# Set of Hausdorff dimension $\alpha$

I would like to construct a set $\Sigma\subset\mathbb{R}^n$ such that $$mr^\alpha\leq\mathscr{H}^\alpha(B(0,r)\cap\Sigma)\leq M r^\alpha,$$ for all $0\leq r\leq1$ and some $0<m\leq M<\infty$. Here $0\leq\alpha\leq n$ and $B(0,r)$ denotes the ball in $\mathbb{R}^n$ of radius $r$ centred at $0$.

The question is very easy for integer dimension (simply take a hiperplane of dimension $\alpha$), so one can also assume that $n-1<\alpha< n$, otherwise the question reduces to a lower dimension, by constructing $\Sigma$ as a subset of a hyperplane of dimension $\lceil \alpha\rceil$.

I should add that it would be very nice if $\Sigma$ is trapped in a given open cone with apex at $0$ (in which case one should impose $\alpha>0$).

Does anyone have any suggestions? Many thanks!

• What is $s$? Do you mean $\alpha$? – Robert Israel Aug 2 '18 at 3:19
• Yes, of course; it's now edited. Sorry for the mistake! – Sobolev Aug 2 '18 at 12:22

Let $\Sigma=\mathbb{R}^{n-1}\times F$ where ${\rm dim}\ F=\gamma$ and $n-1<\alpha =n-1+\gamma <n$. Let $F_1=[0,1]$ and $F_2$ be a union of $n_1$ intervals of length $l_1$ s.t. $n_1l_1^\gamma=1$.

Similarly, $F_3$ has $n_1$ components in a component of $F_2$ s.t. $F_3$ is union of $n_1^2$ intervals of length $l_2$ and $n_1^2l_2^\gamma =1$.

Hence $F_n$ is $n_1^{n-1}$ intervals of length $l_n$ with $n_1^{n-1}l_{n-1}^\gamma=1$. So let $F=\bigcap_{i=1}^\infty\ F_i$.

[Add] In $\mathbb{R}^n$, note that there is a subset $A$ in $\mathbb{S}^{n-1}$ s.t. $[0,1]^n$ and $[r,R]\cdot A$ are bi-Lipschitz.

• If I understand correctly, $F\subset[0,1]$ is a self-similar set that solves the question ($\Sigma=F$) if $n=1$. My question would be: In $n$ dimensions, consider the cone $C=\{(x^\prime,x_n)\colon |x^\prime|^2\leq a^2 x_n^2, |x^\prime|\leq 1\}$, where $a>0$ is a constant and $x^\prime=(x_1,\ldots,x_{n-1})$. If I now place a copy of $F$ on each ray in $C$, i.e., $\Sigma:=\{\theta (x^\prime,a)\colon |x^\prime|\leq 1, \theta\in F\}$, will I then obtain a suitable set $\Sigma\subset C$? This is obviously true if $\alpha=n$, but I'm not sure how can one use "polar coordinates". Thanks! – Sobolev Aug 20 '18 at 14:09