The simplest erasing substitution
TThe simplest erasing substitution
Alessandro Della Corte Stefano Isola Riccardo Piergallini
Scuola di Scienze e Tecnologie – Universit`a di Camerino – Italy
Abstract
In this work, we begin the study of a new class of dynamical systems determined byinterval maps generated by the symbolic action of erasing substitution rules. We do thisby discussing in some detail the geometric, analytical, dynamical and arithmetic propertiesof a particular example, which has the virtue of being arguably the simplest and that atthe same time produces interesting properties and new challenging problems.
Keywords : topological dynamics, Baire class 1 function, erasing substitution, distribu-tional chaos.
AMS Classification : 37B10, 37B05, 37B40, 26A18, 26A21, 54C50, 54H20.
Introduction
Substitutive dynamical systems are a widely studied and quite well-understood class (see forinstance [18, 35]). In this context, it is usual to intend the term substitution in a rather specificsense. Namely, a substitution S is typically understood as a rule replacing every element froma given alphabet by a finite word on the same alphabet, in order to extend S by concatenationto a morphism over all the finite or infinite words. The morphic nature of S means that theaction of the substitution on a certain symbol within a word w is independent of its positionin w . Moreover, most of the times no symbol is mapped by S to the empty word, that is thesubstitution is assumed to be non-erasing .In recent years, erasing substitutions have been taken in some consideration within combina-torics of words (see for instance [16, 36]), mostly in the context of computer science, whilelittle attention seems to be devoted to the analytical and dynamical properties of real mapsgenerated by the symbolic action of erasing substitutions. The shift from symbolic spaces tothe continuous real context, that is from zero-dimensional spaces to one-dimensional ones, hasrelevant consequences on the richness of the properties involved and the problems that seemsnatural to consider. Lately, topological dynamics questions initially considered only in caseof continuous maps, such as the structure of ω -limit sets, topological entropy, the presenceof Devaney, Li-Yorke or distributional chaos and mixing properties, have been addressed in amore general context. For instance: Darboux interval maps of Baire class 1 are considered in[22, 41], general Darboux interval maps are investigated from the point of view of the connec-tions between transitivity, turbulence and topological entropy in [31, 32], interval maps with G δ connected graph are studied in [9, 10], while in [21] an adjustment of the concept of Devaneychaos is proposed for discontinuous maps. In this connection, the study of the dynamics ofreal maps generated by erasing block substitutions appears as a natural development, as theobjects constructed in this way fall into a category which is in a way the direct generalizationof the ones cited above, that is Baire class 1, generally not Darboux functions. The systematicinvestigation of the dynamical properties of these maps has just begun; for instance, genericinterval maps of Baire class 1 are studied in [38, 39, 19]. a r X i v : . [ m a t h . D S ] F e b ur aim is the investigation of an interval map generated by what is arguably the simplesterasing substitution rule, which we will indicate by ρ . We are led to the choice of ρ as follows.We start selecting the smallest non-trivial alphabet, that is { , } . The simplest way to get anerasing substitution is to map one of the symbols, say 0, to the empty word. Then, we mustmake a choice on how to transform the other symbol 1. Clearly, to obtain a non-trivial map,we cannot replace every 1 with a fixed word, independently of its position. Making once againthe simplest choice, we distinguish between odd and even positions, replacing one case with thesymbol 0 and the other with the symbol 1.Even in this model-case, the real map R : [0 , → [0 ,
1] generated by the symbolic action of ρ on binary expansions, presents interesting properties and challenging problems, which involve,among others, measure-theoretic, dimension-theoretic, topological, dynamical and arithmeticalaspects. As we will see better in Section 2, from an operational point of view the map is hardlymanageable, as to obtain the first m digits of the n -th iterate of its action at point x we musttypically provide x with accuracy of order ≈ − mn . This entails that it is almost impossible toinvestigate in a purely numerical fashion objects as ω -limit sets and attractors. Moreover, aswe will see, there is an uncountable and dense subset of [0 ,
1] exhibiting extreme distributionalchaos, which means that the qualitative relative behaviour of sets of points, even arbitrarilyclose, cannot be extrapolated from finite samplings of the dynamics of any time length.It is this combination of simplicity of the object and variety of related properties which in ouropinion confers interest to the proposed subject.The main properties of the map R , defined in Sections 2 and 3, can be summarized as follows.1. The graph of R is a totally disconnected subset of the plane, whose Hausdorff dimensionis between 1 and log R is a singular Borel map of Baire class 1, it is not Darboux, and its mean value is the(admittedly a bit surprising) rational number 3 / R are null-measure Cantor sets, withHausdorff dimension belonging to [1 / , log ϕ ] (Sections 5 and 6).4. There are two rational R -cycles of order two which attract (in a finite time) every rational,moreover R has uncountably many periodic points for any given period, it exhibits Devaneyand (uniform) DC1 chaos, it is topologically mixing, it has infinite topological entropy andevery point in [0 ,
1] is a full entropy point (Section 6).5. The combinatorially simplest periodic points of odd period (including 1) are transcendentalnumbers and there are uncountably many periodic points that are transcendental as wellfor every given period (Section 7).For the reader’s convenience, a list of the most used symbols is given at the end of the paper.
