UNIFORMLY PERFECT ANALYTIC AND CONFORMAL ATTRACTOR SETS

RICH STANKEWITZ

Abstract. Conditions are given which imply that analytic iterated function systems (IFS’s) in the complex plane ℂ have uniformly perfect attractor sets. In particular, it is shown that the attractor set of a finitely generated conformal IFS is uniformly perfect when it contains two or more points. Also, an example of a finitely generated analytic attractor set which is not uniformly perfect is given.

1. Introduction and results

Consider an iterated function system (IFS) G = ⟨g_{i} : i ∈ I⟩, the set of all finite compositions of non-constant generating maps {g_{i} : i ∈ I} for some index set I, where each function maps the open connected set U ⊂ ℂ into a compact set K ⊂ U such that there exists 0 < s < 1 and a metric d on K where d(g_{i}(z), g_{i}(w)) ≤ sd(z, w) for all z, w ∈ K and all i ∈ I. Thus this system is uniformly contracting on the metric space (K, d). We say that G is an IFS on (U, K). We define the attractor set A by A = A(G) = \overline{A′}, the closure of A′ in the Euclidean topology, where A′ = A′(G) = {z : there exists g ∈ G such that g(z) = z} is the set of (attracting) fixed points of G. We will suppress the dependence on G when there is no chance for confusion. We define an IFS and its corresponding attractor set to be analytic (respectively conformal) if all the maps are analytic (respectively conformal) on U. In an analytic IFS, we note that we may, and will, take the metric d to be the hyperbolic metric on U (see section 2).

We note that when the IFS G = ⟨g_{1}, …, g_{N}⟩, we have that A is the unique compact set such that

A = ∪_{i=1}^{N}g_{i}(A) (1.1)

(see [4], p. 724). We also point out that in [6, 7] the limit set J of a conformal IFS is defined in such a way so that J ⊃ A′ and \overline{J} = A. In [6, 7] conformal IFS’s are defined differently, and non equivalently, and so the reader should observe that the definition above does not match exactly the definitions that one may find in the literature.

1991 Mathematics Subject Classification: Primary 30D05, 58F23. Key words and phrases. Attractor sets, Uniformly perfect, iterated function systems.

2 RICH STANKEWITZ

In [11] certain conformal attractor sets have been shown to be uniformly perfect, that is, these sets are uniformly thick near each of their points (see section 2), when the generating maps are Möbius. We note that in [11], the result in Corollary 3.3 is only stated for attractor sets generated by linear maps, but clearly more general cases are handled by Theorem 3.1. In this reference critical use was made of the fact that the generating maps were globally defined (and had globally defined inverses). Such restrictions are not needed in the following theorem.

Theorem 1.1. Let G = ⟨g_{i} : i ∈ I⟩ be an analytic IFS on (U, K) where there exist 0 < δ < diam(A) and C > 0 such that we have the following:

(I) if a ∈ A and i ∈ I, then g_{i} is one-to-one on △(a, δ),

(II) if a ∈ A and g_{i}(a) = a′, then the branch h_{i} of g_{i}^{−1} such that h_{i}(a′) = a is defined on △(a′, δ), (III) if a ∈ A and g_{i}(a) = a′, then the branch h_{i} of g_{i}^{−1} such that h_{i}(a′) = a satisfies |h_{i}(z) − h_{i}(a′)| ≤ C|z − a′| for all z ∈ \overline{△(a′, δ/10)}. Then, if the attractor set A has infinitely many points, then A is uniformly perfect.

Above and throughout we use the following notation. Let q be a metric. For a set F ⊂ ℂ, let diam_{q}(F) = sup{q(z, w) : z, w ∈ F } and F_{ϵ}^{q} = {z : dist_{q}(z, F) < ϵ} where dist_{q}(z, F) = inf{q(z, w) : w ∈ F }. Also let △_{q}(w, r) = {z : q(z, w) < r} and C_{q}(w, r) = {z : q(z, w) = r}. If no metric is noted, then it is assumed that the metric is the Euclidean metric.

Remark 1.1. Suppose δ satisfies (I) in Theorem 1.1. Then there exists (see proof at the end of section 2) δ′ > 0 such that each g ∈ G is one-to-one on △(a, δ′). Hence in the proof of Theorem 1.1 we will replace δ by δ′ and assume that each g ∈ G is one-to-one on △(a, δ) when a ∈ A.

