# Assouad and Lower Dimensions of Some Homogeneous Cantor Sets

Authors: Xiao JiaQing
Journal: Journal of Mathematics
 (2022)

---

## Abstract

We compute the Assouad dimensions and the lower dimensions of a class of homogeneous Cantor sets without the condition that the smallest compression ratioC⋆>0 and find that the lower dimension of a homogeneous Cantor set E may be any a number in the interval 0,dimHE and the Assouad dimension of E may be any a number in the interval dim¯BE,1.


---

## Body

## 1. Introduction

Let us begin with the definition of the Assouad dimension and the lower dimension. The Assouad dimension of a nonempty setF⊆ℝd is defined bydimAF=infα≥0:there exists a constantc>0such that(1)for any0<r<R,andx∈F,NrBx,R∩F≤cRrα.The lower dimension of F is defined bydimLF=supα≥0:there exists a constantc>0such that(2)for any0<r<R≤F,andx∈F,NrBx,R∩F≥cRrα.NrBx,R∩E is the smallest number of open sets with a diameter less than r needed to cover Bx,R∩E.The Assouad dimension which is introduced by Assouad [1] has recently received an enormous interest in the mathematical literature due to its connections with the doubling property. The dual notion of the Assouad dimension is the lower dimension which was introduced by Larman [2]. Just like the Assouad dimension, the lower dimension has also received an enormous interest in the mathematical literature due to its connections with the uniform property of metric spaces. As a result of this, a large number of papers have investigated the Assouad dimension and the lower dimension of different classes of fractal sets. Fraser [3] has a detailed discussion of the Assouad dimension, the lower dimension, and its use in fractal geometry. Olsen [4] computed the Assouad dimension of a graph-directed Moran fractal satisfying the open-set conditions which are Ahlfors regular. However, in general, it is difficult to obtain the Assouad dimensions of sets which are not Ahlfors regular. Mackay [5] calculated the Assouad dimension of the self-affine carpets of Bedford and McMullen, and his main result solved the problem posed by Olsen [4]. For the Moran sets introduced by Wen [6] which are not Ahlfors regular, Li et al. [7] obtained the Assouad dimensions of Moran sets under suitable condition. Li [8] also proved that the Assouad dimensions of some Moran sets coincide with their packing and upper box dimensions under the condition that the smallest compression ratio C⋆>0 and therefore gave a conjecture that the conclusion remains true if the condition C⋆>0 is removed. Xiao [9] proved that the lower dimensions of a class of Moran sets coincide with their Hausdorff dimensions under the condition that the compression sequence and the compression ratio sequence are smoothly changing and found some homogeneous Cantor sets whose Assouad dimension is not equal to their upper box dimensions and packing dimensions and therefore gave a negative answer to the conjecture in the paper [8]. In this paper, we compute the Assouad dimensions and the lower dimensions of a class of homogeneous Cantor sets without the condition that the smallest compression ratio C⋆>0 and find that the lower dimension of a homogeneous Cantor set E may be any a number in the interval 0,dimhE and the Assouad dimension of E may be any a number in the interval dim¯BE,1.


