## Range of the Hausdorff dimension

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?
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 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 $\mathbb{R}^n$.
 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?