Topological and Hausdorff dimension of Koch tetrahedron fractal I am trying to find the topological and Hausdorff dimensions of the fractal on the attached picture (with proofs). It is obtained by similar approach as Koch snowflake, but applied to a tetrahedron instead of a triangle. 
Could you help me how to approach this? or provide me with an article / text-book with the answers?

 A: Paul Lévy, 1938.  This construction is found in Paragraph 14 of [1].  An English translation is Selection 12 in [2].  See my commentary in [2], p. 237—238, "A fractal fairy-tale."  In the final limiting set, from the outside you have just six plane faces of a cube.  Only on the inside do you have complicated fractal structure.  This fractal has been described many times over the years, but (except for Mandelbrot) the new describers were unaware that Lévy had already done it.
"user" already gave the Hausdorff dimension $\log 6/\log 2 \approx 2.6$.  What about the topological dimension?  Since it has plane faces as subsets, it has topological dimension ${}\ge 2$.  And since its Hausdorff dimension is $2.6$, the topological dimension is an integer ${} \le 2.6$.  So the topological dimension is $2$.
[1] Paul Lévy, "Les courbes planes ou gauches et les surfaces composées de parties semblables au tout". Journal de l'École Polytechnique 8 (1938) 18—25
[2] G. A. Edgar, Classics on Fractals (1993) softcover: Westview Press, 2004
A: At each step, every face is divided into 4 subsets, each with side length 1/2 the original side length then 3 more faces constructed on the middle subset so there are 6 times as many faces.  Taking "d" to be the dimension, total area at each step is $\frac{6}{2^d}$.  In order that this be a non-zero number in the limit as the number of steps goes to infinity we must have $\frac{6}{2^d}= 1$ so $2^d= 6$.  $d ln(2)= ln(6)$, $d= \frac{ln(6)}{ln(2)}$.
