How likely can a point be in different Cantor sets
aa r X i v : . [ m a t h . D S ] F e b LARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS
KAN JIANG, DERONG KONG, AND WENXIA LI
Abstract.
Let x ∈ (0 ,
1) and m ∈ N ≥ . We consider the set Λ( x ) of bases λ ∈ (0 , /m ]such that x = P ∞ i =1 d i λ i for some (unique) sequence ( d i ) ∈ { , , . . . , m − } N . In thispaper we show that Λ( x ) is a topological Cantor set; it has zero Lebesgue measure andfull Hausdorff dimension. Furthermore, we show that the intersection Λ( x ) ∩ Λ( y ) has fullHausdorff dimension for any x, y ∈ (0 , Introduction
Given an integer m ≥
2, for λ ∈ (0 , /m ] let K λ be the self-similar set generated by the iterated function system (IFS)Ψ λ := { f i ( x ) = λ ( x + i ) : i = 0 , , . . . , m − } . Then the convex hull of K λ is [0 , ( m − λ − λ ]. In this paper we are interested in the class ofself-similar sets which contain a common point. Fix x ∈ [0 , x ) := { λ ∈ (0 , /m ] : x ∈ K λ } . Observe that the point 0 always belongs to K λ for all λ ∈ (0 , /m ]. Then Λ(0) = (0 , /m ].Furthermore, 1 ∈ K λ if and only if λ = 1 /m . This implies Λ(1) = { /m } . So, in thefollowing we only consider Λ( x ) for x ∈ (0 , x ) comes from unique beta expansions. When λ ∈ (1 /m, λ = { f i ( x ) } m − i =0 has overlaps, that is, for the attractor K λ we have f i ( K λ ) ∩ f i +1 ( K λ ) = ∅ for all 0 ≤ i < m −
1. Denote by U λ the set of x ∈ K λ such that x has a unique coding, i.e., there exists a unique sequence ( d i ) ∈ { , , . . . , m − } N suchthat x = P ∞ i =1 d i λ i . The last two authors and their coauthors recently studied in [9] thetopological and fractal properties of the set U ( x ) := { λ ∈ (1 /m,
1) : x ∈ U λ } . They showedthat U ( x ) has zero Lebesgue measure for all x >
0, and dim H U ( x ) = 1 for any x ∈ (0 , x > U ( x ) contains isolated points. Dajani et al. [5] showedthat for any x ∈ (0 ,
1] the algebraic sum U ( x ) + U ( x ) contains interior points. When x = 1,the set U (1) was extremely studied. Komornik and Loreti [8] showed that the topologicalclosure of U (1) is a topological Cantor set : a nonempty compact set having neither interiornor isolated points. Allaart and the second author studied in [1] the local dimension of Date : 1st March 2021.2010
Mathematics Subject Classification.
Primary:28A78, Secondary: 28A80, 37B10, 11A63.
Key words and phrases.
Self-similar set, univoque base, thickness, Cantor sets, intersection. U (1), and showed that U (1) has more weight close to 1 /m . Bonanno et al. [4] found itsclose connection to the bifurcation set of α -continued fractions and many other dynamicalsystems.Another motivation to study Λ( x ) is from the work of Boes, Darst and Erd˝os [3], wherethey considered a similar class of Cantor sets C λ in [0 ,
1] (in fact, they considered aclass of fat Moran sets with positive Lebesgue measures). They showed that Λ ′ ( x ) := { λ ∈ (0 , /
2) : x ∈ C λ } is of first category for any x ∈ (0 , λ ∈ (0 , /
2) the Cantor set C λ excluding the two endpointscontains only irrational numbers.Now we state our main results on Λ( x ). The first result shows that Λ( x ) is small in thetopological sense. Theorem 1.1.
For any x ∈ (0 , the set Λ( x ) is a topological Cantor set with min Λ( x ) = xm − x and max Λ( x ) = 1 /m .
13 12 ∞ ∞ Figure 1.
The geometrical construction of Λ( x ) with x = 1 / m = 2.See Example 2.6 for more explanation.Observe that the self-similar set K λ is dimensional homogeneous , i.e., for any y ∈ K λ we have lim δ → dim H ( K λ ∩ ( y − δ, y + δ )) = dim H K λ . This means that K λ is distributeduniformly in the dimensional sense. Our next result states that the local dimension of Λ( x )behaves differently. Theorem 1.2.
Let x ∈ (0 , . Then for any λ ∈ Λ( x ) we have lim δ → dim H (Λ( x ) ∩ ( λ − δ, λ + δ )) = dim H K λ = log m − log λ . As a direct consequence of Theorem 1.2 we obtain that Λ( x ) has zero Lebesgue measureand full Hausdorff dimension. Corollary 1.3.
For any x ∈ (0 , the set Λ( x ) is a Lebesgue null set, and dim H Λ( x ) = 1 . ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 3
Another consequence of Theorem 1.2 gives the following result for the segment of Λ( x ). Corollary 1.4.
Let x ∈ (0 , . Then for any open interval ( a, b ) with Λ( x ) ∩ ( a, b ) = ∅ wehave dim H (Λ( x ) ∩ ( a, b )) = sup λ ∈ Λ( x ) ∩ ( a,b ) dim H K λ . Note by Theorem 1.1 that Λ( x ) is a topological Cantor set for any x ∈ (0 , h x : λ → dim H (Λ( x ) ∩ (0 , λ )) is a non-decreasingDevil’s staircase for any x ∈ (0 , x, y ∈ (0 ,
1) there are infinitely many λ ∈ (0 , /m ] such that K λ contains both points x and y . In other words, we show thatΛ( x ) ∩ Λ( y ) contains infinitely many points. Theorem 1.5.
For any x, y ∈ (0 , we have dim H (Λ( x ) ∩ Λ( y )) = 1 . The paper is organized in the following way. In Section 2 we show that Λ( x ) is atopological Cantor set, and prove Theorem 1.1. In Section 3 we study the local dimensionof Λ( x ), and prove Theorem 1.2, Corollaries 1.3 and 1.4. In Section 4 we construct asequence of Cantor subsets ( E k ( x )) of Λ( x ) such that the thickness of E k ( x ) goes to infinityas k → ∞ . Based on this we show that for any x, y ∈ (0 ,
1) the intersection Λ( x ) ∩ Λ( y )has full Hausdorff dimension, and prove Theorem 1.5.2. Topological structure of Λ( x )Note that for λ ∈ (0 , /m ) the IFS Ψ λ satisfies the strong separation condition. So thereexists a natural bijective map from the symbolic space { , , . . . , m − } N to the self-similarset K λ defined by π λ : { , , . . . , m − } N → K λ ; ( d i ) ∞ X i =1 d i λ i . The infinite sequence ( d i ) is called a (unique) coding of π λ (( d i )) with respect to the IFSΨ λ . Now we fix x ∈ (0 , x ) \ { /m } to { , , . . . , m − } N byΦ x : Λ( x ) \ { /m } → { , , . . . , m − } N ; λ π − λ ( x ) . So Φ x ( λ ) is the unique coding of x ∈ K λ with respect to the IFS Ψ λ . For convenience,when λ = 1 /m we define Φ x (1 /m ) as the greedy m -adic expansion of x , i.e., Φ x (1 /m ) isthe lexicographically largest sequence ( d i ) ∈ { , , . . . , m − } N such that x = P ∞ i =1 d i /m i .Now we recall some terminology from symbolic dynamics (cf. [11]). In this paper wefix our alphabet { , , . . . , m − } , and let { , , . . . , m − } N denote the set of all infinitesequences ( d i ) with each element d i ∈ { , , . . . , m − } . For a word w we mean a finitestring of digits, i.e., w = w w . . . w n for some n ≥
0. Here n is called the length of w . In particular, when n = 0 we call w = ǫ the empty word. Let { , , . . . , m − } ∗ denote the set of all finite words. Then { , , . . . , m − } ∗ = S ∞ n =0 { , , . . . , m − } n , K.JIANG, D.KONG, AND W.LI where { , , . . . , m − } n denotes the set of all words of length n . For two words u = u . . . u k , v = v . . . v n we write uv = u . . . u k v . . . v n ∈ { , , . . . , m − } ∗ . Similarly, for asequence a = a a . . . ∈ { , , . . . , m − } N we write ua = u . . . u k a a . . . . For j ∈ N wedenote by u j = u · · · u the j -fold concatenation of u with itself, and by u ∞ = uu · · · theperiod sequence with period block u . For a word u = u . . . u k , if u k < m −
1, then we write u + = u . . . u k − ( u k +1). So u + is also a word. In this paper we will use lexicographical order ≺ , , ≻ and < between sequences in { , , . . . , m − } N . For example, for two sequences( a i ) , ( d i ) we say ( a i ) ≺ ( b i ) if there exists n ∈ N such that a . . . a n − = b . . . b n − and a n < b n . Also, we write ( a i ) ( b i ) if ( a i ) ≺ ( b i ) or ( a i ) = ( b i ). Lemma 2.1.
Let x ∈ (0 , . Then Φ x is strictly decreasing in Λ( x ) with respect to thelexicographical order in { , , . . . , m − } N .Proof. Let λ , λ ∈ Λ( x ) with λ < λ . Write ( a i ) = Φ x ( λ ) and ( b i ) = Φ x ( λ ). Supposeon the contrary that ( a i ) ( b i ). Then x = ∞ X i =1 a i λ i ≤ ∞ X i =1 b i λ i < ∞ X i =1 b i λ i = x, leading to a contradiction. (cid:3) As a direct consequence of Lemma 2.1 we can determine the extreme values of Λ( x ). Lemma 2.2.
