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.

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Is there a way to fix the proof?

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    $\begingroup$ 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


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.


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