Cylinder Covers Of The Middle-Third Cantor Set

Level-by-Level Covers

The middle-third Cantor set is built by repeatedly removing open middle thirds from the intervals that remain. At stage \(n\), the remaining set consists of \(2^n\) closed intervals, each of length \(3^{-n}\). These intervals are useful covers: their sizes and their number are explicit, so their total cost can be computed at any Hausdorff exponent.

Work in the usual metric on \(\mathbb{R}\). Write \(C_n\) for the union of the intervals remaining after stage \(n\), with \(C_0=[0,1]\), and let \(C=\bigcap_{n=0}^{\infty}C_n\) be the middle-third Cantor set. Each interval at stage \(n\) is indexed by a word \(w=(a_1,\ldots,a_n)\) with every \(a_j\) equal to \(0\) or \(2\). It is

There are \(2^n\) such words, hence \(2^n\) intervals. Each interval has diameter \(3^{-n}\), and \(C\subseteq C_n=\bigcup_w I_w\). Thus these intervals form a cover of \(C\) whose maximum diameter tends to zero as \(n\) increases.

The limit defining \(\mathcal{H}^s(E)\) exists in \([0,\infty]\): reducing \(\delta\) restricts the allowed covers, so \(\mathcal{H}^s_\delta(E)\) is nondecreasing as \(\delta\) decreases. The cover cost at stage \(n\), for exponent \(s\), is obtained by multiplying the number of intervals by the \(s\)-th power of their common diameter:

This formula is the central calculation. If \(2/3^s<1\), the stage costs tend to zero; if \(2/3^s=1\), they stay equal to one; and if \(2/3^s>1\), they grow. The equality case defines the critical exponent for this family of covers.

Computing the Critical Exponent

Proof. There are \(2^n\) intervals in the level-\(n\) cover, and each has diameter \(3^{-n}\). Therefore the cost is \(2^n(3^{-n})^s=(2/3^s)^n\). The ratio equals one precisely when \(3^s=2\). Taking logarithms gives \(s=\log 2/\log 3=d\). Substitution gives \(2/3^d=1\), so the cost is \(1^n=1\) at every level. If \(s>d\), then \(3^s>3^d=2\), and the ratio is less than one. If \(0<s<d\), then \(3^s<2\), and the ratio is greater than one. \(\square\)

The Upper Bound for Hausdorff Dimension

Proof. Fix \(s>d\) and \(\delta>0\). Since \(3^{-n}\to0\), there are arbitrarily large \(n\) for which \(3^{-n}\le\delta\). The level-\(n\) intervals are therefore an admissible cover in the definition of \(\mathcal{H}^s_\delta(C)\). By the cost calculation, their cost is \(q^n\), where \(q=2/3^s\). Because \(s>d\), the proposition gives \(0<q<1\), and hence \(q^n\to0\). For every \(\varepsilon>0\), choose such an \(n\) large enough that \(q^n<\varepsilon\). It follows that \(0\le\mathcal{H}^s_\delta(C)\le q^n<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(\mathcal{H}^s_\delta(C)=0\). This is true for every \(\delta>0\), so \(\mathcal{H}^s(C)=0\). Thus every \(s>d\) belongs to the set whose infimum defines \(\dim_H C\), giving \(\dim_H C\le d\).

For \(s=d\) and any \(\delta>0\), choose \(n\) with \(3^{-n}\le\delta\). The level-\(n\) cover is admissible and has cost \(1\), so \(\mathcal{H}^d_\delta(C)\le1\). Taking the limit as \(\delta\downarrow0\) preserves this upper bound, giving \(\mathcal{H}^d(C)\le1\). \(\square\)

What These Covers Do—and Do Not—Show

The argument proves an upper bound: it constructs covers whose costs tend to zero for every \(s>d\). It does not prove that the dimension equals \(d\). In particular, the fact that the level-\(n\) costs grow when \(s<d\) does not show that \(\mathcal{H}^s(C)\) is positive. Hausdorff content takes an infimum over all admissible covers, not just the level-by-level cylinder covers. A lower bound must rule out small-cost covers of every allowed form; that requires an additional argument.

A common error is to use the ordinary interval lengths for every exponent. For example, at \(s=1/2\) and level four, an incorrect calculation might multiply the 16 intervals by \(1/81\), obtaining \(16/81\). That is the cost at \(s=1\), not at \(s=1/2\). The prescribed cost uses the diameter raised to the chosen exponent. Here each diameter contributes \((1/81)^{1/2}=1/9\), so the correct cost is \(16\cdot1/9=16/9\). The general formula confirms it: \(2^4(3^{-4})^{1/2}=16\cdot3^{-2}=16/9\). Thus the exponent changes the contribution of each interval, even though the cover itself is unchanged.

The critical-cost calculation is useful because it identifies exactly where these particular covers stop becoming cheaper with increasing level. At \(s=d\), their cost remains one, so they give a finite upper bound for the critical Hausdorff outer measure. For \(s>d\), increasingly fine cylinder covers have arbitrarily small cost. For \(s<d\), their cost increases—but that observation alone is not a lower-bound proof. Keeping these conclusions separate is essential when using explicit covers to study Hausdorff dimension.