Are all Hausdorff fractals also box counting fractals? The Hausdorff dimension of a set $A$ is always lesser or equal than its boxcounting dimension. 
$$
\mathrm{dim}_\mathcal{H}(A)\leq\mathrm{dim}_\mathcal{B}(A)
$$
More precisely lesser or equal than its lower boxcounting dimension.
If the set $A$ is fulfills the open set condition or is self-similar, the two dimensions are equal.
But if $A$ does not, is it possible that only one of those dimensions is fractal?
For example, $\mathrm{dim}_\mathcal{H}(A)=1$ while $\mathrm{dim}_\mathcal{B}(A)>1$ or $\mathrm{dim}_\mathcal{B}(A)=1$ while $\mathrm{dim}_\mathcal{H}(A)<1$.
 A: Everything, within the bounds $0\le \dim_H\le \dim_B\le n$ (for subsets of $\mathbb{R}^n$) is possible. I'll give two examples in $\mathbb{R}^2$, which can be generalized. 
Example 1: $A=\{(x, y): 0\le y\le 1, x\in \{1/1, 1/2, 1/3, \dots\}\}$. This is the union of countably many line segments $\{1/k\}\times [0,1]$. Since the Hausdorff dimension is countably stable, $\dim_H A = 1$. On the other hand, $\dim_B A = 3/2$. Indeed, if we cover $A$ by boxes of size $\epsilon$, then doubling the size each box we cover the $\epsilon$-neighborhood of $A$. This neighborhood contains the rectangle $[0,\sqrt{\epsilon}]\times [0,1]$ because when $1/k < \sqrt{\epsilon}$, the distance between $1/k$ and $1/(k+1)$ becomes less than $\epsilon$. Since the area of this rectangle is $\sqrt{\epsilon}$, and each $\epsilon$-square has area $\epsilon^2$ it follows that the number of boxes is $\ge C\epsilon^{-3/2}$. A matching upper bound can be shown by using $\epsilon^{-3/2}$ boxes to cover the aforementioned rectangle, and about as many boxes to cover the remaining $\epsilon^{-1/2}$ line segments individually.  
Example 2: pick a Cantor-type set $C\subset [0,1]$ of Hausdorff dimension $1/2$, and let $A=\{(x, y): y\in C, x\in \{1/1, 1/2, 1/3, \dots\}\}$. Again by countable stability, $\dim_H A = 1/2$. On the other hand, $\dim_B A=1$ which is left as an exercise. 
