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?