# Subspaces of metric spaces having prescribed dimension

Let $$(X,d)$$ be a metric space having Hausdorff dimension $$\alpha>0$$ and let $$0<\beta<\alpha$$. Is there a metric subspace of $$X$$ having Hausdorff dimension $$\beta$$?

• Thinking about a fancy way that I don't know if it could work. Take a continous $\gamma$ from [0,1] to the space of compact subsets of $X$ with the hausdorff distance. 1. Is the hausdorff measure of $\gamma(t)$ a continous function of $t$? 2. Is there a compact subset K of X such that it ha s the same hausdorff measure? 3. Is $K$ connected to a point in the space of compact subsets? Of course, this would yield the answer, but I have no idea if it is effective.. Jun 6, 2019 at 15:32
• Welcome Nicola, nice to see you Jun 6, 2019 at 18:55
• For $X=\mathbb R^n$ the answer is positive, see e.g. this question. Jun 7, 2019 at 7:28
• Hi Pietro. Skeeve: the question is about general metric spaces. Locally compact and complete could be a reasonable restriction. Euclidean space is restricting too much. Andrea: thanks for the hint. Jun 7, 2019 at 9:56
• @AndreaMarino: We'd need to look at the Hausdorff dimension, right? And that is certainly not continuous with respect to Hausdorff distance. Consider $X = [0,1]$ and $\gamma(t) = [0,t]$, so $\gamma(t)$ has Hausdorff dimension $1$ for every $t>0$ but dimension zero for $t=0$. Jun 9, 2019 at 2:54

• @NicolaArcozzi: This is an interesting paper but it does not resolve your question what is about existence of subsets of given dimension strictly less than $s=\alpha$. Jun 12, 2019 at 16:48
• @Misha. I think it does: if the dimension of $X$ is $s>\alpha$, then the $\alpha$-Hausdorff measure of $X$ is infinite, by the theorem in the paper I can find a compact subset $K$ of $X$ having finite and positive $\alpha$ measure, hence $K$ has dimension $\alpha$. Or am I missing something? Jun 12, 2019 at 18:20