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It seems that the following statement is pretty simple to prove:

Let $\emptyset \neq M$ be an arbitrary open set in $\mathbb{R}^n$, where n is the smallest integer such that $M \subseteq \mathbb{R}^n$. Then $dim_H(M) = n$, for $M$ contains an n-dimensional ball.

Could you give me hints in order to do so?

Source: Massopoust, Interpolation and Approximation with Splines and Fractals.

user1868607
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    Show that the Hausdorff dimension of an $n$-dimensional ball is $n$. Show that the Hausdorff dimension of $\mathbb R^n$ is $n$. With those two hints, you should be able to finish. – GEdgar Mar 03 '19 at 19:42
  • It would be interesting to investigate if the converse holds true: given $M\subset\Bbb R^n$ such that $\operatorname{dim}_H(M)=n$, then $M$ is open. – Joe Feb 18 '21 at 15:40
  • @Joe unfortunately i left this topic time ago – user1868607 Feb 18 '21 at 17:53

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Hausdorff dimension is monotonic as Why is Hausdorff dimension monotonic? explains. If we show that $dim_H \; \mathbb{R}^n = dim_H \; B(x,r) = n$ then for any open set we have $B(x,r) \subseteq O \subseteq \mathbb{R}^n$. And thus $dim_H \; O = n$.

For the computation of the Hausdorff dimension of a ball, I sketch the following argument. In the plane one may see that the optimal way of filling a squares with squares of side $\epsilon$ requires $\frac{1}{\epsilon^2}$ squares. Generalizing to dimension $n$ we need $\frac{1}{\epsilon^n}$ squares. If we compute $H^s_{\epsilon}$ of the square we then get $\epsilon^{s-n}$. Taking the limit when $\epsilon \to 0$ it is clear that the critical value defining the $0 - \infty$ behaviour is $s = n$.

The computation of the Hausdorff dimension of $\mathbb{R}^n$ is described in Hausdorff dimension of $\mathrm{R}^d$.

user1868607
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