Remark 1.2. Suppose δ satisfies (I) in Theorem 1.1. If there exists ϵ > 0 such that for all a ∈ A and i ∈ I we have g_{i}(△(a, δ)) ⊃ △(g_{i}(a), ϵ), then we see that the branch h_{i} of g_{i}^{−1} such that h_{i}(g_{i}(a)) = a is defined on △(g_{i}(a), ϵ). Hence we may replace δ by min{δ, ϵ} to satisfy both (I) and (II).

Corollary 1.1. Let G = ⟨g_{i} : i ∈ I⟩ be an analytic IFS on (U, K) such that there exists η > 0 where |g′_{i}(a)| ≥ η for all a ∈ A and all i ∈ I. If A has infinitely many points, then A is uniformly perfect.

Remark 1.3. We show in Lemma 2.4 that the hypotheses of Theorem 1.1 and Corollary 1.1 are equivalent.

Corollary 1.2. Let G = ⟨g_{i} : i ∈ I⟩ be a conformal IFS on (U, K) such that there exist η > 0 where |g′_{i}(a)| ≥ η for all a ∈ A and all i ∈ I. If A contains more than one point, then A is uniformly perfect.

UNIFORMLY PERFECT ANALYTIC AND CONFORMAL ATTRACTOR SETS 3

Corollary 1.3. Let G = ⟨g_{1}, …, g_{N}⟩ be a conformal IFS on (U, K). If A contains more than one point, then A is uniformly perfect.

Uniformly perfect sets, which are defined in section 2, were introduced by A. F. Beardon and Ch. Pommerenke in 1978 in [1]. Such sets cannot be separated by annuli that are too “fat” (large ratio of outer to inner radius) and thus near each of its points the set is uniformly “thick” where this thickness is independent of scaling. See [1, 9, 10] for many interesting equivalent definitions. Such sets are known to be regular for the Dirichlet problem (see [9], p. 193) and also have positive Hausdorff dimension (see [5], p. 523). See [6, 7] for a discussion on the Hausdorff dimension, packing dimension, and other properties of limit sets (attractors) for conformal IFS’s.

This paper is organized as follows. Section 2 contains basic lemmas and definitions. Section 3 contains the proofs of Theorem 1.1 and Corollaries 1.1, 1.2 and 1.3. Section 4 contains examples of the following: a uniformly perfect conformal attractor set that is generated by maps of the form z^{2} + c; a uniformly perfect analytic attractor set that is generated by maps of the form z^{2} + c; a non uniformly perfect attractor set which is generated by an infinite number of conformal maps; a (nonconformal) IFS generated by two analytic maps whose attractor set is perfect though not uniformly perfect, yet the generating maps are one-to-one on A; analytic attractor sets whose cardinality is any given integer n.

The author would like to Harold Boas, Marshall Whittlesey, Mohammed Ziane, and especially Aimo Hinkkanen for productive conversations regarding this subject.

2. Definitions and basic facts

Given an open set U ⊂ ℂ such that there exists a non constant analytic map g with g(U) contained in a compact set K ⊂ U, then U is hyperbolic, that is, ℂ \ U contains more than one point. This follows since the image of the plane or the punctured plane under an analytic map is always dense in the plane and hence could not be contained in a compact subset of U. Hence there exists a hyperbolic metric on U (see [2], p. 12).

Lemma 2.1. If the analytic function g maps an open connected set U ⊂ ℂ into a compact set K ⊂ U, then there exists 0 < s < 1, which depends on U and K only, such that d(g(z), g(w)) ≤ sd(z, w) for all z, w ∈ K where d is the hyperbolic metric defined on U.

Proof. If no such s exists then there exist sequences z_{n}, w_{n} ∈ K and analytic maps g_{n} such that g_{n}(U) ⊂ K where d(g_{n}(z_{n}), g_{n}(w_{n}))/d(z_{n}, w_{n}) → 1. By compactness and normality we may assume that z_{n} → z, w_{n} → w and g_{n} → g uniformly on K where g(U) ⊂ K. If z ≠ w then one can

4 RICH STANKEWITZ

show d(g_{n}(z_{n}), g_{n}(w_{n}))/d(z_{n}, w_{n}) → d(g(z), g(w))/d(z, w) < 1 (see [2], p. 12). If z = w, then then d(g_{n}(z_{n}), g_{n}(w_{n}))/d(z_{n}, w_{n}) → |g′(z)|λ(g(z))/λ(z) < 1 where λ(z) denotes the density of the hyperbolic metric of U at the point z of U. Thus we have a contradiction and so the lemma follows.

Lemma 2.2. Let G = ⟨g_{i} : i ∈ I⟩ be an IFS on (U, K). Then for any g ∈ G, we have g(A) ⊂ A.

