Easy to understand proof of Self-similar IFS fractal dimension

Question

I'm preparing for a seminar talk on fractals, the topic is Self-similar IFS fractal dimension, proving the main theorem used:

Given $IFS=\left\{ \mathbb{R}^{n};w_{1},...,w_{N}\right\} $ with Attractor $A$, each mapping is a similitude with scaling factor $0<r_i<1$ if the IFS is totally disconnected then the fractal dimension is the only solution $D(A)$ to the equation: $$ \sum_{i=1}^{N}r_{i}^{D(A)}=1 $$

I've gone over the sketch of the proof given in Barnsley, M.F. (2012) Fractals everywhere, and it's just missing the main concept that make the theorem difficult. It assumes that $D(A)$ which is defined as the limit $D(A)=\lim_{\epsilon\rightarrow 0}{\frac{\ln\left(\mathcal{N}\left(A,\epsilon\right)\right)}{\ln\left(\frac{1}{\epsilon}\right)}}$ exists, and that $\mathcal{N}\left(A,\epsilon\right)\sim C\epsilon^{-D(A)}$ both reasonable assumptions but they make the proof pretty useless besides giving some intuition why it is correct.

I've checked out the proof in a more advanced book Fractal Geometry: Mathematical Foundations and Applications, Falconer (2014). Here the problem is the opposite, the proof is a bit too advance as it uses terms from measure theory that not all of the students know. In addition it proves a stronger theorem where we don't assume the IFS is totally disconnected rather we assume the Open set condition.

I'm looking for a middle ground, not too difficult to understand on one hand, but as extensive and rigorous as possible on the other, and yes I know this is a rather strange question.

Answer 1

I think that the reality is that it's not particularly easy to prove that the similarity dimension agrees with some other notion of dimension under reasonably general hypotheses. You might have a look, though, at Chapter 5 of this text whose main purpose is to prove that the similarity dimension agrees with the box-counting dimension under the appropriate assumptions. Some main features of the text include:

• Self contained descriptions of similarity dimension and box-counting dimension, together with all their properties required for the proof,
• No essential reference to the Hausdorff dimension or measure theory, and
• Use of the strong open set condition as characterization of "non-overlapping".

As far as I know, there's no published proof of the main result that similarity dimension agrees with box-counting dimension under these hypotheses that doesn't use Hausdorff dimension as an intermediate step. So, I wrote this up to teach these ideas to my undergraduate students.

The main complication relative to texts like Barnsley's is that there is no assumption that all functions in the IFS have the same scaling factor. The main idea to deal with this is to construct a tree from the IFS using an algorithm that follows each branch until the terminal nodes generate copies of the whole set with comparable sizes. That appears as Theorem 4.4.2.

The presentation is, I think, more elementary than that in Falconer because it doesn't use measure theory. But, again, I don't think there's any getting around the reality that the theorem is just not particularly easy.