Let's do the real line or Euclidean case of complete metric space?). . Let $X$ be a complete ? metric space with Hausdorff dimension $\alpha$. \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$\beta$. More details to be added 1. J. M. Marstrand, where "The dimension of Cartesian product sets." Proc. Cambridge, Philos. Soc. 50 (1954) 198--202 2. J. Howroyd, "On dimension and the ? are now..existence of sets of finite positive Hausdorff measure." Proc. London Math. Soc. 70 (1995) 581--604 1 I think we use a result of Howroyd? (following Moran? who did the real line or Euclidean space ?). Let$X$be a complete? metric space with Hausdorff dimension$\alpha$. Let$0 < \beta < \alpha$. Then$H^\beta(X) = \infty$, the$\beta$-dimensional Hausdorff measure. By Howroyd's theorem, 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\$.