On the structure of the intersection of two middle third cantor sets

Publicacions Matem`atiques, Vol 39 (1995), 43–60. ON THE STRUCTURE OF THE INTERSECTION OF TWO MIDDLE THIRD CANTOR SETS Gregory J. Davis and Tian-You Hu Abstract Motivated by the study of planar homoclinic bifurcations, in this paper we describe how the intersection of two middle third Cantor sets changes as the sets are translated across each other. The resulting description shows that the intersection is never empty; in fact, the intersection can be either finite or infinite in size. We show that when the intersection is finite then the number of points in the intersection will be either 2nor 3 ·2n. We also explore the Hausdorff dimension of the intersection of two middle third Cantor sets as the sets are translated across one another. We show that the Hausdorff dimension of the intersection can take on any value from 0 to ln 2/ln 3; in addition, we show that for each Hausdorff dimension, between 0 and ln 2/ln 3, there is a dense set of translation parameters for which the intersections have that particular Hausdorff dimension. 1. Dynamical systems and intersecting Cantor sets Our motivation to study the intersection of Cantor sets comes from the discipline of Dynamical Systems. In the late 1800’s, Poincar´e identified a problem common to many nonlinear dynamical systems - how to describe the changes in a dynamical system when a homoclinic bifurcation takes place. This problem is still the center of much work in the theory of dynamical systems (see the recent monograph [9], for example). As a homoclinic bifurcation takes place, the behavior of a deterministic dynamical system can change from being very robust and predictable (with respect to initial conditions) to being completely chaotic. Over the past 15 years, work has accelerated in the area of homoclinic bifurcations. Several majour theories have been explored in conjunction with the creation and destruction of homoclinic bifurcations. Along with the possibility of strange attractors, some of the phenomena 44 G. J. Davis, T.-Y. Hu associated with homoclinic bifurcations include omega explosions [8], infinitely many coexisting sinks [7], [10], [1] and antimonotonicity [4]. At this time, there is no single theory which integrates and predicts the order of occurrence of these phenomena. However, the development of the theories related to each of these phenomena requires the understanding of how certain stable and unstable manifolds intersect as the homoclinic bifurcation takes place. It is known that the intersections of stable and unstable manifolds have the shape of intersecting Cantor sets; because of this fact, it seems that in order to create a theory unifying omega explosions, infinitely many coexisting sinks and antimonotonicity, it is necessary to understand how Cantor sets intersect in general. The theories of infinitely many sinks and antimonotonicity rely heavily on knowing when stable and unstable manifolds cannot be separated as they slide across one another, while the theory of omega explosions requires that the stable and unstable manifolds seldom intersect as they slide across one another. Each of these theories hold for parameter values close to a given homoclinic bifurcation in a dissipative planar diffeomorphism. The criterion used for showing that the stable and unstable manifolds cannot be separated is that the product of the thicknesses of the manifolds is greater than one. Newhouse [6] defines the concept of thickness and show that two Cantor sets which have the product of their thicknesses greater than one cannot be separated. A Cantor set Cin the line is represented as the difference of an interval C0and an infinite collection {Uj}of disjoint open subintervals (also known as gaps) contained in C0. More precisely, C= ∞  i=0 Ci, where C0is the smallest interval containing C, and Ci=C0− i−1  j=0 Uj. Such a sequence of sets {Ci}is called a defining sequence of C. Let Iij for j=1,2 be the two components of Cion either side of the gap Uiand let l(J) be the length of an interval J. The thickness of a defining sequence is defined by: τ({Ci}) = inf{l(Iij)/l(Ui):i≥1,where j=1,2}. The thickness of a Cantor set is then defined by τ(C) = sup{τ({Ci}): {Ci}is a defining sequence for C}. Using this definition, Newhouse proved the