Lipschitz equivalence of fractals generated by nested cubes

$Page 1: Lipschitz equivalence of fractals generated by nested cubes$
Math. Z. (2012) 271:1287–1308DOI 10.1007/s00209-011-0916-5 Mathematische Zeitschrift

Lipschitz equivalence of fractals generatedby nested cubes

Lifeng Xi · Ying Xiong

Received: 25 February 2011 / Accepted: 16 May 2011 / Published online: 1 July 2011© Springer-Verlag 2011

Abstract David and Semmes (Fractured fractals and broken dreams, vol 7, 1997) posed aproblem on the Lipschitz equivalence among fractals generated by nested cubes accordingto various “rules”. For this problem, we show that there are uncountably many Lipschitzequivalence classes if the rules are taken arbitrarily. However, if the same rule is taken ineach step of construction process of fractal, under some suitable restriction, we show thatthere is only a unique Lipschitz equivalence. In particular, we obtain the self-similar set ifonly one rule is used in construction process. Applying the second result to the self-similarsets, we can obtain the following proposition in Xi and Xiong (C R Math Acad Sci Paris348:15–20, 2010): if E A = ⋃

a∈A( 1n E A + a

n ), EB = ⋃b∈B( 1

n EB + bn ) are totally discon-

nected self-similar sets with A, B ⊂ {0, 1, . . . , (n − 1)}d and #A = #B, then E A and EB

are Lipschitz equivalent.

Keywords Fractal · Lipschitz equivalence · Generating rule · Moran set · Self-similar set

Mathematics Subject Classification (2000) 28A80

Supported by National Natural Science of China (Grant Nos 11071224, 11071082), Program for ExcellentTalents in University of China, Research Fund for the Doctoral Program of Higher Education of China(20100172120027), Fundamental Research Funds for the Central Universities, SCUT (D2104900), andMorningside Center of Mathematics.

L. Xi (B)Institute of Mathematics, Zhejiang Wanli University, Ningbo, 315100 Zhejiang,People’s Republic of Chinae-mail: [email protected]

Y. XiongDepartment of Mathematics, South China University of Technology,Guangzhou 510641, People’s Republic of Chinae-mail: [email protected]

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1288 L. Xi, Y. Xiong

1 Introduction

Two metric spaces (X1, d1) and (X2, d2) are called Lipschitz equivalent, denoted by X1 � X2,if there exists a bijection f : X1 → X2 such that for all x, y ∈ X1,

c−1 · d1(x, y) ≤ d2 ( f (x), f (y)) ≤ c · d1(x, y),

where c > 1 is a constant.One important topic in geometric measure theory is to classify fractals under Lipschitz

equivalence. As mentioned in [6], topology may be regarded as the study of equivalenceclasses of sets under homeomorphism, and fractal geometry is sometimes thought of as thestudy of equivalence classes of fractals under bi-Lipschitz mappings. In fact, the Lipschitzequivalence is suitable, since “isometry” leads to a poor and rather boring category and “con-tinuity” takes us out of geometry to the realm of pure topology (see [9]). Another interestingmotivation of studying Lipschitz equivalence of fractal comes from geometry group theory(see [1,7]).

Many works have been devoted to the related topics. For example, Cooper andPignataro [2], Falconer and Marsh [5,6], David and Semmes [3], Xi [20,21] studied the shapeof Cantor set, nearly Lipschitz equivalence, BPI equivalence and quasi-Lipschitz equivalencerespectively. Recently, Mattila et al. [13,14] studied Lipschitz equivalence between subsetsof Ahlfors regular sets or self-conformal sets; while Xi et al. [4,16,18,22–24,26] discussedthe Lipschitz equivalence between self-similar sets.

In Sects. 2.5, 11.7 and 13.1 of [3], David and Semmes studied the geometric structure ofa special fractal family other than self-similar sets made from nested cubes. Some “rules”are used to generated the nested cubes.

Let d ≥ 1, m ≥ 2 and md > n ≥ 2, J0 =[0, 1]d . We call {b1, . . . , bn} ⊂ {0, . . . , m−1}d ,with bi �= b j for i �= j , a rule set.

Definition 1 ([3]) For a rule of type (n, m) with respect to a rule set {b1, . . . , bn}, we referto the manner of replacing any cube S(J0) by n small cubes

⋃nj=1 S(m−1(J0 + b j )), where

S : x → ax + b(a > 0) is a similitude.

According to [3], we can obtain a fractal as follows. At the first step, we use a rule toreplace J0 by n small cubes with side length m−1. Inductively, after k steps, we obtain nk

cubes with side length m−k , and then at (k + 1)th step, we use various rules to replace everycube obtained at kth step by n cubes with side length m−k−1. Again and again, we finallyget the limit set—a fractal.

To be precise, we give the following definition.

Definition 2 For σ = σ1σ2 · · · σk ∈ {1, . . . , n}k and all k ≥ 1, let Sσ = Tσ1 ◦ Tσ1σ2

◦ · · · ◦ Tσ1···σk , where Tσ1···σ j : x → m−1(x + bσ1···σ j ) are similitudes on Rd such that

{bσ1···σ j−11, bσ1···σ j−12, . . . , bσ1···σ j−1n} is a rule set. For k ≥ 1, define

Ek =⋃

σ∈{1,...,n}k

Sσ (J0) and E =⋂

k≥1

Ek .

We use Rd(n, m) to denote the collection of all such sets E . The cube Jσ := Sσ (J0) appearingin the definition of E is called the basic cube of order k for σ ∈ {1, . . . , n}k .

Remark 1 The set Ek+1 is generated from the set Ek with replacing every cube Sσ (J0) bythe rule with respect to the rule set {bσ1, . . . , bσn}.

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Lipschitz equivalence of fractals generated by nested cubes 1289

Fig. 1 The construction process of the first two steps of a set in R2(3, 3)

Remark 2 If only one rule is used in the construction process of a set E ∈ Rd(n, m), thenthe set E is a self-similar set satisfying the open set condition.

Example 1 Figure 1 shows the construction process of the first two steps of a sets in R2(3, 3).Two rule sets

B1 = {(1, 0), (0, 1), (2, 2)}, B2 = {(0, 1), (2, 0), (2, 2)}are used in the construction.

