# Intersections of Translation of a Class of Self-Affine Sets

Authors: Jian Lu; Yuru Zou; Lijing Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

Intersections of translation of a class of self-affine sets are investigated by constructing sofic affine-invariant sets which coincide with these sets and then a scheme for computing the Hausdorff dimensions of these intersections is given.


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## Body

## 1. Introduction

Fix integers2≤m≤n. Let T be the expanding endomorphism of the 2-torus 𝕋2=ℝ2/ℤ2 given by the matrix diag(n,m) and let D⊆I×J be the set of digits with I={0,1,…,n-1}, J={0,1,…,m-1}. The set KT,D is defined as follows:
(1)KT,D={∑k=1∞‍T-kdk:dk∈D∀k≥1},
which can be viewed as the unique invariant set of a family of N(=#D) affine contractions as follows:
(2)Fj(xy)=T-1((xy)+dj),dj∈D,j=1,2,…,N.
If no confusion arises about T, D, we will write K instead of KT,D. Throughout this paper, we let #D denote the cardinality of D. The set K, called McMullen’s self-affine carpet, was first studied by McMullen [1] and Bedford [2], independently, to determine its Hausdorff and box-counting dimensions. From then on, there has been a fast growth in general interest in the study of McMullen’s self-affine carpet K. Its packing and Hausdorff measures were explored in [3, 4]. The results of [1, 2] were extended to the compact subsets of the 2-torus corresponding to shifts of finite type or sofic shifts and to the Sierpinski sponges by Kenyon and Peres [5, 6]. Gui and Li [7] studied a class of subset of Sierpinski carpets for which the allowed digits in the expansions fall into each fiber set with a prescribed frequency. Hua [8] investigated a class of self-affine sets with overlaps and showed that they can be viewed as sofic affine invariant sets without overlaps.Intersection of Cantor sets has been the subject of several studies [9–19], the context and motivation being numerous. Davis and Hu [12] explored the Hausdorff dimension of the intersection of two middle third Cantor sets and showed that the Hausdorff dimension can take any value from 0 to ln2/ln3. Williams [10] produced examples to show that the intersection of two Cantor sets can vary from being one point to containing another Cantor set. Li and Xiao [13] showed that intersection resulting from two middle-α sets being translated across one another is a generalized Moran set if the contraction lies in [0,1/3], whose Hausdorff, packing, upper Box dimension then can be obtained by the results of the Moran set. Nekka and Li [14, 15] studied the properties of intersection of Cantor sets with their translations. Dai and Tian [16, 17] got the related properties of a class of self-similar sets in the plane.In this paper, we develop a new algorithm to explore the structure ofK∩(K+t) for t∈K-K, which extends our previous results in [18] and is different from the listed papers. It is easy to see that K∩(K+t)≠∅ if and only if t∈K-K; so throughout this paper, we always suppose that t∈K-K. The intersection, generally, has a complicated structure which leads the set quite different from McMullen’s self-affine carpet. The main difficulty in the study of the structure of the intersection comes from the fact that the number of affine contractions used in the construction may vary from step to step. However, by defining a suitable equivalence relation on the basic rectangles of K∩(K+t) and partitioning them into equivalence classes, a sofic affine-invariant set coinciding with K∩(K+t) then can be constructed, if the number of equivalence classes is finite. Our idea of the construction of the sofic affine-invariant set is enlightened by [8, 20–22].To state our main result, we need to recall some definitions and results about the sofic affine-invariant set. For details, one can refer to [6, 9].LetG=(𝒱,ℰ) be a finite directed graph in which loops and multiple edges are allowed. Let D0={0,1,…,n-1}×{0,1,…,m-1} be the set of symbols. Suppose that the edges of G are labeled in symbols in D0 in a right resolving fashion: no two edges emanating from the same vertex have the same symbol. Then the symbol sequences which arise from infinite paths in G form a sofic system on #D0 symbols. That is,
(3)Θ={{dk}k≥1:{dk}k≥1fisaninfinitepathinG,wheredk∈D0{dk}k≥1}.
Suppose that Θ⊂D0ℕ is the resulting sofic system. We call
(4)ΓT(Θ)={∑k=1∞T-kdk:{dk}k≥1∈Θ}
a T-invariant sofic set. An adjacency matrix A, according to G=(𝒱,ℰ), is constructed, where, for two vertices v1, v2 in 𝒱, A(v1,v2) is the number of edges in ℰ from v1 to v2.Theorem 1.
Suppose thatt=(p1/q1,p2/q2)∈ℚ2. Then
(5)K∩(K+t)=ΓT(Θ).To compute the Hausdorff dimension ofK∩(K+t), the following result in [6] is needed.Kenyon-Peres’s Theorem(see [6, Theorem  3.2]).  LetG=(𝒱,ℰ)be a finite directed graph with edges labeled byD0and adjacency matrixA. LetΘ⊂D0ℕbe the resulting sofic system. Then the Hausdorff dimension ofΓT(Θ)is given by(6)dimH(ΓT(Θ))=limk→∞logm∑0≤i1,…,ik≤m-1‍∥Aik·Aik-1·…·Ai1∥α,where, for0≤j<mand verticesv1,v2inG,Aj(v1,v2)denotes the number of edges fromv1tov2inGsuch that the second coordinate of their label isj,α=logm/logn.This paper is arranged as follows. A description for equivalence classes onK∩(K+t) is arranged in Section 2. A sofic affine-invariant set coinciding with K∩(K+t) is then constructed in Section 3, if the number of equivalence classes is finite (see Theorem 6). A sufficient condition for the number of equivalence classes to be finite is also established in Section 3 (see Proposition 7). Finally, the proof of the Theorem 1 is completed in this section. Some examples are included in Section 4.