following striking lemma: Lemma. Let C1and C2be two Cantor sets in Rsuch that τ(C1)τ(C2)>1.IfC1is not contained in a gap of C2and C2is not contained in a gap of C1, then C1∩C2=∅. On the intersection of two Cantor sets 45 On the other hand, the theory of omega explosions uses the criterion that the sum of the limit capacities of the stable and unstable manifolds is less than one. The limit capacity, d(C), of a Cantor set Cwill be defined as: lim sup →0 (ln n(C, )/ln(1/)), where n(C, ) is the minimum number of open intervals of radius needed to cover C. The limit capacity of a Cantor set is closely related to its Hausdorff dimension (see [6]). Williams [11], Kraft [5] and Hunt, Kan, and Yorke [3] have explored the size of the intersections resulting from two Cantor sets being translated across one another. All of these explorations have assumed that the Cantor sets in question have product of thickness greater than one. Williams produced examples to show that the intersection of two Cantor sets can vary from being one point to containing another Cantor set. Independently, Kraft and Hunt et. al. further developed these ideas and determined all pairs of thicknesses for which the intersection may be a single point and all pairs of thicknesses which must contain another Cantor set. They also consider the problem of how often —as one Cantor set is being translated over another one— does the intersection contain another Cantor set. In this paper it is our goal to completely describe how the intersection of two middle third Cantor sets change as they are translated across each other. Recall the standard construction for the middle third Cantor set. First we let C0be the closed interval Iof length 1. Although we identify Iwith [0,1], we note that this construction can be carried out in any closed interval of length 1 (or any closed interval in general). C0:01 Now remove from C0the open middle third interval, (1/3,2/3), to obtain C1=[0,1/3] ∪[2/3,1]: C1:01/3 2/31 Next remove from each closed interval in C1the open middle third interval to obtain C2=[0,1/9] ∪[2/9,3/9] ∪[6/9,7/9] ∪[8/9,1]: C2:01/9 2/93/9 6/96/9 8/91 This process is continued indefinitely creating a nested collection of closed nonempty sets C0⊃C1⊃C2⊃C3⊃..., where each Cnis 46 G. J. Davis, T.-Y. Hu the union of 2nintervals, each of length 1/3n. We define the middle third Cantor set to be the infinite intersection: C= ∞  n=0 Cn. Suppose that two middle third Cantor sets are initially positioned on top of each other as in the following figure: C: C:01 Now translate one Cantor set to the right xunits (x∈[0,1]) while keeping the other Cantor set in its initial position: x+C:|←x→| C:01 Hence, our goal is to analyze the intersections (x+C)∩C. At first we will consider the problem of how often x+Cand Chave a nonempty intersection. It can be shown that the thickness of a middle third Cantor set is exactly one, thus Newhouse’s lemma (mentioned above) does not apply directly to intersecting middle third Cantor sets. However, by making small modifications to Newhouse’s lemma, it can be shown [2] that in the middle third Cantor set context, (x+C)∩C=∅for all x∈[0,1]. In this paper we will present an elementary argument, which is independent of Newhouse’s, to explicitly show that (x+C)∩C=∅ for all x∈[0,1]. Our argument, also provides information which allows us to then determine the structure and the size of the intersections of two middle third Cantor sets. In fact, we will completely describe the intersections both in terms of their cardinality and in terms of their Hausdorff dimension. The rest of this paper is organized as follows. Explicit statements of our results are presented in section 2. Section 3 contains proof of the cardinality results and Section 4 contains proofs of the Hausdorff dimension results. 2. Notation and statement of main results The points in the middle third Cantor set Ccan be conveniently represented by ternary decimals. Recall that for any x∈[0,1] we can express On the intersection of two Cantor sets 47 xin a ternary decimal expansion: (2.1) x=(ai)= ∞  n=0 an 3n=a0·a1a2a3... where a0=0,a i=0,1,or 2 for all i>0. For each nonzero x, representation (2.1) is unique except when x= k/3nand kand nare integers. In these exceptional cases xhas two representations: (2.2) x=(ai)=0·a1a2...a n000 ...