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 $\beta$.
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$\begingroup$ The empty set works, as does a 1-point set. $\endgroup$– Igor RivinCommented Nov 15, 2011 at 14:45
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$\begingroup$ (unless the space itself has one point, in which case only the empty set works). $\endgroup$– Igor RivinCommented Nov 15, 2011 at 14:46
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$\begingroup$ (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 RivinCommented Nov 15, 2011 at 14:46
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7$\begingroup$ @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 MajerCommented Nov 15, 2011 at 14:58
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2$\begingroup$ 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 EdgarCommented Nov 15, 2011 at 22:47
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1 Answer
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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 $\beta$.
J. M. Marstrand, "The dimension of Cartesian product sets." Proc. Cambridge, Philos. Soc. 50 (1954) 198--202
J. Howroyd, "On dimension and the existence of sets of finite positive Hausdorff measure." Proc. London Math. Soc. 70 (1995) 581--604
answered Nov 15, 2011 at 15:38
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$\begingroup$ 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? $\endgroup$– EmaCommented Nov 15, 2011 at 18:03
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1$\begingroup$ Howroyd's first two pages can be viewed here: mendeley.com/research/… (see the statement of Corollary 7). $\endgroup$ Commented Nov 15, 2011 at 23:00
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1$\begingroup$ I also note that Howroyd attributes the real & Euclidean cases to Besicovitch and Davies, not to Marstrand. $\endgroup$ Commented Nov 15, 2011 at 23:06