Are Fractals always hollow? If so, how can they have volume or area?

Question

When calculating the dimension of a fractal shape, using the intersecting boxes method (where the number of intersecting boxes at different scales is compared; this is described at the starting at around the 10:30 mark in this video), boxes within an enclosed area of the fractal are not counted.

This makes me think of fractals as structures made of lines of 0 areas. Simplistically, I think of them as hollow structures made out of lines. Does this make mathematical sense? If so, how can some fractals have a finite or infinite area or volume? I can logically comprehend an infinite length/perimeter, however, area and volume make no logical sense to me.

Please note that my math knowledge maxes out at calculus. Would love to hear some logical explanation!

Answer 1

One definition of a fractal (fractional dimension) is an object whose dimension is not an integer. Here the dimension means the number you get using the intersecting boxes method you described.

You said the boxes enclosed by the shape are not counted. Since the boxes outside the shape are not counted either, what this procedure gives you is the dimension of the boundary of the shape rather than the shape itself.

The boundary consists of all points which have enclosed points (shown in yellow) nearby and also exterior points (shown in pink) nearby.

What do I mean by nearby? I mean there will be both types of points within distance $1/2$ to the boundary point, and also (possible different) points within distance $1/4$ to the boundary point, and likewise for $1/8,1/16, \ldots$ and by extension for any small number. I have only drawn the first three dotted circles. In reality there are an infinite number of them with sizes shrinking to zero.

Looking at the picture you can see the boundary cannot have any "area" to it. For then we could fit a small circle inside the boundary, but then there would be a pink point inside that circle, which contradicts how the pink points are outside the shape.

I'm not sure what you get applying the intersecting boxes method to the entire shape rather than just the boundary. But I imagine it says the answer has as big a dimension as possible. For example the "filled mandlebrot" set would have dimension 2 while the boundary has some fractal dimension.

Whether the boundary is made out of lines is a harder question to answer. There are certainly fractal-like subsets of the plane that contain no (curved) lines. For example something called the pseudo-arc. But I'm not sure that guy officially counts as a fractal.

Answer 2

Mitsuhiro Shishikura proved that the boundary of the Mandelbrot set has Hausdorff dimension $2$ and that for generic $c$ on the boundary of the Mandelbrot set, the Julia set of $z \mapsto z^2 + c $ also has Hausdorff dimension $2$.

The Hausdorff Dimension of the Boundary of the Mandelbrot Set and Julia Sets by Mitsuhiro Shishikura. Annals of Mathematics, Vol. 147, No. 2 (Mar., 1998), pp. 225–267

Answer 3

The video describes a technique for computing the box-counting dimension of a bounded subset of a Euclidean space (e.g. $\mathbb{R}^d$ for some nonnegative integer $d$). For a reasonable definition of "hollow", all sets with non-integer box-counting dimension will be hollow. This follows from the observation that if a set $E \subseteq \mathbb{R}^d$ contains an open set $U$, e.g. if there is some point $x \in E$ and some number $r > 0$ such that $$ B(x,r) := \{ y\in\mathbb{R}^d : |x-y| < r \} \subseteq E, $$ then we will have $$ \dim(E) = d, $$ which is an integer. This implies that any set with non-integer box counting dimension must have empty interior, in the sense that it cannot contain any open sets. Having no interior feels like a good notion of "hollowness", which gives an answer to your question.

Alternatively, if a subset of $\mathbb{R}^d$ has non-integer box-counting dimension, then it must have zero $d$-dimensional Lebesgue measure (essentially, the Lebesgue measure generalizes the idea of area or volume). In this sense, a "fractal" has no mass, and is therefore "hollow". Again, this gives an answer to your question.

That being said, your definition of "fractal" is essentially tailor made to get this answer, and leaves out a lot of subtleties. For example, I think that most working mathematicians would regard a fat Cantor set as fractal. However, fat Cantor sets have positive measure, which implies that they have dimension 1 (i.e. the largest possible dimension which they could have). Fat Cantor sets still have empty interior (so maybe you would consider them hollow?), but have positive Lebesgue measure, which means that they do have "mass". Should they be considered hollow?

Finally, consider the following: if $C$ is the usual ternary Cantor set, then $$ E^d := C \times [0,1]^{d-1} \subseteq \mathbb{R}^d$$ (the Cartesian product of $C$ and the unit square in $\mathbb{R}^{d-1}$) has non-integer box-counting dimension, but also contains line segments, squares, cubes, etc. of every dimension less than $d$. Such a set is not made up of lines of zero area.