Let A = N ω be the set of all infinite sequences of non-negative integers, and B = { , } ω ⊂ N ω be the subset of all infinite binary sequences, both endowed with the metric given by d ( s, t ) = 1 / min { k ≥ | s k (cid:54) = t k } , for every s = ( s , s , . . . , s n , . . . ) and t = ( t , t , . . . , t n , . . . ) such that s (cid:54) = t .Then, B splits as the union of two dense subspaces B = C ∪ C (cid:48) , C consists of all the infinite binary sequences which are not eventually 0, while C (cid:48) consistsof all the infinite binary sequences which are not eventually 1.There is a 1-Lipschitz homeomorphism η : A → C defined by a = ( a , a , . . . , a n , . . . ) η (cid:55)−→ η ( a ) = (0 a , , a , , . . . , a n , . . . )= (0 , . . . , (cid:124) (cid:123)(cid:122) (cid:125) a , , , . . . , (cid:124) (cid:123)(cid:122) (cid:125) a , , . . . , , . . . , (cid:124) (cid:123)(cid:122) (cid:125) a n , , . . . ) ∈ C (1)for every a = ( a , a , . . . , a n , . . . ) ∈ A .For the sake of convenience, we introduce the notation (cid:104) a (cid:105) b = η (2 a + η − ( b )) (2)with a ∈ A and b ∈ C . If a = ( a , a , . . . , a n , . . . ) then (cid:104) a (cid:105) b is obtained from b by inserting 0 a k immediately before the k -th occurrence of 1 in b . In particular, (cid:104) a (cid:105) ∞ = (cid:104) a (cid:105) (1 , ,... ) = η (2 a ).There is also a 2-Lipschitz map ξ : B → [0 ,
1] defined by b = ( b , b , . . . , b n , . . . ) ξ (cid:55)−→ ξ ( b ) = ∞ (cid:88) n =1 b n n = 0 .b b . . . b n . . . ∈ [0 ,
1] (3)for every b = ( b , b , . . . , b n , . . . ) ∈ B .The map ξ splits as the union of two bijective maps ξ | C : C → (0 ,
1] and ξ | C (cid:48) : C (cid:48) → [0 , . Then, we have the inverse maps β = ( ξ | C ) − : (0 , → C and β (cid:48) = ( ξ | C (cid:48) ) − : [0 , → C (cid:48) , (4)giving the binary expansions of the real numbers in [0 , binary expression .x x . . . of x ∈ [0 , binary expansion of x which is always aninfinite sequence in B . Namely, for a dyadic rational x ∈ (0 ,
1] admitting the finite binaryexpression 0 .x x . . . x n , we have β (cid:48) ( x ) = ( x , x , . . . , x n , , , . . . ) ∈ C (cid:48) , while we have β ( x ) =( x , x , . . . , x k − , , , , . . . ) ∈ C if k ≤ n is the index of the last occurrence of 1 in 0 .x x . . . x n .Now, let D ⊂ Q denote the set of all dyadic rationals, and put D = D ∩ [0 ,
1] and Q = Q ∩ [0 , . Notice that the two maps β and β (cid:48) coincide and are continuous on (0 , (cid:114) D , while they differon the set D (cid:114) { , } , where β is only left-continuous and β (cid:48) is only right-continuous.Finally, we introduce the further notations b (cid:112)(cid:112) k for the k -th truncation of an infinite binarysequence b , and [ w ] for the cylinder consisting of all the infinite binary sequences having thefinite prefix w . Namely, b (cid:112)(cid:112) k = ( b , b , . . . , b k ) ∈ { , } k for every b = ( b , b , . . . , b n , . . . ) ∈ B , (5)[ w ] = { ( w, b ) | b ∈ B } ⊂ B for every w = ( w , w , . . . , w k ) ∈ { , } k . (6)Then, for every b ∈ B and k ≥ B ( b, / k ) ⊂ B as [ b (cid:112)(cid:112) k ] .In the following, we identify the infinite binary sequence b = ( b , b , . . . ) with the infinite binaryword b = b b . . . , and we write x = 0 .b for (the corresponding binary expression of) x = ξ ( b ).An analogous convention is adopted for finite binary sequences. Thus, the notations introducedin the equations (2), (5) and (6) make sense also in the context of binary words, and hence inthe context of binary expressions/expansions of real numbers in [0 , The substitution rule
Let { , } ∞ be the set of all the finite or infinite binary words, including the empty word (cid:15) .For a word w ∈ { , } ∞ , we indicate by | w | its (possibly infinite) length and by | w | the (possiblyinfinite) number of 1’s occurring in it.Consider the map ρ : { , } ∞ → { , } ∞ digit-wise defined by the alternating substitution rule ρ : (cid:40) (cid:55)→ (cid:15) (cid:55)→ (cid:55)→ . (7)In other words, for any word b = b b . . . ∈ { , } ∞ the word ρ ( b ) is obtained from b by deletingall the b n = 0 and replacing each b n = 1 by 0 if n is odd and by 1 if n is even.We remark that the parity condition in the above definition makes ρ a non-morphic mapwith respect to the concatenation. Indeed, for any v, w ∈ { , } ∞ with v a finite word and | w | >
0, the equality ρ ( vw ) = ρ ( v ) ρ ( w ) holds only if | v | is even, while for | v | odd we have ρ ( vw ) = ρ ( v ) (cid:101) ρ ( w ), where (cid:101) ρ is the complementary substitution rule (cid:101) ρ : (cid:40) (cid:55)→ (cid:15) (cid:55)→ (cid:55)→ . (8)For the sake of convenience, we introduce the notation ρ k = (cid:110) ρ if k is an even integer (cid:101) ρ if k is an odd integer , (9)and we write ρ v to mean ρ | v | . In this way, for any v and w as above we have ρ ( vw ) = ρ ( v ) ρ v ( w ) . (10)We observe that, for binary words of even or infinite length, the action of ρ coincides with thatof the block substitution rule τ : (cid:55)→ (cid:15) (cid:55)→ (cid:55)→ (cid:55)→ . (11)This interpretation of the substitution rule ρ enables us to estimate the vanishing order n (cid:15) ( w ) = min { k > | ρ k ( w ) = (cid:15) } (12)of a finite binary word w under the action of ρ in terms of | w | . Proposition 2.1. If w (cid:54) = (cid:15) is a finite binary word then n (cid:15) ( w ) ≤ (cid:98) log | w |(cid:99) + 2 .Proof. We proceed by induction on | w | ≥
1, based on the trivial case of | w | = 1. So, let | w | > | w | even, because of the obvious equality n (cid:15) ( w ) = n (cid:15) ( w | ρ ( w ) | = | ρ ( τ ( w )) | = | τ ( w ) | . Looking at the contribution to | τ ( w ) | ofeach single pair, according to (14), we immediately get | ρ ( w ) | ≤ | w | /
2. Then, by the inductivehypothesis, n (cid:15) ( w ) ≤ n (cid:15) ( ρ ( w )) + 2 ≤ (cid:98) log | ρ ( w ) |(cid:99) + 4 ≤ (cid:98) log ( | w | / (cid:99) + 4 = 2 (cid:98) log | w |(cid:99) + 2 .
