Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Range of the Hausdorff dimension

  1. Apr 10, 2009 #1

    OB1

    User Avatar

    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?
     
  2. jcsd
  3. Apr 11, 2009 #2
    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].
     
    Last edited: Apr 11, 2009
  4. Apr 13, 2009 #3

    OB1

    User Avatar

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Range of the Hausdorff dimension
  1. Range of Validity (Replies: 1)

  2. Hausdorff Dimension (Replies: 1)

Loading...