Proof. Suppose a ∈ A′(G) and so f(a) = a for some f ∈ G with, say, f = g_{i_{1}} ∘g_{i_{2}} ∘⋯∘g_{i_{k}} where each i_{j} ∈ I. Say g = g_{i′_{1}} ∘g_{i′_{2}} ∘⋯∘g_{i′_{n}} where each i′_{j} ∈ I. Letting Ã = A(⟨g_{i_{1}}, g_{i_{2}}, …, g_{i_{k}}, g_{i′_{1}}, g_{i′_{2}}, …, g_{i′_{n}}⟩) we see by (1.1) that g(a) ∈ g(Ã) ⊂ Ã ⊂ A(G). Hence g(A′(G)) ⊂ A(G) and the lemma follows from the continuity of g.

Lemma 2.3. Let η > 0, r > 0, M > 0, a ∈ ℂ and let F denote the family of maps {f : △(a, r) → △(0, M) such that f is analytic and |f′(a)| ≥ η}. Then there exists ρ > 0 such that each f ∈ F is one-to-one on △(a, ρ).

Remark 2.1. Any ρ such that Mρ(2r − ρ)/(r(r − ρ)^{2}) < η will suffice will satisfy the conclusion of Lemma 2.3.

Proof. Fix ρ such that Mρ(2r − ρ)/(r(r − ρ)^{2}) < η. For |z − a| < r we may obtain by estimates on the Cauchy integral formula that |f′′(z)| ≤ 2Mr/(r − |z − a|)^{3}. Hence for |z − a| ≤ ρ we have |f′(z) − f′(a)| = | ∫_{a}^{z} f′′(s) ds| ≤ 2Mr ∫_{0}^{|z−a|}(r − t)^{−3} dt = Mr[(r − |z − a|)^{−2} − r^{−2}] ≤ Mρ(2r − ρ)/(r(r − ρ)^{2}) < η ≤ |f′(a)| where the integral is over the straight line path. This suffices to show that f is one-to-one on △(a, ρ) (see [3], p. 293).

Lemma 2.4. Let G = ⟨g_{i} : i ∈ I⟩ be an analytic IFS on (U, K). Then there exist 0 < δ < diam(A) and C > 0 such that conditions (I)-(III) of Theorem 1.1 hold if and only if there exists η > 0 such that we have |g′_{i}(a)| ≥ η for all a ∈ A and all i ∈ I.

Proof. Let 0 < δ < diam(A) and C > 0 be such that conditions (I)-(III) hold. If there does not exist η > 0 such that we have |g′_{i}(a)| ≥ η for all a ∈ A and all i ∈ I, then there exist a_{n} ∈ A and g_{n} ∈ {g_{i} : i ∈ I} such that g′_{n}(a_{n}) → 0. Let a′_{n} = g_{n}(a_{n}) and let h_{n} be the branch of the inverse of g_{n} such that h_{n}(a′_{n}) = a_{n}. Conditions (II) and (III) imply, in particular, that h_{n}(△(a′_{n}, δ/10)) ⊂ △(a_{n}, Cδ/10). But by the Koebe 1/4 theorem we have h_{n}(△(a′_{n}, δ/10)) ⊃ △(a_{n}, |h′_{n}(a′_{n})|δ/40) which gives a contradiction for large n since |h′_{n}(a′_{n})| = 1/|g′_{n}(a_{n})| → ∞.

Suppose η > 0 is such that |g′_{i}(a)| ≥ η for all a ∈ A and all i ∈ I. Note that since K is compact there exists M > 0 such that each g_{i} maps U into some △(0, M). Let r > 0 be such that A_{r} ⊂ U.

UNIFORMLY PERFECT ANALYTIC AND CONFORMAL ATTRACTOR SETS 5

By Lemma 2.3 setting δ = ρ we see that condition (I) is satisfied. By the Koebe 1/4 theorem each g_{i}(△(a, δ)) ⊃ △(g_{i}(a), δη/4) and so (see Remark 1.2) we may replace δ by min{δ, δη/4} to satisfy both (I) and (II). Since the branch h_{i} of g_{i}^{−1} such that h_{i}(g_{i}(a)) = a is defined on △(g_{i}(a), δ) we see that |h′_{i}(g_{i}(a))| = 1/|g′_{i}(a)| ≤ 1/η and so by Lemma 2.9 we see that condition (III) is satisfied with C = 100/(81η).

Lemma 2.5. Let G = ⟨g_{i} : i ∈ I⟩ be an IFS such that for each g ∈ G there exists a positive integer N_{g} such that for each a ∈ A the set g^{−1}({a}) has at most N_{g} elements. If A has infinitely many elements, then A is perfect and hence, by the Baire category theorem is uncountable.