## 2. Homogeneous Cantor Set

Firstly, let us recall the definition of Moran sets introduced by Wen [5]. Let nkk≥1⊂ℕ be a sequence of positive integer (we assume nk≥2). Define Do=∅, and for any k≥1, set Dm,k=σmσm+1⋯σk;1≤σj≤nj,m≤j≤k, Dk=D1,k and D=∪k≥0Dk. If σ=σ1σ2⋯σk∈Dk and τ=τ1τ2⋯τm∈Dk+1,m, we define σ∗τ=σ1⋯σkτ1⋯τm.Definition 1.
Suppose thatJ⊂ℝd is a compact set with intJ¯=J. Let Φk be a sequence of positive real vectors with Φk=ck,1,ck,1,…,ck,nk, ∑j=1nkck,j≤1,k∈ℕ. We say the collection ℱ=Jσ;σ∈D of closed subsets of J possesses the Moran structure if it satisfies the following Moran structure conditions (MSC):(1)
Forσ∈D, Jσ is geometrically similar to J; that is, there exists a similarity Sσ:ℝd⟶ℝd such that Jσ=SσJ. For convenience, we write J∅=J(2)
For allk≥0 and σ∈Dk,Jσ∗1,Jσ∗2,…,Jσ∗nk+1 are subsets of Jσ and satisfy that intJσ∗i∩intJσ∗j=∅ whenever i≠j(3)
For anyk≥1 and σ∈Dk−1,1≤j≤nk,(3)Jσ∗jJσ=ck,j,where A denotes the diameter of A.
Suppose thatℱ=Jσ:σ∈D is a collection of closed subsets of J possessing the Moran structure, set(4)Ek=∪σ∈DkJσ,E=∩k≥0Ek.
It is ready to see thatE is a nonempty compact set. The set E:=Eℱ is called the Moran set associated with the collection ℱ.
Letℱk=Jσ;σ∈Dk, and ℱ=∪k≥0ℱk. The elements of ℱk are called kth-level basic sets of E, and the elements of ℱ are called the basic sets of E. Suppose the set J and the sequences nk,Φk are given. We denote by ℳ=ℳJ,nk,Φk the class of the Moran sets satisfying the MSC. We call ℳJ,nk,Φk the Moran class associated with the triplet J,nk,Φk.Definition 2.
SupposeJ is the interval [0,1] and ck,j=ck for any k>0,1≤j≤nk in Definition 1. For all k>0,σ∈Dk,distJσ⋆j,Jσ⋆j+1=distJσ⋆j+1,Jσ⋆j+21≤j≤nk+1−2, and the left endpoint of Jσ⋆1 is the left endpoint of Jσ and the right endpoint of Jσ⋆nk+1 is the right endpoint of Jσ. The set E∈ℳJ,nk,ck is called the homogeneous Cantor set. Write E=CJ,nk,ck.Remark 1.
LetC⋆:=infck. In the present paper, the author obtains formulas for the Assouad dimension and the lower dimension of sets belonging to this class of Moran fractals without assuming the condition C⋆>0.Theorem 1 (see [10]).
Suppose the setE=CJ,nk,ck is a homogeneous Cantor set. Then,(5)dimHE=liminfk⟶∞logn1n2⋯nk−logc1c2⋯ck,dimPE=dim¯BE=limsupk⟶∞logn1n2⋯nk+1−logc1c2···ck+lognk+1,where dimHE, dimPE, and dim¯BE denote the Hausdorff, packing, and upper box dimensions of E, respectively.


