Lipschitz classification of self-similar sets with overlaps

In Rao et al. (Comptes Rendus Acad Sci Paris Ser I(342):191–196, 2006), Rao–Ruan–Xi solved an open question posed by David and Semmes and gave a complete Lipschitz classification of self-similar sets on R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb R$$\end{document} with touching structure. In this short note, by applying a matrix rearrangeable condition introduced in Luo (J Lond Math Soc 99(2):428–446, 2019), we generalize their result onto the self-similar sets with overlapping structure.


Introduction
Two compact metric spaces (E, d 1 ) and (F, d 2 ) are said to be Lipschitz equivalent, denote by E F, if there is a bijection f from E onto F and a constant C > 0 such that for all x, y ∈ E, we have From the definition, it is trivial to know that two Lipschitz equivalent spaces must have the common Hausdorff dimension. But the converse is usually not true in general. Hence except the Hausdorff dimension, Lipschitz equivalence plays another important role for the classification of fractal sets.
Around 1990, with different viewpoints, Cooper-Pignataro [2] and Falconer-Marsh [5] made an initial attempt on the Lipschitz equivalence of Cantor sets. In 2006, Rao-Ruan-Xi [16] achieved a breakthrough on self-similar sets with touching structure. They used a technique of graph directed system to give a solution to an open question of David and Semmes [3]. The result triggered a lot of interest. Since then, there have been fruitful researches on this topic [4,[10][11][12][13]15,17,20,21]. However, very limited results involved the self-similar sets with overlapping structure (in the absence of the open set condition). It is known that the topological structure of self-similar sets with overlaps is quite complicated (see [14]). That may make more difficult to study the Lipschitz equivalence.
Recently, Guo et al. [7] first constructed a Lipschitz equivalence class of self-similar sets with complete overlaps on R. Some other sufficient or necessary conditions for the Lipschitz classification of self-similar sets on R were also concerned in [1,8]. At the same time, Lau and Luo [10,12] introduced the classical Gromov hyperbolic graph into the study of Lipschitz equivalence. They developed a new tractable approach on the problem. In particular, for self-similar sets with overlaps, Luo in [10] provided a very general sufficient condition (later we will call it the matrix rearrangeable condition (MRC)) for the Lipschitz equivalence of self-similar sets.
The main purpose of this note is to apply the MRC to classify the collection of self-similar sets as follows: Let I = [0, 1] be a unit interval, n ≥ 3 be an integer and let 0 < r < 1/n. Denote by C n,r , the collection of self-similar sets K satisfying: We prove that Theorem 1.1 Any two self-similar sets in C n,r , are Lipschitz equivalent.
Note that the self-similar set K ∈ C n,r , is always totally disconnected as nr < 1. Rao-Ruan-Xi [16] proved the case that = 0 (i.e., there are no overlaps on K ). Let C n,r = n−1 =0 C n,r , be the collection of self-similar sets K satisfying the above conditions (1), (2), (3). Then the Lipschitz classification of C n,r only depends on the number of overlaps .

Preliminaries
In this section, we follow the standard notation in [10]. Let (X , E) be a graph, where X is a set of vertices and E is a set of edges, i.e., Moreover, if the path has the minimal length among all possible paths from x to y, then we say that the path is a geodesic and denote by π(x, y). Call X a tree if for any two distinct vertices there is a unique path connecting them. We equip a graph (X , E) with an integer valued metric d(x, y), which is the length of a geodesic π(x, y) from x to y. Let ϑ ∈ X be a fixed vertex and call it the root (or reference point) of the graph. We use |x| to denote d(ϑ, x), and say that x belongs to the n-th level of the graph if |x| = n. We always assume that the graph X is connected, i.e., any two vertices can be joined by a path.
A y)) is the Gromov product [6]. We choose a > 0 with exp(3δa) < √ 2, and define for x, y ∈ X , ρ a (x, y) = δ x,y exp(−a|x ∧ y|), where δ x,y = 0 or 1 according to x = y or x = y. This is not necessarily a metric unless X is a tree. However ρ a is equivalent to a metric, hence we usually regard ρ a as a (visual) metric [19]. Let X be the completion of a hyperbolic graph X under ρ a . The hyperbolic boundary of X is defined by ∂ X := X \ X which is a compact metric space under the metric ρ a . A tree is a hyperbolic graph with δ = 0 and its hyperbolic boundary is a Cantor-type set.
Let K = ∪ n i=1 S i (K ) be a self-similar set of C n,r , as in the introduction. Denote by = {1, 2, . . . , n} and * = ∪ k≥0 k where 0 = ∅ (as a root). For x = i 1 i 2 . . . i k ∈ k , write the length |x| = k and the composition of maps S x = S i 1 • · · · • S i k . Say x, y ∈ * are equivalent, denote by x ∼ y, if S x = S y . Trivially, ∼ defines an equivalence relation on * . Let X k = k / ∼ be the set of equivalence classes on level k. Then X = ∪ ∞ k=0 X k is the quotient space of * under the relation ∼. For convenience, we use [x] to express the equivalence class containing x. Occasionally, we replace [x] by x with no confusion.
There is a natural graph structure on X by the standard concatenation of indices, we denote the (vertical) edge set by E v . That is, (x, y) ∈ E v if and only if there exist x ∈ [x], y ∈ [y] and some z ∈ such that y = x z or x = y z. According to the geometric structure of K , we need to define a horizontal edge set on X : [9], and is a Gromov hyperbolic graph (due to Theorem 1.2 of [18]). Moreover, we have Proposition 2.1 The hyperbolic boundary of (X , E) is Hölder equivalent to K .
Proof According to Theorem 1.3 of [18]. It suffices to prove that the iterated function system {S i } n i=1 of K satisfies the condition (H): There exists a constant c > 0 such that for any integer k > 0 and any x, y ∈ X k , Then the shortest length of the open intervals, say c := min 0≤i≤m |I i |, is the desired constant in condition (H).
We use T D to denote the union of T and its all descendants, that is, Then T D , equipped with the edge set E restricted on T D , is a subgraph of X . For any x ∈ X , we define xT := {x y : y ∈ T }.
Let T ⊂ X k , T ⊂ X m be two horizontal components of X . We say that T and T are equivalent, denote by T ∼ T , if there exists a graph isomorphism g : T D → T D , i.e., the map g and the inverse map g −1 preserve the vertical and horizontal edges of T D and T D . Denote by [T ] the equivalence class and F the family of all horizontal components of X .