For any x ∈ (0 , we have min Λ( x ) = xm − x and max Λ( x ) = 1 /m .Proof. Note that K /m = [0 , x ) it is clear that max Λ( x ) = 1 /m .On the other hand, by Lemma 2.1 it follows that the smallest element λ = min Λ( x ) satisfiesΦ x ( λ ) = ( m − ∞ . This gives that x = ( m − λ − λ , and hence, min Λ( x ) = λ = xm − x . (cid:3) Recall that K λ is the self-similar set generated by the IFS Ψ λ = { f i ( x ) = λ ( x + i ) } m − i =0 .Geometrically, K λ can be constructed via a decreasing sequence of nonempty compact sets.More precisely, K λ = ∞ \ n =1 K λ ( n ) with K λ ( n ) = [ d ...d n ∈{ , ,...,m − } n f d ◦ · · · ◦ f d n (cid:18)h , ( m − λ − λ i(cid:19) . When λ ∈ (0 , /m ), for each n ∈ N the set K λ ( n ) is the union of m n pairwise disjointsubintervals of equal length λ n ( m − λ − λ . Furthermore, K λ ( n + 1) ⊂ K λ ( n ) for any n ≥ d H the Hausdorff metric in the space C ( R ) consisting of all non-empty com-pact sets in R (cf. [6]). In the following lemma we show that the set sequence { K λ ( n ) } ∞ n =1 converges to K λ uniformly. Lemma 2.3. d H ( K λ ( n ) , K λ ) → uniformly as n → ∞ for all λ ∈ (0 , /m ] . ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 5
Proof.
Note that K λ ⊂ K λ ( n ) for any λ ∈ (0 , /m ] and n ∈ N , and K λ ( n ) is the unionof m n pairwise disjoint subintervals of equal length λ n ( m − λ − λ . This implies that for any λ ∈ (0 , /m ] we have d H ( K λ ( n ) , K λ ) ≤ λ n ( m − λ − λ ≤ m n → n → ∞ . (cid:3) Next we show that Λ( x ) is closed. Lemma 2.4.
For any x ∈ (0 , the set Λ( x ) is closed.Proof. Suppose ( λ j ) is a sequence in Λ( x ) with lim j →∞ λ j = λ . We will show that λ ∈ Λ( x ), i.e., x ∈ K λ . By Lemma 2.3 it follows that for each ε > N ∈ N such that for any λ ∈ (0 , /m ] and any n > N we have(2.1) d H ( K λ ( n ) , K λ ) < ε . Take n > N . Since λ j → λ as j → ∞ , there exists J ∈ N such that for any j > J we have(2.2) d H ( K λ j ( n ) , K λ ( n )) < ε . Then by (2.1) and (2.2) it follows that for any j > J and n > N , d H ( K λ j , K λ ) ≤ d H ( K λ j , K λ j ( n )) + d H ( K λ j ( n ) , K λ ( n )) + d H ( K λ ( n ) , K λ ) < ε ε ε ε. Since ε > d H ( K λ j , K λ ) → j → ∞ . Now suppose on the contrary that x / ∈ K λ . Note that K λ is compact. Then dist ( x, K λ ) =inf {| x − y | : y ∈ K λ } >
0, and thus by (2.3) it follows that dist ( x, K λ j ) > j ∈ N sufficiently large. This leads to a contradiction with x ∈ K λ j for all j ≥
1. Hence, x ∈ K λ ,and thus Λ( x ) is closed. (cid:3) To prove Theorem 1.1 we still need the following lemma.
Lemma 2.5.
Let x ∈ (0 , and q ∈ ( xm − x , /m ) . Then there exists a constant C > such that for any two λ , λ ∈ Λ( x ) ∩ (0 , q ] we have | π q (Φ x ( λ )) − π q (Φ x ( λ )) | > C | λ − λ | . Proof.
Note by Lemma 2.2 that min Λ( x ) = xm − x . So Λ( x ) ∩ (0 , q ] = ∅ for any q > xm − x .Take λ , λ ∈ Λ( x ) ∩ (0 , q ] with λ < λ . By Lemma 2.1 we have( a i ) := Φ x ( λ ) ≻ Φ x ( λ ) =: ( b i ) . K.JIANG, D.KONG, AND W.LI
So there exists n ∈ N such that a . . . a n − = b . . . b n − and a n > b n . Then by using λ , λ ∈ Λ( x ) we obtain that n − X i =1 b i λ i ≤ ∞ X i =1 b i λ i = x = ∞ X i =1 a i λ i ≤ n − X i =1 a i λ i + ∞ X i = n ( m − λ i , where for n = 1 we set the sum for the index over an empty set as zero. This implies that xλ λ ( λ − λ ) = xλ − xλ ≤ n − X i =1 a i λ i − − n − X i =1 b i λ i − + ∞ X i = n ( m − λ i − = n − X i =1 a i λ i − − n − X i =1 a i λ i − + m − − λ λ n − ≤ m − − λ λ n − . (2.4)Observe that ( a i ) and ( b i ) are distinct at the first place n . Then(2.5) | π q (( a i )) − π q (( b i )) | ≥ q n − m − − q q n +1 = 1 − mq − q q n . Hence, by (2.4) and (2.5) we conclude that | π q (Φ x ( λ )) − π q (Φ x ( λ )) | = | π q (( a i )) − π q (( b i )) |≥ − mq − q q n ≥ − mq − q λ n ≥ − mq − q × (1 − λ ) x ( m − λ | λ − λ |≥ (1 − mq ) x ( m − q | λ − λ | as desired. (cid:3) Proof of Theorem 1.1.
By Lemmas 2.2 and 2.4 we only need to prove that Λ( x ) has neitherisolated nor interior points. First we prove that Λ( x ) has no isolated points. Let λ ∈ Λ( x ).We split the proof into the following two cases.Case I. λ ∈ (0 , /m ). Then Φ x ( λ ) = ( d i ) is the unique coding of x with respect to theIFS Ψ λ . For n ≥ p n be the unique root in (0 ,
1) of the equation x = n − X i =1 d i p in + d ′ n p nn , ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 7 where d ′ n = d n + 1 (mod m ). Then for all sufficiently large n ∈ N we have p n ∈ (0 , /m )and Φ x ( p n ) = d . . . d n − d ′ n ∞ . So p n ∈ Λ( x ) for all large n ∈ N . Furthermore, by Lemma2.5 it follows that p n → λ as n → ∞ . Thus, λ is not isolated in Λ( x ).Case II. λ = 1 /m . Then Φ x (1 /m ) = ( d i ) is the greedy m -adic expansion of x . Since x ∈ (0 , n ∈ N such that d n < m −
1. For all such n we define p n ∈ (0 ,
1) so that x = P n − i =1 d i p in + ( m − p nn . Then p n ∈ (0 , /m ) andΦ x ( p n ) = d . . . d n − ( m − ∞ . So p n ∈ Λ( x ) for all n ∈ N satisfying d n < m −
1. Again,by Lemma 2.5 it follows that p n converges to 1 /m along a suitable subsequence, and thus1 /m is not isolated in Λ( x ).By Cases I and II we conclude that Λ( x ) has no isolated points. Next we prove that Λ( x )has no interior points. It suffices to prove that for any two points λ , λ ∈ Λ( x ) \ { /m } there must exist λ between λ and λ but not in Λ( x ).Take λ , λ ∈ Λ( x ) \ { /m } with λ < λ . Then by Lemma 2.1 we have( a i ) = Φ x ( λ ) ≻ Φ x ( λ ) = ( b i ) . So there exists n ∈ N such that a . . . a n − = b . . . b n − and a n > b n . We define twosequences ξ = b . . . b n ( m − ∞ , ζ = b . . . b n − ( b n + 1)0 ∞ . Then ( b i ) ξ ≺ ζ ( a i ). So by using 0 < λ < λ < /m it follows that(2.6) π λ ( ξ ) < π λ ( ζ ) ≤ π λ (( a i )) = x = π λ (( b i )) ≤ π λ ( ξ ) < π λ ( ζ ) . Let I λ := ( π λ ( ξ ) , π λ ( ζ )). Then (2.6) implies that x is between two disjoint open intervals I λ and I λ . Observe that the map λ I λ = [ π λ ( ξ ) , π λ ( ζ )] is continuous with respectto the Hausdorff metric d H . So there must exist λ ∈ ( λ , λ ) such that x ∈ I λ . Since I λ ∩ K λ = ∅ , it follows that x / ∈ K λ , i.e., λ ∈ ( λ , λ ) \ Λ( x ). This completes the proof. (cid:3) At the end of this section we describe the geometrical construction of Λ( x ). By Theorem1.1 it follows that Λ( x ) is a Cantor set in R for any x ∈ (0 , conv (Λ( x )) = [ xm − x , m ]. Let( x i ) = Φ x (1 /m ) be the greedy m -adic expansion of x . Since x ∈ (0 , ℓ ∈ N such that x ℓ < m −
1. Note that Φ x ( xm − x ) = ( m − ∞ . We call a word w ∈ { , , . . . , m − } ∗ admissible in Φ x (Λ( x )) if( x i ) ≺ w ∞ ≺ w ( m − ∞ ≺ ( m − ∞ . Since x i = m − i < ℓ , it follows that any admissible word w has length at least ℓ .For each admissible word w we define the associated basic interval J w = [ p w , q w ] byΦ x ( p w ) = w ( m − ∞ and Φ x ( q w ) = w ∞ . Then for two admissible words u , w , if u is a prefix of w , then J w ⊂ J u . Furthermore,for any admissible word w there must exist an admissible word v such that w is a prefixof v , i.e., w has an offspring v . Denote by A ( x ) = S ∞ n = ℓ A n ( x ) the set of all admissiblewords, where A n ( x ) consists of all admissible words of length n . So the basic intervals K.JIANG, D.KONG, AND W.LI { J w : w ∈ A ( x ) } have a tree structure, and the Cantor set Λ( x ) can be constructed geo-metrically as Λ( x ) = ∞ \ n = ℓ [ w ∈ A n ( x ) J w . Example 2.6.