Noticing that all the fractals in Rd(n, m) have the same Hausdorff dimension log n/ log m,David and Semmes [3] inquire a fundamental question concerning the geometric structureof the fractals in Rd(n, m): how to characterize Lipschitz equivalence for these fractals. Asan illustration, they proposed a special question, the so called {1, 3, 5} − {1, 4, 5} problem.Let

E1,3,5 = (E1,3,5/5) ∪ (E1,3,5/5 + 2/5) ∪ (E1,3,5/5 + 4/5),

E1,4,5 = (E1,4,5/5) ∪ (E1,4,5/5 + 3/5) ∪ (E1,4,5/5 + 4/5). (1.1)

It is obvious that E1,3,5, E1,4,5 ∈ R1(3, 5) and the rules of the two sets are {0, 2, 4} and{0, 3, 4}, respectively. David and Semmes [3] asked whether E1,3,5 and E1,4,5 are Lipschitzequivalent or not. It is proved in [16] that they are Lipschitz equivalent. The proof is based onthe clear comprehension of the geometric structure of E1,4,5. Furthermore, Xi et al. [18,23,24]discussed various generalizations. Especially, Xi and Xiong [25] studied the {1, 3, 5}–{1, 4, 5}problem in higher dimensional space R

d(d ≥ 1) and proved that:

Theorem A ([25]) Let A, B ⊂ {0, . . . , m − 1}d and

E A =⋃

a∈A

m−1(E A + a), EB =⋃

b∈B

m−1(EB + b)

be two totally disconnected self-similar sets, then E A � EB if and only if card A = card B.

Remark 3 This theorem is a “high dimensional” version of {1, 3, 5} − {1, 4, 5} problemin [3,16]. In fact, let d = 1, n = 5, A = {0, 2, 4} and B = {0, 3, 4}, then E A = E1,3,5 andEB = E1,4,5. When d = 1, E A is totally disconnected if A �= {0, . . . , n − 1}. But whend ≥ 2, E A may be connected if A �= {0, . . . , n − 1}d (see Fig. 2).

In this paper, we further study the Lipschitz equivalence of the sets in Rd(n, m). Our firstresult (Theorem 1) shows that even in R

1, there are uncountably many Lipschitz equivalenceclasses in R1(n, m), due to the arbitrary choices of the rules. When we focus on a specialcase that the same rule with respect to rule set Bk is used for every basic cube of order k, we

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Fig. 2 Connected self-similar sets with initial constructions in shadow

get a special subclass Sd(n, m) of Rd(n, m) (see Definition 3). Then we obtain the completecharacterization (Theorem 2) for sets in Sd(n, m) to be Lipschitz equivalent to the symbolicspace �(n, m) defined by

�(n, m) = {1, . . . , n}N with distance ρm(ω, ω′) = m− inf{k : ωk �=ω′k }.

Furthermore, in Theorem 3, we generalize the result in Theorem 2 to more general setting—the Moran sets with cubic pattern (see Definition 5).

This paper is organized as follows. In the rest of Sect. 1, we state Theorems 1, 2 and 3,the main results of this paper in Sect. 1.1. In Sect. 1.2, we introduce the Moran sets withcubic pattern and the notion so called “bounded block size”. In Sect. 2, we prove Theorem 1by constructing a collection of uncountably many such fractals and showing that any twofractals in the collection are not Lipschitz equivalent. The proof is based on some ideas ofFalconer and Marsh [6]. Section 3 is devoted to the proof of Theorem 3, which generalizesTheorem 2. We give the proof of necessary part in Sect. 3.1 and point out that Proposition 1can be obtained by a similar argument. The proof of sufficient part in Sect. 3.2 is more diffi-cult. The idea of “equal grouping” (see Lemma 10) in the proof comes from [16], while thetechniques used to realize this idea in any dimension come from [25]. Finally, Corollary 2 isproved in the end of Sect. 3.2.

1.1 Main results

Our first result concerns the cardinality of the Lipschitz equivalence classes in R1(n, m).

Theorem 1 Suppose that m > n ≥ 2, then the cardinality of the Lipschitz equivalenceclasses in the collection R1(n, m) is 2ℵ0 .

Remark 4 It is well known that the cardinality of the closed sets in Euclidean spaces is 2ℵ0 .So the cardinality of the Lipschitz equivalence classes is as many as possible.

We introduce some sub-collections of Rd(n, m) as follows.When the same rule with respect to rule set Bk is used for every basic cube of order k, we

get a special sub-collections Sd(n, m) of Rd(n, m).

Definition 3 Suppose that Bk = {bk,1, bk,2, . . . , bk,n} is a rule set for k ≥ 1. Let Tk, j : x →m−1(x + bk, j ) and Sσ = T1,σ1 ◦ T2,σ2 ◦ · · · ◦ Tk,σk for σ = σ1σ2 · · · σk ∈ {1, . . . , n}k . Fork ≥ 1, define

Ek =⋃

σ∈{1,...,n}k

Sσ (J0) and E =⋂

k≥1

Ek .

We use Sd(n, m) to denote the collection of all such sets E .

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Fig. 3 The construction process of the first two steps of a set in S2(3, 3)

Remark 5 The set Ek+1 is generated from the set Ek by using rule Bk , that means the cubeSσ (J0) is replaced by the smaller cubes

Sσ

⎛

⎝n⋃

j=1

Tk, j (J0)

⎞

⎠ =n⋃

j=1

Sσ

(m−1(J0 + bk, j )

)for all σ ∈ {1, . . . , n}k .

In geometric terms, the sets in Sd(n, m) are exactly the sets in Rd(n, m) such that the samerule is used in each step in the construction process, but the rules used in different steps maybe different.

Example 2 Figure 3 shows the construction process of the first two steps of a sets in S2(3, 3).Two rule sets

B1 = {(1, 0), (0, 1), (2, 2)}, B2 = {(0, 1), (2, 0), (2, 2)}are used in the construction.

Let (�(n, m), ρm) be the metric symbolic space with

�(n, m) = {1, . . . , n}N and ρm(ω, ω′) = m− inf{k : ωk �=ω′k }.

We know that

dimH �(n, m) = log n/ log m = dimH E for all E ∈ Rd(n, m).

It is natural to inquire that

Question 1 What fractals in Sd(n, m) are Lipschitz equivalent to (�(n, m), ρm)?

When we give some restrictions on the connected component in Ek , we obtain two sub-collections BRd(n, m)(⊂ Rd(n, m)) and BSd(n, m)(⊂ Sd(n, m)).

Definition 4 Let E ∈ Rd(n, m). For k ≥ 1, a block B of order k is a sub-collection ofthe basic cubes of order k such that

⋃Jσ ∈B Jσ is a connected component of Ek . Write the

cardinality of a block B as �B. We say that the set E has bounded block size if the cardinalityof blocks is uniformly upper bounded, i.e., there is an integer M > 0 such that �B ≤ M forany block B of E . We use BRd(n, m) and BSd(n, m) to denote the collection of all sets withbounded block size in Rd(n, m) and Sd(n, m), respectively. See Fig. 4.

Remark 6 It follows from the definition that all sets in BRd(n, m) or BSd(n, m) are totallydisconnected.

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Fig. 4 The blocks

The following theorem is not only the answer to Question 1, but also the complete char-acterization of the sets in Sd(n, m) that are Lipschitz equivalent to �(n, m).

