A Covering Workflow For Self-Similar Dimension Problems

A Two-Sided Workflow

For a self-similar set, level-\(n\) cylinders give a natural cover: there are many small copies of the original set, each scaled down by \(n\) iterations. Counting these copies gives an upper bound on dimension. A lower bound requires a different kind of evidence: a measure supported on the set that cannot put too much mass in a small region. The workflow is to calculate both bounds and compare them, rather than treating the cylinder count as an automatic dimension formula.

We work in Euclidean space \(\mathbb{R}^q\) with its usual distance. For \(s\ge0\), the \(s\)-dimensional Hausdorff measure is defined by first taking the infimum of \(\sum_i(\operatorname{diam} U_i)^s\) over countable covers with \(\operatorname{diam}U_i\le\delta\), then letting \(\delta\downarrow0\). The Hausdorff dimension \(\dim_H K\) is the critical exponent separating the values of \(s\) for which this measure is infinite from those for which it is zero. We use the convention that a set of diameter zero contributes zero when \(s>0\).

Definition and upper-bound rule: Suppose \(K\subset\mathbb{R}^q\) is the attractor of \(N\) similarities, all with common contraction ratio \(r\in(0,1)\). A level-\(n\) cylinder is the image of \(K\) under a composition of \(n\) of these similarities. There are at most \(N^n\) such cylinders, and each has diameter \(r^n\operatorname{diam}K\). Consequently, if \(s>\dfrac{\log N}{\log(1/r)}\), then \(\mathcal{H}^s(K)=0\), so \(\dim_H K\le\dfrac{\log N}{\log(1/r)}\). This upper bound does not require the cylinders to be disjoint.

Here is the calculation behind the rule. The level-\(n\) cylinders cover \(K\), their diameters tend to zero, and the sum of their \(s\)-powers is at most

N^n\bigl(r^n\operatorname{diam}K\bigr)^s =(\operatorname{diam}K)^s(Nr^s)^n.

Writing \(d=\log N/\log(1/r)\), we have \(Nr^d=1\). Thus \(s>d\) implies \(Nr^s<1\), and the displayed sum tends to zero. This makes \(\mathcal{H}^s(K)=0\) by the definition of Hausdorff measure.

Worked Example: The Upper Bound for a Separated Two-Map Set

On \(\mathbb{R}\), consider \(S_0(x)=x/4\) and \(S_1(x)=3/4+x/4\), and let \(K\) be their compact attractor. The first-level images of \([0,1]\) are \([0,1/4]\) and \([3/4,1]\), so they lie in \([0,1]\) and are disjoint. The attractor is contained in \([0,1]\), hence \(\operatorname{diam}K\le1\).

At level \(n\), there are \(2^n\) words of length \(n\). Each corresponding cylinder has diameter at most \(4^{-n}\), because each map scales distances by \(1/4\). For any \(s>0\), the total \(s\)-power cost of this cover is at most

2^n(4^{-n})^s=(2\cdot4^{-s})^n.

The exponent at which the base equals one solves \(2\cdot4^{-d}=1\), so \(4^d=2\) and \(d=\log2/\log4=1/2\). For every \(s>1/2\), \(2\cdot4^{-s}<1\), and the cover costs tend to zero. Therefore \(\dim_H K\le1/2\). Check: at \(s=1\) the level-\(n\) cost is \(2^n4^{-n}=2^{-n}\), which tends to zero, in agreement with the upper-bound conclusion.

Turning a Supported Measure into a Lower Bound

A probability measure supported on \(K\) assigns total mass one to \(K\). The useful estimate is a uniform bound at small scales: if the measure of every set \(U\) is at most \(C(\operatorname{diam}U)^s\), then no sufficiently fine cover can have arbitrarily small \(s\)-power cost. In this statement, measure of an arbitrary, possibly nonmeasurable covering set means its outer measure.

