The Moran Formula Under Separation

The Moran Formula and Its Separation Hypothesis

A finite family of contractions has an immediate candidate for the dimension of its self-similar set: choose an exponent so that the number of pieces times the scale factor to that exponent equals one. This equation gives a prediction, not by itself a proof. The open set condition supplies enough separation to show that the predicted exponent is the actual Hausdorff dimension, and that the Hausdorff measure at this exponent is neither zero nor infinite.

Let \(K\subset\mathbb{R}^q\) be the compact attractor of \(N\) similarities \(S_1,\ldots,S_N:\mathbb{R}^q\to\mathbb{R}^q\), each with the same contraction ratio \(r\), where \(0<r<1\). Thus \(|S_i(x)-S_i(y)|=r|x-y|\) and \(K=\bigcup_{i=1}^N S_i(K)\). For a word \(w=(i_1,\ldots,i_n)\), write \(S_w=S_{i_1}\circ\cdots\circ S_{i_n}\) and \(K_w=S_w(K)\); then \(\operatorname{diam}(K_w)=r^n\operatorname{diam}(K)\). We use Hausdorff measure defined from covers by sets of bounded diameter, with cost \(\sum_j(\operatorname{diam} E_j)^s\). Multiplying this convention by a fixed normalization constant would not change the dimension or finiteness and positivity.

The condition requires disjoint images of an open set; it does not require the compact sets \(S_i(K)\) to be separated by a positive distance. The exponent in the Moran formula is the unique \(d\ge0\) satisfying \(Nr^d=1\). When \(N\ge2\), this is \(d=\log(N)/\log(1/r)>0\).

Three Basic Cases

Proof of the Formula

We prove the finite upper bound by covering \(K\) with all its level-\(n\) pieces. Put \(D=\operatorname{diam}(K)\), which is finite because \(K\) is compact. There are at most \(N^n\) such pieces, each of diameter \(r^nD\). Since \(Nr^d=1\), their total \(d\)-dimensional cost is at most

For each diameter bound \(\delta>0\), choose \(n\) large enough that \(r^nD\le\delta\). These pieces then form an admissible cover of \(K\), with cost at most \(D^d\). Taking the infimum over admissible covers and then letting \(\delta\) decrease to zero proves \(\mathcal{H}^d(K)\le D^d<\infty\).

For the lower bound, we construct a probability measure that assigns mass \(N^{-n}\) to each prescribed sequence of \(n\) indices. Give the sequence space \(\{1,\ldots,N\}^{\mathbb{N}}\) the product probability measure for which each index has probability \(1/N\). The coding map sends \((i_1,i_2,\ldots)\) to the unique point in the nested sets \(K_{(i_1,\ldots,i_n)}\). These sets are nonempty compact sets with diameters at most \(r^nD\), so their intersection consists of one point. The coding map is measurable: sequences agreeing in their first \(n\) coordinates have images at distance at most \(r^nD\), which tends to zero. Push the product probability forward by this map and call the resulting Borel probability measure \(\nu\). It satisfies \(\nu(K)=1\) and is supported on \(K\). We use only an upper bound on the mass of a ball; no uniqueness of coding is needed.

Fix \(x_0\in U\). Since \(U\) is open, choose \(a>0\), with \(a\le1\), such that the open Euclidean ball \(B(x_0,a)\) is contained in \(U\). Set \(L=\sup_{y\in K}|x_0-y|<\infty\). At any fixed level \(n\), the open sets \(S_w(U)\), for words \(w\) of length \(n\), are pairwise disjoint. Indeed, two distinct words first differ at some index; their corresponding first-level images of \(U\) are disjoint, and applying their common prefix preserves disjointness. Each \(S_w(B(x_0,a))\) is a ball of radius \(ar^n\) contained in \(S_w(U)\), so these balls are also pairwise disjoint.

