Range of the Hausdorff dimension

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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 [itex]\mathbb{R}^n[/itex], you will not find any set that has a dimension [itex]>n[/itex] !! To construct a set with Hausdorf dimension [itex]0 < d < n[/itex], it suffices to construct a set [itex]E[/itex] that has a [itex]d[/itex]-Hausdorf measure [itex]0 < \mathcal{H}^{(d)}(E) < \infty[/itex]. If you know the example of Cantor-like sets in [itex]\mathbb{R}[/itex], you can adapt the idea in [itex]\mathbb{R}^n[/itex].
 
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OB1
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So, suppose I start with the closed box in [tex] \mathbb{R}^{n}[/tex] and delete everything except the corners of the box with side length [tex] l [/tex], and then repeat this so the ith iteration leaves boxes of side length [tex] l^{i}[/tex]. So I have a Cantor-like set in [tex] \mathbb{R}^{n}[/tex]. I'm still a bit puzzled on finding the dimension of this object, any further hints?
 

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