4s it can be easily realized, the vanishing order n (cid:15) ( w ) of a word of length n ≥ n and 1 n , which vanish exactly after 1 and 2 (cid:98) log n (cid:99) + 2iterations of ρ , respectively.Up to the identification between binary sequences and binary words, we have C ⊂ B ⊂ { , } ∞ .Moreover, it can be easily seen that ρ − ( B ) = C . Then, it makes sense to consider the restriction ρ | C = τ | C : C → B .Now, consider the substitution rule τ : (cid:110) (cid:55)→ (cid:55)→ , (13)and let τ : { , } ∞ → { , } ∞ be the corresponding digit-wise generated map. Then, the image τ ( w ) has even or infinite length for every w ∈ { , } ∞ . Moreover, τ ◦ τ = id { , } ∞ and τ ( w ) ∈ C for every w ∈ B . It immediately follows that the map ρ | C is surjective.Actually, τ ( w ) is not the “simplest” element in ( ρ | C ) − ( w ) for w ∈ B . In fact, τ ( w ) can containpairs of consecutive 0’s and these can be deleted without changing the image under ρ | C . Thereason is that ρ (0) = (cid:15) and that deleting/inserting subwords of even length does not changethe parity of the following positions in the word.In the following we indicate by S ⊂ C the set of the “simplest” infinite binary words, meaningthose which do not contain any pair of consecutive 0’s.In order to directly construct the “simplest” element in ( ρ | C ) − ( w ), the unique one whichbelongs to S , we consider the section σ : { , } ∞ → { , } ∞ digit-wise generated by the rule σ : (cid:110) x (cid:55)→ x is preceded by 1 − x in the word or it is 0 as the first digit x (cid:55)→
01 if x is preceded by x in the word or it is 1 as the first digit . (14)Given w ∈ { , } ∞ , σ ( w ) is actually given by a true substitution rule (in the usual sense)applied to the sequence of first differences of the word 1 w obtained prepending 1 to w . Moreover, σ ( b ) ∈ S ⊂ C for every b ∈ B , and hence σ | B : B → C is a section of ρ | C . Proposition 2.2.
For every b ∈ B the fiber ( ρ | C ) − ( b ) contains uncountably many elements.In fact, ( ρ | C ) − ( b ) = {(cid:104) a (cid:105) σ ( b ) | a ∈ A } , (15) and hence σ ( b ) is the unique element of ( ρ | C ) − ( b ) belonging to S and it is the maximum of ( ρ | C ) − ( b ) with respect to the lexicographic order.Proof. We only need to prove equation (15), since the rest of the statement immediately followsfrom it. A straightforward induction on n gives ρ | C ( σ ( b )) n = b n for every n ≥
1, which impliesthat ρ ( σ ( b )) = b . On the other hand, by deleting pairs of 0’s as discussed above, we get ρ ( (cid:104) a (cid:105) σ ( b ) ) = b for every a ∈ A . Conversely, let any v ∈ ( ρ | C ) − ( b ) be written as v = η ( c ),thanks to (2). The n -th occurrence of 1 in v is in position k n = c + . . . + c n + n . Since theparity of k n is determined by ρ ( v ) n = b n , induction on n shows that also the parity of c n isdetermined for every n ≥
1. So, if σ ( b ) = η ( d ) then we have d n = 0 , c n ≡ d n mod 2for every n ≥
1. This yields c = 2 a + d for a suitable a ∈ A , and hence, according to (2), v = η ( c ) = η (2 a + d ) = (cid:104) a (cid:105) σ ( b ) .We conclude this section by considering some continuity properties of the maps ρ | C : C → B and σ | B : B → C with respect to the metric induced by the inclusions C ⊂ B ⊂ A .5 roposition 2.3. The map ρ | C : C → B is continuous but not uniformly continuous. Moreover, lim v → b ρ ( v ) does not exists for any b ∈ B (cid:114) C . So, ρ cannot be continuously extended to anysubset of B larger than C .Proof. For any b ∈ C and n ≥
1, let k n = | ρ ( b (cid:112)(cid:112) n ) | = | b (cid:112)(cid:112) n | . Then, ρ ( B ( b, / n )) = ρ ([ b (cid:112)(cid:112) n ]) ⊂ [ ρ ( b (cid:112)(cid:112) n )] = [ ρ ( b ) (cid:112)(cid:112) k n )] = B ( ρ ( b ) , / k n ), and hence the continuity of ρ | C at b immediately followsfrom the fact that k n → ∞ for n → ∞ . On the other hand, if ρ were uniformly continuousthen it would be possible to extend it to the whole space { , } ∞ , and this would contradictthe second part of the statement, which we are going to prove.Let b ∈ B (cid:114) C . Then, b is eventually 0, and hence it can be written as b = v ∞ . Consider thesequence ( b n = v n ∞ ) n ≥ ⊂ C . We have, lim n →∞ b n = b , while the limit lim n →∞ ρ ( b n ) cannotexist, being ρ ( b (2 k + 1)) = ρ ( b (1)) (cid:54) = ρ ( b (2)) = ρ ( b (2 k + 2)) for every k ≥ Proposition 2.4.