Lemma 2.2 For any two horizontal components T
Proof Without loss of generality, we may assume that T 1 = {x 1 , . . . , x m } and T 2 = {y 1 , . . . , y m } and φ • S x i = S y i where i = 1, . . . , m. Then for any finite word z ∈ * , we have It follows that for any z, w ∈ * , S x i z = S x i w if and only if S y i z = S y i w and S x i z (I ) ∩ S x i w (I ) = ∅ if and only if S y i z (I ) ∩ S y i w (I ) = ∅. That proving T 1 ∼ T 2 .

Definition 2.3 We call the graph (X , E) simple if the number of equivalence classes in
be the equivalence classes, and let a i j denote the cardinality of the horizontal components of offspring of T ∈ [T i ] that belong to [T j ]. We call A = [a i j ] d×d the incidence matrix of (X , E).
Since S x = S y may happen for x = y, the natural graph (X , E v ) may be not a tree. In that situation, one vertex of X may have multiparents that destroys the tree structure. However, we can reduce it into a tree (X , E * v ) in the following way: for a vertex y ∈ X \ {ϑ}, let x 1 , x 2 , . . . , x k be all the parents of y such that |x i | = |y| − 1 and # c t : c t = b j , t = 1, 2, . . . , p = u i j , where b j is the j-th row vector of B, 1 = [1, . . . , 1] such that the involved matrix product is well defined, and u * = The notion of the matrix rearrangeable condition (MRC) is the most important technical device in constructing the needed bi-Lipschitz map. It was introduced and systematically studied by Lau and Luo (please see a series of papers [4,10,12]). Many matrices, such as primitive matrices, satisfy the MRC.
With the above notion and Proposition 2.1, the main theorems of [10] can be reformulated as follows. A is (B, u)

Theorem 2.6 Suppose the graph (X , E) of a self-similar set K is simple and suppose the incidence matrix
is a tree, its hyperbolic boundary is a Cantor-type set. Then the self-similar set K is Lipschitz equivalent to a Cantor-type set.