Let m = 2 and x = 1 /
2. Then conv (Λ( x )) = [ , ]. Furthermore, Φ x ( ) =1 ∞ and Φ x ( ) = 10 ∞ . Then any admissible word has length at least ℓ = 2. By thedefinition of admissible words it follows that A ( x ) = { , } , A ( x ) = { , , , } , . . . , and in general, for any n ∈ N we have A n +1 ( x ) = { u : u ∈ { , } n } . So, in the first step we remove the open interval H = (0 . , . ∼ (110 ∞ , ∞ )from the convex hull [ , ]; and in the next step we remove two open intervals H = (0 . , . ∼ (1110 ∞ , ∞ ) ,H = (0 . , . ∼ (1010 ∞ , ∞ ) . This procedure can be continued, and after finitely many steps we can get a good approx-imation of Λ( x ) (see Figure 1).3. Fractal properties of Λ( x ) : local dimension In this section we will investigate the local dimension of Λ( x ), and prove Theorem 1.2.Our proof will be split into the following two cases: (I) local dimension of Λ( x ) at λ = 1 /m ;(II) local dimension of Λ( x ) at λ ∈ (0 , /m ).3.1. Local dimension of Λ( x ) at λ = 1 /m . Observe that 1 /m ∈ Λ( x ) for any x ∈ (0 , x ) at 1 /m is one. Proposition 3.1.
For any x ∈ (0 , we have lim δ → dim H (cid:18) Λ( x ) ∩ (cid:16) m − δ, m + δ (cid:17)(cid:19) = 1 = dim H K /m . Our strategy to prove Proposition 3.1 is to construct a large subset of Λ( x ) ∩ (1 /m − δ, /m + δ ) with its Hausdorff dimension arbitrarily close to one. Let x ∈ (0 , k ∈ N we setΛ k ( x ) := (cid:8) λ ∈ Λ( x ) : ∃ N such that σ N (Φ x ( λ )) does not contain k consecutive zeros (cid:9) , where σ is the left-shift map on { , , . . . , m − } N . Note that for any λ ∈ Λ( x ) \ S ∞ k =1 Λ k ( x )the coding Φ x ( λ ) must end with 0 ∞ . Thus the difference between Λ( x ) and S ∞ k =1 Λ k ( x ) is ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 9 at most countable. So, by the countable stability of Hausdorff dimension it follows thatdim H Λ( x ) = dim H ∞ [ k =1 Λ k ( x ) = sup k ≥ dim H Λ k ( x ) . In the following we will give a lower bound for the Hausdorff dimension of Λ k ( x ). Let( x i ) = Φ x (1 /m ) be the greedy m -adic expansion of x . Since x ∈ (0 , x i < m −
1. So there exists a subsequence ( n i ) ⊂ N such that x n i < m − i ≥
1. For a large integer j , let γ j be the unique root in (0 , /m ) of the equation x = n j − X i =1 x i γ ij + ( x n j + 1) γ n j j + ∞ X i = n j +1 ( m − γ ij . Then Φ x ( γ j ) = x . . . x + n j ( m − ∞ , and γ j ր /m as j → ∞ . DefineΓ k,j ( x ) := n x . . . x + n j d d . . . ∈ { , , . . . , m − } N : ( d i ) does not contain k consecutive zeros o . Lemma 3.2.
Let x ∈ (0 , and k ∈ N . Then for a large j ∈ N we have Γ k,j ( x ) ⊆ Φ x (Λ k ( x ) ∩ [ γ j , /m )) . Proof.
Take a sequence ( c i ) ∈ Γ k,j ( x ). Then the equation x = P ∞ i =1 c i λ i determines aunique λ ∈ (0 , /m ), i.e., Φ x ( λ ) = ( c i ). Note thatΦ x (1 /m ) = ( x i ) ≺ Φ x ( λ ) = ( c i ) x . . . x + n j ( m − ∞ = Φ x ( γ j ) . By Lemma 2.1 we conclude that λ ∈ [ γ j , /m ). Furthermore, by the definition of Γ k,j ( x ) itfollows that σ n j (( c i )) does not contain k consecutive zeros. So, λ ∈ Λ k ( x ). This completesthe proof. (cid:3) To give a lower bound of dim H (Λ k ( x ) ∩ [ γ j , /m )) we still need the following lemma. Lemma 3.3.
Let x ∈ (0 , and k ∈ N . Then for a large j ∈ N there exists C > suchthat for any λ , λ ∈ Φ − x (Γ k,j ( x )) we have | π γ j (Φ x ( λ )) − π γ j (Φ x ( λ )) | ≤ C | λ − λ | . Proof.
Let λ , λ ∈ Φ − x (Γ k,j ( x )) with λ < λ . Then by Lemma 3.2 we have λ , λ ∈ Λ k ( x ) ∩ [ γ j , /m ). Furthermore, ( a i ) = Φ x ( λ ) and ( b i ) = Φ x ( λ ) have a common prefixof length at least n j . Since λ < λ , by Lemma 2.1 we have ( a i ) ≻ ( b i ). So there exists n > n j such that a . . . a n − = b . . . b n − and a n > b n . Note that σ n j (( a i )) does not contain k consecutive zeros. Then λ n + k + n X i =1 a i λ i < ∞ X i =1 a i λ i = x = ∞ X i =1 b i λ i < n X i =1 a i λ i . Rearranging the above equation it gives that λ n + k < n X i =1 a i ( λ i − λ i ) < ∞ X i =1 ( m − λ i − λ i ) = m − − λ )(1 − λ ) ( λ − λ ) . Since λ , λ ∈ [ γ j , /m ), this implies that(3.1) λ n < m − λ k (1 − λ )(1 − λ ) ( λ − λ ) < m γ kj ( m −
1) ( λ − λ ) . Note that ( a i ) and ( b i ) have a common prefix of length n −
1. It follows that | π γ j (( a i )) − π γ j (( b i )) | ≤ γ n − j ( m − γ j − γ j = m − − γ j γ nj . Therefore, by (3.1) it follows that | π γ j (Φ x ( λ )) − π γ j (Φ x ( λ )) | = | π γ j (( a i )) − π γ j (( b i )) |≤ m − − γ j γ nj ≤ m − − γ j λ n < m − − γ j × m γ kj ( m − | λ − λ | = m (1 − γ j ) γ kj | λ − λ | as desired. (cid:3) Lemma 3.4.
Let x ∈ (0 , and k ∈ N . Then for a large j ∈ N we have dim H (Λ k ( x ) ∩ [ γ j , /m )) ≥ ( k −
1) log m + log( m − − k log γ j . Proof.
This follows by Lemmas 3.2 and 3.3 thatdim H (Λ k ( x ) ∩ [ γ j , /m )) ≥ dim H Φ − x (Γ k,j ( x )) ≥ dim H π γ j (Γ k,j ( x )) ≥ ( k −
1) log m + log( m − − k log γ j , where the last inequality follows from that σ n j (Γ k,j ( x )) is a sub-shift of finite type withthe forbidden block 0 k . (cid:3) Proof of Proposition 3.1.
Note that S ∞ k =1 Λ k ( x ) ⊆ Λ( x ) and the difference set Λ( x ) \ S ∞ k =1 Λ k ( x ) is at most countable. Furthermore, by the definition of Λ k ( x ) we know thatΛ k ( x ) ⊂ Λ k ( x ) for any k < k . So by Lemma 3.4 it follows that for any large j ∈ N ,dim H (Λ( x ) ∩ [ γ j , /m )) = dim H ∞ [ k =1 (Λ k ( x ) ∩ [ γ j , /m ))= lim k →∞ dim H (Λ k ( x ) ∩ [ γ j , /m )) ≥ lim k →∞ ( k −
1) log m + log( m − − k log γ j = log m − log γ j . ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 11
Note that γ j → /m as j → ∞ . This implies thatlim δ → dim H (Λ( x ) ∩ (1 /m − δ, /m + δ )) ≥ lim j →∞ dim H (Λ( x ) ∩ [ γ j , /m )) ≥ lim j →∞ log m − log γ j = 1 , completing the proof. (cid:3) Local dimension of Λ( x ) at λ ∈ Λ( x ) \ { /m } . In this part we will prove Theorem1.2 for λ ∈ Λ( x ) \ { /m } . Fix x ∈ (0 ,
1) and λ ∈ Λ( x ) \ { /m } . Let ( x i ) = Φ x ( λ ) ∈{ , , . . . , m − } N be the unique coding of x ∈ K λ with respect to the IFS Ψ λ . Let n ∈ N large enough such that x . . . x n = 0 n , and then let β n and γ n be the unique roots in (0 , x = n X i =1 x i β in + ∞ X i = n +1 ( m − β in and x = n X i =1 x i γ in . In fact, by choosing n ∈ N sufficiently large we can even require that(3.2) 0 < β n ≤ λ ≤ γ n < /m. Then Φ x ( β n ) = x . . . x n ( m − ∞ and Φ x ( γ n ) = x . . . x n ∞ . Furthermore, lim n →∞ β n =lim n →∞ γ n = λ .Recall from the previous subsection that Λ k ( x ) consists of all q ∈ Λ( x ) such that the tailsequence of Φ x ( q ) does not contain k consecutive zeros. Now for a large n ∈ N , we alsodefineΓ λk,n ( x ) := n x . . . x n d d . . . ∈ { , , . . . , m − } N : ( d i ) does not contain k consecutive zeros o . Lemma 3.5.