The map σ | B : B → C is uniformly continuous, in fact it is -Lipschitz.Proof. Since | σ ( v ) | ≥ k for every v ∈ { , } k , we have that d ( b, b (cid:48) ) = 1 / k ⇒ b (cid:112)(cid:112) k = b (cid:48) (cid:112)(cid:112) k ⇒ σ ( b ) (cid:112)(cid:112) k = σ ( b (cid:48) ) (cid:112)(cid:112) k ⇒ d ( σ ( b ) , σ ( b (cid:48) )) ≤ / k , for every b, b (cid:48) ∈ B . We define the map R : [0 , → [0 ,
1] as follows R ( x ) = (cid:110) / x = 0 ξ ( ρ ( β ( x ))) if x ∈ (0 , . (16)For x (cid:54) = 0 this amounts to say that R ( x ) admits a binary expansion which is the transformunder ρ of the unique binary expansion of x in C . In formulas, if x = 0 .b with b ∈ C then R ( x ) = R (0 .b ) = 0 .ρ ( b ) . The image of 0 could be chosen somewhat arbitrarily. This particular choice is convenient forlater purposes.The continuity properties we have seen in the previous section for the maps involved in thedefinition of R give us the next proposition. Proposition 3.1.
The map R is continuous everywhere but on the set D (cid:114) { } , where it isonly left-continuous. Hence, R is a Borel function of Baire class , i.e. it is a point-wise limitof continuous functions.Proof. The first assertion immediately derives from the continuity of ξ and ρ | C (Proposition2.3), and the fact that β is continuous everywhere but on the set D (cid:114) { } , where it is onlyleft-continuous. Then, the second assertion follows by the fact that R has countably manydiscontinuities (see [43, Theorem 11.8] or [23, Chapter Three, Section 34, Paragraph VII]).It is worth remarking that the right-discontinuity of R on D (cid:114) { } is not simply due to theright-discontinuity of β on that set, but it is instead also related to the fact that ρ | C is notcontinuously extendable, as stated in Proposition 2.3. In fact, as we will see in Proposition 4.2,for x ∈ D (cid:114) { , } the right limit lim y (cid:38) x R ( y ) does not exist and even more R ( x ) / ∈ (cid:2) lim inf y (cid:38) x R ( y ) , lim sup y (cid:38) x R ( y ) (cid:3) . R (1 /
2) = 2 / , while lim inf x (cid:38) / R ( x ) = 0 and lim sup x (cid:38) / R ( x ) = 1 / ,R (1 /
4) = 1 / , while lim inf x (cid:38) / R ( x ) = 1 / x (cid:38) / R ( x ) = 1 . In particular, R is not a Darboux function, meaning that it does not satisfy the intermediatevalue property. Actually, it is not Darboux from the right for every x ∈ D (cid:114) { } (see Corollary4.4). Thus, by Darboux’s theorem, the integral function (cid:82) x R ( t ) dt is differentiable everywherebut on the set D (cid:114) { } , where it is only left-differentiable.By a classical characterization of Baire class 1 functions given by Lebesgue [24] (see also [5,Section 4.4] or [23, Chapter Two, Section 31, Paragraph II]), we know that for every ε > { C n } n ≥ of [0 ,
1] such that the oscillation of R on each C n is less than ε . We recall that the oscillation of a real function f on a subset S of its domainis defined as O f ( S ) = sup x,y ∈ S | f ( x ) − f ( y ) | .Before going on, we want to provide an explicit construction of a covering { C n } n ≥ as above.First, we need to estimate the oscillation of R on the dyadic intervals as follows.For every x ∈ { , } n and n ≥
1, we consider the cylinder [ x ] ⊂ B and the interval ξ ([ x ]) =[0 .x ∞ , .x ∞ ] ⊂ [0 , R ( ξ ([ x ])) = (cid:26) [0 ,
1] if x = 0 n ξ ( { ρ ( x (cid:112)(cid:112) k − ∞ ) } ∪ [ ρ ( x )]) if x (cid:54) = 0 n , (17)where k is the index of the last occurrence of 1 in x . In particular, in the latter case we have R ( ξ ([ x ])) ⊂ ξ ([ ρ ( x (cid:112)(cid:112) k − )]), which implies that O R ( ξ ([ x ])) ≤ / h with h = | ρ ( x (cid:112)(cid:112) k − ) | = | x | − . (18)Now we can start with our construction. Given ε >
