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The Hausdorff dimension $\dim(A)$ of $A \subseteq \mathbb{R}^d$ is the unique number $\alpha$ such that the $\beta$-dimensional Hausdorff measure $H_\beta(A)=0$ for $\beta>\alpha$ and $H_\beta(A)=\infty$ for $\beta<\alpha$.

What is $\dim(\mathbb{R}^n)$?

For every $\delta,\beta>0$ and every (countable) cover $\mathfrak{A}$ of $\mathbb{R}^n$ where $\forall A\in\mathfrak{A}. \text{diam}(A)>\delta$ we get $\sum\limits_{A\in\mathfrak{A}}\text{diam}(A)^\beta=\infty$ which would imply $\dim(\mathbb{R}^n)=0$ while Wikipedia states $\dim(\mathbb{R}^n)=n$.

So where did i go wrong?

Thanks, Takirion

max_zorn
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Takirion
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  • Although the diameters may shrink they have to be covered from below. Note that $H^\alpha (A)=\lim\limits_{\delta\rightarrow 0}\inf\lbrace\sum_{A'\in A} diam(A')^\alpha\vert A\text{ is countable covering with }diam(A)>\delta\rbrace$ – Takirion Feb 14 '16 at 11:26
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    I don't see that in the definition on the wiki page. If it were the case, then clearly every unbounded subset of $R^n$ would have dimension zero, as you say. – Mr. Cooperman Feb 14 '16 at 11:30
  • See my comment on the first answer. The definition of the hausdorff measure can be found here https://en.wikipedia.org/wiki/Hausdorff_measure – Takirion Feb 14 '16 at 11:33
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    I think you flipped the inequality – Mr. Cooperman Feb 14 '16 at 11:35
  • Oh my... what a shame. You are right, sorry. But thanks for your answer. – Takirion Feb 14 '16 at 11:40
  • In fact, for every sequence $(a_i){i\in\mathbb{N}}$ with $a_i > \delta$, the series $\sum_i a_i$ must necessarily diverge, since then $\sum{i=1}^n a_i \geq n\cdot\delta$. So putting a lower bound on the diameters of the covering sets in the definition of the Hausdorff measure makes no sense. – fgp Feb 14 '16 at 11:42
  • Sure, No worries :) – Mr. Cooperman Feb 14 '16 at 11:42
  • The definition of Hausdorff measure should have $\text{diam}(A) < \delta$. But that "Hausdorff dimension" web page defines Hausdorff content (not Hausdorff measure), and in that case there is no restriction except $\text{diam}(A) > 0$ – GEdgar Feb 14 '16 at 12:43
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    Possible duplicate of Hausdorff dimension of $\mathrm{R}^d$ – user1868607 Mar 04 '19 at 21:18

1 Answers1

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I'm not an expert on this but I think I figured it out for $\mathbb R^1$ at least. Consider Covering $\mathbb R^1$ with balls (intervals) $I_n$ of length $\frac{1}{n}$. We know this is possible because $\sum_{n=1}^\infty\frac{1}{n}$ diverges so the intervals can cover the whole line. Now, by the $p$-series test or the integral test, $\sum_{n=1}^\infty \text{diam}(I_n)^\beta=\sum_{n=1}^\infty\frac{1}{n^\beta}$ converges for all $\beta\leq1$ and diverges for $\beta\geq 1$.

Can you finish the argument from here? And generalize it to $n$ using boxes of volume $1/n$ instead?

Arkady
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  • Unfortunately this doesn't work, because the cover with the $I_n$ doesn't satisfy $diam(I_n)>\delta$ for any $\delta$ and all $n$. – Takirion Feb 14 '16 at 11:20
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    The diameters don't have to be bounded away from zero – Mr. Cooperman Feb 14 '16 at 11:21
  • Hm, ok i noticed that in the article about hausdorff-dimension they defined the dimension via the hausdorff content which is "just" a content and may differ from the hausdorff measure. Are the definitions equvalent? – Takirion Feb 14 '16 at 11:30