Let x ∈ (0 , , λ ∈ Λ( x ) \ (cid:8) m (cid:9) and k ∈ N . Then for any δ ∈ (0 , min (cid:8) λ, m − λ (cid:9) ) there exists a large N ∈ N such that Γ λk,n ( x ) ⊂ Φ x (Λ k ( x ) ∩ ( λ − δ, λ + δ )) for any n ≥ N .Proof. By (3.2) there exists N ∈ N such that for any n > N we have 0 < β n ≤ λ ≤ γ n < /m . Take ( c i ) ∈ Γ λk,n ( x ). Then the equation x = P ∞ i =1 c i q i determines a unique q ∈ (0 , /m ), i.e., Φ x ( q ) = ( c i ). Observe thatΦ x ( γ n ) = x . . . x n ∞ ( c i ) = Φ x ( q ) x . . . x n ( m − ∞ = Φ x ( β n ) . By Lemma 2.1 we have β n ≤ q ≤ γ n . Note that lim n →∞ β n = lim n →∞ γ n = λ . Then for δ ∈ (0 , min { λ, /m − λ } ) there exists an integer N > N such that for any n > N we have q ∈ [ β n , γ n ] ⊂ ( λ − δ, λ + δ ) ⊂ (0 , /m ) . Clearly, by the definition of Γ λk,n ( x ) the tail sequence σ n (( c i )) does not contain k consecutivezeros. This proves q ∈ Λ k ( x ), and hence completes the proof. (cid:3) Now by Lemma 3.5 and the same argument as in the proof of Lemma 3.4 we obtain thefollowing lower bound for the dimension of Λ k ( x ) ∩ ( λ − δ, λ + δ ). Lemma 3.6.
Let x ∈ (0 , , λ ∈ Λ( x ) \ (cid:8) m (cid:9) and k ∈ N . Then for any δ ∈ (0 , min (cid:8) λ, m − λ (cid:9) ) we have dim H (Λ k ( x ) ∩ ( λ − δ, λ + δ )) ≥ dim H π λ − δ (Γ λk,n ( x )) ≥ ( k −
1) log m + log( m − − k log( λ − δ ) . Proof of Theorem 1.2.
By Proposition 3.1 it suffices to prove the theorem for λ ∈ Λ( x ) \{ /m } . By Lemma 3.6 it follows that for any δ ∈ (0 , min (cid:8) λ, m − λ (cid:9) ) we havedim H (Λ( x ) ∩ ( λ − δ, λ + δ )) = dim H ∞ [ k =1 (Λ k ( x ) ∩ ( λ − δ, λ + δ ))= lim k →∞ dim H Λ k ( x ) ∩ ( λ − δ, λ + δ ) ≥ lim k →∞ ( k −
1) log m + log( m − − k log( λ − δ ) = log m − log( λ − δ ) . Letting δ → δ → dim H (Λ( x ) ∩ ( λ − δ, λ + δ )) ≥ log m − log λ . On the other hand, by using Lemma 2.5 it follows thatdim H (Λ( x ) ∩ ( λ − δ, λ + δ )) ≤ dim H π λ + δ ◦ Φ x (Λ( x ) ∩ ( λ − δ, λ + δ )) ≤ dim H K λ + δ = log m − log( λ + δ ) . (3.4)Letting δ → δ → dim H (Λ( x ) ∩ ( λ − δ, λ + δ )) = log m − log λ . (cid:3) Proof of Corollary 1.3.
By Theorem 1.2 it follows thatdim H Λ( x ) ≥ lim n →∞ dim H (Λ( x ) ∩ (1 /m − / n , /m + 1 / n )) = dim H K /m = 1 . So Λ( x ) has full Hausdorff dimension. In the following it suffices to prove that Λ( x ) haszero Lebesgue measure for any x ∈ (0 , λ ∈ Λ( x ) \ { /m } there exists δ λ > H (Λ( x ) ∩ ( λ − δ λ , λ + δ λ )) <
1, which implies that Λ( x ) ∩ ( λ − δ λ , λ + δ λ ) is a Lebesgue nullset. Note by Theorem 1.1 that for any γ ∈ (0 , /m ) the segment Λ( x ) ∩ (0 , γ ] is compact,and then it can be covered by a finite number of open intervals { ( λ i − δ λ i , λ i + δ λ i ) } Ni =1 forsome λ i ∈ Λ( x ) with 1 ≤ i ≤ N . Since each set Λ( x ) ∩ ( λ i − δ λ i , λ i + δ λ i ) is a Lebesgue nullset, so is Λ( x ) ∩ (0 , γ ] for any γ < /m . Let ( γ n ) be a sequence in (0 , /m ) with γ n ր /m ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 13 as n → ∞ . Then we conclude that Λ( x ) = { /m } ∪ S ∞ n =1 (Λ( x ) ∩ (0 , γ n ]) has zero Lebesguemeasure. (cid:3) Proof of Corollary 1.4.
Note that the function D : λ dim H K λ = log m − log λ is continuousand strictly increasing in (0 , /m ]. Let ( a, b ) be an open interval such that ( a, b ) ∩ Λ( x ) = ∅ .We consider the following two cases.Case (I). There exists a λ ∗ ∈ Λ( x ) ∩ ( a, b ) such that D ( λ ∗ ) = sup λ ∈ Λ( x ) ∩ ( a,b ) D ( λ ). ThenΛ( x ) ∩ ( λ ∗ , b ) = ∅ . Take ε >
0. By Theorem 1.2 it follows that(3.5) dim H (Λ( x ) ∩ ( a, b )) ≥ dim H (Λ( x ) ∩ ( λ ∗ − δ, λ ∗ + δ )) ≥ D ( λ ∗ ) − ε for δ > λ ∈ Λ( x ) ∩ [ a, b ) = Λ( x ) ∩ [ a, λ ∗ ] there must exist a δ λ > H (Λ( x ) ∩ ( λ − δ λ , λ + δ λ )) < D ( λ ) + ε. Observe that the union of ( λ − δ λ , λ + δ λ ) with λ ∈ Λ( x ) ∩ [ a, b ) is an open cover of thecompact set Λ( x ) ∩ [ a, b ) = Λ( x ) ∩ [ a, λ ∗ ]. So there exists a subcover of Λ( x ) ∩ [ a, b ), say { ( λ i − δ λ i , λ i + δ λ i ) : 1 ≤ i ≤ N } , such that Λ( x ) ∩ [ a, b ) ⊂ S Ni =1 (Λ( x ) ∩ ( λ i − δ λ i , λ i + δ λ i )).Therefore, by (3.6) we conclude thatdim H (Λ( x ) ∩ ( a, b )) = dim H (Λ( x ) ∩ [ a, b )) = max ≤ i ≤ N dim H (Λ( x ) ∩ ( λ i − δ λ i , λ i + δ λ i )) ≤ max ≤ i ≤ N ( D ( λ i ) + ε ) ≤ D ( λ ∗ ) + ε. Since ε > H (Λ( x ) ∩ ( a, b )) = dim H (Λ( x ) ∩ ( a, λ ∗ ]) = D ( λ ∗ ) = sup λ ∈ Λ( x ) ∩ ( a,b ) dim H K λ as desired.Case (II). There is no λ ∗ ∈ Λ( x ) ∩ ( a, b ) such that D ( λ ∗ ) = sup λ ∈ Λ( x ) ∩ ( a,b ) D ( λ ). Thenthere exists a sequence ( λ n ) ⊂ Λ( x ) ∩ ( a, b ) such that λ n ր b as n → ∞ , and sup n ≥ D ( λ n ) =sup λ ∈ Λ( x ) ∩ ( a,b ) D ( λ ). By Case (I) we know that for any n ≥ H (Λ( x ) ∩ ( a, λ n ]) = D ( λ n ) . By the countable stability of Hausdorff dimension we conclude thatdim H (Λ( x ) ∩ ( a, b )) = dim H ∞ [ n =1 (Λ( x ) ∩ ( a, λ n ]) = sup n ≥ D ( λ n ) = sup λ ∈ Λ( x ) ∩ ( a,b ) D ( λ ) . This completes the proof. (cid:3) Large intersection of Λ( x ) and Λ( y )In this section we will show that the intersection of any two sets Λ( x ) and Λ( y ) contains aCantor subset of Hausdorff dimension arbitrarily close to one, and then prove Theorem 1.5.Our strategy is to show that Λ( x ) contains a Cantor subset with its thickness arbitrarilylarge. Construction of Cantor subsets of Λ( x ) with large thickness. First we recallfrom [12] the definition of thickness for a Cantor set in R . Suppose E ⊂ R is a Cantorset. Then E can be obtained by successively removing countably many pairwise disjointopen intervals { U i } ∞ i =1 from the closed interval I = conv ( E ), where conv ( E ) denotes theconvex hull of E . In the first step, we remove the open interval U from I , and we obtaintwo closed subintervals L and R . In this case we call I the father interval of U , andcall L , R the generating intervals of U . In the second step, we remove U . Withoutloss of generality we may assume U ⊂ L . Then we obtain two subintervals L and R from L . This procedure can be continued. Suppose in the n -th step we remove U n fromsome closed interval L j for some 1 ≤ j ≤ n −