(an=0) and (2.3) x=(ai)=0·a1a2...a n−1(an−1)222 ...(an=0). In order for there to be a 1 −1 correspondence between the x’s in the interval [0,1] and their ternary representations (ai), we will use representation (2.3) if and only if an=1. The points in the middle third Cantor set Care characterized by the set of ternary representations which contain no 1’s as digits. That is, x∈Cif and only if each aiin the representation of xis either a 0 or 2. If there is an index nsuch that an=1,ai= 0 or 2 for i<nand ai=0 (or ai= 2) for all i>n, then we will still consider it is a point in C, since obviously anand ai,i>n, can be rewritten as either 0 or 2. To simplify our notation we will make the following definitions. The segment of digits arar+1 ...a r+k, where arand ar+k= 1, and for all i=r+1,... ,r +k−1, ai= 0 or 2 is called a (1,1)-string of x. Furthermore, the (1,1)-string sequence of xis defined to be the sequence of all mutually disjoint (1,1)-strings of xstarting with the first 1 digit of the ternary representation (ai). If there are an odd number of 1’s in the ternary representation of x, then we will make the assumption that there is a 1 at the infinity of the ternary representation to form a final (1,1)-string of x. For each x∈[0,1] we will let T=T(x) denote the number of 2’s in the (1,1)-string sequence of x. Similarly we will let Z=Z(x) denote the number of 0’s (a0does not count) in the complement of the (1,1)-string sequence of x. Finally, for each x∈[0,1], we will define the set Cxto be: Cx={y∈C|x+y∈C}. 48 G. J. Davis, T.-Y. Hu Notice that the collection of points in the set Cxcorresponds exactly to the points in (x+C)∩C. Hence all results stated in terms of Cxapply directly to the intersection (x+C)∩C. The phrase Cxcontains a Cantor set is to mean that Cx=Cor that Cxcontains a self similar copy of C. Since for x= 0 and for x= 1, the structure of (x+C)∩Cis clear, to avoid triviality, we always assume that x∈(0,1) for discussion. Result A: The cardinality of middle third Cantor set intersections. 1. Let x∈(0,1) have the expansion x=0·a1a2a3..., then Cx contains a Cantor set if and only if either the number of 1’s in the expansion is even and all but finitely many of the remaining digits are 0’s, or the number of 1’s is odd and all but finitely many of the remaining digits are 2’s. All the x’s which satisfy these conditions form a countable, dense subset of (0,1). 2. The number of elements in Cx, denoted by |Cx|, satisfies: i) |Cx|=3·2T+Z−1if and only if the number of 1’s in the ternary representation of xis finite and T+Z<∞. All the x’s which satisfy these conditions form a countable, dense subset of (0,1). ii) |Cx|=2 T+Zif and only if the number of 1’s in the ternary representation of xis infinite and T+Z<∞. All the x’s which satisfy these conditions form an uncountable, dense subset of (0,1) having Lebesgue measure zero. iii) |Cx|is uncountably infinite in all other cases. All x’s, with the exception of an uncountable dense subset of (0,1) having Lebesgue measure zero, satisfy these conditions. Result B: The Hausdorff dimension of middle third Cantor set intersections. 1. For every 0 ≤α≤1 there exists an x∈(0,1) such that the Hausdorff dimension of (x+C)∩Csatisfies: dim[(x+C)∩C]=(1−α)ln 2 ln 3. 2. Let Dαbe the set of translation parameter x’s where all of the Hausdorff dimensions, dim[(x+C)∩C], have the following common value: Dα=x∈[0,1] : dim[(x+C)∩C]=(1−α)ln 2 ln 3. On the intersection of two Cantor sets 49 Then for every 0 ≤α≤1 we have that: i) Dαis a dense subset of (0,1). ii) Either mDα=0ormDα= 1,where mDαis the Lebesgue measure of Dα. 3. Proof of Result A: The cardinality of the intersection Throughout the paper, xwill represent an arbitrary number in the interval (0,1). We will identify xwith its ternary representation (ai) where each aiis a 0, 1, or 2 digit for all positive integers i. Similarly, y will be any number in the interval [0,1] which is also a point of the middle third Cantor set C. We will identify ywith its ternary representation (bi) where each biis a 0 or 2 digit for each positive integer i. We define the indices Ikfor each positive integer kby: Ik= max{i|0≤i<k, and ai+bi= 0 or 3},k=1,2,3,... . Notice that Ikalways exists due to the fact that a0+b0=0+0=0. The translate of yin Cby a distance of xwill be given by t=x+y. We will identify twith its ternary