The proof of Theorem 1.1
By Theorem 2.6, to prove Theorem 1.1, we only need to show that the graph (X , E) induced by any self-similar set K ∈ C n,r , where ≥ 1 is simple and the incidence matrix A is (B, u)-rearrangeable. Lemma 3.1 Let K ∈ C n,r , , ≥ 1. Then the hyperbolic graph (X , E) is simple.
Proof Obviously, 0, 1 ∈ K and K ⊂ I . Let X = ∪ ∞ k=0 X k and E = E v ∪ E h be defined as in Sect. 2. From the conditions (1), (2), (3) of K , it can be seen that any horizontal component of X is generated by either one vertex (see Fig. 1b where 1T 0 is generated by the vertex 1) or two vertices that are joined by one horizontal edge in the previous level (see Fig. 1b where Write From the condition (2) of K , there are at least two components in the first iteration ∪ i∈ S i (I ) (see Fig. 1a). Hence there are at least two horizontal components in the first level X 1 of the hyperbolic graph (X , E). We denote them by T 0 , T 1 , . . . , T k , where k ≥ 1. Without loss of generality, we may assume that 1 ∈ T 0 and n ∈ T 1 . Let For any two vertices x, y ∈ T i with x < y, if (x, y) ∈ E h , then we need to consider the following two cases: It can be seen that #T k+1 = m k+1 + k+1 + 1 where m k+1 := #(T k+1 ∩ V 1 ) and (ii) the edge (x, y) is of Type B. In the offsprings of {x, y}, we can find a new horizontal component T k+2 := xT 1 ∪ yT 0 with #T k+2 = #T 0 + #T 1 (see Figure 1(c)). All the horizontal components generated by {x, y} are xT 0 , xT 2 , . . . , xT k , T k+2 , yT 2 , . . . , yT k , yT 1 .
By similarity and Lemma 2.2, it is easy to check that yT i ∼ T i for i = 0, . . . , k. We mention that it is possible that [T i ] = [T j ] for i = j ∈ {0, 1, . . . , k + 2}, but that will not affect the finiteness of the equivalence classes.
In the third level X 3 , for i = 0, 1, . . . , k + 2, we have #T i = m i + i + 1 with m i = #(T i ∩ V 1 ) and i = #(T i ∩ V 2 ). It follows that, in T i , there are m i pairs of {x, y} so that (x, y) is of Type A and i pairs of {x, y} so that (x, y) is of Type B. By using the similar arguments as above and Lemma 2. Based on the previous arguments, there will be no new classes appearing in the offsprings of any component T ∈ [T i ] for i = 0, 1, . . . , k + 2. Therefore, (X , E) is simple and the incidence matrix A is of the form Proof By using the similar arguments as in the proof of Lemma 3.1, we shall find the incidence matrix B. For any pair of indices x, y ∈ * with x < y. If |S x (I )∩ S y (I )| = r |x|+1 then we must have S xn = S y1 and xn, y1 ∈ * are equivalent (i.e., xn ∼ y1), hence {xn, y1} forms an identifying vertex of X (= * / ∼). On the other hand, from the conditions (1), (2), (3) of the self-similar set K , we can see that any identifying vertex of X has at most two members of the form {xn, y1}. Let t 1 = ϑ be the root of X and [t 1 ] be the equivalence class containing ϑ. Due to condition (4) of K , we may assume that i 1 , . . . , i are the indices in the first level X 1 such that |S i j (I )∩ S i j +1 (I )| = r 2 , j = 1, . . . , . Hence there are identifying vertices in the second level X 2 , say x j := {i j n, (i j + 1)1}, j = 1, . . . , . By the construction of the reduced tree (X , E * v ), we need to remove the vertical edges (i j + 1, x j ) and keep (i j , x j ) in the graph (X , E v ) as i j < i j + 1. Then the number of offsprings of vertex i j + 1 is n − 1 for each j. We denote by [t 2 ] the equivalence class containing i j + 1. If we continue this progress until the third level X 3 , it can be seen that there are only two classes of vertices in (X , , and the incidence matrix follows. is (B, u)
Proof of Theorem 1.1 Let K , K ∈ C n,r , , and let X , Y be their hyperbolic graphs, respectively. By Lemma 3.2, their reduced trees are both simple with the common incidence matrix B as in (3.5), hence are graph isomorphic by Theorem 2.4. Lemma 3.1 implies that both X and Y are simple. We denote their incidence matrices by A and A , respectively. According to Lemma 3.3, A is (B, u)-rearrangeable and A is (B, u )rearrangeable where u, u are the corresponding representations. Then K and K are Lipschitz equivalent to the common Cantor-type set by Theorem 2.6. Therefore K K .
Proof of Corollary 1. 2 If K ∈ C n,r , for ≥ 1. Let (X , E) be the hyperbolic graph induced by K . Lemma 3.2 implies that the reduced tree (X , E * v ) is simple with the incidence matrix B as in (3.5). The spectral radius of B is ρ := n+ √ n 2 −4 2 . Let e 1 = [1, 0], 1 = [1, 1] and 1 t be the transpose of 1. Then the number of vertices of k-th level can be calculated by # X k = e 1 B k 1 t and lim k→∞ (e 1 B k 1 t ) 1/k = lim k→∞ (||B k || ∞ ) 1/k = ρ by Gelfand's formula. Moreover, # X k also counts the number of different pieces of K in the k-th iteration. Since the corresponding IFS {S i } n i=1 of K satisfies the weak separation property [18], by Theorem 2 in [22], we obtain dim H K = lim k→∞ log(# X k ) −k log r = log ρ − log r .
If K ∈ C n,r ,0 , there are no overlaps, then dim H K = log n − log r . Suppose K , K ∈ C n,r (= n−1 =0 C n,r , ) and K K , then they have the same Hausdorff dimension. Thus, K , K belong to the common collection C n,r , for some . The converse follows from Theorem 1.1.
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