1, and we obtain two subintervals L n , R n .So L j is the father interval of U n , and L n , R n are the two generating subintervals of U n .Continuing this procedure indefinitely we obtain the Cantor set E . Then the thickness of E introduced by Newhouse [12] is defined as(4.1) τ N ( E ) := sup inf n ∈ N (cid:26) | L n || U n | , | R n || U n | (cid:27) , where L n and R n are the generating intervals of U n in the procedure, and | U | denotesthe length of a subinterval U ⊂ R . Here the supermum is taken over all permutations of { U i } ∞ i =1 . It is worth to mention that the supermum in (4.1) can be achieved by orderingthe lengths of the open intervals { U i } ∞ i =1 in a decreasing order (cf. [2]).Let x ∈ (0 , x i ) = Φ x (1 /m ) be the greedy m -adic expansion of x . Thenthere exist infinitely many i ≥ x i < m −
1. Denote by ( n j ) ⊂ N the set ofall indices i ≥ x i < m −
1. Then for any j ≥ x n j < m − x n j +1 . . . x n j +1 − = ( m − n j +1 − n j − . For j ≥ b ∈ (cid:8) x n j + 1 , x n j + 2 , . . . , m − (cid:9) let p j,b , q j,b ∈ (0 , /m ) be defined by(4.2) Φ x ( p j,b ) = x . . . x n j − b ( m − ∞ and Φ x ( q j,b ) = x . . . x n j − b ∞ . Then by Lemma 2.1 it follows that0 < p j,m − < q j,m − < p j,m − < q j,m − < · · · < p j,x nj +1 < q j,x nj +1 < p j +1 ,m − < q j +1 ,m − < · · · < p j +1 ,x nj +1 +1 < q j +1 ,x nj +1 +1 < · · · < /m (4.3)for all j ≥
1, and p j,m − ր /m as j → ∞ . Note that the intervals I j,b = [ p j,b , q j,b ] with j ≥ b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) are pairwise disjoint. Then we can rename theseintervals in an increasing order: I = [ p , q ] , I = [ p , q ] , · · · , I k = [ p k , q k ] , I k +1 = [ p k +1 , q k +1 ] · · · such that q k < p k +1 for all k ≥
1. So for each interval I k there exist a unique j ∈ N anda unique b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) such that I k = I j,b . Since we have only finitely manychoices for the index b , it follows that k → ∞ is equivalent to j → ∞ . ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 15
Let k ≥
1. Then I k = I j,b for some j ≥ b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) . Now for a word w ∈ { , , . . . , m − } ∗ we define the basic interval associated to w by I k ( w ) = [ p k ( w ) , q k ( w )] , where(4.4) Φ x ( p k ( w )) = x . . . x n j − b w ( m − ∞ , Φ x ( q k ( w )) = x . . . x n j − b w ∞ . Then for the empty word ǫ we have I k ( ǫ ) = [ p k , q k ]. Observe that for any two words u and v , if u is a prefix of v , then I k ( v ) ⊂ I k ( u ). So, for any w ∈ { , , . . . , m − } ∗ we have I k ( w ) ⊃ S m − d =0 I k ( w d ) with the union pairwise disjoint. Furthermore, the leftendpoint of I k ( w ( m − I k ( w ), and the right endpointof I k ( w
0) coincides with the right endpoint of I k ( w ). In other words, these intervals I k ( w ), w ∈ { , , . . . , m − } ∗ have a tree structure. Therefore,(4.5) E k ( x ) := ∞ \ n =0 [ w ∈{ , ,...,m − } n I k ( w )is a Cantor subset of Λ( x ). Here the inclusion E k ( x ) ⊂ Λ( x ) follows by our construction of E k ( x ) that each λ ∈ E k ( x ) corresponds to a unique coding ( d i ) ∈ { , , . . . , m − } N suchthat Φ x ( λ ) = ( d i ), which implies that λ ∈ Λ( x ). Since E k ( x ) ⊂ Λ( x ) for all k ≥
1, we thenconstruct a sequence of Cantor subsets of Λ( x ). Furthermore, by (4.3) it follows that theseCantor subsets E k ( x ) , k ≥ d H ( E k ( x ) , { /m } ) → k → ∞ ,where d H is the Hausdorff metric. I k ( w + ) λ = p k ( w + ) λ = q k ( w + ) G k ( w ) I k ( w ) λ = p k ( w ) λ = q k ( w ) Figure 2.
The geometrical structure of the basic intervals I k ( w + ) =[ p k ( w + ) , q k ( w + )] = [ λ , λ ] and I k ( w ) = [ p k ( w ) , q k ( w )] = [ λ , λ ], and thegap G k ( w ) = ( q k ( w + ) , p k ( w )) = ( λ , λ ).Let k ≥
1. Recall that for a word w = w . . . w n with w n < m − w + = w . . . w n − ( w n + 1). For two neighboring basic intervals I k ( w + ) = [ p k ( w + ) , q k ( w + )] and I k ( w ) = [ p k ( w ) , q k ( w )] of the same level, we call the open interval G k ( w ) := ( q k ( w + ) , p k ( w ))the gap between them (see Figure 2). Now we write τ ( E k ( x )) := inf n ≥ min w , w + ∈{ , ,...,m − } n (cid:26) | I k ( w + ) || G k ( w ) | , | I k ( w ) || G k ( w ) | (cid:27) = inf n ≥ min w , w + ∈{ , ,...,m − } n (cid:26) q k ( w + ) − p k ( w + ) p k ( w ) − q k ( w + ) , q k ( w ) − p k ( w ) p k ( w ) − q k ( w + ) (cid:27) , (4.6)In the next lemma we show that τ N ( E k ( x )) ≥ τ ( E k ( x )). Lemma 4.1.
Let x ∈ (0 , . Then for any k ≥ we have τ N ( E k ( x )) ≥ τ ( E k ( x )) .Proof. Let x ∈ (0 ,
1) and k ≥
1. We remove the open intervals (gaps) G k ( w ) with w ∈{ , , . . . , m − } ∗ from I k ( ǫ ) = [ p k , q k ] in the following way. First we remove from I k ( ǫ ) theopen intervals G k (0), and then the open interval G k (1), and so on, and in the ( m − G k ( m − ≤ d ≤ m − (cid:26) | L k ( d ) || G k ( d ) | , | R k ( d ) || G k ( d ) | (cid:27) ≥ min ≤ d ≤ m − (cid:26) | I k ( d + 1) || G k ( d ) | , | I k ( d ) || G k ( d ) | (cid:27) , where L k ( d ) and R k ( d ) are the generating intervals of G k ( d ) in the construction of E k ( x )by Newhouse. After removing the ( m −
1) open intervals G k (0) , G k (1) , . . . , G k ( m − m basic intervals I k (0) , I k (1) , . . . , I k ( m − I k ( d )we remove from I k ( d ) the open intervals G k ( d , G k ( d , . . . , G k ( d ( m − m basic subintervals I k ( d , I k ( d , . . . , I k ( d ( m − I k ( w ) for some w of length n . Thenwe successively remove from I k ( w ) the open intervals G k ( w , G k ( w , . . . , G k ( w ( m − m basic subintervals I k ( w , I k ( w , . . . , I k ( w ( m − ≤ d ≤ m − (cid:26) | L k ( w d ) || G k ( w d ) | , | R k ( w d ) || G k ( w d ) | (cid:27) ≥ min ≤ d ≤ m − (cid:26) | I k ( w ( d + 1)) || G k ( w d ) | , | I k ( w d ) || G k ( w d ) | (cid:27) , where L k ( w d ) and R k ( w d ) are the generating intervals of G k ( w d ). By induction it followsthat min w , w + ∈{ , ,...,m − } n (cid:26) | L k ( w ) || G k ( w ) | , | R k ( w ) || G k ( w ) | (cid:27) ≥ min w , w + ∈{ , ,...,m − } n (cid:26) | I k ( w + ) || G k ( w ) | , | I k ( w ) || G k ( w ) | (cid:27) for all n ≥
1. Therefore, by (4.1) and (4.6) we conclude that τ N ( E k ( x )) ≥ inf n ≥ min w , w + ∈{ , ,...,m − } n (cid:26) | L k ( w ) || G k ( w ) | , | R k ( w ) || G k ( w ) | (cid:27) ≥ inf n ≥ min w , w + ∈{ , ,...,m − } n (cid:26) | I k ( w + ) || G k ( w ) | , | I k ( w ) || G k ( w ) | (cid:27) = τ ( E k ( x )) . (cid:3) In the following we show that the thickness of E k ( x ) goes to infinitey as k → ∞ . Proposition 4.2.