Theorem (Mass estimate gives a dimension lower bound): Let \(K\subset\mathbb{R}^q\), and let \(\mu\) be a Borel probability measure on \(\mathbb{R}^q\) with \(\mu(K)=1\). Suppose that for some \(s>0\) and \(C<\infty\), \(\mu^*(U)\le C(\operatorname{diam}U)^s\) for every set \(U\subset\mathbb{R}^q\) of sufficiently small diameter. Then \(\mathcal{H}^s(K)\ge1/C\), and consequently \(\dim_H K\ge s\).

Proof. Choose a diameter threshold \(\delta_0>0\) for which the assumed estimate holds. Let \((U_i)\) be any countable cover of \(K\) with \(\operatorname{diam}U_i\le\delta\le\delta_0\). By countable subadditivity of the outer measure, and because \(\mu(K)=1\),

1=\mu^*(K)\le\sum_i\mu^*(U_i) \le C\sum_i(\operatorname{diam}U_i)^s.

Thus every such cover has cost at least \(1/C\). Taking the infimum over covers preserves this lower bound, and letting \(\delta\downarrow0\) gives \(\mathcal{H}^s(K)\ge1/C\). A positive \(s\)-dimensional Hausdorff measure implies \(\dim_H K\ge s\) by the definition of Hausdorff dimension. \(\square\)

Example: Completing the Two-Map Calculation

Worked Example: A Measure Bound at Every Small Scale

Continue with \(S_0(x)=x/4\), \(S_1(x)=3/4+x/4\), and their attractor \(K\subset[0,1]\). Give each of the two choices probability \(1/2\) at every stage. The resulting Bernoulli probability measure \(\mu\) is supported on \(K\); each level-\(n\) cylinder has measure \(2^{-n}\). We show that \(\mu^*(U)\le2(\operatorname{diam}U)^{1/2}\) for every sufficiently small \(U\), which is enough for the lower-bound theorem.

Write \(\ell=\operatorname{diam}U\), with \(0<\ell<1\), and choose \(n\ge1\) so that \(4^{-n}\le\ell<4^{-(n-1)}\). Put \(L=4^{-n}\), so \(\ell<4L\). Distinct level-\(n\) cylinder intervals are separated by a gap of at least \(2L\): if their words first differ after \(j-1\) common choices, their gap is \(\tfrac12\,4^{-(j-1)}\), which is at least \(\tfrac12\,4^{-(n-1)}=2L\). An interval of length less than \(4L\) cannot meet three cylinders. Indeed, to meet three in order it must span the middle cylinder, of length \(L\), and two gaps, each at least \(2L\); that requires length at least \(5L\).

If \(U\) is nonempty, it is contained in the interval between its infimum and supremum, of length \(\ell\). That interval meets at most two level-\(n\) cylinders. Since each has measure \(2^{-n}\),

\mu^*(U)\le2\cdot2^{-n} =2(4^{-n})^{1/2} \le2\ell^{1/2}.

The empty set has outer measure zero, so the bound holds for it as well. The lower-bound theorem with \(s=1/2\) and \(C=2\) gives \(\mathcal{H}^{1/2}(K)\ge1/2\), and hence \(\dim_H K\ge1/2\). Combined with the cylinder-cover upper bound, this proves \(\dim_H K=1/2\). Check: for \(\ell=4^{-n}\), the estimate reads \(2^{-n}\le2\cdot2^{-n}\), since \((4^{-n})^{1/2}=2^{-n}\); the cylinder mass used in the calculation is exactly \(2^{-n}\).

Example: A Four-Corner Set in the Plane

Worked Example: Matching Bounds for a Square Construction

In \(\mathbb{R}^2\), take the four similarities of ratio \(1/3\) that map the unit square to its four corner squares: the lower-left coordinates of those squares are \((0,0),(2/3,0),(0,2/3),(2/3,2/3)\). Let \(K\) be the attractor. Level \(n\) consists of \(4^n\) squares of side \(3^{-n}\), so each has diameter \(\sqrt{2}\,3^{-n}\). The upper-bound calculation gives the candidate exponent \(d=\log4/\log3\): for \(s>d\), \(4\cdot3^{-s}<1\), and the level-\(n\) cost is at most \((\sqrt{2})^s(4\cdot3^{-s})^n\), which tends to zero. Therefore \(\dim_H K\le d\).

