# Hausdorff Dimension

• Jan 25th 2009, 11:58 AM
Lighterning
Hausdorff Dimension
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.

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.