Theorem 2 Suppose that E ∈ Sd(n, m). Then

E � �(n, m) if and only if E ∈ BSd(n, m).

Remark 7 Theorem 2 generalizes Theorem A. In fact, it is proved in [25] that all self-similarsets in Theorem A have bounded block size, so-called “finite type” in [25].

Especially, in the case of the dimension d = 1, we have BS1(n, m) = S1(n, m), since�B ≤ 2n for all block B due to m > n. So the following corollary holds.

Corollary 1 For all sets E, F ∈ S1(n, m), we have E � F � �(n, m).

If E /∈ Sd(n, m) and is totally disconnected, it is difficult to determine whether E isLipschitz equivalent to �(n, m) or not. The following is a necessary condition.

Proposition 1 Suppose that E ∈ Rd(n, m). If E � �(n, m), then we have E ∈ BRd(n, m).

In fact, the results in Theorem 2 and Corollary 1 hold in more general setting. With thetechnical notations and definitions of Moran sets with cubic pattern in Sect. 1.2, we can statethe generalized results.

Given a sequence of positive integers {nk}k≥1 and a sequence of positive numbers {ck}k≥1

satisfying nk ≥ 2 and ck < 1 for all k ≥ 1, let Md({nk}, {ck}) and BMd({nk}, {ck}) be thecollections of “Moran sets with cubic pattern” defined in Definitions 5 and 7, respectively.

Let � = ∏∞k=1{1, 2, . . . , nk} be a “symbolic space” which has no shift operator, equipped

with metric

ρ(ω, ω′) = c1c2 · · · ck, where k = inf{i : ωi �= ω′i },

for any distinct points ω,ω′ ∈ � with ω = ω1ω2 · · · and ω′ = ω′1ω

′2 · · ·.

Theorem 3 Suppose that E ∈ Md({nk}, {ck}). Then E � � if and only if E ∈BMd({nk}, {ck}).Remark 8 From Remarks 9, 12 and 13, we know that Theorem 3 generalizes Theorem 2.

When the dimension d = 1, we have the following corollary which generalizesCorollary 1.

Corollary 2 If infk{1 − nkck} > 0, then BM1({nk}, {ck}) = M1({nk}, {ck}). That meansE � � for all E ∈ M1({nk}, {ck}).

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1.2 The Moran sets with cubic pattern

Some special cases of Moran sets were first studied by Moran [15]. The later works [8,10–12,17,19] developed the theory on the geometrical structure and dimensions of Moran setssystematically.

We recall some notations of Moran set as follows.Let {nk}k≥1 be a sequence of positive integers and {ck}k≥1 a sequence of positive numbers

satisfying nk ≥ 2 and ck < 1 for all k ≥ 1.Let �0 = {∅}, where ∅ is the empty word. For any integers k, let �k = ∏k

j=1{1, . . . , n j },the collection of words with length k. Define �∗ = ⋃

k≥0 �k , the collection of all finitewords.

For any integers l, k with l > k ≥ 1, let �k,l = ∏lj=k+1{1, . . . , n j }. If σ = σ1 · · · σk ∈

�k, τ = τk+1 · · · τl ∈ �k,l , set

σ ∗ τ = σ1 · · · σkτk+1 · · · τl ∈ �l .

For convenience, write Nk = ∏kj=1 n j , Ck = ∏k

j=1 ck, Nk,l = ∏lj=k+1 n j and Ck,l =

∏lj=k+1 c j for l > k ≥ 1. Let N0 = C0 = 1 for k = 0.

Definition 5 (Moran sets with cubic pattern) Suppose that d ≥ 1, bk, j ∈ [0, 1)d for k ≥ 1and 1 ≤ j ≤ nk . Let Tk, j : x → ck(x + bk, j ) and Sσ = T1,σ1 ◦ T2,σ2 ◦ · · · ◦ Tk,σk for σ ∈ �k .Let J0 = [0, 1]d be the unit cube of R

d . For all k ≥ 1 and σ ∈ �k , define

Jσ = Sσ (J0) = T1,σ1 ◦ · · · ◦ Tk,σk (J0) and J∅ = J0.

For σ ∈ �k with k ≥ 0, we call the cube Jσ a basic cube of order k.If c∗ = infk ck > 0 and

(i) for all k ≥ 0 and σ ∈ �k, Jσ∗1, Jσ∗2, . . . , Jσ∗nk+1 are subsets of Jσ ;(ii) for all k ≥ 1, int Jσ ∩ int Jσ ′ = ∅, whenever σ, σ ′ ∈ �k with σ �= σ ′;

we call the set E = ⋂k≥1

⋃σ∈�k

Jσ a Moran set with cubic pattern and use Md({nk}, {ck})to denote the collection of all such sets. Md(n, c) is used when nk ≡ n and ck ≡ c.

Remark 9 It is natural that Sd(n, m) ⊂ Md(n, m−1). In fact, the notion of Moran setswith cubic pattern generalizes the fractal collection Sd(n, m) and it is quite different fromSd(n, m) in the following points:

• the placements of the basic cubes at each step of the construction can be more flexible;• the contraction ratios can be different at different steps;• the cardinality of the cubes used in replacement at different steps can be different.

Remark 10 The Moran sets with cubic pattern are a special case of Moran sets. It is knownthat all the Moran sets in Md({nk}, {ck}) have the same Hausdorff and packing dimensiondue to c∗ > 0. For the general definition and the results on the dimensions of Moran sets,we refer to [10–12,19].

Remark 11 We notice that ck ≤ 1/2 for all k in the setting of Definition 5. In fact, in anycube Jσ of side length Ck−1, there are nk(≥ 2) sub-cubes Jσ∗1, . . . , Jσ∗nk of side length Ck

such that int Jσ∗1, . . . , int Jσ∗nk are pairwise disjoint, which implies ck ≤ 1/2.

We will adhere to the following conventions. The metric in Rd is taken as

|x − y| = max1≤i≤d

|xi − yi |.

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Fig. 5 δ-blocks for δ = 1/2 in M1(3, 1/5) and M2(4, 1/5)

For X, Y ⊂ Rd , the shortest distance dist(X, Y ) is defined by

dist(X, Y ) = infx∈X,y∈Y

|x − y|.

Let diam(X) denote the diameter of the set X .

Definition 6 (δ-block) Let E be a Moran set with cubic pattern. For δ > 0, a δ-block B oforder k related to σ ∈ �k is a collection of basic cubes of order k defined by

B ={

Jσ ′ : σ ′ = σ or there is a sequence σ 1(= σ), σ 2, . . . , σ j (= σ ′) ∈ �k

such that dist(Jσ i , Jσ i+1) ≤ δ · Ck for every 1 ≤ i < j}.

By omitting σ , we always say B is a δ-block of order k. The collection of all the δ-blocks oforder k is denoted by Bk . Write the order of B as ord B.

