Overlaps Can Lower Similarity Dimension

Two Contractions with Overlapping Images

For a family of contractions, the equation used to define its similarity dimension counts the contraction ratios, but it does not record how the images are positioned. If those images overlap, the actual attractor can be much smaller in dimension than the similarity dimension suggests. The maps here make the contrast explicit: their attractor is an interval in \(\mathbb{R}\), while their similarity dimension is greater than one.

Throughout, the ambient space is \(\mathbb{R}\) with its usual distance. The maps \(S_1,S_2:\mathbb{R}\to\mathbb{R}\) are similarities with common contraction ratio \(r=3/4\). A compact invariant set for these maps means a nonempty compact set \(K\) satisfying \(K=S_1(K)\cup S_2(K)\); the attractor is the unique such set. The Lebesgue measure space used below is \((\mathbb{R},\mathcal{L},m)\), which is complete and \(\sigma\)-finite.

For clarity, for \(t\ge0\) and \(\delta>0\), the \(t\)-dimensional Hausdorff content at scale \(\delta\) is the infimum of \(\sum_j(\operatorname{diam}U_j)^t\) over countable covers by sets \(U_j\subset\mathbb{R}\) with \(\operatorname{diam}U_j\le\delta\). The Hausdorff measure is obtained by letting \(\delta\) decrease to zero; normalization conventions do not affect Hausdorff dimension. The Hausdorff dimension is the critical exponent separating zero Hausdorff measure from nonzero Hausdorff measure.

Finding and Verifying the Attractor

To conclude that this invariant interval is the attractor, we verify uniqueness of a nonempty compact invariant set rather than relying only on the endpoint calculation.

Proof. We have already shown that \(I\) is nonempty, compact, and satisfies the invariance equation. Let \(A\) and \(B\) be any two nonempty compact invariant sets. Their Hausdorff distance is \(d_H(A,B)=\max\{\sup_{a\in A}\inf_{b\in B}|a-b|,\sup_{b\in B}\inf_{a\in A}|b-a|\}\), which is finite because both sets are compact. For each \(a\in A\), invariance gives \(a=S_i(a_0)\) for some \(i\in\{1,2\}\) and \(a_0\in A\). Compactness of \(B\) ensures there is \(b_0\in B\) with \(|a_0-b_0|\le d_H(A,B)\). Then \(S_i(b_0)\in B\) by invariance, and \[ |a-S_i(b_0)|=\tfrac34|a_0-b_0|\le\tfrac34d_H(A,B). \] The same argument with \(A\) and \(B\) interchanged bounds both directed distances, so \(d_H(A,B)\le(3/4)d_H(A,B)\). Since this distance is finite and nonnegative, it must be zero. A zero Hausdorff distance between compact sets implies \(A=B\). Therefore the invariant compact set \(I\) is unique. \(\square\)

The Two Dimensions

The attractor \(I=[0,2/5]\) has Hausdorff dimension one. Here is a brief verification using covers, consistent with the earlier results on Hausdorff measures of intervals. For \(t>1\), divide \(I\) into \(N\) equal closed subintervals, each of diameter \(2/(5N)\). Their total \(t\)-power diameter is \[ N\left(\frac{2}{5N}\right)^t=\left(\frac25\right)^tN^{1-t}, \] which tends to zero as \(N\to\infty\). Hence the \(t\)-dimensional Hausdorff measure is zero for every \(t>1\).

For \(0\le t<1\), take any countable cover of \(I\) by sets of diameters \(d_j\le\delta\). Each set is contained in an interval of length arbitrarily close to \(d_j\); countable subadditivity of Lebesgue outer measure therefore gives \(2/5\le\sum_jd_j\). For \(0<d_j\le\delta\), \(d_j^t\ge\delta^{t-1}d_j\); zero diameters contribute nothing to \(\sum_jd_j\). Consequently the sum of \(t\)-power diameters is at least \((2/5)\delta^{t-1}\), which tends to infinity as \(\delta\downarrow0\). At \(t=0\), any cover of a nondegenerate interval has at least one nonempty member, so its zero-dimensional covering sum cannot tend to zero either. Thus the dimension is exactly one.

This is the essential difference between similarity dimension and Hausdorff dimension in this example. The similarity-dimension equation records two copies scaled by \(3/4\) as if their sizes contributed separately. But the actual images overlap in an interval of positive length, and the invariant set fills only the interval \([0,2/5]\). Since that interval lies in \(\mathbb{R}\), its Hausdorff dimension is one, not approximately \(2.409\).

A Common Mistake: Using the Similarity Dimension as the Answer

A tempting but incorrect calculation is to solve \(2(3/4)^s=1\) and then report \(s\approx2.409\) as the Hausdorff dimension of the attractor. The equation is solved correctly, but it defines the similarity dimension; it does not by itself establish the Hausdorff dimension. In this case that proposed answer cannot be right: the attractor is a subset of the line, and its dimension is exactly one.

The correction is to check how the images sit before applying a dimension formula that requires separation. Here \(S_1(I)\cap S_2(I)=[1/10,3/10]\), so the two pieces are not disjoint, and their overlap has positive length \(1/5\). The separated-piece conclusion from the Moran formula is therefore not available. The verified invariant set and its interval dimension give the actual answer: Hausdorff dimension one, strictly less than the similarity dimension.