# Issue with the definition of the Hausdorff dimension

1. May 26, 2010

### mikfig

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 $$\textbf{D}(\textbf{A})$$ of A is the infimum of d>=0 such that the d-dimensional Hausdorff measure of A is 0

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

The $$\textbf{d}$$-dimensional Hausdorff measure of $$\textbf{A}$$, $$\textbf{H}^\textbf{d}(\textbf{A})$$, is the infimum of positive numbers $$\textbf{y}$$ such that for every $$\textbf{r}>0$$, $$\textbf{A}$$ can be covered by a countable family of closed sets, each of diameter less than $$\textbf{r}$$, such that the sum of the $$\textbf{d}$$th powers of their diameters is less than $$\textbf{y}$$. Note that $$\textbf{H}^\textbf{d}(\textbf{A})$$ may be infinite, and $$\textbf{d}$$ 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. May 26, 2010

### eok20

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.