Range of the Hausdorff dimension

[OB1]

My analysis textbook mentioned in passing that the range of the Hausdorff dimension is all nonnegative real numbers, i.e. for any nonnegative real number a, there's some compact subset of R^n whose Hausdorff dimension is exactly a. The problem is that I don't see how to prove this (and my oh-so-concise book doesn't bother proving it). Does anyone know how to go about proving this?

 

[alekk]

in $\mathbb{R}^n$, you will not find any set that has a dimension $>n$ ! To construct a set with Hausdorf dimension $0 < d < n$, it suffices to construct a set $E$ that has a $d$-Hausdorf measure $0 < \mathcal{H}^{(d)}(E) < \infty$. If you know the example of Cantor-like sets in $\mathbb{R}$, you can adapt the idea in $\mathbb{R}^n$.

 

[OB1]

So, suppose I start with the closed box in $$\mathbb{R}^{n}$$ and delete everything except the corners of the box with side length $$l$$, and then repeat this so the ith iteration leaves boxes of side length $$l^{i}$$. So I have a Cantor-like set in $$\mathbb{R}^{n}$$. I'm still a bit puzzled on finding the dimension of this object, any further hints?