Question on geometric measure theory

2011-11-15T14:39:23Z

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 $\beta$.

Answer by Gerald Edgar for Question on geometric measure theory

2011-11-15T15:38:06Z

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

Question on geometric measure theory - MathOverflow

Question on geometric measure theory

Answer by Gerald Edgar for Question on geometric measure theory