Given a real number $d , (1<d<2)$, is there a fractal with fractal dimension $d$?

Question

There are fractals with fractal dimension $\log2/\log3$ and $\log5/\log3$. My question is that given any real number $d$ between one and two is it possible to construct a fractal with fractal dimension $d$?

Answer 1

Yes, though it depends upon what, exactly, you mean by "fractal dimension." The answer will be yes in basically any reasonable sense of the word "fractal dimension", but the details will be different.

On the assumption that you mean the Hausdorff dimension, there are many ways to build such objects. If you want to construct a set of Hausdorff dimension $D > 1$, one cheap and simple way is to build a Cantor set dimension $\{D\}$ (the fractional part of $D$), then take the Cartesian product of that set with a cube in $\mathbb{R}^{\lfloor D \rfloor}$. The basic idea is as follows:

Fix some $\alpha \in (0,1)$. Define two maps $\mathbb{R}\to\mathbb{R}$ by $$ \varphi_1(x) = \alpha x \qquad\text{and}\qquad \varphi_2(x) = \alpha x + (1-\alpha). $$ By some abstract nonsense (essentially the Banach fixed point theorem), there is a unique nonempty compact set $K$ such that $$ K = \varphi_1(K) \cup \varphi_2(K). $$ It can be shown that the Hausdorff dimension of $K$ is given by $$ \dim_H(K) = \min\left\{ 1, \frac{\log(2)}{-\log(\alpha)} \right\} $$ (for details, there is a seminal paper by Hutchinson that was published in 1981 that you should really consider reading). As long as $\alpha < \frac{1}{2}$, the Hausdorff dimension of $K$ will be less than 1. Indeed, we can solve $$ \{D\} = \frac{\log(2)}{-\log(\alpha)} $$ for $\alpha$ in order to figure out what $\alpha$ needs to be in order to construct a Cantor set of dimension $\{D\}$. If we do this, we get $$ -\log(\alpha) = \frac{\log(2)}{\{D\}} \implies \alpha = \exp\left( -\frac{\log(2)}{\{D\}} \right).$$ It then follows by other abstract nonsense that $$ \dim_H(K \times [0,1]^D) = \{D\} + \lfloor D \rfloor = D. $$ Do be a little careful, however: the Hausdorff dimension is not stable under Cartesian products. It is not true in general that $\dim_H(A\times B) = \dim_H(A) + \dim_H(B)$. I've just chosen my sets to be well enough behaved that everything works out nicely. If you want to see some counterexamples, I think that Falconer has a few.

You can also construct sets more directly that have arbitrary dimension, using a similar approach. We say that $\varphi : \mathbb{R}^n \to \mathbb{R}^n$ is a contracting similitude with contraction ratio $c \in (0,1)$ if $d(\varphi(x),\varphi(y)) = c d(x,y)$ for all $x,y\in\mathbb{R}^n$ (where $d(\cdot,\cdot)$ is the Euclidean distance). Suppose that $\{ \varphi_{j} : \mathbb{R}^n \to \mathbb{R}^n \}_{j=1}^{J}$ is a collection of contracting similitudes, with contraction ratio $c_j$ corresponding to $\varphi_j$. As above, there is a unique nonempty compact set $K$ such that $$ K = \bigcup_{j=1}^{J} \varphi_j(K). $$

If this collection satisfies the open set condition, that is, if there is an open set $U$ such that

1. $\varphi_j(U) \subseteq U$ for all $j$, and
2. $\varphi_j(U) \cap \varphi_k(U) = \emptyset$ for all $j\ne k$,

then the Hausdorff dimension of $K$ will be $s$, where $s$ is the unique real solution to the Moran equation $$ 1 = \sum_{j=1}^{J} c_j^s. $$ If we assume that there is some $c$ such that $c_j = c$ for all $j$, then this reduces to $$ 1 = Jc^s \implies s \log(c) = -\log(J) \implies s = -\frac{\log(J)}{\log(c)}. $$ By choosing $J$ and $c$ appropriately, and ensuring that the ambient space is large enough to provide room for a system that satisfies the open set condition with those parameters, we can construct many self-similar sets with any desired Hausdorff dimension $s$.