I want to know the following is wellknown 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 $\beta$.
I want to know the following is wellknown 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 $\beta$. 


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 $\beta$.


