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$\Rightarrow$ is proved here.

I suspect that if a subset $M\subset\Bbb R^n$ has Hausdorff dimension $n$, then it must contain an open ball. Is that true?

Joe
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2 Answers2

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This is false - in fact, there are compact subsets of the real line containing no open intervals but which have positive Lebesgue measure - in particular fat Cantor sets have this property.

The usual Cantor set construction starts with an interval and removes the middle third of that interval, then the middle thirds of the two resulting intervals and so on - but this construction can be modified so that instead of removing a third each time we remove, say, a proportion of $\frac{1}{4^n}$ from the middle of each line at the $n^{th}$ step of the construction, which leaves an area of $$\prod_{n=1}^{\infty}\left(1 - 4^{-n}\right) > 0$$ in the set. However, after the $n^{th}$ step, the remaining set is the union of $2^n$ intervals of length at most $2^{-n}$ - hence the final result, which is the intersection of all of these steps, does not contain any open interval.

Milo Brandt
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  • Thanks for the answer. It's funny that I am actually working on Cantor sets (of the plane, meaning they are subsets of $\Bbb R^2$ homemorphic to the standard Cantor set). But I was asking about Hausdorff measure, not Lebesgue; my question arose from the fact that I'm wondering if such Cantor sets could have full Hausdorff measure; and I suspect that it's not the case, that is $\operatorname{dim}_H(C)<n$. Is this true? – Joe Feb 18 '21 at 16:45
  • @Joe The $n$-dimensional Hausdorff measure on $\mathbb R^n$ is a scalar multiple of Lebesgue measure - so if a set has positive Lebesgue measure, it has positive $n$-dimensional Hausdorff measure, hence Hausdorff dimension $n$. – Milo Brandt Feb 18 '21 at 16:48
  • The positivity of the $n$-dimensional Hausdorff measure implies the Hausdorff dimension is $\ge n$. If you say that it is exactly $n$ I suspect $\dim_H(M)\le n$ holds true for any $M\subset\Bbb R^n$, is this right? – Joe Feb 18 '21 at 22:37
  • @Joe Yes, that's right - for any $d>n$, it's possible to cover all of $\mathbb R^n$ with a sequence of balls of radii $r_i$ so that $\sum r_i^{d}$ is as small as you like. You can get most of the way to this result by noting that for any $k$ you can cover all the points in the hypercube $[0,1]^n$ by placing a ball of radius $\frac{\sqrt{n}}k$ at all the $(k+1)^n$ lattice points from $(\frac{1}k\mathbb Z)^n$ within the hypercube. But if $k$ is big $(k+1)^n\cdot \left(\frac{\sqrt{n}}{k}\right)^d$ will be small. – Milo Brandt Feb 18 '21 at 23:08
  • That sounds reasonable also thinking at the measure of the $r$-balls in $\Bbb R^k$ is proportional to $r^k$. Thank you very much again Milo – Joe Feb 18 '21 at 23:32
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No, it is not necessarily true that if $M$ has Hausdorff dimension $n,$ then $M$ contains an open ball. I will provide you with a construction of such a set. Consider for simplicity the case $n = 2$ (you can construct similar examples in higher dimensions) by the same technique. Consider the square $[-2, 2] \times [-2, 2].$ We will fill it with disjoint scaled Cantor dusts with slightly tweaked parameters such that its dimension is $2,$ but it doesn't contain any open balls by construction.

In $[0,1] \times [0,1]$, we construct a Cantor dust $D_1$ with parameter $f_1 = \frac13.$ Its dimension will be $\dim_H D_1 = \log_3 4.$ In $[1 + \frac12, 1 + \frac12 + \frac14] \times [1 + \frac12, 1 + \frac12 + \frac14],$ we construct a Cantor dust $D_2$ with parameter $f_2 := \frac{1}{2 + \frac12}$ and its dimension will be $\dim_H D_2 = - \log_{\frac{2}{5}} 4.$ Inductively, in $[\sum_{0 \leq l \leq 2k - 1} \frac{1}{2^l}, \sum_{1 \leq l \leq 2k} \frac{1}{2^l}] \times [\sum_{0 \leq l \leq 2k - 1} \frac{1}{2^l}, \sum_{1 \leq l \leq 2k} \frac{1}{2^l}]$ we construct a Cantor dust $D_k$ with parameter $f_k := \frac{1}{2 + \frac1k}$ and $\dim_H D_k = - \log_{f_k} 4.$ Denote $D := \bigcup_{1 \leq k} D_k.$ Then $\dim_H D = \sup_{k \geq 1} \dim_H D_k = 2,$ but it doesn't contain any open ball by construction. I hope this helps. :)

Actually Fritz
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