I know that for bounded non-empty subsets of $\mathbb{R}^n$ the following inequality holds, where $D_T$ is the Lebesgue covering dimension, $D_H$ the Hausdorff-Besicovitch dimension and $D_F$ the box counting dimension:

$0\leq D_T \leq D_H \leq D_F \leq n$

Mandelbrot's definition of fractal states:

"A fractal is by definition a set for which the Hausdorff Besicovitch dimension strictly exceeds the topological dimension." (from: The Fractal Geometry of Nature. Updated and Augmented. 3rd edition, 1983.)

Sadly, $D_F$ is easier to calculate for many sets than $D_H$. Furthermore, I know that many sets (including non overlapping IFS fractals) $D_F$ and $D_H$ are equal. So I wonder If the definition could not as well use $D_F$ instead of $D_H$. This reduces to the following question:

Is there a subset of $\mathbb{R}^n$, such that $D_T=D_H< D_F$?

  • $\begingroup$ Do you mean a strict inequality in the last line? $\endgroup$
    – Del
    Commented Jan 2, 2018 at 23:42
  • $\begingroup$ Yes, of course. Thank you, I will edit it right now. $\endgroup$ Commented Jan 2, 2018 at 23:43
  • $\begingroup$ It should be noted that Mandelbrot backed off of that definition pretty quickly. I'll try to remember to check the exact reference when I get back into the office next week, by my recollection is that in the appendix to the second edition of The Fractal Geometry of Nature, he discusses the problem with that characterization of a fractal set. You seem to have the third edition, so you might be able to check for yourself. $\endgroup$
    – Xander Henderson
    Commented Jan 3, 2018 at 1:00

1 Answer 1


The set of points with rational coordinates is countable but dense. Hence, it has Hausdorff dimension 0 and box-counting dimension n


Let the first coordinate be all reals and the other coordinates be rational. Then topological and Hausdorff are 1 and box counting is n


Let the first coordinate be the standard Cantor set obtained by removing middle thirds from $[0,1]$ and the other coordinates be rational. Then topological is 0, Hausdorff is 0.63... and box counting is n-1+0.63 (you need to prove it)

  • $\begingroup$ Is there also an example when the topological and hausdorff dimension is not 0? $\endgroup$ Commented Jan 3, 2018 at 0:22
  • $\begingroup$ Yes, but it’s kind of cheating. See edit. $\endgroup$
    – user334639
    Commented Jan 3, 2018 at 0:33
  • $\begingroup$ Nice construction! $\endgroup$ Commented Jan 3, 2018 at 0:40
  • $\begingroup$ The example of $\mathbb{Q}^n \subseteq \mathbb{R}^n$ is nice, in that it shows the danger of defining fractality in terms of a strict inequality involving the box dimension. I don't think that anyone would argue that the set of rational points are fractal (or maybe they would?), but @MinecraftShamrock's definition would consider it to be fractal. $\endgroup$
    – Xander Henderson
    Commented Jan 3, 2018 at 1:05

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