representation (di). Our initial goal is to understand when tis a point of the middle third Cantor set Cand when it is not. In the following series of lemmas we will present four conditions which guarantee that a translated of a point from Cdoes not fall on another point of C; that is, conditions which imply that t=x+y is not an element of C. Lemma 3.1. If ak+bk=4for some k≥1,ai=1for all Ik<i<k, and aj+bj=4for some j>I k, then t=(di)/∈C. Proof: By the definition of Ikwe see that our assumption of ai= 1 for all Ik<i<kimplies that ai+bi= 2 or 4 for all Ik<i<k. Further, our assumption that ak+bk= 4 implies that di=ai+bi+ 1 (mod 3) for all Ik<i<k. Hence, we may conclude that dIk= 1. We also have that dk= 1 or 2 since ak+bk= 4. If we can show that di<2 for some i>I k, then (di)/∈C. The index j>I kfor which aj+bj= 4 can be divided into two cases: Case 1: j<k. As before, we have aj+bj= 2, hence dj=0. Case 2: j>k. Without loss of generality, assume that jis the smallest such index greater than k.Ifaj+bj= 3, then dj= 0 or 1. If aj+bj<2, then dj−1=aj−1+bj−1(mod 3) = 4 (mod 3) = 1. Suppose that aj+ bj= 2. Then dj−1=1ifdj= 2. Otherwise, dj=aj+bj+1 (mod 3) = 0. In either case we have t=(di)/∈C. 50 G. J. Davis, T.-Y. Hu Lemma 3.2. If ak+bk=3for some k≥1,ai=1for all Ik<i<k, and aj+bj>0for some j>k, then t=(di)/∈C. Proof: Repeating the first part of the proof of lemma 3.1 (with ak+ bk= 3 replacing ak+bk= 4), we can again conclude that dIk= 1. Here the hypothesis that ak+bk= 3 implies that dk= 0 or 1. Without loss of generality, let j>kbe the smallest index such that aj+bj>0. If dj>0, then clearly (di)/∈C.Ifdj= 0, since aj+bj>0, so dj−1=aj−1+bj−1+ 1 (mod 3) = 0 + 1 = 1. Hence t=(di)/∈C. Lemma 3.3. If aj+bj=ak+bk=1for some j<k,ai=1for all j<i<k, and al+bl=4for some l>k, then t=(di)/∈C. Proof: First we will consider the case in which ai+bi≤2 for all j<i<k. It follows that dj= 1. Without loss of generality, assume that l>kis the smallest index such that al+bl= 4. Notice that dk=1 or 2. By considering the cases of al+blequals 3 or 2, or less than 2, as discussed in the proof of lemma 3.1, it will imply that dl=0,1or dl−1=1. Sot=(di)/∈C. Now suppose that ai+bi≥3 for some j<i<k. We define i0to be the maximum index between jand ksuch that ai+bi≥3. Due to the fact that ai0= 1, we see that ai0+bi0= 4 which in turn implies that di0=1. Now,ai+bi≤2 for all i0<i<k. At this point we appeal to the first part of this lemma’s proof to conclude that t=(di)/∈C. Lemma 3.4. If aj+bj=1,ak+bk=0for some k>j,ai=1, for j<i<k, and al+bl>0for some l>k(or if ak−1+bk−1=2), then t=(di)/∈C. Proof: Assume at first that ai+bi≤2 for all j<i<k. This assumption implies that dj= 1. Note that dk= 0 or 1. Thus if ak−1+ bk−1= 2, then dk−1= 2, hence (di)/∈C.Ifal+bl>0 for some l>k, without loss of generality, assume that lis the smallest one of such index. If dl>0, then (di)/∈C.Ifdl= 0, as in the proof of lemma 3.2, we have dl−1=1,so(di)/∈C. Now suppose that ai+bi≥3 for some j<i<k. Define i0to be the largest index ibetween jand ksuch that ai+bi≥3, then di0= 1 (as in the proof of lemma 3.3). Note that ai+bi≤2 for all i0<i<k, implying t=(di)/∈C. Proposition 3.5. Let xbe any real number in the interval (0,1) with an even number of 1’s in its ternary representation (including no 1’s at all in which case x∈C). If Z<∞, then |Cx|=3·2T+Z−1. On the intersection of two Cantor sets 51 Proof: Let x=(ai) and let ai= 1 for i=p1,q 1,... ,p n,q n, where p1<q 1<··· <p n<q n. Assume that ai= 0 or 2 for all i>q nand i<p 1. By conventions (2.2) and (2.3), there is an i>q nsuch that ai=0,soZ>0. Now let iZbe the largest index such that aiZ=0. Clearly, qn<i Z<∞since Z<∞, and ai= 2 for all i>i Z. We will first show that a necessary condition for x+y∈C, where y=(bi) and each biis a 0 or 2 is that (3.1) bpi= 0 and bqi= 2 for i=1,... ,n. To see that bp1= 0 we notice that if bp1= 0, then ap1+bp1= 3; and then by lemma 3.2, we would have that x+y=(di)/∈C. To see that bq1= 2 we notice that if bq1= 2, then bq1=0,ap1+bp1=aq1+bq1=1 and aiZ+biZ= 4; so by lemma 3.3, we would have that (di)/∈C. Hence bp1= 0 and bq1=2. Now suppose that (3.1) is true for i≤k−1. We first show that bpk=0. If bpk= 0, then apk+bpk= 3. Since aqk−1+bqk−1=3,soIpk≥qk−1and ai= 1 for all Ipk<i<p k. By lemma 3.2, we have (di)/∈C.Sobpk=0. Next, we show that bqk= 2. Suppose not, then bqk= 0 and