0, we choose a positive integer (cid:96) suchthat 1 / (cid:96) − < ε , and consider the set D (cid:96) ⊂ D consisting of all dyadic rational x whose binaryexpansion β (cid:48) ( x ) contains at most (cid:96) | β (cid:48) ( x ) | ≤ (cid:96) . Then, the ordered set ( D (cid:96) , ≥ ) isisomorphic to the countable ordinal ω (cid:96) + 1, and hence D (cid:96) has Cantor-Bendixson rank (cid:96) + 1.Moreover, considering D (cid:96) with the standard order in R , we have min D (cid:96) = 0 and max D (cid:96) =0 . (cid:96) = ξ (1 (cid:96) ), and for every x ∈ D (cid:96) (cid:114) { } , the immediate predecessor of x in D (cid:96) , that is thelargest element of D (cid:96) smaller than x , is p ( x ) = ξ ( β ( x ) (cid:112)(cid:112) k ∞ ), where k denotes the index of the (cid:96) -th 1 in β ( x ). Keeping the same notations, we put C x = ξ ([1 (cid:96) ]) = [ ξ (1 (cid:96) ∞ ) , ξ (1 (cid:96) ∞ )] = [0 . (cid:96) ,
1] if x = 1 ξ ([ β ( x ) (cid:112)(cid:112) k ]) = [ ξ ( β ( x ) (cid:112)(cid:112) k ∞ ) , ξ ( β ( x ) (cid:112)(cid:112) k ∞ )] = [ p ( x ) , x ] if x ∈ D (cid:96) (cid:114) { }{ } if x = 0 . Then, { C x } x ∈ D (cid:96) ∪{ } is a countable closed covering of [0 ,
1] with the wanted oscillation bound.Indeed, in all cases, we have O R ( C x ) ≤ / (cid:96) − < ε by (18).Concerning the fibers, it immediately follows from (16) that R − ( y ) ∩ (0 ,
1] = β − ( ρ − ( ξ − ( y ))) = ξ ( ρ − { β ( y ) , β (cid:48) ( y ) } )) (19)for every y ∈ [0 , R − ( y ) if y (cid:54) = 2 /
3, while 0 must be added toget R − ( y ) if y = 2 /
3. 7aking into account equation (15) and the properties of the maps β and β (cid:48) we have seen at theend of Section 1, we have R − ( y ) = ξ ( (cid:104)(cid:104) σ ( β ( y )) (cid:105)(cid:105) ) if y ∈ [0 , (cid:114) ( D ∪ { / } ) or y = 1 ξ ( (cid:104)(cid:104) σ ( β ( y )) (cid:105)(cid:105) ) ∪ { } if y = 2 / ξ ( (cid:104)(cid:104) σ ( β ( y )) (cid:105)(cid:105) ) ∪ ξ ( (cid:104)(cid:104) σ ( β (cid:48) ( y )) (cid:105)(cid:105) ) if y ∈ D (cid:114) { , } ξ ( (cid:104)(cid:104) σ ( β (cid:48) ( y )) (cid:105)(cid:105) ) if y = 0 , (20)where the following notation is used for b ∈ C (cid:104)(cid:104) b (cid:105)(cid:105) = {(cid:104) a (cid:105) b | a ∈ A } . (21)Since ξ | C is injective, the fiber R − ( y ) contains uncountably many elements for every y ∈ [0 , R − ( y ) admits a maximum element S ( y ) = (cid:40) ξ ( σ ( β ( y ))) if y ∈ [0 , (cid:114) D or y = 1max { ξ ( σ ( β ( y ))) , ξ ( σ ( β (cid:48) ( y ))) } if y ∈ D (cid:114) { , } ξ ( σ ( β (cid:48) ( y ))) if y = 0 . (22)The map S : [0 , → [0 ,
1] defined by the equation above is a null measure section for R , aswe will see in Proposition 5.1. Here, we limit ourselves to consider the following continuityproperties of it. Proposition 3.2.
The map S is continuous everywhere but on the set D (cid:114) { , } , where it iseither left- or right-continuous, with different existing left- and right-limits. Hence, S is a Borelfunction of Baire class .Proof. The first assertion easily follows from the continuity properties of ξ , σ | B (Proposition2.4), β and β (cid:48) , and the fact that, for every y = (2 k + 1) / n ∈ D (cid:114) { , } , we have | ξ ( σ ( β ( y ))) − ξ ( σ ( β (cid:48) ( y ))) | ≤ d ( σ ( β ( y )) , σ ( β (cid:48) ( y ))) ≤ d ( β ( y ) , β (cid:48) ( y )) = 1 / n − . Then, we can apply [43] as above to get the second assertion.Recalling how ρ and (cid:101) ρ act on the binary digits of x we readily deduce the next proposition. Proposition 3.3.