## 3. Statement and Proof of Results

Now, consider four conditions thatnk and ck satisfy.Condition 1.
liminfknkck>0 and there are two strictly increasing sequence lii and kii of positive integers with li≤ki such that(a)
limicli+1⋯cki=0(b)
The following limit(6)s¯=limi⟶∞lognli+1⋯nki−logcli+1⋯cki,exists (note thats¯ is defined as the value of the limit in (6))(c)
For eachε>0, there is an r0 (only depending on ε) such that(7)lognl+1⋯nk−logcl+1⋯ck>s¯−ε,for each pair of positive integers l and k with l≤k and cl+1⋯ck<r0.Condition 2.
liminfknkck=0.Condition 3.
limsupknk<+∞ and there are two strictly increasing sequence lii and kii of positive integers with li≤ki such that(a)
limicli+1⋯cki=0(b)
The following limit(8)s¯=limi⟶∞lognli+1…nki−logcli+1…cki,exists (note thats¯ is defined as the value of the limit in (8))(c)
For eachε>0, there is an r0 (only depending on ε) such that(9)lognl+1⋯nk−logcl+1⋯ck<s¯+ε,for each pair of positive integersl and k with l≤k and cl+1⋯ck<r0.Condition 4.
limsupknk=+∞Theorem 2.
Suppose the setE=CJ,nk,ck is a homogeneous Cantor set. Then,(10)dimLE=s¯,if Condition1is satisfied;0,if Condition2is satisfied.dimAE=s¯,if Condition3is satisfied;1,if Condition4is satisfied.Proof .
of Theorem2. For two positive numbers r and R with r<R≤1, there exist l,k∈ℕ and l≤k such that(11)c1c2⋯cl+1≤R<c1c2⋯cl,c1c2⋯ck+1≤r<c1c2⋯ck.
Forσ∈Dk, we consider the relationship between r and NrJσ∩E. Notice that(12)c1c2⋯ck+1+c1c2⋯ck1−nk+1ck+1nk+1−1=c1c2⋯ck1−ck+1nk+1−1,where c1c2⋯ck+1 is the length of the basic elements of order k+1 of the homogeneous Cantor set E, and c1c2⋯ck1−nk+1ck+1/nk+1−1 is the length of the interval among the basic elements of order k+1 of the homogeneous Cantor set E. By a simple calculation, we obtain the relationship between r and NrJσ∩E. The results are shown in Table 1. where 1≤q≤nk+1−1.
Secondly, letx∈E, we consider the relationship between R and #τ∈Dl+1|Bx,R∩E∩Jτ≠∅. Denote #τ∈Dl+1|Bx,R∩E∩Jτ≠∅ by Bτ. By a simple calculation, we obtain the relationship between R and Bτ. The results are shown in Table 2. where 1≤p≤nl+1−1.
Next, we consider the relationship betweenNrBx,R∩E and R/r. Notice that(13)14BτNrJσ∩Enl+2⋯nk≤NrBx,R∩E≤BτNrJσ∩Enl+2⋯nk.
By a simple calculation, we obtain the relationship betweenNrBx,R∩E and R/r. The results are shown in Tables 3 and 4.
Using (13) and Tables 1–3, we obtain(14)14pnk+1qnl+2⋯nk≤NrBx,R∩E≤2pnk+1+1qnl+2⋯nk≤4pnk+1qnl+2⋯nk,
For entry 1 in Table3, by a simple calculation, we obtain(15)−2lg2+logp−logq+lognl+2⋯nk+1logp−11−cl+1/nl+1−1+cl+1−logq−11−ck+1/nk+1−1+ck+1−logcl+1⋯ck≤logNrBx,R∩ElogR/r≤2lg2+logp−logq+lognl+2⋯nk+1logp−11−cl+1/nl+1−1+cl+1−logq−11−ck+1/nk+1−1+ck+1−logcl+1⋯ck,for q≠0 and p≠0.
In order to simplify expression, the notationA≍B is introduced. If there exist two constant positive numbers a1,a2 such that a1A≤B≤a2A we remark A≍B. Denote logNrBx,R∩E/logR/r by α.
Using (13) and Tables 1 and 2, one gets(16)14pnk+1q+1nl+2⋯nk≤NrBx,R∩E≤4p+1nk+1qnl+2⋯nk,for 1≤p≤nl+1−1 and 1≤q≤nk+1−1. It follows that NrBx,R∩E≍pnk+1/qnl+2⋯nk where the constants a1=1/8 and a2=8 be chosen uniformly in r,R and x. By a simple calculation that is similar to entry 1 in Table 3, we obtain Table 4.