Let x ∈ (0 , . Then τ ( E k ( x )) → + ∞ as k → ∞ .Proof. Let x ∈ (0 , x i ) = Φ x (1 /m ) be the greedy m -adic expansion of x . Thenthere exists a smallest integer ℓ ≥ x ℓ >
0. Take k large enough, so thereexist j ≥ , b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) such that I k = I j,b and n j > ℓ . For n ≥ ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 17 w ∈ { , , . . . , m − } n such that w + ∈ { , , . . . , m − } n . In view of (4.6) we need toestimate the lower bounds for the two quotients q k ( w + ) − p k ( w + ) p k ( w ) − q k ( w + ) , q k ( w ) − p k ( w ) p k ( w ) − q k ( w + ) . For simplicity we write λ = p k ( w + ) , λ = q k ( w + ) , λ = p k ( w ) and λ = q k ( w ). Then λ < λ < λ < λ (see Figure 2). So we need to estimate the lower bounds for λ − λ and λ − λ , and the upper bounds for λ − λ . A lower bound for λ − λ . Note by (4.4) that π λ ( x . . . x n j − b w + ( m − ∞ ) = x = π λ ( x . . . x n j − b w + ∞ ) . Rearranging the above equation we obtain π λ (0 n j + n ( m − ∞ ) = π λ ( x . . . x n j − b w + ( m − ∞ ) − π λ ( x . . . x n j − b w + ( m − ∞ ) ≤ π λ (( m − ∞ ) − π λ (( m − ∞ )= m − − λ λ − m − − λ λ = m − − λ )(1 − λ ) ( λ − λ ) . This implies that(4.7) λ − λ ≥ (1 − λ )(1 − λ ) m − π λ (0 n j + n ( m − ∞ ) = (1 − λ ) λ n j + n +12 . A lower bound for λ − λ . Note by (4.4) that π λ ( x . . . x n j − b w ( m − ∞ ) = x = π λ ( x . . . x n j − b w ∞ ) . Then by the same argument as in the estimate for λ − λ one can verify that(4.8) λ − λ ≥ (1 − λ ) λ n j + n +14 . An upper bound for λ − λ . Note by (4.4) that π λ ( x . . . x n j − b w + ∞ ) = x = π λ ( x . . . x n j − b w ( m − ∞ ) . This implies that π λ (0 n j + n − ∞ ) − π λ (0 n j + n ( m − ∞ )= π λ ( x . . . x n j − b w ∞ ) − π λ ( x . . . x n j − b w ∞ ) ≥ λ ℓ − λ ℓ ≥ λ ℓ − ( λ − λ ) , where the first inequality follows by the definition of ℓ that ℓ < n j and x ℓ >
0. So, byusing λ > λ it follows that λ − λ ≤ λ − ℓ (cid:18) λ n j + n − m − − λ λ n j + n +13 (cid:19) ≤ λ n j − ℓ + n +12 (cid:18) − m − − λ λ (cid:19) = λ n j − ℓ + n +12 m − λ (cid:18) m − λ (cid:19) . (4.9)From this we still need to estimate 1 /m − λ . Note that Φ x (1 /m ) = ( x i ). Then π λ ( x . . . x n j − b w ( m − ∞ ) = x = π /m ( x x . . . ) . This yields that π λ (0 n j − ( b − x n j ) w ( m − ∞ ) − π /m (0 n j x n j +1 x n j +2 . . . )= π /m ( x . . . x n j ∞ ) − π λ ( x . . . x n j ∞ ) ≥ (1 /m ) ℓ − λ ℓ ≥ λ ℓ − (1 /m − λ ) . So, by using b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) it follows that1 /m − λ ≤ λ − ℓ π λ (0 n j − ( b − x n j ) w ( m − ∞ ) ≤ λ − ℓ π λ (0 n j − ( m − ∞ ) < λ − ℓ λ n j − = λ n j − ℓ . (4.10)Substituting (4.10) into (4.9) we obtain an upper bound for λ − λ :(4.11) λ − λ ≤ mλ n j − ℓ − λ λ n j − ℓ + n +12 . Therefore, by (4.7) and (4.11) it follows that(4.12) q k ( w + ) − p k ( w + ) p k ( w ) − q k ( w + ) = λ − λ λ − λ ≥ (1 − λ )(1 − λ ) λ ℓ mλ n j − ℓ → + ∞ as k → ∞ . Here we emphasize that ℓ ∈ N depends only on x , and k → ∞ implies j → ∞ . Note that λ > λ . Then by (4.8) and (4.11) we obtain that(4.13) q k ( w ) − p k ( w ) p k ( w ) − q k ( w + ) = λ − λ λ − λ ≥ (1 − λ ) λ ℓ mλ n j − ℓ → + ∞ as k → ∞ . Hence, according to (4.6), and by (4.12) and (4.13) we conclude that τ ( E k ( x )) → + ∞ as k → ∞ . This completes the proof. (cid:3) Remark . (1) Note by [13] that for any Cantor set E we have(4.14) dim H E ≥ log 2log(2 + τ N ( E ) ) . ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 19
So, by Lemma 4.1, Proposition 4.2 and (4.14) it follows that for any x ∈ (0 , H Λ( x ) ≥ dim H E k ( x ) ≥ log 2log(2 + τ N ( E k ( x )) ) ≥ log 2log(2 + τ ( E k ( x )) ) → k → ∞ . This provides an alternative proof of dim H Λ( x ) = 1.(2) It is well-known that for a Cantor sets E ⊂ R , if τ N ( E ) ≥
1, then the algebraicsum E + E := { a + b : a, b ∈ E } contains an interval (cf. [12]). So, Proposition 4.2suggests that the algebraic sum Λ( x ) + Λ( x ) contains an interval for any x ∈ (0 , E ⊂ R is said to contain arbitrarily long arithmetic progressions iffor any n ∈ N there exist a, b ∈ R such that { a + b, a + 2 b, a + 3 b, . . . , a + nb } ⊂ E .Recently, Yavicoli [14, Remark of Theorem 4] proved that for any n ∈ N thereexists a (finite) constant τ n such that if the thickness of E is larger than τ n , then E contains arithmetic progressions of length n . This, together with Proposition 4.2,implies that Λ( x ) contains arbitrarily long arithmetic progressions for any x ∈ (0 , Large intersections of Λ( x ) and Λ( y ) . In order to prove Theorem 1.5 we recall thefollowing results from Kraft [10, Theorem 1.1] and Hunt et al. [7, Theorem 1]. Two Cantorsets F and F are called interleaved if F ∩ conv ( F ) = ∅ and conv ( F ) ∩ F = ∅ . Lemma 4.4.
Let x, y ∈ (0 , . If there exist i, j ∈ N such that E i ( x ) and E j ( y ) areinterleaved with τ i,j ( x, y ) := min { τ N ( E i ( x )) , τ N ( E j ( y )) } > √ , then E i ( x ) ∩ E j ( y ) contains a Cantor subset of thickness at least of order p τ i,j ( x, y ) . By Lemma 4.1 and the proof of Proposition 4.2 it follows that for any x ∈ (0 ,
1) wehave τ N ( E i ( x )) → ∞ as i → ∞ . So, if we can show that there exist infinitely many pairs( i, j ) ∈ N × N such that E i ( x ) and E j ( y ) are interleaved Cantor sets, then Theorem 1.5can be deduced from Lemma 4.4 and (4.14). In the following we will show that there existinfinitely many pairs ( i, j ) such that E i ( x ) and E j ( y ) are interleaved.For x ∈ (0 ,
1) we recall from (4.5) that E k ( x ) is a Cantor subset of Λ( x ) for any k ≥ I k = conv ( E k ( x )) = [ p k , q k ] there exist a unique j ≥ b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) such that I k = I j,b . Then the endpoints p k , q k are determined byΦ x ( p k ) = x . . . x n j − b ( m − ∞ and Φ x ( q k ) = x . . . x n j − b ∞ . First we define the thickness of the interval sequence { I k : k ≥ } by θ ( x ) := lim inf k →∞ θ k ( x ) , where(4.15) θ k ( x ) := min (cid:26) | I k || G k | , | I k +1 || G k | (cid:27) = min (cid:26) q k − p k p k +1 − q k , q k +1 − p k +1 p k +1 − q k (cid:27) . Here G k = ( q k , p k +1 ) is the gap between I k and I k +1 (see Figure 3).In the following lemma we show that the thickness of the interval sequence { I k : k ≥ } is infinity for any x ∈ (0 , I k p k q k I k +1 p k +1 q k +1 1 m Figure 3.
The geometrical structure of the basic intervals I k = [ p k , q k ] = conv ( E k ( x )) and I k +1 = [ p k +1 , q k +1 ] = conv ( E k +1 ( x )). They converge to { /m } in the Hausdorff metric d H as k → ∞ . Lemma 4.5.