The map R satisfies the functional equations for every x ∈ (0 , R (cid:16) x (cid:17) = 1 − R ( x ) , (23) R (cid:16) x + 12 (cid:17) = 1 − R ( x )2 . (24) Proof.
For every x ∈ (0 , R (cid:16) x (cid:17) = ξ (cid:16) ρ (cid:16) β (cid:16) x (cid:17)(cid:17)(cid:17) = ξ ( ρ (0 β ( x ))) = ξ ( (cid:101) ρ ( β ( x ))) = 1 − ξρ ( β ( x ))) = 1 − R ( x ) ,R (cid:16) x + 12 (cid:17) = ξ (cid:16) ρ (cid:16) β (cid:16) x + 12 (cid:17)(cid:17)(cid:17) = ξ ( ρ (1 β ( x ))) = ξ (0 (cid:101) ρ ( β ( x ))) = 1 − ξρ ( β ( x )))2 = 1 − R ( x )2 . Notice that the relations (23) and (24) are verified by any map defined by means of a generalizedsubstitution of type (7), when odd-indexed 1’s go into a word w and even-indexed 1’s go intoits complementary word (cid:101) w . 8y using the relations (23) and (24), one can get a countable family of functional equations,one for each word w ∈ { , } ∞ . Namely, if w is a finite binary word such that | w | = n , ρ ( w ) = v and | v | = m , for every x ∈ (0 ,
1] we have R (cid:16) x n + n (cid:88) i =1 w i i (cid:17) = m (cid:88) i =1 v i i + 12 m (cid:16) − n +1 − n R ( x ) (cid:17) . Moreover, from (23) and (24) it easily follows that R ( x ) = R (cid:16) x + 12 (cid:17) if x ∈ (0 , / R (cid:16) x − (cid:17) if x ∈ (1 / , . (25) In this section we consider some properties of the map R related to the geometry of its graph G = { ( x, R ( x )) | x ∈ [0 , } ⊂ [0 , , which is depicted in Figure 1 below.We remark that there is a priori no reason to think that the dots in the picture represent anumerical approximation of points of the graph of R , as numerical computation is of coursebased on (finite) binary expressions of dyadic rational numbers, whereas ρ acts always onthe infinite representation and is discontinuous precisely on dyadic rationals. However, thepicture still represents a numerical approximation of the graph of R for a somewhat deeperreason, that is because we can interpret it as the plot of an element of a sequence of stepfunctions, constant on cylinder sets sharing a common finite prefix, converging to R . That this“numerical” convergence is mathematically meaningful is indeed ensured by Proposition 3.1,which establishes that R can be point-wise limit of continuous functions.Figure 1: The graph of the map R (blue) and of its integral function (orange).The functional relations (23) and (24) induce an interesting recursive structure on G .9onsider, the transformations T , T : [0 , → [0 , defined by the equations T ( x, y ) = (cid:16) x , − y (cid:17) and T ( x, y ) = (cid:16) x + 12 , − y (cid:17) . Let T be the operator which associates to each subset E ⊂ [0 , the subset T ( E ) = T ( E ) ∪ T ( E ) . Of course, T preserves compactness. Moreover, (23) and (24) imply the equality T ( H ) = H for H = G − { (0 , / } . Indeed, for every x ∈ (0 , T ( x, R ( x )) = (cid:16) x , − R ( x ) (cid:17) = (cid:16) x , R (cid:16) x (cid:17)(cid:17) and T ( x, R ( x )) = (cid:16) x + 12 , − R ( x )2 (cid:17) = (cid:16) x + 12 , R (cid:16) x + 12 (cid:17)(cid:17) . Therefore, T ( H ) = H ∩ ([0 , / × [0 , T ( H ) = H ∩ ([1 / , × [0 , , which gives T ( H ) = H .We inductively define K n ⊂ [0 , for n ≥
0, by putting K = [0 , and K n +1 = T ( K n ) = T ( K n ) ∪ T ( K n ) , where T ( K n ) = K n +1 ∩ ([0 , / × [0 , T ( K n ) = K n +1 ∩ ([1 / , × [0 , . (26)The trivial inclusion K ⊂ K implies that K n +1 ⊂ K n for every n ≥
