Hausdorff Dimension

Lighterning · #1 $Old$ January 25th, 2009, 12:58 PM

Hi all,

I would like to prove rigorously that the fractal dimension of the unit interval is one. Indeed, it feels obvious from the definition of the Hausdorff Measure that at dimension 1, it measures length.

It is relatively easy to prove that the Hausdorff measure is zero for dimensions greater than 1, but showing the Hausdorff Measure is infinity for dimension less than 1 and indeed 1 when looking at dimension equal to one is proving more difficult.

Thanks in advance!

Rebesques · #2 $Old$ June 5th, 2009, 11:41 AM

Quote:

but showing the Hausdorff Measure is infinity for dimension less than 1...

Consider just the half interval $I=(0,1)$ .
Let's suppose that $\mathcal{H}^{1-\lambda}(I)$ is finite for some $0<\lambda<1$ . Choose a covering $(S)$ of $I$ costisting of intervals of individual length $\delta>0$ . Their number must be no less than $\left[1/\delta\right]$ (as $(S)$ is also a covering for the Lebesgue measure of I in $\mathbb{R}$ ). This means that the related Hausdorff sum will be equal to $C\delta^{1-\lambda}\left[1/\delta\right]$ , where $C$ is a constant not depending on $\delta$ or $(S)$ . So, the infimum $m(\delta)$ of all such partitions (which we assumed exists) satisfies $m(\delta)\geq C\delta^{1-\lambda}\left[1/\delta\right]$ . Now in the last relation, the limit of the right hand side as $\delta\rightarrow0$ behaves like the limit of $\delta^{-\lambda}$ as $\delta\rightarrow0$ , which is infinite. A contradiction.

alunw · #3 $Old$ June 10th, 2009, 10:07 AM

The Hausdorff d-measure of a set is non-zero and finite for at most one value of d and if so that d is the dimension. So to prove the Hausdorff dimension of the unit interval is 1 it is only necessary to show the 1-measure is finite and non-zero. So really it would be more use to learn how to prove my first sentence in the general case (I must admit I'd need to refer to a textbook for the proof myself), as then if you can have a lucky guess at the right answer for d, and the d-measure happens to be non-zero and finite you don't have to worry about any other possible dimensions.
Sorry if I am stating the obvious, as I probably am.