## 2. Equivalence Classes onK∩(K+t)

In this section we consider the geometrical structure ofK∩(K+t)(≠∅). First we define a suitable equivalence relation on the basic rectangles of K∩(K+t) and partition them into equivalence classes.Let{Fj}j=1N be defined as in (2); simple computation yields
(7)K+t=⋃j=1N‍(T-1(K+t)+T-1dj+t-T-1t).
Then define
(8)Sj(xy)=T-1((xy)+dj+Tt-t),j=1,…,N,
such that K+t can be viewed as the unique invariant nonempty compact set under mappings Sj,j=1,…,N. Let O=t+(0,1)2, {Sj}j=1N, obviously, satisfy the OSC. Set
(9)D*={dj+Tt-t,dj∈D}.In order to analyze the structure ofK∩(K+t), we get the following standard notations.LetΣ={1,…,N} and Σ*=∪p≥0Σp be the set of all finite words, where Σp is the set of all words of length p, with Σ0 containing only the empty word ∅. Σℕ={i1i2⋯:in∈Σforall1≤n<∞}. For J∈Σp, let |J|=p denote the length of J. For I∈Σp and J∈Σq, let IJ be the concatenation of I and J. For I=i1i2⋯∈Σℕ and a positive integer k, let I∣k=i1⋯ik denote the truncation of I to the kth place.We define the iterated mappings using these notations. LetJ=(j1,…,jk)∈Σk. Then
(10)FJ=Fj1∘Fj2∘⋯∘Fjk,SJ=Sj1∘Sj2∘⋯∘Sjk
with F∅=S∅=I, the identity map on ℝ2. Thus we get
(11)K=⋂k=1∞‍⋃J∈Σk‍FJ(Ω),K+t=⋂k=1∞⋃J∈Σk‍SJ(Ωt),
where Ω=[0,1]2 and Ωt=t+[0,1]2. We call FJ(Ω), SJ(Ωt) the basic rectangles of rank k, respectively. For simplicity without confusion, we further call thembasick-rectangles and we write FJ(Ω)=ΩJ and SJ(Ωt)=ΩJt in this paper.Suppose that there existI∈Σk, J∈Σk+1,  and ℓ∈Σ1 such that
(12)τ1=ΩI,τ2=ΩJ,J=Iℓ.
Then we connect a directed edge ℓ from τ1 to τ2 and denote it by τ1→ℓτ2. We call τ1 the parent of τ2 and τ2 an offspring of τ1. Definitions of parent and offspring can be parallel extended to K+t.Following we investigate the relationships between the basick-rectangles of K and K+t.Suppose thatΩJ, J∈Σk, k≥0; the neighborhoodof ΩJ with respect to the basic k-rectangles of K+t is defined as
(13)N(ΩJ)={ΩIt:I∈Σk,ΩJ∩ΩIt≠∅}.
Let Λk={J∈Σk:N(ΩJ)≠∅,k≥0}. Note that, for some ΩJ, J∈Σk, N(ΩJ) may be empty; namely, these basic k-rectangles do not meet any basic k-rectangles of K+t, and then they have nothing to do with the intersection K∩(K+t), which will be deleted. So let
(14)ℳk={ΩJ,J∈Λk}.The basick-rectangles of K and K+t are completely determined by their left-lower corner points. For this point, if τ∈ℳk we define
(15)Rk(τ)={Tk(δ^-τ^):δ∈N(τ)},(16)R={Rk(τ):τ∈ℳk,k≥0},
where δ^, τ^ are the left-lower corner points of δ, τ, respectively. For τ1∈ℳk1 and τ2∈ℳk2, we say that τ1 and τ2 are equivalent, if Rk1(τ1)=Rk2(τ2). Next we prove a recurrent relationship on R.Letτ1=ΩI and τ2=ΩJ, where I∈Σk1, J∈Σk2. Suppose that Rk1(τ1)=Rk2(τ2) and set
(17)N(τ1)={v1,…,vt},N(τ2)={u1,…,ut},
such that Tk1(v^i-τ^1)=Tk2(u^i-τ^2), 1≤i≤t. Let σj1, σj2 be the offsprings of τ1 and τ2 connected by edge j∈Σ1, respectively, and denote them by τ1→jσj1, τ2→jσj2. Similarly, vki, uki are the offsprings of vi and ui connected by edge k∈Σ1, respectively. That is vi→kvki, ui→kuki.Lemma 2.
