Sierpinski triangle, Hausdorff-dimension and Roughness I'm currently trying to understand fractals better and I'm not really a mathematician, so please excuse if I express myself not very professional or scientifically.
I was watching this video that explains fractal dimension: https://www.youtube.com/watch?v=gB9n2gHsHN4
I have a question regarding the roughness that can be measured by the Hausdorff-dimension, so the video says. It's hard to explain what I mean without images, so I created this graphic:

When you would just fill all the "holes" in the Sierpinski triangle except for the big one in the middle, you would get exactly the same Hausdorff-dimension log2(3) ≈ 1.585, even though this reduced triangle is (at least from my current perspective) obviously less rough than the original Sierpinski mesh. 
How can the Hausdorff-dimension be a measurement for roughness in this case? 
Thanks in advance!
 A: When you fill in all of the holes (other than the big one), the Hausdorff dimension of the new object is not the same as the Hausdorff dimension of the Sierpinski gasket.  Indeed, the new object has Hausdorff dimension 2.
To see this, note that the area (or mass) of an equilateral triangle is given by
$$ \text{Area} = \frac{1}{2} bh = \frac{1}{2} \cdot s \cdot \frac{\sqrt{3}s}{2} = \frac{\sqrt{3}}{4}s^2,$$
where $s$ is the length of each side.  Suppose that we start with a "filled-in" Sierpinski gasket with sides of length 2.  This filled in gasket is composed of three identical equilateral triangles of side length 1 each, thus the area of the original object is
$$ A_o = 3 \cdot \frac{\sqrt{3}}{4} \cdot 1^2 = \frac{3\sqrt{3}}{4}, $$
(where I am using $A_o$ to denote the original area).  If we scale this down by a factor of $\frac{1}{2}$, then the scaled down version will consist of three equilateral triangles, each of side length $\frac{1}{2}$.  The new area will be
$$ A_n = 3 \cdot \frac{\sqrt{3}}{4} \cdot \left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2 \cdot \frac{3\sqrt{3}}{4} = \left(\frac{1}{2}\right)^2 A_o. $$
In this case, the area (or mass) scales like the square of the length, hence the Hausdorff dimension is 2.
The source of confusion, I think, is that the "filled-in" gasket cannot actually be covered by exactly three copies of itself scaled by $\frac{1}{2}$---there are always going to be some areas that don't get covered.  This is in contrast to the actual gasket, which can be exactly covered by three copies of itself scaled by $\frac{1}{2}$.  However, the mass of the Sierpinski gasket does not quite correspond to an area in the way that the mass of a triangle does, which further clouds the issue.
