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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?).

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Isometric, unlikely. Show that the hyperspace fails the parallelogram law. –  Gerald Edgar Nov 3 '12 at 20:12
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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. –  Gerald Edgar Nov 4 '12 at 13:03

2 Answers 2

up vote 5 down vote accepted

The hyperspace of any Peano continuum (locally connected metric continuum) is homeomorphic to the Hilbert cube; this is a result of Curtis and Schori, see here. I learnt about this result in a paper of Torunczyk, where a different proof is given (that paper is also available online)

When it comes to asking about an isometry, I'm not sure which metric you put on the Hilbert cube...

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@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. –  Marcus Nov 3 '12 at 21:10
    
@Marcus You're welcome. I think for the isometric version Gerald Edgar suggestion looks good. –  Julien Melleray Nov 3 '12 at 21:53

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.

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