Proof. If a ∈ A = \overline{A′} were isolated it would have to be an attracting fixed point for some element f ∈ G. Suppose △(a, r) ∩ A = {a}. For some positive integer n, we have f^{n}(A) ⊂ △(a, r) since a ∈ f^{n}(A) and diam_{d}(f^{n}(A)) → 0 as n → ∞. By Lemma 2.2 we see that f^{n}(A) = {a} and this contradicts the hypothesis on the integer N_{f^{n}}.

Remark 2.2. If G is an analytic IFS, then the proof of Lemma 2.5 and the identity theorem for non-constant analytic maps imply that the attractor set A is perfect when A has infinitely many elements.

Lemma 2.6. Let G = ⟨g_{i} : i ∈ I⟩ be an analytic IFS such that each g′_{i} never takes the value zero on A. If a ∈ A′, then there exists two sequences of maps F_{1} = g_{i_{1}}, F_{2} = g_{i_{1}} ∘ g_{i_{2}}, …, F_{m} = g_{i_{1}} ∘ g_{i_{2}} ∘ ⋯ ∘ g_{i_{m}}, … where each i_{j} ∈ I and H_{1} = h_{i_{1}}, H_{2} = h_{i_{2}} ∘ h_{i_{1}}, …, H_{m} = h_{i_{m}} ∘ ⋯ ∘ h_{i_{1}}, … where each h_{i_{j}} is a branch of g_{i_{j}}^{−1}, such that for each positive integer m we have (H_{m} ∘ F_{m})(z) = z (on a small neighborhood of H_{m}(a)) and H_{m}(a) ∈ A.

Proof. Suppose g(a) = a for g ∈ G and so g = g_{i_{1}} ∘ g_{i_{2}} ∘ ⋯ ∘ g_{i_{k}} for some choice of i_{j} ∈ I. Let h_{i_{1}} be the branch of g_{i_{1}}^{−1} such that h_{i_{1}}(a) = (g_{i_{2}} ∘ ⋯ ∘ g_{i_{k}})(a) ∈ A by Lemma 2.2. Similarly choose h_{i_{2}}, …, h_{i_{k}} to be branches of g_{i_{2}}^{−1}, …, g_{i_{k}}^{−1}, respectively, so that h_{i_{2}}(h_{i_{1}}(a)) = (g_{i_{3}} ∘ ⋯ ∘ g_{i_{k}})(a) ∈ A, …, h_{i_{j}}((h_{i_{j−1}} ∘ ⋯ ∘ h_{i_{1}})(a)) = (g_{i_{j+1}} ∘ ⋯ ∘ g_{i_{k}})(a) ∈ A, …, h_{i_{k}}((h_{i_{k−1}} ∘ ⋯ ∘ h_{i_{1}})(a)) = a ∈ A.

We define F_{1} = g_{i_{1}}, F_{2} = g_{i_{1}} ∘ g_{i_{2}}, …, F_{k} = g_{i_{1}} ∘ g_{i_{2}} ∘ ⋯ ∘ g_{i_{k}}, F_{k+1} = g_{i_{1}} ∘ g_{i_{2}} ∘ ⋯ ∘ g_{i_{k}} ∘ g_{i_{1}}, . . .. For F_{m} = g_{i_{1}} ∘ g_{i_{2}} ∘ ⋯ ∘ g_{i_{m}} we define H_{m} = h_{i_{m}} ∘ ⋯ ∘ h_{i_{1}} and we are done.

Definition 2.1. A true annulus A = Ann(w; r, R) = {z : r < |z − w| < R} is said to separate a set F if F intersects both components of \overline{ℂ} \ A and F ∩ A = ∅.

Definition 2.2. [[9], p. 192] A compact subset F ⊂ ℂ is uniformly perfect if there exists M > 1 such that whenever w ∈ F, 0 < R < diam(F), R/r > M, then Ann(w; r, R) ∩ F ≠ ∅.

Lemma 2.7. [[10], p. 315] Let F be a perfect compact subset of ℂ and let A_{n} = Ann(z_{n}; r_{n}, R_{n}) be annuli which separate F, have z_{n} ∈ F and R_{n}/r_{n} → ∞. Then R_{n} → 0.

Proof. Since F is compact we may assume that z_{n} → w ∈ F. We also see that R_{n} ≤ diam(F) and hence r_{n} → 0. If there exists δ > 0 such that R_{n} > δ along some subsequence of R_{n} then one can show that △(w, δ) ∩ F = {w} since △(w, δ) \ {w} is contained in the “limit” of this subsequence of A_{n}. This contradicts the hypothesis that F is perfect.

