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 $\beta$?

  • $\begingroup$ 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.. $\endgroup$ Jun 6, 2019 at 15:32
  • $\begingroup$ Welcome Nicola, nice to see you $\endgroup$ Jun 6, 2019 at 18:55
  • $\begingroup$ For $X=\mathbb R^n$ the answer is positive, see e.g. this question. $\endgroup$
    – Skeeve
    Jun 7, 2019 at 7:28
  • $\begingroup$ 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. $\endgroup$ Jun 7, 2019 at 9:56
  • $\begingroup$ @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 $t=0$. $\endgroup$ Jun 9, 2019 at 2:54

1 Answer 1


By Corollary 7 in [How95], every analytic subset of a complete separable metric space which has positive (or infinite) Hausdorff measure of dimension s contains a compact set which has finite and positive Hausdorff measure of dimension s.

[How95] J.D. Howroyd. On dimension and on the existence of sets of finite positive Hausdorff measure. Proc. London Math. Soc. (3), 70:581–604, 1995.

  • $\begingroup$ Thank you, Yuval: this completely answers my question. $\endgroup$ Jun 9, 2019 at 21:20
  • $\begingroup$ Hi Nicola, In that case the standard procedure is to upvote and officially accept the answer See meta.stackexchange.com/questions/5234/… $\endgroup$ Jun 10, 2019 at 1:25
  • $\begingroup$ Thanks also for this tip, Yuval: I'm new here and still learning. I hope I did it. $\endgroup$ Jun 10, 2019 at 6:27
  • $\begingroup$ @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 $s=\alpha$. $\endgroup$
    – Misha
    Jun 12, 2019 at 16:48
  • $\begingroup$ @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? $\endgroup$ Jun 12, 2019 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.