This is regarding the area enclosed by the Sierspinski curve as shown white(!) in the added image. I get a Hausdorff dimension of this shape equal to 2, which is the same as the topological dimension. But I thought this was a fractal, is it not?
Here's some background (and my understanding):
The Hausdorff dimension is equal to the similarity dimension when the self-similar object satisfies the open set condition (perfect self-similarity) E.g. source. The object in the image above seems to not be perfectly self-similar when the iteration goes to infinity, as it has 4 copies of itself connected through a central box. However, the area of this central box rapidly goes towards zero, making it ~perfectly self-similar in the infinite fractal iteration limit. This would give a Hausdorff dimension equal to $log_2(4) = 2$.
The topological dimension is also 2.
However, Mandelbrot writes in his The Fractal Geometry of Nature (1982).
A fractal is by definition a set for which the Hausdorff Besicovitch dimension strictly exceeds the topological dimension.
So I'm a bit confused. The areas enclosed by e.g. the Sierpinski square or the Sierpinski curve have Hausdorff dimensionality equal to 2. So are these areas not fractals?
Kind regards, Fredrik
EDIT 1: Maybe it not valid to ignore the contribution from the central square? The same type of central square is repeated in all the smaller copies as well. Maybe it is not negligible?
EDIT 2: The open set condition formulation used is: The set of self similiar objects $\{S_0, \dots S_N\}$ constituting e.g. a fractal,satisfies the open set condition if there exists a non-empty open set $O$ such that $Si(O)$ are pairwise disjoint and contained in $O$. Explicitly written \begin{equation} \begin{split} &S_i(O) \subset O \;\; \forall \; i \in\left[0, N\right], \;\; \text{and}\\ &S_i(O) \cap S_i(O) = \emptyset \;\; \text{if} \;\; i = j. \end{split} \end{equation} As formulated here