, is not.

$\endgroup$
1
  • $\begingroup$ Brilliant, this answers my question perfectly. Thanks! $\endgroup$
    – K. Strong
    Commented Nov 14, 2021 at 21:18

You must log in to answer this question.

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