Question on geometric measure theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T06:07:09Z http://mathoverflow.net/feeds/question/80986 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80986/question-on-geometric-measure-theory Question on geometric measure theory Ema 2011-11-15T14:39:23Z 2011-11-15T17:54:07Z <p>I want to know the following is well-known or not:</p> <p>Let X be a metric space with Hausdorff dimension $\alpha$. Then for any $\beta &lt; \alpha$, X contains a closed subset whose Hausdorff dimension is $\beta$.</p> http://mathoverflow.net/questions/80986/question-on-geometric-measure-theory/80987#80987 Answer by Gerald Edgar for Question on geometric measure theory Gerald Edgar 2011-11-15T15:38:06Z 2011-11-15T15:56:09Z <p>Let's do the case of complete metric space. Let $X$ be a complete metric space with Hausdorff dimension $\alpha &lt; \infty$. Then of course $X$ is separable, as well. </p> <p>We use a result of Howroyd [2] (following Marstrand [1] who did the real line). Let $0 &lt; \beta &lt; \alpha$. 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 $0 &lt; H^\beta(A) &lt; \infty$. Then since a finite Borel measure is regular, there is a Cantor set $B \subseteq A$ with $0 &lt; H^\beta(B) &lt; \infty$, so of course $B$ has Hausdorff dimension $\beta$.</p> <ol> <li><p>J. M. Marstrand, "The dimension of Cartesian product sets." Proc. Cambridge, Philos. Soc. 50 (1954) 198--202</p></li> <li><p>J. Howroyd, "On dimension and the existence of sets of finite positive Hausdorff measure." Proc. London Math. Soc. 70 (1995) 581--604</p></li> </ol>