aqk+bqk=1. Recall that apk+bpk= 1 and that aiZ+biZ=biZ= 4, where iZ>q k. Applying lemma 3.3 we see that (di)/∈C. The contradiction shows that bqk= 2. So (3.1) is true by induction. Via the following three claims we will examine the remaining choices for the bi’s such that x+y=(di)∈C: Claim 1. Suppose that iis an index such that 0 <i<i Zand aiis in the complement of the (1,1)-string sequece of x.Ifai= 0, then bican be either a 0 or 2. If ai= 2, then bi=0. Claim 2. Suppose that iis an index such that pj<i<q j, where j=1,... ,n.Ifai= 0, then bi=2. Ifai= 2, then bican be either a 0 or 2. Claim 3. There are exactly three permissible choices for the sequence of bi’s for i≥iZ. For index iin Claim 1, we have that ai= 1 therefore ai+bi=0,2or 4. Since al+bl= 3 for l=qj,j=1,... ,n, we have Ii≥     0if0<i<p 1 qjif qj<i<p j+1,j=1,... ,n−1 qnif qn<i<i Z. 58 G. J. Davis, T.-Y. Hu x∈Cbe chosen so that in the first 10ndigits of its ternary representation there are exactly 1... ndigits which are 2’s and the rest of the digits are 0’s. From the proof of proposition 3.5 we know that the set (x+C)∩C consists precisely of those (bi)∈Csuch that bi=0ifai= 2 and bi=0 or 2 otherwise. It now follows from proposition 4.1 and the construction of xthat x∈Dαand therefore Dα=∅. Furthermore, if yis obtained by adding a finite number of (1,1)-strings to the tail digits of x, then corollary 4.2 implies that y∈Dα. Clearly these y’s form a dense subset of (0,1). Proposition 4.4. For every 0≤α≤1, either mDα=0or mDα= 1. Here mDαis the Lebesgue measure of Dα. Proof: Let β∈(0,1); we will show that if mDα>0 then mDα>β and therefore mDα= 1. Notice that if mDα>0, then there exists a ternary interval Iof length 3−nsuch that: (4.2) 1 ≥m(Dα∩I) m(I)>β. Proposition 4.4 will be proven if we can show that the interval Iin expression (4.2) can be replaced by any ternary interval of length 3−n from [0,1]. Let Jbe an arbitrarily chosen ternary interval of length 3−n from [0,1]. Recall that all points in I(or J, respectively) have the same beginning ndigits in their ternary expansion. For any given pair of points (ai), (bi) with (ai)∈Iand (bi)∈J, either (ai) and (bi) will both have an even (odd) number of 1’s in their first n digits or one will have an even number of 1’s and the other will have an odd number of 1’s in their first ndigits. Suppose that both a1...a nand b1...b nhave an even (or odd, respectively) number of 1’s. Let Adenote the translation of I∩Dαto J; that is, A={yx∈J|yx=0·b1...b nan+1an+2 ... , for all x=(ai)∈(I∩Dα)}. The points xand yxsatisfy the hypotheses of corollary 4.2, thus dim[(y+ C)∩C]=αfor all y∈A. Therefore A⊆(J∩Dα) and we now see that m(I∩Dα)=m(A)≤m(J∩Dα). Hence expression (4.2) remains true when Iis replaced by J. Now suppose that one of the sequences a1...a nor b1...b nhas an odd number of 1’s and the other sequence has an even number of 1’s. Let B be the union of the similitudes of I∩Dαon every middle third ternary On the intersection of two Cantor sets 59 interval of J\C; that is, B={yx∈J|yx=0·b1...b n+kan+1an+2,... , for all x=(ai)∈(I∩Dα), where bn+k=1,b i= 0 or 2 for i=n+1,... ,n+k−1,k=1,2,...}. It is straightforward to verify that mB =m(I∩Dα). Now by corollary 4.2 we have dim[(y+C)∩C]=αfor all y∈B.SoB⊆(J∩Dα) and expression (4.2) remains true when Iis replaced by J. By combining the results in this section we obtain a proof of Result B on the Hausdorff dimension of middle third Cantor set intersections. Acknowledgements. The authors would like to thank the editor and an anonymous reviewer for their careful reading and constructive remarks, which have made this manuscript a stronger contribution. References 1. Davis G. J., Infinitely many coexisting sinks from degenerate homoclinic tangencies, A.M.S. Trans. 323(2) (1991), 727–748. 2. Davis G. J., Intersections of middle αCantor sets, preprint, 1993. 3. Hunt B. 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F., How big is the intersection of two thick Cantor sets, in “Contemporary Mathematics,” M. Brown, ed., Proc. of the 60 G. J. Davis, T.-Y. Hu 1989 Joint Summer Research Conference on Continua and Dynamical Systems, A.M.S., Providence R. I., 1991. Department of Mathematics University of Wisconsin - Green Bay Green Bay, WI 54311-7001 U.S.A. Primera versi´o rebuda el 25 d’Agost de 1993, darrera versi´o rebuda el 7 de Setembre de 1994