Roughly speaking, a δ-block of order k is the collection of basic cubes which make up anapproximate connected component of the k-th stage construction of E (Fig. 5).

Remark 12 Note that the blocks defined in Definition 4 are the δ-blocks with any δ < 1.

Definition 7 We say that E has bounded block size if there are δ > 0 and M ∈ N such thatthe cardinality of any δ-block of E is not greater than M . Write the cardinality of a δ-blockB as �B. Denote by BMd({nk}, {ck}) the collection of Moran sets with bounded block sizein Md({nk}, {ck}). BMd(n, c) is used when nk ≡ n and ck ≡ c.

Remark 13 We know that BSd(n, m) ⊂ BMd(n, m−1) from Remarks 9 and 12.

2 Uncountably many Lipschitz equivalent classes in R1(n, m)

We will construct a sub-collection {Eλ : λ ∈ (0, 1)} of R1(n, m) such that

Eλ �� Eλ′ whenever λ �= λ′.

2.1 Preliminary

For any collection A of sets, let⋃ A = ⋃

A∈A A denote the union of all sets in A. ForE ∈ R1(n, m), we denote the collection of basic cubes of order k by Jk . If U is a connectedcomponent of

⋃ Jk = ⋃{J : J ∈ Jk}, we say U is of order k, write ord U = k. Denotethe set of connected components of

⋃ Jk by Uk . For another E ′ ∈ R1(n, m), we use thecorresponding notations J ′

k and U ′k .

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Lemma 1 Let E, E ′ ∈ R1(n, m) and f : E → E ′ be a bi-Lipschitz bijection. Then there isan integer K0 such that

(a) for all k ≥ 1 and for every connected component U ∈ Uk , there exists a subset U ′U of

U ′k+K0

satisfying card U ′U ≥2 and

f (E ∩ U ) = E ′ ∩⋃

U ′U = E ′ ∩

⋃{U ′ : U ′ ∈ U ′

U };(b) for all k ≥ 1 and U ′ ∈ U ′

k , there exists a subset UU ′ of Uk+K0 satisfying card UU ′ ≥ 2and

f −1(E ′ ∩ U ′) = E ∩⋃

UU ′ .

Proof Let β > 1 be the Lipschitz constant of f , i.e.,

β−1|x − y| ≤ | f (x) − f (y)| ≤ β|x − y| for all x, y ∈ E .

Let K0 be an integer large enough such that

mK0−2 > 2nβ.

Let

U ′U = {U ′ ∈ U ′

k+K0: U ′ ∩ f (E ∩ U ) �= ∅}.

By the structure of sets in R1(n, m), we know that

(1) U ′ contains at most 2n basic intervals of order k + K0;(2) dist(U ′, E ′\U ′) ≥ m−k−K0 ;(3) dist(U, E\U ) ≥ m−k .

For conclusion (a), we first show that card U ′U ≥ 2. If U ′

U contains only one connectedcomponent U ′, then

2nm−k−K0 ≥ diam U ′ ≥ diam f (U ∩ E) ≥ β−1 diam(U ∩ E) > β−1m−k−2.

For the last inequality, notice that U contains at least four basic cubes of order k + 2, sodiam(U ∩ E) > m−k−2. The above estimation implies that 2nβ > mK0−2, contradicting thechoice of K0.

It remains to show that if U ′ ∈ U ′U , then E ′ ∩ U ′ ⊂ f (E ∩ U ). Suppose on the contrary

that there are U ′ ∈ U ′U and x ′, y′ ∈ U ′ ∩ E ′ with x ′ ∈ f (E ∩ U ) and y′ /∈ f (E ∩ U ). So we

have

β−1m−k ≤ β−1| f −1(x ′) − f −1(y′)| ≤ |x ′ − y′| ≤ diam(U ′) ≤ 2nm−k−K0 .

As a result, mK0 ≤ 2nβ. This is contradictory to the choice of K0.A similar argument proves the conclusion (b). ��As in Lemma 1, let E, E ′ ∈ R1(n, m) and f : E → E ′ is a bi-Lipschitz bijection. For

U ∈ Uk , define

φ(U ) = card{J ∈ Jk : J ⊂ U },φ̃(U ) = card{J ′ ∈ J ′

k+K0: J ′ ∩ E ′ ⊂ f (U ∩ E)}, (2.1)

where K0 is as in Lemma 1. Similarly, for U ′ ∈ U ′k , we can also define

φ(U ′) = card{J ′ ∈ J ′k : J ′ ⊂ U ′},

φ̃(U ′) = card{J ∈ Jk+K0 : J ∩ E ⊂ f −1(U ′ ∩ E ′)}. (2.2)

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Since φ(U ) is a positive integer not greater than 2n, there exists an U0 ∈ ⋃k≥1 Uk such

that

φ̃(U0)/φ(U0) = min

⎧⎨

⎩φ̃(U )/φ(U ) : U ∈

⋃

k≥1

Uk

⎫⎬

⎭. (2.3)

For such U0, we have the following two lemmas which are the key to the proof of Theorem 1.

Lemma 2 Suppose that the connected component U0 is as in (2.3). Then for all U ∈ ⋃k≥1 Uk

with U ⊂ U0, we have

φ̃(U )/φ(U ) = φ̃(U0)/φ(U0).

Proof Assume that U0 ∈ Uk and U ∈ Uk+l with U ⊂ U0. By Lemma 1,

nl φ̃(U0) = card{J ′ ∈ J ′k+l+K0

: J ′ ∩ E ′ ⊂ f (U0 ∩ E)} =∑

U∈Uk+l , U⊂U0

φ̃(U )

=∑

U∈Uk+l , U⊂U0

φ(U ) · φ̃(U )/φ(U ) ≥ φ̃(U0)/φ(U0) ·∑

U∈Uk+l , U⊂U0

φ(U )

= φ̃(U0)/φ(U0) · nlφ(U0) = nl φ̃(U0).

The computation above shows that φ̃(U )/φ(U ) = φ̃(U0)/φ(U0) for all U ∈ Uk+l withU ⊂ U0. This completes the proof. ��Lemma 3 Suppose that the connected component U0 ∈ Uk is as in (2.3). let

U ′U0

= {U ′ ∈ U ′k+K0

: U ′ ∩ E ′ ⊂ f (U0 ∩ E)}.For U ′ ∈ U ′

U0, let

UU ′ = {U ∈ Uk+2K0 : U ∩ E ⊂ f −1(U ′ ∩ E ′)}.Then for any U ′, V ′ ∈ U ′

U0, we have

∑U∈UU ′ φ(U )

∑U∈UV ′ φ(U )

= φ(U ′)/φ(V ′).