Now use the probability measure that chooses each of the four maps with probability \(1/4\) at every stage. It is supported on \(K\), and each level-\(n\) square has mass \(4^{-n}\). Let \(U\) be a nonempty set with diameter \(\ell\), where \(0<\ell<1\). Choose \(n\) with \(3^{-n}\le\ell<3^{-(n-1)}\), and set \(L=3^{-n}\). The coordinates of the lower-left corners of level-\(n\) squares are integer multiples of \(L\). If a level-\(n\) square meets \(U\), its lower-left coordinate in either direction lies within distance \(L\) of the corresponding coordinate range of \(U\). That range has length at most \(\ell\), so the possible lower-left coordinates lie in an interval of length at most \(\ell+2L<5L\). There are at most five integer multiples of \(L\) in any interval of length less than \(5L\). Thus at most \(25\) level-\(n\) squares meet \(U\), and

\mu^*(U)\le25\cdot4^{-n} =25(3^{-n})^d \le25\ell^d.

The empty set again has zero outer measure. The mass estimate theorem yields \(\dim_H K\ge d\), so \(\dim_H K=\log4/\log3\). Check: the equality \((3^{-n})^d=4^{-n}\) follows by substituting \(d=\log4/\log3\): \(3^d=4\), hence \(3^{-nd}=4^{-n}\). The lower estimate and upper cover use the same exponent.

Compare the Bounds, Do Not Guess from the Count

The workflow is useful because its two halves have different hypotheses and different roles. The cylinder cover always supplies an upper bound when there are \(N\) maps of common ratio \(r\), even if some cylinders overlap. A lower bound needs a measure estimate that applies uniformly to small sets, not merely a statement of the mass of each cylinder. In the examples, separation made it possible to control how many cylinders a small interval or box could meet.

A common mistake is to calculate \(d=\log N/\log(1/r)\) from the number of maps and declare that it is the dimension. That calculation only shows the upper-bound exponent unless a valid lower bound is also established. For a concrete warning, take two identical maps on \(\mathbb{R}\), \(S_0(x)=x/2\) and \(S_1(x)=x/2\). Their attractor is the singleton \(\{0\}\): a point in the attractor must satisfy \(x=x/2\), so \(x=0\). Counting the two maps gives the candidate exponent \(\log2/\log2=1\), but the set has Hausdorff dimension zero. Indeed, for every \(s>0\), the singleton can be covered by an interval of arbitrarily small diameter, so its \(s\)-dimensional Hausdorff measure is zero. The repeated map supplies no second distinct piece and no measure estimate that could force dimension one.

Takeaway: Use level-\(n\) cylinders to prove an upper bound, and a supported measure with a checked small-set estimate to prove a lower bound. Equality of the two bounds certifies the dimension; a cylinder count by itself does not.

Check Your Understanding

For each question, identify which part of the covering workflow applies and state the hypotheses needed for the conclusion.

For \(N=3\) similarities of common ratio \(r=1/5\), what exponent does the level-\(n\) cover suggest, and for which \(s\) does its cost tend to zero?
Why does a probability measure satisfying \(\mu^*(U)\le C(\operatorname{diam}U)^s\) imply \(\mathcal{H}^s(K)\ge1/C\)?
In the two-map example, how many level-\(n\) cylinders can an interval of length less than \(4\cdot4^{-n}\) meet, and what separation estimate justifies the count?
What additional work is needed before interpreting the similarity exponent as the dimension when cylinders overlap?
For the four-corner construction, verify directly that \(3^d=4\) when \(d=\log4/\log3\).