The Hausdorff Dimension Of The Cantor Set

Two Bounds Determine the Dimension

The middle-third Cantor set is produced by repeatedly removing open middle thirds from the interval \([0,1]\). At level \(n\), the remaining set is a union of \(2^n\) closed intervals, each of length \(3^{-n}\). These covers suggest the critical exponent: their total \(s\)-power diameter is \(2^n3^{-ns}\), which changes from growing to shrinking when \(2=3^s\). This gives an upper bound for the dimension. To prove equality, however, a cover calculation alone is not enough: we also need a lower bound that applies to every sufficiently fine cover.

Work in \(\mathbb{R}\) with its usual distance. Write \(C_0=[0,1]\), and obtain \(C_n\) from \(C_{n-1}\) by removing the open middle third of each interval in \(C_{n-1}\). The middle-third Cantor set is the compact Borel set \(C=\bigcap_{n=0}^{\infty}C_n\). A level-\(n\) cylinder interval is one of the \(2^n\) closed intervals making up \(C_n\).

The critical exponent in this case is \(s_0=\log(2)/\log(3)\), so that \(3^{s_0}=2\). The measure used below is a Borel probability measure on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), and it assigns total mass one to \(C\). The estimates will also apply to arbitrary, possibly nonmeasurable, cover sets by using the measure's outer measure.

The Cylinder Covers Give the Upper Bound

For any \(s>s_0\), \(3^s>2\), so \(2\cdot3^{-s}<1\). The level-\(n\) cylinders cover \(C\), and their diameters \(3^{-n}\) tend to zero. Their total \(s\)-cost is \((2\cdot3^{-s})^n\), which tends to zero. Thus, for every \(\delta>0\), arbitrarily large \(n\) give covers with diameters at most \(\delta\) and costs tending to zero. It follows that \(\mathcal{H}_{\delta}^{s}(C)=0\) for every \(\delta>0\), hence \(\mathcal{H}^{s}(C)=0\). This proves \(\dim_H C\le s_0\).

The Equal-Weight Cantor Probability Measure

We now construct a measure that gives each level-\(n\) cylinder mass \(2^{-n}\). For \(t\in[0,1)\), define binary digits $$ \xi_k(t)=\lfloor 2^kt\rfloor-2\lfloor 2^{k-1}t\rfloor,\qquad k\ge1, $$ and set \(\xi_k(1)=1\) for every \(k\). Each digit takes values in \(\{0,1\}\). Define \(\Phi:[0,1]\to[0,1]\) by $$ \Phi(t)=\sum_{k=1}^{\infty}\frac{2\xi_k(t)}{3^k}. $$ The finite partial sums are measurable, and their pointwise limit \(\Phi\) is measurable. Its values lie in \(C\): after fixing the first \(n\) digits, the remaining sum is between \(0\) and \(\sum_{k>n}2/3^k=3^{-n}\), so \(\Phi(t)\) lies in the corresponding level-\(n\) cylinder.

Let \(m\) denote Lebesgue measure on \([0,1]\), and define \(\mu(A)=m(\Phi^{-1}(A))\) for Borel \(A\subseteq\mathbb{R}\). This is the pushforward probability measure: countable additivity follows because inverse images preserve disjoint unions and \(m\) is countably additive. Also \(\mu(\mathbb{R})=m([0,1])=1\), and \(\mu(C)=1\). For a fixed word \(w=(w_1,\ldots,w_n)\in\{0,1\}^n\), the corresponding cylinder interval is $$ \left[\sum_{k=1}^n\frac{2w_k}{3^k},\ \sum_{k=1}^n\frac{2w_k}{3^k}+3^{-n}\right]. $$ The set of parameters with those first \(n\) binary digits is a dyadic interval of Lebesgue measure \(2^{-n}\), up to endpoints. The endpoints have Lebesgue measure zero. Further, distinct \(0,2\)-ternary digit sequences give distinct points: if their first differing digit is \(k\), the difference contributed there is \(2/3^k\), while the total possible later difference is at most \(\sum_{j>k}2/3^j=1/3^k\). Consequently, the preimage of that cylinder has measure \(2^{-n}\), and $$ \mu(\text{each level-}n\text{ cylinder})=2^{-n}. $$

A Uniform Bound for Short Sets

The essential geometric fact is that distinct level-\(n\) cylinders are separated by gaps of length at least \(3^{-n}\). Indeed, if two cylinders first branch at stage \(k\le n\), the gap created at that stage has length \(3^{-k}\ge3^{-n}\). This separation lets a short interval meet at most one level-\(n\) cylinder.