Lemma 2.8. Let δ > 0 and let g be analytic and univalent on △(0, δ) with g(0) = 0 such that the there exists z with |z| ≤ δ/10 and |g(z)| = R. Then g(△(0, δ)) ⊃ △(0, 2R) and hence g^{−1} is defined on △(0, 2R).

Proof. Defining the univalent map K(z) = \frac{1}{δg′(0)}g(δz) on △(0, 1) we have K(0) = 0 and K′(0) = 1. By hypothesis there exists a point z′ = z/δ such that |z′| ≤ 1/10 and |K(z′)| = \frac{R}{δ|g′(0)|}. By the Koebe distortion theorem and the maximum modulus theorem (see [2], p. 3) we see that for |z| ≤ 1/10 we have

|K(z)| ≤ \frac{1/10}{(1 − 1/10)^{2}}

and so

\frac{R}{δ|g′(0)|} ≤ \frac{10}{81}.

By the Koebe 1/4 theorem (see [2], p. 2) we have K(△(0, 1)) ⊃ △(0, 1/4) and so we conclude that g(△(0, δ)) ⊃ △(0, δ|g′(0)|/4) ⊃ △(0, 2R).

Lemma 2.9. Let η > 0 and let g be analytic and univalent on △(a, δ) with |g′(a)| ≤ 1/η. If |z − a| ≤ δ/10, then |g(z) − g(a)| ≤ 100|z − a|/(81η).

Proof. This follows from the Koebe distortion theorem in much the same manner as the proof of Lemma 2.8 and therefore we omit the details.

Lemma 2.10. Let g be univalent on △(0, 2R) with g(0) = 0. If R > 9r, then g(Ann(0; r, R)) ⊃ Ann(0; r′, R′) for some R′ > r′ > 0 such that R′/r′ ≥ R/(9r).

Proof. The univalent map h(z) = \frac{g(2Rz)}{2Rg′(0)} defined in △(0, 1) is such that h(0) = 0 and h′(0) = 1. By the Koebe distortion theorem we have

|h(z)| ≤ \frac{r/(2R)}{[1 − (r/(2R))]^{2}} on |z| = r/(2R)  and  |h(z)| ≥ \frac{1/2}{[1 + (1/2)]^{2}} on |z| = 1/2.

UNIFORMLY PERFECT ANALYTIC AND CONFORMAL ATTRACTOR SETS 7

So

|g(z)| ≤ \frac{r|g′(0)|}{[1 − (r/(2R))]^{2}} = r′ on |z| = r  and  |g(z)| ≥ \frac{R|g′(0)|}{[1 + (1/2)]^{2}} = R′ on |z| = R.

Hence g(Ann(0; r, R)) ⊃ Ann(0; r′, R′) where \frac{R′}{r′} = \frac{R[1−(r/(2R))]^{2}}{r[1+(1/2)]^{2}} ≥ R/(9r) since r < R.

Lemma 2.11. Let ρ > 0. If p and q are metrics that induce the same topology on a compact topological space F, then there exists r > 0 such that △_{p}(a, r) ⊂ △_{q}(a, ρ) for all a ∈ F.

Proof. Consider the open cover {△_{q}(a, ρ/2) : a ∈ F } of F. By the Lebesgue number lemma (see [8], p. 179) applied to the compact metric space (F, p), there exists R > 0 such that for each a ∈ F there exists a′ ∈ F such that △_{p}(a, R/3) ⊂ △_{q}(a′, ρ/2) ⊂ △_{q}(a, ρ). Hence we see the result follows with r = R/3.

Proof of Remark 1.1. Let d be the hyperbolic metric on U and recall that the hyperbolic and Euclidean metrics induce the same topology on U. Since A is compact one may apply Lemma 2.11 twice to obtain δ′, r > 0 such that △(a, δ′) ⊂ △_{d}(a, r) ⊂ △(a, δ) for all a ∈ A. Hence for any f ∈ G we have that f(△(a, δ′)) ⊂ f(△_{d}(a, r)) ⊂ △_{d}(f(a), r) ⊂ △(f(a), δ). Note that we used the fact that f(A) ⊂ A.

