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 $D_T=D_H< D_F$?
The set of points with rational coordinates is countable but dense. Hence, it has Hausdorff dimension 0 and box-counting dimension n
Also:
Let the first coordinate be all reals and the other coordinates be rational. Then topological and Hausdorff are 1 and box counting is n
Finally:
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)