1. Are there examples of homeomorphic compact metric spaces of different Hausdorff dimension?

  2. If yes, are there sufficient conditions on the spaces which would imply the equality of Hausdorff dimensions?

In my case the spaces are the so called multiple conic singularities (MCS-) spaces. They are defined inductively: MCS-spaces of dimension -1 are empty. Any point of an $n$-dimensional MCS-space has an open neighborhood homeomorphic to an open cone over its boundary which is a compact $(n-1)$-dimensional MCS-space. (Such examples come from the theory of Alexandrov spaces.)

  • 18
    $\begingroup$ 1) Yes, of course. All Cantor sets are homeomorphic, and their Hausdorff dimension can be anything. $\endgroup$ Nov 8, 2015 at 15:24
  • $\begingroup$ It beats me why this question got migrated from MO... $\endgroup$
    – Alex M.
    Nov 15, 2015 at 23:00
  • $\begingroup$ Somewhat related: Behavior of Hausdorff dimension under homeomorphisms $\endgroup$
    – user147263
    Nov 15, 2015 at 23:03

2 Answers 2


Concerning (1), one can say more: for any compact metric space $X$ of positive Hausdorff dimension $d$, and for any $d'>d$, there exists a compact metric space $Y$ homeomorphic to $X$ and with $\dim_H Y=d'$.

Indeed, let $Y$ be the same set as $X$ but with the metric raised to power $d/d'$; that is $\rho_Y(a,b) = \rho_X(a,b)^{d/d'}$. Since the function $t\mapsto t^{d/d'}$ is concave and increasing, the triangle inequality holds for $\rho Y$. The identity map is a homeomorphism between $X$ and $Y$. And it's easy to show that $\dim_H Y = \frac{d'}{d}\dim_HX = d'$.

In view of the above, any sufficient condition in (2) must involve both spaces, not just one of them. E.g., we can assume both are Alexandrov spaces. Then it's true that homeomorphisms preserve Hausdorff dimension. This follows from a topological characterization of Hausdorff dimension of such spaces (due to Burago, Gromov, and Perelman): an Alexandrov space $X$ contains an open dense subset $U$ that is a topological manifold, and the Hausdorff dimension of $X$ is the topological dimension of $U$. Reference: The Riemannian Structure of Alexandrov Spaces by Otsu and Shioya.


An $\alpha$-Holder map increases the dimension by a factor of at most $1/\alpha$. So, if for any $\alpha<1$ there are surgective $\alpha$-Holder maps both ways, then the dimensions are equal.


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