Proof From Lemma 1, we have U0 ⊃ ⋃U ′∈U ′

U0

⋃U∈UU ′ U . So by Lemma 2,

φ̃(U )/φ(U ) = φ̃(U0)/φ(U0),

for all U ∈ UU ′ and all U ′ ∈ U ′U0

. Therefore, we have

φ(U ′)/φ(V ′) = card{J ′ ∈ J ′k+3K0

: J ′ ⊂ U ′}card{J ′ ∈ J ′

k+3K0: J ′ ⊂ V ′} =

∑U∈UU ′ φ̃(U )

∑U∈UV ′ φ̃(U )

=∑

U∈UU ′ φ(U ) · φ̃(U )/φ(U )∑

U∈UV ′ φ(U ) · φ̃(U )/φ(U )=

∑U∈UU ′ φ(U )

∑U∈UV ′ φ(U )

.

��

123


Fig. 6 Method A for n = 2 and m = 3

2.2 Construction of Eλ

In this subsection, we will give the construction of Eλ ∈ R1(n, m) for all λ ∈ (0, 1) suchthat Eλ �� Eλ′ .

We use the notations Jk and Uk as in Sect. 2.1. In our construction of Eλ, there will beonly three types of connected components in Uk for all k ≥ 1:

I1 = [0, 1], I2 = [0, n] and I3 = [0, n + 1].

This means for all k ≥ 1 and all U ∈ Uk , there are SU and i ∈ {1, 2, 3} such that U = SU (Ii ),where SU (x) = m−k x + b.

In the progress of construction of Eλ, we will use two iteration methods, which involven + 1 rules. Let

A0 = {m − n, m − n + 1, . . . , m − 1},A1 = {0, m − n + 1, m − n + 2, . . . , m − 1},

. . . ,

Ai = {0, 1, . . . , i − 1, m − n + i, m − n + i + 1, . . . , m − 1},. . . ,

An = {0, 1, . . . , n − 1}.

Method A Replace all intervals J = SJ ([0, 1]) by

⋃

i∈A0

SJ(m−1([0, 1] + i)

),

which is a connected component of type I2 (see Fig. 6).

Method B There are three cases according to the three types of connected components (seeFig. 7).

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Fig. 7 Method B for n = 2 and m = 3

(1) For each connected component U = SU ([0, 1]) of type I1, replace it by⋃

i∈A0

SU(m−1([0, 1] + i)

),

which is a connected component of type I2.(2) For each connected component U = SU ([0, n]) of type I2, replace it by

n−1⋃

i=0

⋃

j∈Ai

SU(m−1([0, 1] + j) + i

),

which consists of n−1 connected components of type I3 and one connected componentof type I1.

(3) For each connected component U = SU ([0, n + 1]) of type I3, replace it by

n⋃

i=0

⋃

j∈Ai

SU(m−1([0, 1] + j) + i

),

which consists of n connected components of type I3.

For each λ ∈ (0, 1), we pick a infinite sequence ς = ς0ς1 · · · ∈ ∏∞i=0{0, 1} such that

limk→∞ k−1

k−1∑

i=0

ςi = λ.

Let

Nλ(A) =⋃

ςi =0

[2i , 2i+1) ∩ N and Nλ(B) =⋃

ςi =1

[2i , 2i+1) ∩ N.

Then Nλ(A) ∪ Nλ(B) = N.For Eλ, recall that Jk(k ≥ 0) is the collection of all the basic cubes of order k. We shall

determine {Jk} by induction on k. This completes the construction of Eλ. Let J0 = {[0, 1]}.Assume that Jk−1 has been determined, then

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Fig. 8 The construction process of the set Eλ

(a) Jk is derived from Jk−1 by Method A if k ∈ Nλ(A);(b) Jk is derived from Jk−1 by Method B if k ∈ Nλ(B).

Now we give an example to show how the construction works.

Example 3 Suppose that n = 2 and m = 3. Let λ = 1/2. Pick ς = 0101 . . . = (01)∞ ∈∏∞i=0{0, 1}. Then

Nς (A) ={

1, 4, 5, 6, 7, . . . , 22k, . . . , 22k+1 − 1, . . .},

Nς (B) ={

2, 3, 8, 9, . . . , 15, . . . , 22k+1, . . . , 22(k+1) − 1, . . .}.

The first 3 steps of the construction process of the set Eλ is showed in Fig. 8.

By the construction, we can check the following lemma easily.

Lemma 4 Suppose that U ∈ Uk is of type I1 or I2, then for all l ≥ k, there exists at leastone V ∈ Ul such that V ⊂ U and V is of type I1 or I2.

2.3 Proof of Eλ �� Eλ′ for λ �= λ′

We use the notations as in Sect. 2.1 except for replacing E, E ′ by Eλ, Eλ′ . In other words, Jk

and Uk denote the collections of basic cubes and connected components of order k for Eλ,while J ′

k and U ′k denote the corresponding collections for Eλ′ .

We will prove Eλ �� Eλ′ by reduction to absurdity. Assume on the contrary that thereexists a bi-Lipschitz bijection f : Eλ → Eλ′ such that

β−1|x − y| ≤ | f (x) − f (y)| ≤ β|x − y| for all x, y ∈ Eλ.

Let K0 be as in Lemma 1, φ, φ̃ as in (2.1) and (2.2), and U0 as in (2.3). According to Lemma 2and the construction of Eλ, we can further assume that ord U0 = k0 and Jk0 is derived fromJk0−1 by Method A. As a result, U0 is of type I2.

New suppose that

limk→∞ k−1

k−1∑

i=0

ςi = λ > λ′ = limk→∞ k−1

k−1∑

i=0

ς ′i .

So we can find i0 ≥ 1 large enough such that

ςi0 = 1, ς ′i0

= 0, 2i0 − 1 > k0 and 2i0 > 2K0. (2.4)

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By Lemma 4, we can find U1 ∈ U2i0 −1 such that U1 ⊂ U0 and U1 is of type I1 or I2. ByLemma 2,

φ̃(U1)/φ(U1) = min

⎧⎨

⎩φ̃(U )/φ(U ) : U ∈

⋃

k≥1

Uk

⎫⎬

⎭. (2.5)

According to the construction of Eλ, Eλ′ and (2.4), we know that J2i0 , J2i0 +1, . . . ,

J2i0 −1+2K0are all derived by Method B and J ′

2i0 −1+K0is derived by Method A. Conse-

quently, all connected components in {U ∈ U2i0 −1+2K0: U ⊂ U1} are of type I3 except for

one V which is of type I1 or I2 since U1 is of type I1 or I2. And all connected componentsin U ′

2i0 −1+K0are of type I2. In other words, we can write

{U ∈ U2i0 −1+2K0: U ⊂ U1} = U∗ ∪ {V }

such that

φ(V ) = 1 or n and φ(U ) = n + 1 for all U ∈ U∗. (2.6)

And

φ(U ′) = n for all U ∈ U ′2i0 −1+K0

. (2.7)

By Lemma 1, we have the following decomposition.

