There are at least one continuous fractal from dimension 0 to 1? (I do not speak English. Please, excuse my bad grammar.)
This is about fractals over the real line, composed of points or segments inside the real line.
If the integers set is a fractal with dimension D=0, and the real line is a fractal with dimension D=1, then is possible to continuously vary the dimension D from the integer set to the real line set? (making some continuous figure in the plane)
There is a drawing of all the fractals side by side? (something like this image)
I do not mean that all the fractals make one larger fractal, but just that for each value D there is one fractal over the real line with that dimension, and all the fractals "touch each neighbor" continuously.
In other words, I want to continuously transform a fractal, from the integer set to the real line, by continuously changing his dimension D from 0 to 1.
 A: This can be done by generalizing the Cantor set. In particular, fix some $\alpha$ to be the dimension and let $\beta=2^{-1/\alpha}$. Define two functions $f_1(x)=\beta x$ and $f_2(x)=1-\beta+\beta x$, which are contractions towards $0$ and $1$ respectively. One may prove that there is a unique compact set $S_{\alpha}$ such that $S=f_1(S_{\alpha})\cup f_2(S_{\alpha})$. We have that $S_{1/
\log_2(3)}$ is the Cantor set and $S_1=[0,1]$ and in the limit as $\alpha$ goes to $0$, we get the set $\{0,1\}$. Additionally, it is fairly clear that $S_{\alpha}$ has dimension $\alpha$, since, being composed of two copies of it scaled by $\beta$, its dimension satisfies $\beta^{d}=1/2$.
One may prove that $S_{\alpha}$ consists exactly of points $x$ of the form, where $s_i$ is a sequence in $\{0,1\}$:
$$x=\sum_{i=0}^{\infty}s_i(1-\beta)\beta^i$$
which clearly varies continuously when $s_i$ is fixed. A plot of these fractals follows, where the cross-section of the graph at $x=d$ is a fractal of dimension $d$:

