Fractal obtained from its "number of dimensions''.

Question

Does the most regular fractal of a given dimension $r\in\mathbb{R}$ exist? Could I make one from its number $r$?

Answer 1

It's not clear what the "most regular" fractal of a given dimension is.

If $0<r<n$ it's not hard to construct a subset of $\mathbb R^n$, analogous to a Cantor set, with Hausdorff dimension $r$.

How? Like so.

Suppose first that $0<r<1$. Choose $\alpha>0$ with $$\alpha^r=1/2,$$and note that $$\alpha<1/2.$$

If $I$ is an interval let $|I|$ denote the length of $I$. Given $I=[a,b]$, define $$I_L=[a,a+\alpha|I|],\quad I_R=[b-\alpha|I|,b].$$Then

If $I$ is a compact interval then $I_L\subset I$, $I_R\subset I$, $I_L\cap I_R=\emptyset$ and $|I_L|^r+|I_R|^r=|I|^r$.

Now construct a Cantor set $K$ as usual, starting with $I=[0,1]$, except: Instead of deleting the middle third of each interval at each step, replace each interval $I$ by the two intervals $I_L$, $I_R$. Note that

If $I_1,\dots I_{2^n}$ are the intervals remaining after the $n$th step then $\sum_{j=1}^{2^n}|I_j|^r=1$.

If $h_r$ is $r$-dimensional Hausdorff measure it's clear from the above that $$h_r(K)\le1.$$It's very plausible that $$h_r(K)>0;$$probably the cleanest way to prove this is using the trivial direction of Frostman's Lemmma, with the "uniform" measure on $K$. So $$\dim(K)=r.$$

So we're done if $0<r<1$. Suppose that $0<r<n$. The construction above gives $K\subset\mathbb R$ with $\dim(K)=r/n$, and hence $\dim(K^n)=r$.