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Issue with the definition of the Hausdorff dimension

  1. May 26, 2010 #1
    Issue with the definition of the Hausdorff dimension

    1. The problem statement, all variables and given/known data
    http://mathworld.wolfram.com/HausdorffDimension.html" [Broken] involves a n-dimensional Hausdorff measure of 0. I'm having trouble understanding cases that would give such a value.


    2. Relevant equations
    the Hausdorff dimension [tex]\textbf{D}(\textbf{A})[/tex] of A is the infimum of d>=0 such that the d-dimensional Hausdorff measure of A is 0

    Let [tex]\textbf{X}[/tex] be a metric space, [tex]\textbf{A}[/tex] be a subset of [tex]\textbf{X}[/tex], and [tex]\textbf{d}[/tex] a number [tex]>=0[/tex].

    The [tex]\textbf{d}[/tex]-dimensional Hausdorff measure of [tex]\textbf{A}[/tex], [tex]\textbf{H}^\textbf{d}(\textbf{A})[/tex], is the infimum of positive numbers [tex]\textbf{y}[/tex] such that for every [tex]\textbf{r}>0[/tex], [tex]\textbf{A}[/tex] can be covered by a countable family of closed sets, each of diameter less than [tex]\textbf{r}[/tex], such that the sum of the [tex]\textbf{d}[/tex]th powers of their diameters is less than [tex]\textbf{y}[/tex]. Note that [tex]\textbf{H}^\textbf{d}(\textbf{A})[/tex] may be infinite, and [tex]\textbf{d}[/tex] need not be an integer.

    3. The attempt at a solution
    I read in some wikipedia article that a 2-dimensional Hausdorff measure of planar Brownian motion would be 0. So, I'm thinking that it may be that if one forms subsets out of said set A in which each subset contains only one isolated point, then the diameter of each of those subsets would be 0 and thus the Hausdorff measure would be 0. However, I don't think this is the explanation I'm looking for because then that set A would have a Hausdorff measure of 0 for any given dimension d.

    Another issue I have is related to the definition of the Hausdorff measure.
    It is the "infimum of positive numbers y such that..." How can the measure be 0 if it has to be a member of a set of positive numbers?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 26, 2010 #2
    The infimum of a set of positive numbers can be 0, e.g. the infimum of the open interval (0,1) is 0 even though every number in (0,1) is positive.

    For example, look at the d = 1 Hausdorff measure of the natural numbers (so this is just the Lebesgue measure). For any y>0 we can cover 0 by the closed interval [-y/4, y/4], 1 by the closed interval [1-y/8,1+y/8], 2 by [2-y/16, 2+y/16] and so forth. The total length of these intervals is y/2 + y/4 + y/8 + y/16 + ... = y. Since this worked for all positive y, the inf over all these y is 0 so that the d = 1 Hausdorff measure of the natural numbers is 0.
     
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