Let g ∈ G and say it can be written as g = g_{i_{k}} ∘ ⋯ ∘ g_{i_{2}} ∘ g_{i_{1}}. Note that g_{i_{1}} is one-to-one on △(a, δ′). Since from above g_{i_{1}}(△(a, δ′)) ⊂ △(g_{i_{1}}(a), δ) we use (property (I)) the fact that g_{i_{2}} is one-to-one on △(g_{i_{1}}(a), δ) to conclude that g_{i_{2}} ∘ g_{i_{1}} is one-to-one on △(a, δ′) and again from above (g_{i_{2}} ∘ g_{i_{1}})(△(a, δ′)) ⊂ △((g_{i_{2}} ∘ g_{i_{1}})(a), δ) . We then proceed inductively to show that each map g_{i_{m}} ∘ ⋯ ∘ g_{i_{2}} ∘ g_{i_{1}} including g is one-to-one on △(a, δ′).

3. Proof of the Theorem and Corollaries

Proof of Theorem 1.1. Let the hypotheses of the theorem be satisfied. Recall that d is the hyperbolic metric on U. Since K is a compact subset of U there exists ϵ > 0 such that \overline{K_{ϵ}^{d}} ⊂ U. Let s̃ < 1 be the uniform contraction coefficient on \overline{K_{ϵ}^{d}} for the generating maps as given by Lemma 2.1. Since we may shrink δ and still satisfy conditions (I)-(III) of the theorem we will assume (see Lemma 2.11) that for each a ∈ A we have △(a, δ/(10C)) ⊂ △_{d}(a, ϵ).

Suppose A contains infinitely many points, but is not uniformly perfect. Hence there exist annuli A_{n} = Ann(a_{n}; r_{n}, R_{n}) that separate A where R_{n}/r_{n} → ∞ and each a_{n} ∈ A. Since A′ is dense in A a simple geometric argument allows one to assume that each a_{n} ∈ A′. Since A is perfect (see Remark 2.2) we must have both r_{n} and R_{n} tending to zero by Lemma 2.7. We assume all R_{n} < δ/(10C) and R_{n} > 9r_{n}.

8 RICH STANKEWITZ

The method of proof will be to show that the A_{n} can be expanded by certain locally defined inverses of maps in G given in Lemma 2.6, such that the expanded (conformal) annuli contain true annuli of large ratio of outer to inner radius. The Koebe 1/4 theorem (Lemma 2.8) and the hypothesis on δ is used to show that the chosen inverse maps are well defined on △(a_{n}, 2R_{n}). The deformation that occurs since the maps are not linear (similitudes) is controlled by the use of the Koebe distortion theorem (see Lemma 2.10). The expanded true annuli are constructed so that the outer radii do not tend to zero, yet these annuli do separate A. This contradicts the fact that A is perfect (see Lemma 2.7). We now carry out this program.

Since a_{n} ∈ A′ we may construct sequences of maps F_{m} and H_{m} for m = 1, 2, … as given by Lemma 2.6. Note that the maps F_{m} and H_{m} depend on n. There exists m such that

F_{m}(△(H_{m}(a_{n}), δ/(10C))) ⊂ F_{m}(△_{d}(H_{m}(a_{n}), ϵ)) ⊂ △_{d}(a_{n}, s̃^{m}ϵ) ⊂ △(a_{n}, R_{n}). (3.1)

The first inclusion follows from the choice of δ as stated at the beginning of the proof and the last inclusion follows from the fact that the Euclidean and the hyperbolic metrics induce the same topology on U.

We define f_{n} = F_{m^{∗}} where m^{∗} is the smallest integer m such that F_{m}(△(H_{m}(a_{n}), δ/(10C))) ⊂ △(a_{n}, R_{n}) and define its “inverse” by k_{n} = H_{m^{∗}}. Therefore F_{m^{∗}−1}(△(H_{m^{∗}−1}(a_{n}), δ/(10C))) ∩ C(a_{n}, R_{n}) ≠ ∅. Choose z ∈ △(H_{m^{∗}−1}(a_{n}), δ/(10C)) so that we have |F_{m^{∗}−1}(z) − a_{n}| = R_{n}. By the choice of δ we see that h_{i_{m^{∗}}} (z) is defined and since |f_{n}(h_{i_{m^{∗}}} (z)) − a_{n}| = R_{n} we also have δ/(10C) ≤ |h_{i_{m^{∗}}} (z) − k_{n}(a_{n})| by the definition of f_{n}. By the choice of C and property (III) we conclude

δ/(10C) ≤ |h_{i_{m^{∗}}} (z) − k_{n}(a_{n})| ≤ δ/10.