Take \(0<\rho<1\) and choose \(n\ge1\) so that \(r^n\le\rho<r^{n-1}\). The choice is possible because \(r^n\) decreases to zero, and this strict range is used only for \(\rho<1\). Let \(M\) be the number of words \(w\) of length \(n\) for which \(K_w\cap\overline B(x,\rho)\ne\varnothing\). If \(z\) is in this intersection, then \(z=S_w(y)\) for some \(y\in K\), and \[ |S_w(x_0)-x|\le |S_w(x_0)-z|+|z-x|\le r^nL+\rho. \] Consequently every ball \(S_w(B(x_0,a))\) counted by \(M\) lies inside \(B(x,\rho+(L+a)r^n)\). Comparing their ordinary \(q\)-dimensional volumes, using disjointness and the fact that a radius-\(t\) ball has volume \(v_qt^q\), gives \[ M v_q(ar^n)^q\le v_q\bigl(\rho+(L+a)r^n\bigr)^q. \] Cancel \(v_q>0\), divide by \((ar^n)^q\), and use \(\rho/r^n<1/r\). This yields \[ M\le\left(\frac{1/r+L+a}{a}\right)^q=:A. \] If a coded point lies in \(\overline B(x,\rho)\), its first \(n\) indices must be one of these \(M\) words. Each such prefix has product probability \(N^{-n}\). Therefore \[ \nu(\overline B(x,\rho))\le MN^{-n}\le A N^{-n} =A(r^n)^d\le A\rho^d. \] Here \(N^{-n}=(r^n)^d\) follows from \(Nr^d=1\), and the last inequality uses \(r^n\le\rho\).

The endpoint \(\rho=1\) is not covered by the choice \(r^n\le\rho<r^{n-1}\) with \(n\ge1\), so it must not be included in that argument. Instead, \(\nu(\overline B(x,1))\le1\le A\), since \(A>1\). For every \(\rho\ge1\), the same probability bound gives \(\nu(\overline B(x,\rho))\le1\le A\rho^d\). We have thus proved, for all \(x\in\mathbb{R}^q\) and all \(\rho>0\), that \(\nu(\overline B(x,\rho))\le A\rho^d\).

Now let \((E_j)\) be any countable cover of \(K\) by sets of diameter at most \(\delta\). For each \(E_j\) that meets \(K\) and has positive diameter \(t_j\), choose \(x_j\in E_j\cap K\). Then \(E_j\cap K\subset\overline B(x_j,t_j)\), so its outer measure with respect to \(\nu\) is at most \(At_j^d\). If \(E_j\) has diameter zero, its intersection with \(K\) is empty or a singleton. Each singleton has \(\nu\)-measure zero, because it is contained in closed balls of every positive radius and the ball bound tends to zero as the radius tends to zero; here \(d>0\). Countable subadditivity of outer measure and \(\nu(K)=1\) now imply \[ 1\le\sum_j\nu^*(E_j\cap K)\le A\sum_j(\operatorname{diam}E_j)^d. \] Taking the infimum over covers gives \(\mathcal{H}^d_\delta(K)\ge1/A\), and letting \(\delta\) decrease to zero proves \(\mathcal{H}^d(K)\ge1/A>0\).

Finally, the critical-exponent comparison identifies the dimension. If \(s>d\), then for every cover with diameters at most \(\delta\), each cost term satisfies \((\operatorname{diam}E_j)^s\le\delta^{s-d}(\operatorname{diam}E_j)^d\). Since \(\mathcal{H}^d(K)<\infty\), this implies \(\mathcal{H}^s(K)=0\). If \(0\le t<d\), the corresponding inequality is \((\operatorname{diam}E_j)^d\le\delta^{d-t}(\operatorname{diam}E_j)^t\). The lower bound \(\mathcal{H}^d_\delta(K)\ge1/A\) therefore forces \(\mathcal{H}^t_\delta(K)\ge A^{-1}\delta^{-(d-t)}\), which tends to infinity as \(\delta\) decreases to zero. Thus \(\mathcal{H}^t(K)=\infty\). By the definition of Hausdorff dimension as the critical exponent separating infinite measure from zero measure, \(\dim_H K=d\). \(\square\)

The Single-Map Case and a Common Pitfall

Common mistake: replacing disjoint open images with a positive gap between compact pieces. Take \(S_1(x)=x/2\) and \(S_2(x)=x/2+1/2\). For \(U=(0,1)\), the images \((0,1/2)\) and \((1/2,1)\) are disjoint, so the open set condition holds. But the attractor is \([0,1]\), and its compact pieces \([0,1/2]\) and \([1/2,1]\) meet at \(1/2\); their distance is zero. The incorrect inference “disjoint open images, therefore the compact pieces have positive distance” fails in this concrete case. The proof above instead packs disjoint balls inside the open images. That argument remains valid even when compact pieces touch.