Proof. First suppose \(0<r<1\). Choose the integer \(n\ge0\) for which \(3^{-(n+1)}\le r<3^{-n}\). The set \(A\) is contained in the closed interval between its infimum and supremum, of length \(r\). That interval cannot meet two level-\(n\) cylinders: if it met two, points in its intersections with the two cylinders would be separated by at least the gap \(3^{-n}\), contradicting its length \(r<3^{-n}\). Since \(\mu\) is supported on \(C\), the interval therefore has measure at most \(2^{-n}\). By the choice of \(n\), \(r^{s_0}\ge3^{-(n+1)s_0}=2^{-(n+1)}\). Thus \(\mu^*(A)\le2^{-n}\le2r^{s_0}\). If \(r\ge1\), then \(\mu^*(A)\le\mu(\mathbb{R})=1\le2r^{s_0}\). If \(r=0\), then \(A\) is empty or a singleton. A singleton lies in a level-\(n\) cylinder for every \(n\), so its outer measure is at most \(2^{-n}\) for every \(n\), and hence is zero. This proves both claims. \(\square\)

The Mass Distribution Argument and the Dimension

We use the mass distribution principle in the form needed here: if a probability measure supported on a set \(E\) satisfies \(\mu^*(A)\le K(\operatorname{diam}A)^s\) for every set \(A\), with \(K<\infty\), then every countable cover of \(E\) has total \(s\)-cost at least \(1/K\). In this instance, the proof is direct. For any countable cover \(C\subseteq\bigcup_j U_j\), countable subadditivity of outer measure gives $$ 1=\mu(C)\le\sum_j\mu^*(U_j)\le2\sum_j(\operatorname{diam}U_j)^{s_0}. $$ Therefore every such cover has \(s_0\)-cost at least \(1/2\). The estimate applies to arbitrary cover sets, not just measurable ones.

To establish the dimension lower bound, fix \(t\) with \(0<t<s_0\), and consider a cover with \(\operatorname{diam}U_j\le\delta\). For each positive diameter, \((\operatorname{diam}U_j)^{s_0}\le\delta^{s_0-t}(\operatorname{diam}U_j)^t\); zero-diameter sets have outer measure zero by the proposition. The preceding cover estimate now yields $$ \sum_j(\operatorname{diam}U_j)^t\ge\frac12\delta^{t-s_0}. $$ As \(\delta\downarrow0\), the right-hand side tends to \(+\infty\), since \(t-s_0<0\). Thus \(\mathcal{H}^t(C)=\infty\) for every \(0<t<s_0\), proving \(\dim_H C\ge s_0\).

Proof. The cylinder covers prove \(\mathcal{H}^s(C)=0\) for every \(s>s_0\), so \(\dim_H C\le s_0\). The probability measure estimate proves \(\mathcal{H}^t(C)=\infty\) for every \(0<t<s_0\), so \(\dim_H C\ge s_0\). Combining the inequalities gives \(\dim_H C=s_0=\log(2)/\log(3)\). \(\square\)

Why Lebesgue Measure Alone Is Not Enough

A common mistake is to observe that the total length of the level-\(n\) cylinders is \(2^n3^{-n}=(2/3)^n\), which tends to zero, and then conclude that the Cantor set has dimension zero. That calculation proves that its one-dimensional Lebesgue measure is zero; it does not determine its Hausdorff dimension. For example, the same cylinders have total \(s_0\)-cost \(2^n3^{-ns_0}=2^n2^{-n}=1\), not a cost tending to zero. More importantly, the Cantor probability measure shows that no cover can have arbitrarily small total \(s_0\)-cost, and that every smaller positive exponent has Hausdorff measure infinite.

The two parts of the dimension proof have different jobs. Cylinder covers give the upper bound because they are a convenient explicit family of covers. The measure estimate gives the lower bound because it controls all cover sets at a common scale. Together they prove the dimension without identifying the exact value of the critical Hausdorff measure.