U∗ ∪ {V } = {U ∈ U2i0 −1+2K0: U ⊂ U1} =

⋃

U ′∈U ′U1

UU ′ . (2.8)

By Lemma 3, (2.5) and (2.7), for any U ′, V ′ ∈ U ′U1

,∑

U∈UU ′φ(U ) =

∑

U∈UV ′φ(U ). (2.9)

Also by Lemma 1, we know that card U ′U1

≥ 2. Therefore, (2.9) contradicts (2.6). Thiscontradiction shows that Eλ �� Eλ′ .

3 The Moran sets with cubic pattern

3.1 The sets with bounded block size

This subsection is devoted to the proofs of the necessary part of Theorem 3 and Proposition 1.Since the two proofs make use of the same idea and the arguments are also similar, we onlygive the proof of the necessary part of Theorem 3.

Suppose that E ∈ Md({nk}, {ck}) is a Moran set with cubic pattern such that E � �. Letf : � → E be a bi-Lipschitz bijection with Lipschitz constant β > 1, i.e.,

β−1| f (ω) − f (ω′)| ≤ ρ(ω, ω′) ≤ β| f (ω) − f (ω′)| for all ω,ω′ ∈ �.

Take a positive integer K1 large enough such that

2K1 > 3β. (3.1)

Let Jk be the collection of basic cubes of order k of E for all k ≥ 1. We begin with a lemmasimilar to Lemma 1.

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Lemma 5 For all k ≥ 1 and σ ∈ �k , there is a subset Jσ of Jk+K1 satisfying

f [σ ] = E ∩⋃

Jσ .

Here K1 is as in (3.1), [σ ] = {ω ∈ � : ωi = σi for 1 ≤ i ≤ k} and⋃ Jσ = ⋃

J∈JσJ .

Moreover, for all distinct σ, σ ′ ∈ �k , we have

dist(J, J ′) > Ck+K1 for any J ∈ Jσ and J ′ ∈ Jσ ′ .

Proof Let Jσ = {J ∈ Jk+K1 : J ∩ f [σ ] �= ∅} for σ ∈ �k . We first show that f [σ ] =E∩⋃ Jσ . Suppose on the contrary that there are J ∈ Jσ and x, y ∈ J ∩E such that x ∈ f [σ ]and y /∈ f [σ ]. Recall that the metric in R

d is taken as |x − y| = max1≤i≤d |xi − yi |. Then

Ck ≤ ρ(

f −1(x), f −1(y)) ≤ β|x − y| ≤ β diam J = βCk+K1 ≤ 2−K1βCk < Ck .

For inequality Ck+K1 ≤ 2K1 Ck , notice that ci ≤ 1/2 for all i ≥ 1 (see Remark 11). Thecontradiction in above computation proves that f [σ ] = E ∩ ⋃ Jσ .

It remains to show that dist(J, J ′) > Ck+K1 . In fact,

dist(J, J ′) ≥ dist( f [σ ], f [σ ′]) − diam J − diam J ′

≥ β−1Ck − 2Ck+K1 ≥ (2K1β−1 − 2)Ck+K1 > Ck+K1 .

Thus the proof is completed. ��Proof of the necessary part of Theorem 3 Let E and f : � → E be as above. Take δ = 1.Let B be a 1-block of order k + K1 for some k ≥ 1. From Lemma 5, we know that there is aσ ∈ �k such that B ⊂ Jσ . So to show that E has bounded block size, we only need to findan upper bound for card Jσ for all σ .

For this, we notice that for all σ ∈ �k and all k ≥ 1,

diam f [σ ] ≤ βCk ≤ βc−K1∗ Ck+K1 .

Together with the fact that diam J = Ck+K1 for all J ∈ Jk+K1 , we know that

diam⋃

Jσ ≤ (βc−K1∗ + 2)Ck+K1 .

Denote by L the Lebesgue measure of Rd . Noting that int J ∩ int J ′ = ∅ for all distinct J ,

J ′ ∈ Jσ , we have

card Jσ · Cdk+K1

=∑

J∈Jσ

L(J ) = L(⋃

Jσ

)≤ 2d(βc−K1∗ + 2)dCd

k+K1.

Thus card Jσ ≤ 2d(βc−K1∗ + 2)d for all σ , and so the proof is completed. ��3.2 The geometric structure of sets in BMd({nk}, {ck})

In this subsection, we will prove the sufficient part of Theorem 3 and Corollary 2. We beginwith some notations.

Fix a Moran set E ∈ BMd({nk}, {ck}). We assume that, for some δ > 0, the cardinalityof any δ-block of E is not greater than M . Let Jk be the collection of basic cubes of order kof E for all k ≥ 1. Suppose Jσ ∈ Jk is a basic cube of E with σ ∈ �k . For an integer l > kand a subset P of �k,l , let

Jσ (P) =⋃

τ∈P

Jσ∗τ .

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For any collection A of sets, let⋃ A = ⋃

A∈A A denote the union of all sets in A.Considering the faces of [0, 1]d of dimension d − 1, they are

{F01 , . . . , F0

d } ∪ {F11 , . . . , F1

d },where F0

i = {x ∈ [0, 1]d : xi = 0} and F1i = {x ∈ [0, 1]d : xi = 1}. Given a set of diameter

less than 1/3, then for each 0 ≤ i ≤ d , either F0i or F1

i is far away from this set, with adistance greater than 1/3. We use this simple property to describe the position of subsetsin [0, 1]d . A collection F of faces is called good if {F0

i , F1i } �⊂ F for all 1 ≤ i ≤ d , i.e., at

most one faces in {F0i , F1

i } is contained in F . Let F denote all the good collections of faces.Notice that ∅ ∈ F .

Take an integer K2 so large that

Md(2δ + 1)2−K2 < 1/3. (3.2)

Recall that the metric in Rd is taken as |x − y| = max1≤i≤d |xi − yi |.

Lemma 6 For all k ≥ 1, there is a partition

�k,k+K2 =⋃

F∈F

PFk ,

which is a disjoint union such that for all σ ∈ �k ,

(a) dist (x, Sσ (F)) ≤ M(δ + 1)Ck+K2 , for all x ∈ Jσ (PFk ) and all F ∈ F;

(b) dist(Jσ (PF

k ), Sσ (F))

> δCk+K2 , for all F /∈ F;

(c) dist(

Jσ (PFk ), Jσ (PF ′

k ))

> δCk+K2 whenever F �= F ′.

