TITLE: Intuitive meaning of fractal dimension.
QUESTION [7 upvotes]: I'm studying M. Barnsley's book 'Fractals Everywhere', but I'm stuck in the chapter 'Fractal Dimension'.
Suppose $(X, d)$ is a complete metric space and let $A \in \mathcal{H}(X)$ be a nonempty compact subset of $X$. Write $\mathcal{N}(A, \varepsilon)$ for the smallest amount of closed balls with radius $\varepsilon$ needed to cover $A$. Barnsley states the following:

The intuitive idea behind fractal dimension is that the set $A$ has fractal dimension $D$ if $\mathcal{N}(A, \varepsilon) \approx C \varepsilon^{-D}$ for some positive constant $C$, where $f(x) \approx g(x)$ if $\lim_{\varepsilon \to 0} \frac{\ln(f(x))}{\ln(g(x))}=1$.

I don't understand the intuition behind this definition. Can you explain this a little bit better?

REPLY [3 votes]: Imagine a Sierpinski triangle of side length $1$. And imagine one of its Sierpinski sub-triangles of side length $1/2^n$. If it helps, think about one of the sub-triangles of side length $1/4$.
Now suppose $\epsilon$ just the right size so your disc of radius $\epsilon$ just barely covers such a s sub-triangle. You need $3^n$ such discs to cover the whole thing. (In the $1/4$ case, there are $9$ sub-triangles of that size that comprise the full triangle. So you need $9=3^2$ such discs.)
So $\mathcal{N}(A, 1/4) =9$. More generally, $\mathcal{N}(A, 1/2^n) =3^n$. So there is an exponential relationship between $\epsilon$ and $\mathcal{N}$. You can rewrite this as $\mathcal{N}(A, \epsilon) =\epsilon^{\log_{1/2}(3)}=\epsilon^{-\ln(3)/\ln(2)}$. The fact that the $\ln(3)/\ln(2)$ doesn't depend on $\epsilon$ is what is neat, and makes that quantity worthy of being given a name: the fractal dimension of the Sierpinski triangle.
Added later: 
Oh, and the Sierpinski triangle is a "perfect" fractal because of the equality in the relationship $\mathcal{N}(A, 1/2^n) =3^n$. If you had some sort of shape that was less "perfect", but still had the asymptotic behavior defined here, it gets to have a fractal dimension too.