Notice thatlogp/logp1−cl+1/nl+1−1+cl+1 is decreasing for 1≤p≤cl+1−1 and logq/logq1−ck+1/nk+1−1 is decreasing for 1≤q≤ck+1−1, using (16) and Table 4, we obtain that(17)−3log2−lognk+1−1+lognl+2⋯nk+1log1−cl+1−lognl+1−1−logcl+1⋯ck≤α≤3log2+lognl+1−1+lognl+2⋯nk+1−log1−ck+1+lognk+1−1−logcl+1⋯ck.
If Condition1 is satisfied, we prove dimLE=s¯. Using (c) in Condition 1 and (17) and notice that liminfknkck>0, it follows that for all small ε>0, there exists r0>0 such that(18)NrBx,R∩E=Rrα≥Rr−3log2−lognk+1−1+lognl+2⋯nk+1/log1−cl+1−lognl+1−1−logcl+1⋯ck>Rrs¯−ε,for cl⋯ck+1<r0. If cl⋯ck+1≥r0, notice that R/r≤1/cl⋯ck+1≤1/r0,(19)NrBx,R∩E≥1=Rr−s¯Rrs¯≥1r0−s¯Rrs¯.
By combining (18) and (19), we obtain that dimLE≥s¯. To prove that dimLE≤s¯, it suffices to find a constant c>0 and a sequence of points xi∈E and scales 0<ri<Ri such that Ri/ri⟶∞ and for all i(20)NriBxi,R∩E≤cRiris¯+ε,for all small ε>0. Recall that there exists a sequence lognli+1⋯nki/−logcli+1⋯ckii whose limit is s¯. Take xi=x∈E,Ri=c1⋯cli+1 and ri=c1⋯cki+1, there exists i0∈ℕ such that(21)NriBx,Ri∩E≤2nli+2⋯nki+1=2Ririlognli+2⋯nki+1/−logcli+2⋯cki+1≤2Riris¯+ε,for all i>i0. The proof of dimLE=s¯ is finished.
If Condition2 is satisfied, there exists a subsequence ki such that limi⟶+∞nkicki=0. Let xi be the left endpoint of some basic elements of order ki, take Ri=c1c2···cki+1+c1c2···cki1−nki+1cki+1/nki+1−1 and ri=c1c2···cki+1. It is obvious that(22)NrBxi,Ri∩E≤4=4Riri0.
Notice thatRi/ri=1−cki+1/nki+1−1cki+1⟶∞ when i⟶∞; therefore, dimLE=0.
Next, we provedimAE=s¯. Using (c) in Condition 3 and (17), and notice that limsupknk<+∞, it follows that for all small ε>0, there exists r0>0 such that(23)NrBx,R∩E=Rrα≤Rr3log2+lognl+1−1+lognl+2⋯nk+1/−log1−ck+1+lognk+1−1−logcl+1⋯ck<Rrs¯+ε,for cl⋯ck+1<r0. If cl⋯ck+1≥r0. Notice that R/r≤1/cl⋯ck+1≤1/r0,(24)NrBx,R∩E≤Rr=Rr1−s¯Rrs¯≤1r01−s¯Rrs¯.
By combining (23) and (24), we obtain that dimAE≤s¯. To prove that dimAE≥s¯, it suffices to find a constant c>0 and a sequence of points xi∈E and scales 0<ri<Ri such that Ri/ri⟶∞ and for all i(25)NriBxi,R∩E≥cRiris¯−ε,for all small ε>0. Recall that there exists a sequence lognli+1⋯nki/−logcli+1⋯ckii whose limit is s¯. Take xi=x∈E,Ri=c1⋯cli+1 and ri=c1⋯cki+1, there exists i0∈ℕ such that(26)NriBx,Ri∩E≥nli+2⋯nki+1=Ririlognli+2⋯nki+1/−logcli+2⋯cki+1≥Riris¯−ε,for all i>i0. The proof of dimAE=s¯ is finished.
Finally, we provedimAE=1 if limsupknk<+∞. There exist a subsequence ki such that limi⟶+∞nki=+∞. Let xi be the left endpoint of some basic elements of order ki, take Ri=c1c2···cki and ri=c1c2···cki1−cki+1/nki+1−1. It is obvious that(27)NrBxi,Ri∩E≥nki+1≥1−cki+1nki+1nki+1−1Riri≥12Riri.
Notice thatRi/ri=nki+1−1/1−cki+1⟶∞ when i⟶∞; therefore, dimAE=1.Table 1
The calculation ofNrJσ∩E.
rNrJσ∩Ec1c2⋯ck+1,c1c2⋯ck1−ck+1/nk+1−12nk+1+1≥NrJσ∩E≥nk+1c1c2⋯ck1−ck+1/nk+1−1,c1c2⋯ck1−ck+1/nk+1−1+c1c2⋯ck+12nk+1+1≥NrJσ∩E≥1/2nk+1c1c2⋯ck1−ck+1/nk+1−1+c1c2⋯ck+1,2c1c2⋯ck1−ck+1/nk+1−121/2nk+1+1≥NrJσ∩E≥1/2nk+1⋯⋯q−1c1c2⋯ck1−ck+1/nk+1−1+c1c2⋯ck+1,qc1c2⋯ck1−ck+1/nk+1−121/qnk+1+1≥NrJσ∩E≥1/qnk+1qc1c2⋯ck1−ck+1/nk+1−1,qc1c2⋯ck1−ck+1/nk+1−1+c1c2⋯ck+121/qnk+1+1≥NrJσ∩E≥1/q+1nk+1⋯⋯[nk+1−1c1c2⋯ck1−ck+1/nk+1−1,c1c2⋯ck)2Table 2
