# Dimension of fractals

I would like to know is it possible to generate a fractal in the plane with dimension higher than 2? If that is possible, please could you explain the intuition behind that? If it is not possible, is there some proof for that? Thank you in advance.

Best regards,

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Meaning of "dimension"? If you mean Hausdorff dimension, then NO. If $A \subset B$, then $\dim A \le \dim B$. And $\dim \mathbb R^2 = 2$.

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Thank you for the answer. Yes, it is Hausdorff dimension. There are a few examples that puzzle me. Please let me introduce one of them, the one at kaziprst.com/fractal.gif. Assume $A$ is divided into two parts, and each of the parts is replaced by $B$ -- that is a fractal rule. If my understanding is correct, according to en.wikipedia.org/wiki/Fractal_dimension#Specific_definitions, $N(l) = 8$, and $l = 2$, implying $D = 3$. Please, could you point where I am making a mistake in reasoning? –  Slobodan Jun 22 '11 at 19:10
Does that wiki mention the open set condition? That is what fails in your example. Without it, the simple dimension formula fails. Or, stated another way, the 8 parts that result will have considerable overlap. –  GEdgar Jun 23 '11 at 0:13
I think I got what you mean. Thank you. Would you consider the following as a reasonable explanation why $\dim_H(X) \le d$ for $X \subset \mathbb{R}^d$: Hausdorff dimension does not count overlaps. Consider two fractals in $\mathbb{R}^2$ that produce the same drawings: the first one $F_1$ without overlaps, and the second one $F_2$ with overlaps. Then $\dim_H(F_1) = \dim_H(F_2)$, but clearly $\dim_H(F_1) \le 2$. Please, could you tell me what would be dimension of fractal that I gave as the example? Thank you. –  Slobodan Jun 27 '11 at 1:21
Oh, I have forgotten to ask -- in order to computer Hausdorff dimension by using $N(l)$ stuff, we can only do that when fractal drawing does not overlap itself? –  Slobodan Jun 27 '11 at 1:39
Or when there is minimal overlap, as specified in the "open set condition". –  GEdgar Aug 1 '11 at 23:32