Proof Fix a word σ ∈ �k . Suppose that J ∈ Jk+K2 and J ⊂ Jσ . We say that Jis close to a face F , if there is a sequence J 1(= J ), J 2, . . . , J t ∈ Jk+K2 such thatJ i ⊂ Jσ , dist(J i , J i+1) ≤ δCk+K2 for each i and dist(J t , Sσ (F)) ≤ δCk+K2 . Recall that Ehas bounded block size with constants δ and M , so t ≤ M . Together with (3.2) and ci ≤ 1/2(Remark 11), we have

dist (J, Sσ (F)) ≤t∑

i=2

(dist(J i−1, J i ) + diam J i

)+ dist

(J t , Sσ (F)

)

≤ (Mδ + M − 1)Ck+K2 < Ck/3 (3.3)

if J is close to F . The inequality (3.3) implies that all the faces that are close to a such cubeJ form a good face collection.

For each good face collection F , let

JF = {J ∈ Jk+K2 : J ⊂ Jσ and J is close to F if and only if F ∈ F}.Then {JF }F∈F is a partition of the collection {J ∈ Jk+K2 : J ⊂ Jσ } and

dist(⋃

JF ,⋃

JF ′)

> δCk+K0 whenever F �= F ′. (3.4)

Now for each good face collection F , select PFk ⊂ �k,k+K2 , which may be empty, such that

Jσ (PFk ) = ⋃ JF . Then {PF

k }F∈F is a partition of �k,k+K0 .We will show that the conclusions (a), (b) and (c) hold. If x ∈ Jσ (PF

k ), then x ∈ J forsome J ∈ JF . Thus for all F ∈ F , by (3.3), we have

dist(x, Sσ (F)) ≤ diam J + dist (J, Sσ (F)) ≤ M(δ + 1)Ck+K2 .

123


Therefore, (a) holds for this fixed σ . It follows from the definition of JF and (3.4) that (b)and (c) also hold for this fixed σ . By the definition of Moran sets with cubic pattern, for anyσ ′ ∈ �k and any face F ,

Jσ ′(PFk ) = Sσ ′ ◦ S−1

σ

(Jσ (PF

k ))

and Sσ ′(F) = Sσ ′ ◦ S−1σ (Sσ (F)),

where Sσ ′ ◦ S−1σ is a translation. That means the partition {PF

k }F∈F is as desired for allσ ′ ∈ �k . ��

For each basic cube J and each good face collection F , define

dF (J ) = minx∈J

∑

F∈Fdist(x, F),

where d∅(J ) = 0 for all J when F = ∅.

Lemma 7 Let B ∈ Bk be a δ-block of order k. For each good face collection F , pick abasic cube JF ∈ B such that

dF (JF ) = minJ∈B

dF (J ).

Then for all basic cube J ′ of order k + K2 with JF (PFk ) ∩ J ′ = ∅, we have

dist(JF (PF

k ), J ′) > δCk+K2 .

Proof If F = ∅, then the distance between JF (PFk ) and any face of JF is greater than

δCk+K2 by (b) of Lemma 6. So the lemma holds in this case.Suppose that F �= ∅. By symmetry, we can assume that

F = {F01 , F0

2 , . . . , F0t } for some 1 ≤ t ≤ d.

On the contrary, suppose that J ′ is a basic cube of order k + K2 such that

dist(JF (PF

k ), J ′) ≤ δCk+K2 . (3.5)

By (c) of Lemma 6, J ′ �⊂ JF . On the other hand, we have J ′ ⊂ ⋃ B since B is a δ-block.So J ′ is contained in some cube J ∈ B \ {JF }. Assuming that JF = Sσ ([0, 1]d), we get twocubes [0, 1]d = S−1

σ (JF ) and J ∗ = ∏di=1 I ∗

i = S−1σ (J ) with side length 1 such that

int[0, 1]d ∩ int J ∗ = ∅. (3.6)

By (3.5), we can find x∗ = (x∗1 , . . . , x∗

d ) ∈ S−1σ

(JF (PF

k )) ⊂ [0, 1]d and y∗ =

(y∗1 , . . . , y∗

d ) ∈ S−1σ (J ′) ⊂ J ∗ such that

max1≤i≤d

|x∗i − y∗

i | ≤ δck+1 · · · ck+K2 = δCk,k+K2 . (3.7)

By (a) and (b) of Lemma 6, we also have

x∗i = C−1

k dist(Sσ (x∗), Sσ (F0

i )) ≤ M(δ + 1)Ck,k+K2 for 1 ≤ i ≤ t; (3.8)

δCk,k+K2 < x∗i ≤ 1 − δCk,k+K2 for t + 1 ≤ i ≤ d. (3.9)

This implies

t∑

i=1

|y∗i | ≤ Md(2δ + 1)Ck,k+K2 ≤ Md(2δ + 1) · 2−K2 < 1/3, (3.10)

123


where we use (3.2) and the fact ci ≤ 1/2 (Remark 11). It follows from (3.6) and (3.9) thatthere is an index 1 ≤ j ≤ t such that I ∗

j = [a, a + 1] satisfying

(a, a + 1) ∩ (0, 1) = ∅.

Notice that y∗j ∈ [a, a + 1] and y∗

j < 1/3. So we have a ≤ −1. Let z∗ ∈ J ∗ be defined byz∗

i = y∗i for i �= j and z∗

j = a ≤ −1. Then

dF (J ∗) ≤t∑

i=1

z∗i =

t∑

i=1

y∗i − y∗

j + a ≤t∑

i=1

|y∗i | − 1 ≤ −2/3 < 0 = dF ([0, 1]d).

This contradicts dF (JF ) ≤ dF J . ��For k, l ≥ 1, let B be a δ-block of order k and

Bl(B) ={B′ : B′is a δ-block of order k + l and

⋃B′ ⊂

⋃B}.

It is plain to see that∑

B′∈Bl (B)

�B′ = card{

J ∈ Jk+l : J ⊂⋃

B}

= nk+1nk+2 · · · nk+l · �B = Nk,l · �B.

Lemma 8 For each δ-block B of order k, there is a decomposition BK2(B) = B′K2

(B) ∪B′′

K2(B), which is a disjoint union such that

∑

B′∈B′K2

(B)

�B′ = Nk,k+K2 and∑

B′∈B′′K2

(B)

�B′ = Nk,k+K2 · (�B − 1).

Proof For each good face collection F , let JF (PFk ) be as in Lemma 7. It follows from

Lemma 7 that the collection

BF = {J ∈ Jk+K2 : J ⊂ JF (PF

k )}

is a union of some δ-blocks of order k + K2.Let

B′K2

(B) =⋃

F∈F

{B′ : B′ ⊂ BF } and B′′K2

(B) = BK2(B)\B′K2

(B).

By Lemma 6, we have∑

B′∈B′K2

(B)

�B′ =∑

F∈F

card BF =∑

F∈F

card PFk = card �k,k+K2 = Nk,k+K2 .

So B′K2

(B) and B′′K2

(B) are as desired. ��The following lemma is a combinatorial fact comes from [25].