The calculation ofBτ.
RBτc1c2⋯cl+1,c1c2⋯cn1−cl+1/nl+1−11≤Bτ≤2c1c2⋯cl1−cl+1/nl+1−1,c1c2⋯cl1−cl+1/nl+1−1+c1c2⋯cl+12≤Bτ≤4c1c2⋯cl1−cl+1/nl+1−1+c1c2⋯cl+1,2c1c2⋯cl1−cl+1/nk+1−13≤Bτ≤6⋯⋯p−1c1c2···cl1−cl+1/nl+1−1+c1c2⋯cl+1,pc1c2⋯cl1−cl+1/nl+1−1p≤Bτ≤2ppc1c2⋯cl1−cl+1/nl+1−1,pc1c2⋯cl1−cl+1/nl+1−1+c1c2⋯cl+1p+1≤Bτ≤2p+1⋯⋯nl+1−1c1c2⋯cl1−cl+1/nl+1−1,c1c2⋯clnl+1≤Bτ≤2nl+1Table 3
The calculation ofR/r.
RrR/r1p−1c1⋯cl1−cl+1/nl+1−1+c1⋯cl+1q−1c1⋯ck1−ck+1/nk+1−1+c1⋯ck+1p−11−cl+1/nl+1−1+cl+1/q−11−cl+1/nk+1−1+ck+1cl+1⋯ck2p−1c1⋯cl1−cl+1nl+1−1+c1⋯cl+1qc1⋯ck1−ck+1nk+1−1p−11−cl+1/nl+1−1+cl+1/q1−ck+1/nk+1−1cl+1⋯ck3pc1⋯cl1−cl+1/nl+1−1q−1c1⋯ck1−ck+1/nk+1−1+c1⋯ck+1p1−cl+1/nl+1−1/q−11−ck+1/nk+1−1+ck+1cl+1⋯ck4pc1⋯cl1−cl+1/nl+1−1qc1⋯ck1−ck+1/nk+1−1p1−cl+1/nl+1−1/q1−ck+1/nk+1−1cl+1⋯ckTable 4
The calculation ofα
NumberNrBx,R∩E≍α≍1pnk+1/qnl+2⋯nklogp−logq+lognl+2⋯nk+1/logp−11−cl+1/nl+1−1+cl+1−logq−11−ck+1/nk+1−1+ck+1−logcl+1⋯ckifp≠1,q≠1−logq+lognl+2⋯nk+1/logcl+1−logq−11−ck+1/nk+1−1+ck+1−logcl+1⋯ck,ifp=1,q≠1logp+lognl+2⋯nk+1/logp−11−cl+1/nl+1−1+cl+1−logck+1−logcl+1⋯ck,ifp≠1,q=1lognl+2⋯nk+1/logcl+1−logck+1−logcl+1⋯ckifp=1,q=12pnk+1/qnl+2⋯nklogp−logq+lognl+2⋯nk+1/logp−11−cl+1/nl+1−1+cl+1−logq1−ck+1/nk+1−1−logcl+1⋯ckifp≠1−logq+lognl+2⋯nk+1/logcl+1−logq1−ck+1/nk+1−1−logcl+1⋯ck,ifp=13pnk+1/qnl+2⋯nklogp−logq+lognl+2⋯nk+1/logp1−cl+1/nl+1−1−logq−11−ck+1/nk+1−1+ck+1−logcl+1⋯ckifq≠1logp+lognl+2⋯nk+1/logp1−cl+1/nl+1−1−logck+1−logcl+1⋯ckifq=14pnk+1/qnl+2⋯nklogp−logq+lognl+2⋯nk+1/logp1−cl+1/nl+1−1−logq1−ck+1/nk+1−1−logcl+1⋯ckCorollary 1.
Give a numberμ and 0<μ<1, for any number λ which satisfies that 0≤λ≤μ there exists a homogeneous Cantor set E such that(28)dimHE=μ,dimLE=λ.Proof .
of Corollary1.Case 1.
0<λ≤μ.
LetTkk≥1 be a sequence of integers such that(29)T1=1,T2=2,T2k+1=T2k+k2,T2k+2=T2k+1+k.Define the family of parameters ni,ci as follows: n1=3,n2=4,(30)ni=3,ifT2k<i≤T2k+1;4,ifT2k+1<i≤T2k+2.(31)c1=131/μ,c2=141/λci=131/μ,ifT2k<i≤T2k+1;141/λ,ifT2k+1<i≤T2k+2.By Theorem 1 and Theorem 2, it is obvious that dimHE=μ,dimLE=λ.Case 2.
λ=0.
LetTkk≥1 be a sequence of integers such that(32)T1=1,T2=2,T2k+1=T2k+k2,T2k+2=T2k+1+1.
Define the family of parametersni,ci as follows: n1=3,n2=2,(33)ni=3,ifT2k<i≤T2k+1;k+1,ifT2k+1<i≤T2k+2.(34)c1=131/μ,c2=13,ci=131/μ,ifT2k<i≤T2k+1;13k+1,ifT2k+1<i≤T2k+2.
By Theorem1 and Theorem 2, it is obvious that dimHE=μ,dimLE=0.Corollary 2.
Give a numberμ and 0<μ<1, for any number λ which satisfies that μ≤λ≤1 there exists a homogeneous Cantor set E such that(35)dimPE=dim¯BE=μ,dimAE=λ.Proof .
of Corollary2.Case 3.
μ≤λ<1.
LetTkk≥1 be a sequence of integers such that(36)T1=1,T2=2,T2k+1=T2k+k2,T2k+2=T2k+1+k.
Define the family of parametersni,ci as follows: n1=3,n2=4,(37)ni=3,ifT2k<i≤T2k+1;4,ifT2k+1<i≤T2k+2.(38)c1=131/μ,c2=141/λci=131/μ,ifT2k<i≤T2k+1;141/λ,ifT2k+1<i≤T2k+2.
By Theorem1 and Theorem 2, it is obvious that dimPE=dim¯BE=μ,dimAE=λ.Case 4.
λ=1.
For Case2 in Corollary 1, notice that limsupknk=+∞, it is obvious that dimPE=dim¯BE=μ,dimAE=0.



---

*Source: 3501228-2022-05-29.xml*