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Given any metric space M, Hausdorff defined a new metric space h(M) whose "points" are the non-empty closed and bounded subsets of M. The hierarchy emerges from the following iteration process. Let H(0,M)=M and for each non-negative integer n, let H(n+1,M)=h(H(n,M)). If M is (for example) a finite dimensional Euclidean space or Hilbert space, does this process ever reach a fixed point-in the sense that there exists a non-negative integer k for which the spaces H(k,M) and H(k+1,M) are homeomorphic (or even isometric)? If there is no fixed point in the case of some particular metric space M, is it possible to continue this iteration process into the transfinite?

flag	asked Jan 2 at 19:16 Garabed Gulbenkian 3,003●6●13

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Assume you start with Euclidean space. The maximal number of points on distance $>\varepsilon$ in the unit ball of $H(n,M)$ grows with $n$. So you will not reach the roof in finite number of steps. – Anton Petrunin Jan 2 at 19:55

I assume Hausdorff used what is now known as the Hausdorff metric on sets to get metrics on the derived spaces? – Kevin R. Vixie Jan 2 at 20:14

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If the discrete metric is used and you start with a space containing an infinite number of points, you get a transfinite sequence that never converges to a fixed point (since the power set has a higher cardinality than the original set and closed bounded sets are all subsets of the space when the metric is discrete). – Kevin R. Vixie Jan 2 at 20:41

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@Kevin, usually $h(M)$ is defined as set of all compact subsets. In this case $M=h(M)$ for countable discrete metric space $M$. – Anton Petrunin Jan 2 at 21:04

@Anton, Thanks for the clarification. – Kevin R. Vixie Jan 2 at 21:36

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A question about a hierarchy of metric spaces arising from an operation defined by Hausdorff.

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