# Is the hyperspace of the Hilbert cube homeomorphic to the Hilbert cube

Question: Is the hyperspace of the Hilbert cube $H=[0,1]^\mathbb {N}$ homeomorphic to $H$?

Remarks and definitions:

1) The Hilbert cube $H$ is a compact metric space, where the metric is given by the $\ell_2$-norm of sequences. A classical theorem on metric spaces says that every compact metric space is isometric to a closed subspace of $H$.

2) The hyperspace of a metric space $X$ is the metric space of all non-empty compact subsets of $X$ given by the Hausdorff metric. Another classical theorem on metric spaces says that the hyperspace of a compact metric space is again a compact metric space.

Combining 1) and 2) shows that the hyperspace of the Hilbert cube is isometric to a closed subspace of the Hilbert cube. So my question asks whether we also can get a homeomorphism (can we even get both spaces isometric?).

• Isometric, unlikely. Show that the hyperspace fails the parallelogram law. Nov 3 '12 at 20:12
• Actually, the parallelogram law involves the affine structure as well as the metric, so we cannot do that directly. Maybe try this: there are two points in the hyperspace with non-unique midpoint. Nov 4 '12 at 13:03

• @Julien Thanks for the reference. I'm not officially asking the isometric version (using the $\ell_2$-norm, also see the comment of Gerald above), but I wanted to know the topological type of the hyperspace of $H$ in the first place. Nov 3 '12 at 21:10
Rule out isometry ... In the hyperspace of $\mathbb R$, let $A=\{0,1,2\}$, $B=\{0\}$, $C=\{2\}$ and $M = \{1\}$. Then points $A, B, C$ all have distance $2$ from each other; they are the vertices of an equilateral triangle. But $M$ has distance $1$ from each of $A, B, C$, so $M$ is the midpoint of each of the three sides of that triangle. Not possible in Euclidean geometry.