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Issue with the definition of the Hausdorff dimension
http://mathworld.wolfram.com/HausdorffDimension.html" involves a n-dimensional Hausdorff measure of 0. I'm having trouble understanding cases that would give such a value.
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.
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?
Homework Statement
http://mathworld.wolfram.com/HausdorffDimension.html" involves a n-dimensional Hausdorff measure of 0. I'm having trouble understanding cases that would give such a value.
Homework 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.
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?
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