With above notations, letg:ℝ2→ℝ2 be an affine mapping,
(18)g(x)=Tk1-k2x+τ^2-Tk1-k2τ^1,
where τ^1, τ^2 are the left-lower corner points of τ1 and τ2, respectively. Then σj2=g(σj1). If Rk1(τ1)=Rk2(τ2), then ui=g(vi), i=1,…,t. Furthermore uki=g(vki).Proof.
Note that(19)σj1=T-k1-1(Ω)+τ^1+djT-k1-1,σj2=T-k2-1(Ω)+τ^2+djT-k2-1,
where dj∈D. Then
(20)g(σj1)=T-k2-1(Ω)+Tk1-k2τ^1+djT-k2-1+τ^2-Tk1-k2τ^1=σj2.
We now prove that ui=g(vi), i=1,…,t. τ^2-Tk1-k2τ^1=u^i-Tk1-k2v^i is the direct conclusion from Rk1(τ1)=Rk2(τ2), which yields ui=g(vi), i=1,…,m. uki=g(vki) can be easily obtained by the same argument as σj2=g(σj1).Proposition 3.
With above notations, ifRk1(τ1)=Rk2(τ2), then, for j∈Σ1, σj1∈ℳk1+1 if and only if σj2∈ℳk2+1. If σj1∈ℳk1+1, then Rk1+1(σj1)=Rk2+1(σj2).Proof.
By Lemma2, we have
(21)vki∩σj1≠∅⟺g(vki)∩g(σj1)=uki∩σj2≠∅,
so σj1∈ℳk1+1⇔σj2∈ℳk2+1.
One can check that(22)N(σj1)={vki:vi∈N(τ1),vki∩σj1≠∅},N(σj2)={uki:ui∈N(τ2),uki∩σj2≠∅}.
By Lemma 2 and (21), (22) can be rewritten as
(23)N(σj2)={g(vki):ui∈N(τ2),uki∩σj2≠∅}={g(vki):vi∈N(τ1),vki∩σj1≠∅}={g(vki):vki∈N(σj1)}.
Observe that
(24)σ^j1=τ^1+djT-k1-1,σ^j2=τ^2+djT-k2-1,v^ki=v^i+ckT-k1-1+T-k1t(T-1-1),u^ki=u^i+ckT-k2-1+T-k2t(T-1-1),
where ck∈D* (defined as in (9)). By condition Rk1(τ1)=Rk2(τ2), we obtain
(25)Tk2+1(u^ki-σ^j2)=Tk1+1(v^ki-σ^j1).LetR be defined as in (16); if R is a finite set, we then call that the types of the basic rectangles of K∩(K+t) is finite.We remark that, fort∈K-K, if the types of the basic rectangles of K∩(K+t) is recursive, such as type A produces types B, and C, B produces D, C produces B and finally D produces A, then by Proposition 3, one can claim that the types of the basic rectangles of K∩(K+t) are finite.From (11), and the definition of Λk, we have
(26)K∩(K+t)=(⋂k=1∞⋃J∈Σk‍ΩJ)⋂‍(⋂k=1∞⋃I∈Σk‍ΩIt)=⋂k=1∞‍Πk,
where Πk=(⋃J∈Λk‍ΩJ)⋂‍(⋃I∈Σk‍ΩIt).Proposition 4.
(27)
K
∩
(
K
+
t
)
=
⋂
k
=
1
∞
⋃
J
∈
Λ
k
‍
Ω
J
.Proof.
It is easy to see thatΠk⊂⋃J∈Λk‍ΩJ; therefore
(28)K∩(K+t)⊆⋂k=1∞‍⋃J∈Λk‍ΩJ.
To prove the reverse inclusion, we let ε/2>0; then for k1, k2 large enough, such that |ΩJ|≤ε/2 for J∈Λk1, and Πk2⊆[K∩(K+t)]ε/2 (|A| denotes the diameter of A and [A]ε the open ε-neighborhood of A). Let k=max{k1,k2}, then max{|ΩJ′|,J′∈Λk}≤ε/2 (J′ always exists; otherwise K∩(K+t)=∅). On the other hand ΩJ′∩Πk≠∅, then ΩJ′⊆[K∩(K+t)]ε. Hence
(29)⋂k=1∞‍⋃J′∈Λk‍ΩJ′⊆[K⋂‍(K+t)]ε.
The desired result is then obtained by letting ε tend to 0.For someJ∈Λk, it is possible that, for all i∈Σ1, Ji∉Λk+1; namely, all offsprings of ΩJ do not belong to ℳk+1. So set
(30)Λ={J=(j1,j2,…,jk,…):fforanyk≥0,J∣k=(j1,j2,…,jk)∈Λk}.Remark 5.
We remark that, by Proposition4, K∩(K+t) can also be expressed in radix expansions as follows:
(31)K∩(K+t)=limk→∞∑i=1k‍T-idli,
where dli∈D, I=(l1,…,lk,…)∈Λ.


