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 $\beta$?
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$\begingroup$ 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.. $\endgroup$– Andrea MarinoCommented Jun 6, 2019 at 15:32
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$\begingroup$ Welcome Nicola, nice to see you $\endgroup$– Pietro MajerCommented Jun 6, 2019 at 18:55
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$\begingroup$ For $X=\mathbb R^n$ the answer is positive, see e.g. this question. $\endgroup$– SkeeveCommented Jun 7, 2019 at 7:28
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$\begingroup$ 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. $\endgroup$– Nicola ArcozziCommented Jun 7, 2019 at 9:56
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$\begingroup$ @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 $t=0$. $\endgroup$– Nate EldredgeCommented Jun 9, 2019 at 2:54
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1 Answer
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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.
answered Jun 9, 2019 at 2:45
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$\begingroup$ Thank you, Yuval: this completely answers my question. $\endgroup$ Commented Jun 9, 2019 at 21:20
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$\begingroup$ Hi Nicola, In that case the standard procedure is to upvote and officially accept the answer See meta.stackexchange.com/questions/5234/… $\endgroup$ Commented Jun 10, 2019 at 1:25
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$\begingroup$ Thanks also for this tip, Yuval: I'm new here and still learning. I hope I did it. $\endgroup$ Commented Jun 10, 2019 at 6:27
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$\begingroup$ @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 $s=\alpha$. $\endgroup$– MishaCommented Jun 12, 2019 at 16:48
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$\begingroup$ @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? $\endgroup$ Commented Jun 12, 2019 at 18:20