We note that (see Remark 1.1) f_{n} is one-to-one on △(k_{n}(a_{n}), δ). By Lemma 2.8 we have that k_{n} is defined on △(a_{n}, 2R_{n}) and Lemma 2.10 shows that k_{n}(A_{n}) contains the annulus

B_{n} = Ann(k_{n}(a_{n}); r′_{n}, R′_{n})

where R′_{n} = dist(k_{n}(C(a_{n}, R_{n})), k_{n}(a_{n})) and r′_{n} = max{|k_{n}(z) − k_{n}(a_{n})| : |z − a_{n}| = r_{n}} satisfy R′_{n}/r′_{n} ≥ \frac{R_{n}}{9r_{n}} → ∞. By construction each δ/(10C) ≤ R′_{n} ≤ δ/10 (the lower bound follows from the condition that f_{n}(△(k_{n}(a_{n}), δ/(10C))) ⊂ △(a_{n}, R_{n})). Each B_{n} separates A since f_{n}(B_{n}) ⊂ A_{n} and f_{n}(A) ⊂ A (Lemma 2.2). Thus we see by Lemma 2.7 that we have obtained the desired contradiction.

Proof of Corollary 1.1. The corollary follows directly from Lemma 2.4 and Theorem 1.1.

UNIFORMLY PERFECT ANALYTIC AND CONFORMAL ATTRACTOR SETS 9

Proof of Corollary 1.2. Suppose A contains two or more points. Let z ∈ A′ with, say, f(z) = z and let w ∈ A \ {z}. Since f^{n}(w) are distinct points in A (which tend to z) we see that A contains infinitely many points. The result follows from Corollary 1.1.

Proof of Corollary 1.3. Since each g_{i} for i ∈ I is conformal, we have that each g′_{i} is never zero on A. Since A is compact η = min_{i=1...N} inf_{a∈A} |g′_{i}(a)| > 0 and so the result follows from Corollary 1.2.

4. Examples

Example 4.1. Let G = ⟨g_{1}, …, g_{N}⟩ where g_{i}(z) = z^{2} + c_{i} and each c_{i} is chosen as follows. Fix 0 < ϵ < 1/2 and choose the c_{i} ∈ Ann(0; ϵ^{2}, ϵ − ϵ^{2}) ⊂ △(0, 1/4) so that there exists a half plane H with the origin on its boundary which contains ∪_{i=1}^{N}△(c_{i}, ϵ^{2}). Choose a connected open set U which is contained H such that U ⊃ ∪_{i=1}^{N}△(c_{i}, ϵ^{2}) and \overline{U} ⊂ △(0, ϵ). Since each g_{i}(△(0, ϵ)) ⊂ U we see that each g_{i}(\overline{U}) ⊂ U. With this U and K = ∪_{i=1}^{N}g_{i}(\overline{U}) the IFS G is conformal (even though the generating maps are not globally conformal) and by Corollary 1.3 the resulting attractor set is uniformly perfect.

Example 4.2. Let G = ⟨g_{i} : i ∈ I⟩ where each g_{i}(z) = z^{2} + c_{i} and each c_{i} is chosen as follows. Fix 0 < ϵ < 1/2 and ρ > 0 such that 2ρ < ϵ − 2ϵ^{2}. Choose c_{i} ∈ Ann(0; ρ + ϵ^{2}, ϵ − ϵ^{2} − ρ) ⊂ △(0, 1/4). Hence for U = △(0, ϵ) and K = \overline{Ann(0; ρ, ϵ − ρ)} we have g_{i}(U) = △(c_{i}, ϵ^{2}) ⊂ K. Hence A ⊂ K and we see that for η = 2ρ and a ∈ A each |g′_{i}(a)| = 2|a| ≥ η for i ∈ I. By Corollary 1.1, if A has infinitely many points, then A is uniformly perfect.

If A has three or more points, then A has infinitely many points thus A is uniformly perfect. This follows since g_{1}(A) ⊂ △(c_{1}, ϵ^{2}) and so there exists a point a ∈ A ∩ △(c_{1}, ϵ^{2}) other than the attracting fixed point of g_{1}. Hence the points of the form g_{1}^{n}(a) are distinct since g_{1} is one-to-one on △(c_{1}, ϵ^{2}). We see that A has three or more points if it is not the case that G = ⟨g_{1}, g_{2}⟩ where, denoting the attracting fixed points of g_{i} by a_{i}, a_{2} = ±a_{1}. (If G were such an IFS, then we would have A = \{a_{1}, a_{2}\}.) For example, one can check that if c_{1} > 0 then c_{2} ≠ c_{1} and c_{2} ≠ c_{1}−1+\sqrt{1 − 4c_{1}} implies A is uniformly perfect. Also, if three of the maps g_{i} are distinct, then A is uniformly perfect.