## 3. Sofic Affine-Invariant Set

Suppose thatR is a finite set; let 𝒱=R={Rk(τ):τ∈ℳkk≥0}. Set 𝒱={ω,ω1,…,ωQ} and R0(Ω)=ω. We now define the directed edges ℰ={Ei,j}i,j=1Q on 𝒱. Fix i (1≤i≤Q) and a vertex τ=ΩI∈ℳk such that Rk(τ)=ωi. Let δℓ=ΩIℓ in ℳk+1 be an offspring of τ connected by edge ℓ∈Σ1. Consider
(32)Ei,j={dℓ:τ⟶ℓδℓ,Rk+1(δℓ)=ωj,dℓ∈D}.
Note that if none of the offsprings of τ belongs to ℳk+1, then Ei,j=∅, for all j. We discard those ωi’s that there is no path going out and discard all the paths going to such discarded vertices. Hence, without loss of generality, we assume, that for each i,
(33)∑j=1Q‍#Ei,j≥1.
A finite directed graph G=(𝒱,ℰ) then is obtained. Let
(34)Θ={{dk}k≥1:{dk}k≥1fisaninfinitepathinG,wheredk∈D{dk}k≥1}.
Then Θ⊂Dℓ is the resulting sofic system and
(35)ΓT(Θ)={∑k=1∞‍T-kdk:{dk}k≥1∈Θ}
is a sofic invariant set.Theorem 6.
IfR is a finite set, then
(36)K∩(K+t)=ΓT(Θ).Proof.
By Remark5, K∩(K+t) can be expressed as
(37)K∩(K+t)=limk→∞∑i=1k‍dliT-i,
where dli∈D, I=(l1,…,lk,…)∈Λ. By the definition of Λ in (30) and the construction of G=(𝒱,ℰ), one can claim that dl1⋯dlk⋯ is an infinite path in G and then belongs to Θ. Thus K∩(K+t)⊆ΓT(Θ). The inverse inclusion is left for the readers.Following we give some sufficient condition for the setR (defined as in (16)) to be finite.Proposition 7.
Suppose thatt=(p1/q1,p2/q2)∈ℚ2. Then the set R is finite.Proof.
Suppose thatτ∈Mk, δ∈N(τ). τ^, δ^ denote the left-lower corner points of τ, δ respectively; then
(38)τ^=∑i=1k‍dℓiT-i,
where dℓi∈D;
(39)δ^=∑i=1k‍cℓiT-i+T-kt,
where cℓi∈D*. Consider
(40)Tk(δ^-τ^)=Tk(∑i=1k‍(cℓi-dℓi)T-i+T-kt)=1q(∑i=1k‍(eℓi+pT-p)Tk-i+p),
where p=(qp1/q1,qp2/q2); q is the least common multiple of q1 and q2. eℓi∈q(D-D), qD:={qdj,dj∈D}. So the coordinates of qTk(δ^-τ^) are integers. On the other hand τ∩δ≠∅ implies that
(41)Tk(δ^-τ^)∈[-n,n]×[-m,m].
Consequently,
(42)qTk(δ^-τ^)∈[-qn,qn]×[-qm,qm].
Then, for any τ∈Mk, Rk(τ)⊂(([-qn,qn]×[-qm,qm])∩ℤ2)/q. So the set R is finite.Proof of Theorem1.
It is a direct conclusion from Theorem6 and Proposition 7.