For any x ∈ (0 , we have θ ( x ) = + ∞ .Proof. The idea to prove this lemma is similar to that for Proposition 4.2. By the definitionof θ ( x ) we need to estimate the lower bounds θ k ( x ) for k large enough. Take k ∈ N suffi-ciently large. Then there exist a unique j ∈ N and a unique b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) such that I k = I j,b . In view of (4.15) we consider the following two cases: (I) b ∈ (cid:8) x n j + 2 , . . . , m − (cid:9) ; (II) b = x n j + 1.Case (I). b ∈ (cid:8) x n j + 2 , . . . , m − (cid:9) . Then I k = I j,b and I k +1 = I j,b − . We need toestimate the lower bounds of | I k | = q k − p k and | I k +1 | = q k +1 − p k +1 , and the upper boundsof | G k | = p k +1 − q k . Observe that π q k ( x . . . x n j − b ∞ ) = x = π p k ( x . . . x n j − b ( m − ∞ ) . This implies that π q k (0 n j ( m − ∞ ) = π q k ( x . . . x n j − b ( m − ∞ ) − π p k ( x . . . x n j − b ( m − ∞ ) ≤ m − − q k q k − m − − p k p k = m − − p k )(1 − q k ) ( q k − p k ) . Whence,(4.16) q k − p k ≥ (1 − p k )(1 − q k ) m − × m − − q k q n j +1 k = (1 − p k ) q n j +1 k . Similarly, one can prove that(4.17) q k +1 − p k +1 ≥ (1 − p k +1 ) q n j +1 k +1 . Now we turn to the upper bounds of p k +1 − q k . Note that π q k ( x . . . x n j − b ∞ ) = x = π p k +1 ( x . . . x n j − ( b − m − ∞ ) . Then π q k (0 n j − ∞ ) − π p k +1 (0 n j ( m − ∞ )= π p k +1 ( x . . . x n j − ( b − ∞ ) − π q k ( x . . . x n j − ( b − ∞ ) ≥ p ℓk +1 − q ℓk ≥ q ℓ − k ( p k +1 − q k ) , ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 21 where ℓ < n j is the smallest integer such that x ℓ >
0. Therefore, p k +1 − q k ≤ q − ℓk (cid:18) q n j k − m − − p k +1 p n j +1 k +1 (cid:19) ≤ q n j − ℓ +1 k (cid:18) − m − − p k +1 p k +1 (cid:19) = q n j − ℓ +1 k m − p k +1 (cid:18) m − p k +1 (cid:19) . (4.18)So we still need to estimate 1 /m − p k +1 . Observe that π p k +1 ( x . . . x n j − ( b − m − ∞ ) = x = π /m ( x . . . x n j x n j +1 . . . ) . Then π p k +1 (0 n j − ( b − − x n j )( m − ∞ ) ≥ π /m ( x . . . x n j ∞ ) − π p k +1 ( x . . . x n j ∞ ) ≥ (1 /m ) ℓ − p ℓk +1 ≥ p ℓ − k +1 (1 /m − p k +1 ) , which implies that(4.19) 1 /m − p k +1 ≤ p − ℓk +1 p n j − k +1 = p n j − ℓk +1 . Substituting (4.19) into (4.18) we obtain an upper bound for p k +1 − q k :(4.20) p k +1 − q k ≤ q n j − ℓ +1 k m − p k +1 p n j − ℓk +1 . Hence, by (4.16), (4.17) and (4.20) it follows that θ k ( x ) = min (cid:26) q k − p k p k +1 − q k , q k +1 − p k +1 p k +1 − q k (cid:27) ≥ min ( (1 − p k )(1 − p k +1 ) q ℓk mp n j − ℓk +1 , (1 − p k +1 ) q ℓk +1 mp n j − ℓk +1 ) → + ∞ as k → ∞ .Case (II). b = x n j + 1. Then I k = I j,b = [ p k , q k ] and I k +1 = I j +1 ,m − = [ p k +1 , q k +1 ]. Inthis case we haveΦ x ( p k ) = x . . . x + n j ( m − ∞ , Φ x ( q k ) = x . . . x + n j ∞ ;Φ x ( p k +1 ) = x . . . x n j +1 − ( m − ∞ = x . . . x n j ( m − ∞ , Φ x ( q k +1 ) = x . . . x n j +1 − ( m − ∞ = x . . . x n j ( m − n j +1 − n j ∞ . Here we emphasize that by the definition of ( n j ) it follows that x i = m − n j < i < n j +1 . By the same argument as in Case I one can prove that(4.21) q k − p k ≥ (1 − p k ) q n j +1 k , q k +1 − p k +1 ≥ (1 − p k +1 ) q n j +1 +1 k +1 , and(4.22) p k +1 − q k ≤ q n j − ℓ +1 k m − p k +1 p n j +1 − ℓk +1 . Therefore, by (4.21) and (4.22) it follows that θ k ( x ) = min (cid:26) q k − p k p k +1 − q k , q k +1 − p k +1 p k +1 − q k (cid:27) → + ∞ as k → ∞ .Hence, by Cases (I) and (II) we conclude that θ ( x ) = + ∞ . (cid:3) In order to prove that E i ( x ) and E j ( y ) are interleaved for infinitely many pairs ( i, j ) westill need the following estimate. Lemma 4.6.
Let x ∈ (0 , . Then there exist a constant η ( x ) > and a large integer N such that for any k ≥ N we have | I k +1 || I k | = q k +1 − p k +1 q k − p k < η ( x ) . Proof.
Suppose I k = I j,b = [ p k , q k ] for some j ≥ b ∈ (cid:8) x n j + 1 , . . . , m − (cid:9) . We splitthe proof into the following two cases: (I) b ∈ (cid:8) x n j + 2 , . . . , m − (cid:9) ; (II) b = x n j + 1.Case (I). b ∈ (cid:8) x n j + 2 , . . . , m − (cid:9) . Then I k +1 = I j,b − = [ p k +1 , q k +1 ]. Note by (4.16)that q k − p k ≥ (1 − p k ) q n j +1 k . In the following it suffices to estimate the upper bounds of q k +1 − p k +1 . Observe that π q k +1 ( x . . . x n j − ( b − ∞ ) = x = π p k +1 ( x . . . x n j − ( b − m − ∞ ) . Then π p k +1 (0 n j ( m − ∞ ) = π q k +1 ( x . . . x n j − ( b − ∞ ) − π p k +1 ( x . . . x n j − ( b − ∞ ) ≥ q ℓk +1 − p ℓk +1 ≥ p ℓ − k +1 ( q k +1 − p k +1 ) , where ℓ < n j is the smallest integer such that x ℓ >
0. This implies that(4.23) q k +1 − p k +1 ≤ p − ℓk +1 m − − p k +1 p n j +1 k +1 < p n j − ℓ +1 k +1 , where the last inequality follows by p k +1 < /m . Therefore, by (4.16) and (4.23) it followsthat(4.24) q k +1 − p k +1 q k − p k < p n j − ℓ +1 k +1 (1 − p k ) q n j +1 k = 1(1 − p k ) p ℓk +1 (cid:18) p k +1 q k (cid:19) n j +1 . Since p k ր /m as k → ∞ , there exists a large integer N such that p k +1 > m for any k > N . Then (4.24) suggests that(4.25) q k +1 − p k +1 q k − p k < − /m )(1 / m ) ℓ (cid:18) p k +1 q k (cid:19) n j +1 = m (2 m ) ℓ m − (cid:18) p k +1 q k (cid:19) n j +1 ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 23 for any k > N . Note that k → ∞ is equivalent to j → ∞ . So, to finish the proof itsuffices to prove that lim k →∞ ( p k +1 /q k ) n j +1 = 1. This follows by (4.20) thatlim sup k →∞ (cid:18) p k +1 q k (cid:19) n j +1 = lim sup k →∞ (cid:18) p k +1 − q k q k (cid:19) n j +1 ≤ lim sup k →∞ (cid:18) q n j − ℓk m − p k +1 p n j − ℓk +1 (cid:19) n j +1 = exp (cid:20) lim sup k →∞ (cid:18) ( n j + 1) q n j − ℓk m − p k +1 p n j − ℓk +1 (cid:19)(cid:21) = 1 , and that ( p k +1 /q k ) n j +1 ≥ k ≥
1. So, by (4.25) there must exist an integer
N > N such that for any k > N , q k +1 − p k +1 q k − p k < (2 m ) ℓ +1 m − b = x n j + 1. Then I k +1 = I j +1 ,m − = [ p k +1 , q k +1 ]. Note by (4.21) that q k − p k ≥ (1 − p k ) q n j +1 k . In the following it suffices to estimate the upper bounds for q k +1 − p k +1 . Observe that π q k +1 ( x . . . x n j ( m − n j +1 − n j ∞ ) = x = π p k +1 ( x . . . x n j ( m − ∞ ) . Then by the same argument as in Case (I) one can show that q k +1 − p k +1 ≤ p n j +1 − ℓ +1 k +1 . This together with (4.21) implies that(4.26) q k +1 − p k +1 q k − p k ≤ p n j +1 − ℓ +1 k +1 (1 − p k ) q n j +1 k = q n j +1 − n j k (1 − p k ) p ℓk +1 (cid:18) p k +1 q k (cid:19) n j +1 +1 . Since p k ր /m as k → ∞ , and n j +1 − n j ≥
1, there exists a large integer N such that p k +1 > / (2 m ) for any k > N . Then (4.26) suggests that(4.27) q k +1 − p k +1 q k − p k ≤ − /m )(1 / m ) ℓ (cid:18) p k +1 q k (cid:19) n j +1 +1 = m (2 m ) ℓ m − (cid:18) p k +1 q k (cid:19) n j +1 +1 for any k > N . Note by (4.22) thatlim sup k →∞ (cid:18) p k +1 q k (cid:19) n j +1 +1 = lim sup k →∞ (cid:18) p k +1 − q k q k (cid:19) n j +1 +1 ≤ lim sup k →∞ (cid:18) q n j − ℓk m − p k +1 p n j +1 − ℓk +1 (cid:19) n j +1 +1 = exp (cid:20) lim sup k →∞ (cid:18) ( n j +1 + 1) p n j +1 − ℓk +1 q n j − ℓk m − p k +1 (cid:19)(cid:21) = 1 . So, by (4.27) there must exist an integer
N > N such that for any k > N , q k +1 − p k +1 q k − p k < (2 m ) ℓ +1 m − . This completes the proof. (cid:3)
Based on Lemmas 4.5 and 4.6 we are ready to show that E i ( x ) and E j ( y ) are interleavedfor infinitely many pairs ( i, j ) ∈ N × N . Lemma 4.7.
For any x, y ∈ (0 , there exists a sequence of pairs ( i k , j k ) ∈ N × N suchthat for any k ≥ we have i k < i k +1 , j k < j k +1 , and the two Cantor sets E i k ( x ) , E j k ( y ) areinterleaved.Proof. Take x, y ∈ (0 , x = y . So in the followingwe assume x = y . To emphasize the dependence on x we write I xi := conv ( E i ( x )) =[ p xi , q xi ]. We also denote by G xi := ( q xi , p xi +1 ) the gap between the two intervals I xi and I xi +1 .Observe that the closed intervals { I xi : i ≥ } are pairwise disjoint, and for each i ≥ I xi +1 is on the righthand side of I xi (see Figure 3). Furthermore, p xi ր /m as i → ∞ . Let η ( x ) > η ( y ) > η := max { η ( x ) , η ( y ) } . Then by Proposition 4.2, Lemma 4.5 and Lemma 4.6 there exist N ∈ N such that for any i, j > N we have(4.28) τ ( E i ( x )) > η, τ ( E j ( y )) > η ;(4.29) θ i ( x ) = min (cid:26) | I xi || G xi | , | I xi +1 || G xi | (cid:27) > η, θ j ( y ) = min (cid:26) | I yj || G yj | , | I yj +1 || G yj | (cid:27) > η ;and(4.30) | I xi +1 || I xi | < η ( x ) , | I yj +1 || I yj | < η ( y ) . Then (4.29) suggests that for each i > N the length of the gap G xi is strictly smaller thanthe length of each of its neighboring intervals I xi and I xi +1 . Also, for each j > N the lengthof the gap G yj is strictly smaller than the length of each of its neighboring intervals I yj and I yj +1 . Claim 1: there must exist a pair ( i, j ) ∈ N × N with i, j > N such that I xi ∩ I yj = ∅ . Proof of Claim 1.