Example 4.3. We construct a conformal IFS G such that A is not uniformly perfect. Choose sequences a_{n} and b_{n} of positive real numbers such that 1 > b_{1} > a_{1} > b_{2} > a_{2} > … where a_{n}↘0, b_{n}↘0 and a_{n}/b_{n+1} → ∞. Then the maps g_{n}(z) = (b_{n} − a_{n})z + a_{n} are such that the conformal IFS G = ⟨g_{n} : n ≥ 1⟩ (where U is an appropriately defined neighborhood of the interval [0, 1]) has an attractor set A that is separated by each of the annuli Ann(0; b_{n+1}, a_{n}). Thus this attractor set is not uniformly perfect. We note that any attempt to apply Theorem 1.1 breaks down since neither a choice of δ nor a choice of C can be found to satisfy its hypotheses.

Example 4.4. We construct an example of a finitely generated (polynomial) IFS which has an attractor set which is not uniformly perfect. Let f(z) = z^{23} and g(z) = (z − 1/2)^{2} + 1/2. Let I = [0, 3/4], I_{1/10} = {z : |z − w| ≤ 1/10 for some w ∈ I} and let U = I_{1/10} ∪ △(1/2, 0.37). Since \overline{U} ⊂ △(0, 0.9) ∩ △(1/2, 0.601) we see that f(\overline{U}) ⊂ f(△(0, 0.9)) ⊂ △(0, 1/10) ⊂ U and g(\overline{U}) ⊂ g(△(1/2, 0.601)) ⊂ △(1/2, 0.37) ⊂ U. Hence G = ⟨f, g⟩ is an IFS defined on (U, K) where K = f(\overline{U}) ∪ g(\overline{U}). Since f(I) ∪ g(I) ⊂ I we see that A ⊂ f(I) ∪ g(I). Let V be the open interval ((3/4)^{23}, 1/2) and note that V ∩ A = ∅. We claim that f^{n}(V) ∩ A = ∅ for all n = 1, 2, . . ..

Proof. We first show that f(V) ∩ A = ∅. Suppose a ∈ f(V) ∩ A. Since A = f(A) ∪ g(A) and f(V) ∩ g(A) ⊂ f(I) ∩ g(I) = ∅ we conclude that a ∈ f(V) ∩ f(A). Hence there exists a_{1} ∈ A such that f(a_{1}) = a. Since f is one-to-one on I ⊃ A ∪ V we must then have a_{1} ∈ V. This contradiction shows that f(V) ∩ A = ∅. One can then use this argument in an inductive manner to conclude f^{n}(V) ∩ A = ∅ for all n = 1, 2, . . ..

Since f^{n}(V) = ((3/4)^{23^{n+1}}, 2^{−23^{n}}) we see that Ann(0; (3/4)^{23^{n+1}}, 2^{−23^{n}}) separates A and thus A is not uniformly perfect. We note, however, that since each map in G is a polynomial and so has finite degree, Lemma 2.5 implies A is perfect since A contains the infinitely many points of the form f^{n}(1/2).

Remark 4.1. We note that in Example 4.4 it can be shown that both f and g are one-to-one on A, but not on any open set containing A and thus we may not apply Corollary 1.3. To see that g is one-to-one on A ⊂ f(I) ∪ g(I) we note that g(g(I)) ∩ g(f(I)) = ∅ and both g|_{g(I)} and g|_{f(I)} are one-to-one.

Example 4.5. We give an example that demonstrates that it is possible for the attractor set of a (non conformal) IFS which is generated by analytic maps to have cardinality any finite number.

Fix a positive integer n. Consider the complex polynomial f_{0}(z) = c + (z^{n} − c^{n})^{2} which has a super-attracting fixed point at c, i.e., f_{0}(c) = c and f′_{0}(c) = 0. For α = e^{2πi/n} we see that f_{0}(α^{k}c) = c for k = 0, …, n − 1. Hence f_{k}(z) = α^{k}f_{0}(z) has a super-attracting fixed point at α^{k}c and maps each α^{j}c to α^{k}c for j = 0, …, n − 1. It is elementary to verify that for small values of c and for a suitable choice of r > 0 we get that f_{k}(△(0, r)) ⊂ K for k = 0, …, n − 1 where K is a suitably chosen compact subset of △(0, r). Hence these functions generate an iterated function system on (U, K) where U = △(0, r).

One can easily see by (1.1) that A = {α^{k}c : 0 ≤ k ≤ n − 1}. So for any n there exists an attractor set A such that card(A) = n.

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Department of Mathematics, Texas A&M University, College Station, TX 77843, richs@math.tamu.edu Current address: School of Science, Penn State University at Erie, Station Road, Erie, PA 16563, rls42@psu.edu