## 4. Examples

Example 1.
LetT=(4003) and D={d1,d2,d3,d4}, where d1=(00), d2=(30), d3=(11), and d4=(32).t=(1/3,1/4).By Proposition7, the set R is finite. The types of the basic rectangles can be determined by (15). It is clear that R0(Ω)=t, denoted by A. For k=1, among four offsprings of Ω, only Ω3 and Ω4 have nonempty neighborhood, which are connected by edges (1,1) and (3,2) from their parent Ω, respectively. By (15), they are of different types; denote them symbolically as
(43)A⟶(1,1)B,A⟶(3,2)C.
For k=2, direct computation yields
(44)Ω32,Ω41,Ω43∈ℳ2.
They are of new types and Ω32, Ω43 are of the same type. So we have
(45)B⟶(3,0)D,C⟶(1,1)D,C⟶(0,0)E.
Above computation process can also be seen from Figure 1. Upon the third iteration, none of the offsprings of Ω41 belongs to ℳ3; then type E and all paths going to E are discarded.Figure 1
Construction ofK∩K+t. Red and blue rectangles denote the construction of K and K+t, respectively.By the same argument as above, we have(46)D⟶(0,0)F,D⟶(1,1)C.
However none of the offsprings of Ω321, whose type is of F, belongs to ℳ4. Then type F and all paths going to F are deleted. By Proposition 3, the above process exhausts all possible types and yields the adjacency matrix
(47)S=(0110000100010010),
and by Kenyon-Peres’s Theorem, we have
(48)S0=(0000000100000000),S1=(0100000000010010),S2=(0010000000000000).Example 2.
LetT=(5004) and D={d1,d2,d3,d4}, where d1=(00), d2=(40), d3=(21), and d4=(32). t=(12/25,5/16).Again by Proposition7, the set R is finite. First, R0(Ω)=t; denote it by A. For k=1, Ω3,Ω4∈ℳ1, and N(Ω3)={Ω1t}, N(Ω4)={Ω1t}. They are of different types and then denoted by
(49)A⟶(2,1)B,A⟶(3,2)C.
For k=2, Ω33,Ω34,Ω41∈ℳ2, and N(Ω33)={Ω11t}, N(Ω34)={Ω11t,Ω13t}, N(Ω41)={Ω14t}. Their types are of D, E, and F, respectively. Thus we have
(50)B⟶(2,1)D,B⟶(3,2)E,C⟶(0,0)F.
Upon the third iteration, Ω33 produces four offsprings in ℳ3, which yields
(51)D⟶(0,0)D,D⟶(2,1)G,D⟶(3,2)H,D⟶(4,0)D.
Only one offspring of Ω34, whose type is of E, belongs to ℳ3; then we get
(52)E⟶(4,0)I.
None of the offsprings of Ω41, whose type is F, belongs to ℳ3; then type F and all paths going to F are discarded. For k=4, by the same argument as above, we have
(53)G⟶(0,0)D,G⟶(2,1)G,G⟶(3,2)H,G⟶(4,0)D,H⟶(0,0)D,H⟶(2,1)D,H⟶(3,2)H,H⟶(4,0)D,I⟶(4,0)I.
Figure 2 can avail to understand the above computation process. By Proposition 3, these are all the types of the intersection and the adjacency matrix is
(54)S=(0100000001100000201100000001002011000201100000001).Figure 2
Construction ofK∩K+t. Red and blue rectangles denote the construction of K and K+t, respectively.Also by Kenyon-Peres’s Theorem, we get(55)S0=(0000000000000000200000000001002000000200000000001),S1=(0100000001000000001000000000000010000001000000000),S2=(0000000000100000000100000000000001000000100000000),S3=(0000000000000000000000000000000000000000000000000).



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