Lemma 9 ([25, Lemma 2.7]) Let p, q, and { ı }ı∈I be positive integers with∑

ı∈I ı = pq.Suppose that there exists an integer r < p such that ı ≤ r for all ı ∈ I and card{ı : ı =1} ≥ rq. Then there is a decomposition I = ⋃q

j=1 I j such that for every 1 ≤ j ≤ q,∑

ı∈I j ı=p.

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Take a positive integer K3 large enough such that

2K3 ≥ M2 + 1 > M. (3.11)

Let K4 = K2 + K3 where K2 is defined by (3.2).The following Lemma, which shows the δ-block collection BK4(B) can be divided into

“equal” parts, is the key in the proof of the sufficient part of Theorem 3.

Lemma 10 For each δ-block B of order k, there is a decomposition

BK4(B) =�B⋃

j=1

BjK4

(B)

such that∑

B′∈BjK4

(B)

�B′ = nk+1nk+2 · · · nk+K4 = Nk,k+K4 for all 1 ≤ j ≤ �B.

Proof Since K4 = K2 + K3, we have

BK4(B) =⋃

C∈BK3 (B)

BK2(C) =⋃

C∈BK3 (B)

(B′

K2(C) ∪ B′′

K2(C)

).

Now we will apply Lemma 9. Let

I = {B′

K2(C) : C ∈ BK3(B)

} ∪ {B′′

K2(C) : C ∈ BK3(B)

}.

For ı ∈ I, let ı = ∑B′∈ı �B′/Nk+K3,k+K4 . Thus ı = 1 if ı = B′

K2(C) for some C and

ı = �C − 1 if ı = B′′K2

(C) for some C. Let p = Nk,k+K3 , q = �B and r = M . Then ı < �C ≤ M = r and

card{ı : ı = 1} ≥ card BK3(B) ≥∑

C∈BK3 (B) �Cmax{�C : C ∈ BK3(B)}

≥ Nk,k+K3 · �BM

≥ 2K3 · �B/M > M · �B = rq.

So the conditions of Lemma 9 hold, we have a decomposition I = ⋃�Bj=1 I j with

∑ı∈I j

ı =Nk,k+K3 . For 1 ≤ j ≤ �B, put

BjK4

(B) =⋃

ı∈I j

ı.

Then∑

B′∈BjK4

(B)

�B′ =∑

ı∈I j

∑

B′∈ı

�B′ =∑

ı∈I j

ı · Nk+K3,k+K4

= Nk,k+K3 · Nk+K3,k+K4 = Nk,k+K4 .

and BK4(B) = ⋃�Bj=1 B

jK4

(B). ��For W ⊂ �k , write

[W ] =⋃

σ∈W

[σ ].

Recall that [σ ] = {ω ∈ � : ω j = σ j for1 ≤ j ≤ k} for σ ∈ �k .

123


Lemma 11 For all k ≥ 0 and all B ∈ BkK4 , there is

W B = {σB,1, σB,1, . . . , σB,�B} ⊂ �kK4

such that

W B ∩ W B′ = ∅ when B �= B′;and for B′ ∈ BkK4 , B ∈ B(k+1)K4 with

⋃ B ⊂ ⋃ B′, there is some σ ′ ∈ W B′satisfying

[W B] ⊂ [σ ′].Proof We prove this lemma by induction on k.

When k = 0, we have B0 = {{[0, 1]d}} and �0 = {∅}. Let W {[0,1]d } = {∅}.By induction, suppose that for all B′ ∈ B(k−1)K4 , we have found the desired W B′ ⊂

�(k−1)K4 . Write

W B′, j ={σ ∈ �kK4 : [σ ] ⊂ [σB′, j ]

}for all σB′, j ∈ W B′

.

Note that for all 1 ≤ j ≤ �B′,

card W B′, j = n(k−1)K4+1 · · · n(k−1)K4+2 · · · nkK4 = N(k−1)K4,kK4 .

By Lemma 10, we have the decomposition

BK4(B′) =�B′⋃

j=1

BjK4

(B′)

with∑

B∈BjK4

(B′)

�B = N(k−1)K4,kK4 for 1 ≤ j ≤ �B′.

Therefore, we can find a decomposition

W B′, j =⋃

B∈BjK4

(B′)

{σB,1, σB,2, . . . , σB,�B} =⋃

B∈BjK4

(B′)

W B.

Such W B’s are as desired and the induction is completed. ��Proof of the sufficient part of Theorem 3 For all x ∈ E , there are δ-blocks Bk ∈ BkK4 forall k ≥ 1 such that

{x} =⋂

k≥1

⋃Bk .

For each δ-block B, let the collection W B be as in Lemma 11. A bijection f : E → � isdefined by

{ f (x)} =⋂

k≥1

[W Bk ].

Then f (x) is well defined. In fact, for some j, [W Bk ] ⊂ [σBk−1, j ] ⊂ [W Bk−1 ], and so

diam[W Bk ] ≤ diam[σBk−1, j ] ≤ C(k−1)K4 → 0 as k → ∞.

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That means⋂

k≥1[W Bk ] is a singleton.To complete the proof, it remains to show that f is bi-Lipschitz.For any two distinct points x, y ∈ E , there are k ≥ 0, Bx,y ∈ BkK4 and distinct Bx , By ∈

B(k+1)K4 such that

x ∈⋃

Bx , y ∈⋃

By and x, y ∈ Bx,y .

By the definition of δ-block and �Bx,y ≤ M , we have

δC(k+1)K4 ≤ dist(⋃

Bx ,⋃

By

)≤ dist(x, y) ≤ diam

⋃Bx,y ≤ M(1 + δ)CkK4 .

(3.12)

On the other hand, by the definition of f and W B, we have

f (x), f (y) ∈ [W Bx,y ] ⊂ [σ x,y] for some σ x,y ∈ �(k−1)K4

and

f (x) ∈ [σ x ] ⊂ [W Bx ] and f (y) ∈ [σ y] ⊂ [W By ]for distinct σ x , σ y ∈ �(k+1)K4 . And so

C(k+1)K4 ≤ ρ([σ x ], [σ y]) ≤ ρ ( f (x), f (y)) ≤ diam[σ x,y] ≤ C(k−1)K4 . (3.13)

From (3.12) and (3.13) and the fact that c∗ ≤ c j ≤ 1/2 for all j , we know that f isbi-Lipschitz. ��Proof of Corollary 2 Suppose that E ∈ M1({nk}, {ck}) with infk{1 − cknk} > 0. ByTheorem 3, we only need to show that E has bounded δ-block size for some constant δ > 0.

Since c∗ = infk ck > 0, we have n∗ = supk nk < ∞. Let

δ = infk{1 − nkck}2(n∗ + 1)

> 0 and M = 2n∗.

By the construction of Moran sets with cubic pattern on the line, we can see that �B ≤ Mfor all the δ-block of E . This completes the proof. ��

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Lipschitz equivalence of fractals generated by nested cubes