# Question on geometric measure theory

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 $\beta$.

• The empty set works, as does a 1-point set. – Igor Rivin Nov 15 '11 at 14:45
• (unless the space itself has one point, in which case only the empty set works). – Igor Rivin Nov 15 '11 at 14:46
• (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). – Igor Rivin Nov 15 '11 at 14:46
• @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. – Pietro Majer Nov 15 '11 at 14:58
• 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. – 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 $\beta$.