Take δ > N > N such that I xN ∩ (1 /m − δ, /m ) = ∅ , and there exists a smallest integer N > N such that I yN ∩ (1 /m − δ, /m ) = ∅ . Then I xi ⊂ (1 /m − δ, /m ) for any i > N , and I yj ⊂ (1 /m − δ, /m ) for any j > N . So to prove Claim 1 it suffices to prove that thereexist i > N and j > N such that I xi ∩ I yj = ∅ . Suppose this is not true. Then each basicinterval I xi with i > N must belong to a gap of the interval sequence (cid:8) I yj : j ≥ N (cid:9) . Inother words, for each i > N there must exists j ≥ N such that I xi ⊂ G yj . Observe that p xi ր /m as i → ∞ , and p yj ր /m as j → ∞ . So there must exist i > N such that I xi ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 25 and I xi +1 belong to two different gaps of the interval sequence (cid:8) I yj : j ≥ N (cid:9) , say I xi ⊂ G yj and I xi +1 ⊂ G yj ′ for some j ′ > j ≥ N (see Figure 4). Since θ i ( x ) > I yj I xi I xi +1 I yj +1 I yj ′ I yj ′ +1 Figure 4.
The geometrical structure of the closed intervals I xi = [ p xi , q xi ] = conv ( E i ( x )) and I xi +1 = [ p xi +1 , q xi +1 ] = conv ( E i +1 ( x )). They belong to thegaps G yj and G yj ′ respectively. | I yj ′ | ≤ | G xi | < min (cid:8) | I xi | , | I xi +1 | (cid:9) . This implies that θ j ′ ( y ) ≤ | I yj ′ || G yj ′ | ≤ | I yj ′ || I xi +1 | < , which leads to a contradiction with (4.29) that θ j ( y ) > η > j > N . Hence, theremust exist i, j > N such that I xi ∩ I yj = ∅ , proving Claim 1. Claim 2: there exists a pair ( i , j ) ∈ N × N with i , j > N such that the two Cantorsets E i ( x ) and E j ( y ) are interleaved. Proof of Claim 2.
By Claim 1 we can choose i, j > N such that I xi ∩ I yj = ∅ . If I xi * I yj and I xi + I yj , then the two Cantor sets E i ( x ) and E j ( y ) are interleaved, and thenwe are done by setting i = i and j = j . Otherwise, without loss of generality we mayassume I xi ⊂ I yj = conv ( E j ( y )). If I xi is not contained in a gap of E j ( y ), then E i ( x ) and E j ( y ) are interleaved Cantor sets, and again we are done by setting i = i and j = j .In the following we assume that I xi is contained in a gap of E j ( y ), say I xi ⊂ G yj ( w ) forsome word w of smallest length. Here G yj ( w ) is the gap between the two basic intervals I yj ( w +1 ) and I yj ( w ) which are defined as in (4.6), see Figure 5. We set s = i and j = j . I yj ( w +1 ) I yj ( w ) I xs I xs − Figure 5.
The geometrical structure of the closed intervals I yj ( w +1 ) , I xs , I xs − and I yj ( w ). The gap G yj ( w ) is the open intervalbetween the two basic intervals I yj ( w +1 ) and I yj ( w ).Note by (4.28) that the thickness of E j ( y ) is strictly larger than one. Then the lengthof the interval I yj ( w ) is strictly larger than the length of the gap G yj ( w ). Furthermore, note by (4.29) that the thickness of the interval sequence { I xi : i > N } is larger than one,and p xi ր /m as i → ∞ , we claim that there must exist a smallest integer s > s suchthat(4.31) I xs ∩ I yj ( w ) = ∅ . Suppose on the contrary that there is no such s > s satisfying I xs ∩ I yj ( w ) = ∅ . Thenthere is an s > s such that I yj ( w ) is contained in the gap G xs between I xs and I xs +1 . Thisimplies that θ s ( x ) ≤ | I xs || G xs | ≤ | I xs || I yj ( w ) | < , where the last inequality follows by that I xs ⊂ G yj ( w ) and | G yj ( w ) | < | I yj ( w ) | . However,this leads to a contradiction with (4.29) that θ i ( x ) > i > N . Hence, we prove theexistence of s such that (4.31) holds.Note by (4.28) and (4.30) that(4.32) | I xs | ≤ η ( x ) | I xs − | ≤ η | G yj ( w ) | < | I yj ( w ) | . Since I xs ∩ I yj ( w ) = ∅ , by (4.32) it follows that if I xs is not contained in a gap of E j ( y ) ∩ I yj ( w ), then E s ( x ) and E j ( y ) are interleaved, and then we are done by setting i = s .Otherwise, suppose I xs is contained in a gap of E j ( y ) ∩ I yj ( w ). Then there exists a word w of smallest length such that I xs ⊂ G yj ( w w ). Note that the thickness of E j ( y ) andthe thickness of { I xi : i ≥ N } are both larger than one. Then by the same argument as in(4.31) there exists a smallest integer s > s such that I xs ∩ I yj ( w w ) = ∅ . Again, by the same reason as in the proof of (4.32) we can prove that | I xs | < | I yj ( w w ) | .So, if I xs is not contained in a gap of E j ( y ) ∩ I yj ( w w ), then E s ( x ) and E j ( y ) areinterleaved, and then we set i = s . Otherwise, there exists a word w of smallest lengthsuch that I xs ⊂ G yj ( w w w ).Repeating the above argument, and we claim that our procedure must stop at somefinite time n , i.e., there exist an integer s n and words w , w , . . . , w n such that I xs n ∩ I yj ( w w · · · w n ) = ∅ , | I xs n | < | I yj ( w w . . . w n ) | , and I xs n is not contained in a gap of E j ( y ) ∩ I yj ( w w · · · w n ). If this claim is true, thenwe are done by setting i = s n . Suppose it is not true. Then I xi ⊂ I yj for all large integers i > s , which implies that q xi ≤ q yj < /m for all i ≥
1. This leads to a contradiction withlim i →∞ q xi = 1 /m . Therefore, we have find a pair ( i , j ) ∈ N × N with i , j > N such thatthe two Cantor sets E i ( x ) and E j ( y ) are interleaved. This proves Claim 2.Now we take δ > /m − δ > max (cid:8) q xi , q yj (cid:9) , and repeating theargument as in Claim 1 and Claim 2. Then we can find another pair ( i , j ) ∈ N × N with i > i , j > j such that E i ( x ) and E j ( y ) are interleaved. We can proceed this argumentindefinitely, and then we find a sequence of pairs ( i k , j k ) ∈ N × N such that for any k ≥ ARGE INTERSECTION OF UNIVOQUE BASES OF REAL NUMBERS 27 we have i k < i k +1 , j k < j k +1 , and the two Cantor sets E i k ( x ) and E j k ( y ) are interleaved.This completes the proof. (cid:3) Proof of Theorem 1.5.
By Lemmas 4.4 and 4.7 it follows that Λ( x ) ∩ Λ( y ) contains a Cantorset of thickness arbitrarily large. More precisely, for the infinite sequence of pairs ( i k , j k ) ∈ N × N defined as in Lemma 4.7 it follows that for any k ≥ x ) ∩ Λ( y ) ⊃ E i k ( x ) ∩ E j k ( y )and the intersection E i k ( x ) ∩ E j k ( y ) contains a Cantor set with its thickness at least oforder p min { τ N ( E i k ( x )) , τ N ( E j k ( y )) } . Note by Lemma 4.1 and Proposition 4.2 thatlim k →∞ τ N ( E i k ( x )) = lim k →∞ τ N ( E j k ( y )) = + ∞ . Then by (4.14) and (4.33) we conclude thatdim H (Λ( x ) ∩ Λ( y )) ≥ dim H ( E i k ( x ) ∩ E j k ( y )) → k → ∞ . This completes the proof. (cid:3)
Acknowledgements
The first author was supported by NSFC No. 11701302, Zhejiang Provincial NaturalScience Foundation of China with No.LY20A010009, and the K.C. Wong Magna Fund inNingbo University. The second author was supported by NSFC No. 11971079 and theFundamental and Frontier Research Project of Chongqing No. cstc2019jcyj-msxmX0338and No. cx2019067. The third author was supported by NSFC No. 12071148 and Scienceand Technology Commission of Shanghai Municipality (STCSM) No. 18dz2271000.
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Department of Mathematics, Ningbo University, Ningbo, Zhejiang, People’sRepublic of China
Email address : [email protected] (D. Kong) College of Mathematics and Statistics, Chongqing University, 401331, Chongqing,P.R.China
Email address : [email protected] (W. Li) School of Mathematical Sciences, Shanghai Key Laboratory of PMMP, EastChina Normal University, Shanghai 200062, People’s Republic of China
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