For a Polish space $(X,d)$ its hyperspace $(K(X),d\_H)$ is also a Polish space. (Here $K(X)$ denotes the set of all nonempty compact subsets of $X$, and the Hausdorff metric $d\_H$ is defined by $d\_H(K,L)=\inf \{\varepsilon>0:\ K\subseteq L\_\varepsilon,\ L\subseteq K\_\varepsilon\}$, where $A\_\varepsilon$ is the open $\varepsilon$-neighborhood of $A$).


**Question:** Is there any nontrivial Polish space $(X,d)$ that is isometric to its hyperspace $(K(X),d\_H)$?


What I know so far:


* There is [this MO question](https://mathoverflow.net/questions/117893/a-question-about-a-hierarchy-of-metric-spaces-arising-from-an-operation-defined) with a couple of comments but no real answer.
* Trivial examples: the empty space, a singleton space, a countably infinite discrete space.
* $X$ cannot be totally bounded. In particular, it cannot be compact. (Proof: if $X$ has at least 2 points and it is totally bounded, then for some $\varepsilon>0$ we have that $N\_X(\varepsilon)=\max\{|F|:\ F\subseteq X \text{ is finite and } d(x,y)>\varepsilon \text{ for any two distinct points }x,y\in F\}$
is finite but at least 2, thus $N\_{K(X)}(\varepsilon)>N\_X(\varepsilon)$, which distinguishes the two spaces.
* A natural candidate would be [the Urysohn universal space](https://en.wikipedia.org/wiki/Urysohn_universal_space) $U$. Unfortunately, the above property fails for $U$ because $K(U)$ is not homogeneous. In fact, there exists a four-point metric space $A$, a point $a\in A$, and an isometric embedding $f: A\setminus{a}\to K(U)$ that does not extend to $A$.
An example: $A=\{a\_1,a\_2,a\_3,a\_4\}$, the pairwise distances are $1,1,1,1,1,2$ with only $d\_A(a\_1,a\_4)=2$. Pick two points $x,y\in U$ with $d\_U(x,y)=1$ and let $K\_1=\{x\}$, $K\_2=\{y\}$, $K\_3=\{x,y\}$. Let $f(a\_i)=K\_i$ for every $i\leq 3$, which is an isometric embedding of $A\setminus \{a\_4\}$ into $K(U)$. It is easy to check that no suitable image exists for $a\_4$.
* If we replace *isometry* with *homeomorphism*, then there are many examples including the Cantor space, the Baire space, and the Hilbert cube. In this case, the interesting problem seems to be classifying such Polish spaces. This is probably too much to ask. A more reasonable goal is classifying some subclass (e.g. compact 0-dimensional Polish spaces).



Polish space isometric to its hyperspace
It is much easier to deal with zero dimensional spaces and ultrametrics than connected metrics. $\omega^\omega$ with the standard ultrametric is isometrically isomorphic to its own hyperspace.


In particular, we give $\omega^\omega$ the ultrametric where if $f,g:\omega\rightarrow\omega$, then let $d(f,g)=2^{-n}$ where $n$ is the least natural number with $f(n)\neq g(n)$ and where $d(f,f)=0$.


We observe that a metric space $(X,d)$ is isometrically isomorphic to $\omega^\omega$ if and only if $(X,d)$ is a Polish ultrametric space where
$d[X\times X]=\{0\}\cup\{2^{-n}\mid n\in\omega\}$ and $x\in X,n\in\omega$ implies that there is an infinite family $(x\_r)\_{r\in\omega}$ where
$d(x\_r,x\_s)=2^{-n}$ for $r<s$ and where $x\_0=x$.


It is easy to see that $d\_H[K(\omega^\omega)\times K(\omega^\omega)]=\{0\}\cup\{2^{-n}\mid n\in\omega\}$.


Now suppose that $C\in K(\omega^\omega)$ is non-empty compact and $n\in\omega$. Let $f\in C$. Then let $C\_0=C$. By compactness, the set $\pi\_n[C]$ is finite, so let $v$ be an natural number with $v>\pi\_n(c)=c(n)$ for each $c\in C$. If $r>0$, then let $f\_r\in\omega^\omega$ be a function with $f\_r(n)=v+r$ and where $f\_r(m)=f(m)$ for each $m<n$. Then let $C\_r=C\cup\{f\_r\}$. Then we have $d\_H(C\_r,C\_s)=2^{-n}$ whenever $r,s\in\omega,r\neq s$. Therefore, $\omega^\omega$ is isometrically isomorphic to $K(\omega^\omega)$.


