I want to know the following is well-known or not:

Let X be a metric space with Hausdorff dimension $\alpha$. Then for any $\beta < \alpha$, X contains a closed subset whose Hausdorff dimension is $\beta$.

  • $\begingroup$ The empty set works, as does a 1-point set. $\endgroup$ – Igor Rivin Nov 15 '11 at 14:45
  • $\begingroup$ (unless the space itself has one point, in which case only the empty set works). $\endgroup$ – Igor Rivin Nov 15 '11 at 14:46
  • $\begingroup$ (unless you consider the dimension of the empty set to be undefined, in which case a one point set is a counterexample to the claim). $\endgroup$ – Igor Rivin Nov 15 '11 at 14:46
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    $\begingroup$ @Igor: I don't see how what you say implies, for instance, the existence of a subset of $[0,1]$ of Hausdorff dimension say 1/2. $\endgroup$ – Pietro Majer Nov 15 '11 at 14:58
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    $\begingroup$ A counterexample when $\alpha=\infty$. Let $X$ be uncountable with the discrete metric. Then a subset has either dimension $\infty$ (if uncountable) or $0$ (if countable). The only place finiteness of $\alpha$ is used in my answer is to get $X$ separable. $\endgroup$ – Gerald Edgar Nov 15 '11 at 22:47

Let's do the case of complete metric space. Let $X$ be a complete metric space with Hausdorff dimension $\alpha < \infty$. Then of course $X$ is separable, as well.

We use a result of Howroyd [2] (following Marstrand [1] who did the real line). Let $0 < \beta < \alpha$. Then $H^\beta(X) = \infty$, the $\beta$-dimensional Hausdorff measure. By Howroyd's theorem ($H^\beta$ is semifinite), there is a Borel subset $A \subset X$ with $0 < H^\beta(A) < \infty$. Then since a finite Borel measure is regular, there is a Cantor set $B \subseteq A$ with $0 < H^\beta(B) < \infty$, so of course $B$ has Hausdorff dimension $\beta$.

  1. J. M. Marstrand, "The dimension of Cartesian product sets." Proc. Cambridge, Philos. Soc. 50 (1954) 198--202

  2. J. Howroyd, "On dimension and the existence of sets of finite positive Hausdorff measure." Proc. London Math. Soc. 70 (1995) 581--604

  • $\begingroup$ Thank you very much. I do not know where you use the finitness assumption of \alpha in your argument. The finiteness seems to be needed only for \beta. If it is possible, would you please tell me which statement you used in Howroyd's paper? $\endgroup$ – Ema Nov 15 '11 at 18:03
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    $\begingroup$ Howroyd's first two pages can be viewed here: mendeley.com/research/… (see the statement of Corollary 7). $\endgroup$ – Gerald Edgar Nov 15 '11 at 23:00
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    $\begingroup$ I also note that Howroyd attributes the real & Euclidean cases to Besicovitch and Davies, not to Marstrand. $\endgroup$ – Gerald Edgar Nov 15 '11 at 23:06

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