5
$\begingroup$

In Falconer's Geometry of Fractal Sets, he establishes a lower bound of $s=\log(2)/\log(3)$ on the Hausdorff dimension of the Cantor set $E$ as follows:

We show that if $\mathscr{I}$ is any collection of intervals covering $E$, then $$ 1 \leq \sum_{I \in \mathscr{I}}|I|^s \quad(1.21). $$ By expanding each interval slightly and using the compactness of $E$, it is enough to prove $(1.21)$ when $\mathscr{I}$ is a finite collection of closed intervals. By a further reduction we may take each $I \in \mathscr{I}$ to be the smallest interval that contains some pair of net intervals, $J$ and $J^{\prime}$, that occur in the construction of $E$. ($J$ and $J^{\prime}$ need not be intervals of the same $E_j$.) If $J$ and $J^{\prime}$ are the largest such intervals, then $I$ is made up of $J$, followed by an interval $K$ in the complement of $E$, followed by $J^{\prime}$ [emphasis mine]. From the construction of the $E_j$ we see that $$ |J|,\left|J^{\prime}\right| \leq|K| . $$ Then $$ \begin{aligned} |I|^s &=\left(|J|+|K|+\left|J^{\prime}\right|\right)^s \\ & \geq\left(\frac{3}{2}\left(|J|+\left|J^{\prime}\right|\right)\right)^s=2\left(\frac{1}{2}|J|^s+\frac{1}{2}\left|J^{\prime}\right|^s\right) \geq|J|^s+\left|J^{\prime}\right|^s, \end{aligned} $$ using the concavity of the function $t^s$ and the fact that $3^s=2$. Thus replacing $I$ by the two subintervals $J$ and $J^{\prime}$ does not increase the sum in $(1.21)$. We proceed in this way until, after a finite number of steps, we reach a covering of $E$ by equal intervals of length $3^{-j}$, say. These must include all the intervals of $E_j$, so as $(1.21)$ holds for this covering it holds for the original covering $\mathscr{I} .$

In this proof, $E$ is the Cantor set and $E_j$ are the intervals that occur in its construction (i.e., $E_1=[0,1]$, $E_2=[0,1/3]\cup [2/3,1]$, etc.). However, I do not think this proof is correct, as I believe the part in bold is untrue. It is not true that $I=J \cup K \cup J'$. Here is a counterexample.

Consider $E_5$. Let $I$ be an interval covering the first 11 sub-intervals. Then there is no way to split $I$ up into $J \cup K \cup J'$. A picture might be helpful here.

enter image description here

Is there a way to fix the proof?

$\endgroup$
2
  • 1
    $\begingroup$ In your example is $I$ the smallest such interval? $\endgroup$ Sep 8 at 22:32
  • $\begingroup$ @LorenoHeer Yes, it is, so I suppose I should specify that it is closed. $\endgroup$ Sep 8 at 22:55

1 Answer 1

4
$\begingroup$

When you choose the "largest such intervals", I don't believe it's necessary for them to be net intervals (although that certainly seems to be what Falconer is taking them to be).

First note that by taking $\ I\ $ to be the "smallest [closed] interval that contains some pair of net intervals" you're ensuring that the end points of $\ I\ $ belong to $\ E\ $.

Now let $\ k_\max\ $ be the largest value of $\ k\ $ such that $\ I\subseteq E_k\ $. Then there's a net interval $\ M\ $ of $\ E_{k_\max}\ $ and two net intervals $\ L\ $ and $\ L'\ $ of $\ E_{k_\max+1}\ $ such that $\ M=L\cup K\cup L'\ $, where $\ K\ $ is an interval in the complement of $\ E\ $, and $\ K\subseteq I\subseteq M\ $. Thus, $\ I=(L\cap I)\,\cup$$\,K\,\cup$$\,(L'\cap I)\ $, is made up of an interval $\ J=I\cap L\ $ followed by the interval $\ K\ $ in the complement of $\ E\ $ followed by an interval $\ J'=I\cap L'\ $, and you still have $$ |J|,|J'|\le|K|\ , $$ so the rest of the proof will still work.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.