# Error in Falconer's proof of Hausdorff dimension of Cantor set

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.

Is there a way to fix the proof?

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

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.