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Menger sponge is a 3d fractal however it's fractal dimension is still less than 3. In fact most of the natural objects like coast lines have fractal dimensions between 2 and 3. This might be because we are calculating fractal surface dimension. But how does it make sense to have a 3d fractal (menger sponge) and get 2.727d value?

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  • $\begingroup$ This is just a matter of expressing the generating equation in $\mathbb{R}^n$, and i dont see any impedimeny to formulate them in major dimensions. Besides one thing is the space dimension, and other is the dimension metric associated with the set. $\endgroup$ – Brethlosze Sep 13 '17 at 4:35
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That would of course depend on the exact definition for the fractal dimension, but if we're using box-counting definition then there's nothing hindering it, you just have to create it in a dimension larger than $3$ to start with.

For the box-counting dimension the dimension of the space the fractal is a subset of is a hard limit.

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  • $\begingroup$ Indeed, every notion of dimension that I am aware of is monotone, in the sense that if $A \subseteq B$, then $\dim(A) \le \dim(B)$. Also, every notion of dimension that I am aware of has $\dim(\mathbb{R}^n) = n$ (with respect to the usual Euclidean metric). Thus no subspace of $\mathbb{R}^n$ can ever have dimension greater than $n$. $\endgroup$ – Xander Henderson Sep 14 '17 at 13:54
  • $\begingroup$ " you just have to create it in a dimension larger than 3 to start with." Is there any software for that? $\endgroup$ – quantised Sep 16 '17 at 6:21
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As skyking pointed out, there are various ways to characterize the "dimension" of a set. However, when we are talking about fractals, we are generally trying to connect metric properties with properties of measures. Roughly speaking, a metric is a way of measuring distances between points in an abstract set, while a measure is a way of measuring the "volume" (or length, or area, or content) of a set.

For example, if we shrink a cube by a factor of $\frac{1}{2}$, we are reducing all of the distances by a factor of $\frac{1}{2}$. Thus a cube that is 1 unit on each side shrinks down to a cube that is $\frac{1}{2}$ of a unit on each side. However, the measure of a cube is the product of its length, width, and height. Thus the original cube has a measure (or volume) of 1 unit$^3$, while the smaller cube has a measure of $\frac{1}{2^3}$ units$^3$. In general, if we compare the scaling of a unit cube and the volume of a scaled cube, we get the relation $$ \text{volume} = \text{scaling}^3.$$ The exponent on the length tells us about the dimension of a cube; a cube is three-dimensional.

Generalizing this construction to a Menger sponge, suppose that the volume of some Menger sponge is 1. If we shrink the sponge down by a factor of $\frac{1}{3}$, what is the volume? This is a little hard to get at directly, but notice that I can put 20 of the scaled sponges together to get the original sponge. Since each of these 20 little sponges are identical, it is reasonable to assume that they all have the same volume, thus if we let $v$ represent the volume of the little sponges, we have $$ 1 = 20v \implies v = \frac{1}{20}. $$ But we also know that the volume of the little sponges should behave like side length raised to some power, thus we have $$ v = \left( \frac{1}{3} \right)^s. $$ But now we have $$ \frac{1}{20} = \left(\frac{1}{3}\right)^s \implies -\log(20) = -s \log(3) \implies s = \frac{-\log(20)}{-\log(3)} = \frac{\log(20)}{\log(3)} \approx 2.7268. $$ Thus the Menger sponge is about 2.7-dimensional.

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  • $\begingroup$ but hansdoff dimension should exceed topological dimension of 3? $\endgroup$ – quantised Sep 16 '17 at 6:20
  • $\begingroup$ If by "topological dimension" you mean Lebesgue covering dimension, my recollection is that the Lebesgue covering dimension of the sponge is 1. I'm not sure I understand the problem... $\endgroup$ – Xander Henderson Sep 16 '17 at 13:48
  • $\begingroup$ Sponge occupies space so it's topological dimension is 3 bu then it's hansdoff dimension is 2.72 $\endgroup$ – quantised Sep 16 '17 at 16:44
  • $\begingroup$ @quantised What do you mean that it "occupies space?" It is a subset of $\mathbb{R}^3$, but so is a smooth curve, which is very much one-dimensional. The sponge has zero Lebesgue measure, so in that sense, it does not occupy space. And, again, its Lebesgue covering dimension (the usual notion of topological dimension) is 1. $\endgroup$ – Xander Henderson Sep 16 '17 at 18:30

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