0. So, the K n ’s form adecreasing sequence of compact subspaces of [0 , converging to K ∞ = ∩ n ≥ K n ⊂ [0 , . Figure 2: K n for n = 0 , . . . , K n ’s from a different point of view, we consider themaps t , t : [0 , → [0 ,
1] defined by t ( y ) = 1 − y and t ( y ) = 1 − y , which give the action of T and T on the second coordinate.10oreover, we consider the intervals I = [0 ,
1] and I x = t x ◦ t x ◦ · · · ◦ t x n ([0 , x ∈ D (cid:114) { } where 0 .x x . . . x n is a finite binary expression of x . Since the terminal 0’s in the binary word x x . . . x n do not affect the interval I x , this is well-defined, depending only on x .We point out that the interval I x satisfies the identities t ( I x ) = I x/ and t ( I x ) = I ( x +1) / . (27)Then, by reasoning inductively as above, the interval I x for x = 0 .x . . . x n can be shown tocoincide with the cylinder generated by the word ρ ( x . . . x n ), that is I x = [ ρ ( x . . . x n )] . (28)For every n ≥
0, we can decompose K n as a union of rectangles as follows K n = ∪ k =0 ,..., n − [ k/ n , ( k + 1) / n ] × I k/ n . This equality is trivially true for n = 0, while it can be proved by induction for n > T ( K n ) = ∪ k =0 ,..., n − [ k/ n +1 , ( k + 1) / n +1 ] × I k/ n +1 and T ( K n ) = ∪ k =2 n ,..., n +1 − [ k/ n +1 , ( k + 1) / n +1 ] × I k/ n +1 . Then, we can obtain K n +1 from K n by replacing each rectangle[ k/ n , ( k + 1) / n ] × I k/ n by the union([ k/ n , (2 k + 1) / n +1 ] × I k/ n ) ∪ ([(2 k + 1) / n +1 , ( k + 1) / n ] × I (2 k +1) / n +1 ) . Remark 4.1.
Passing from K n to K n +1 amounts to split each rectangle [ k/ n , ( k +1) / n ] × I k/ n into four congruent closed sub-rectangles and remove the right-top or right-bottom one, depend-ing on n being even or odd respectively, while keeping the union of the other three sub-rectangles. Now, come back to H and G . The trivial inclusion H ⊂ K and the equality T ( H ) = H implythat H ⊂ K n for every n ≥
0. On the other hand, as { } × [0 ,
1] = T ( { } × [0 , ⊂ T ( { } × [0 , , we also have { } × [0 , ⊂ K n for every n ≥
0. Hence, G ⊂ ( { } × [0 , ∪ H implies G ⊂ K ∞ ,and therefore Cl G ⊂ K ∞ . We want to show that the reversed inclusion holds as well. In fact,we have the following slightly stronger result. Proposition 4.2.
The following equalities hold K ∞ = Cl G = G ∪ ( ∪ x ∈ D (cid:114) { } ( { x } × I x )) . Moreover, for every x ∈ D (cid:114) { , } , R ( x ) / ∈ I x = (cid:2) lim inf ξ (cid:38) x R ( ξ ) , lim sup ξ (cid:38) x R ( ξ ) (cid:3) .
11n order to prove the proposition a technical lemma is needed.
Lemma 4.3.
For every x ∈ (0 , we have K ∞ ∩ ( { x } × [0 , (cid:26) { x } × ( { R ( x ) } ∪ I x ) with R ( x ) / ∈ I x if x ∈ D { ( x, R ( x )) } if x / ∈ D . Proof.
The case when x ∈ D immediately follows from the definition of K ∞ and the followingstatement: if 0 .x . . . x k is the shortest binary expression of x , then for every n ≥ kK n ∩ ( { x } × [0 , { x } × ( I x ∪ J x,n )where ( J x,n ) n ≥ k is a non-increasing sequence of intervals such that ∩ n ≥ k J x,n = { R ( x ) } with R ( x ) (cid:54)∈ I x . We prove this statement by induction on k ≥ k = 1 then x = 1 /
2, and for every n >
1, we have K n ∩ ( { / } × [0 , T ( K n − ∩ ( { } × [0 , ∪ T ( K n − ∩ ( { } × [0 , , so that, by recalling equation (26), K n ∩ ( { / } × [0 , T ( { } × t n − ([0 , ∪ T ( { } × [0 , { / } × ( t ◦ t n − ([0 , ∪ [0 , / . Then, J / ,n = ( t ◦ t n − ([0 , n ≥ is a non-increasing sequence of intervals whose intersectionis t ( R (1)) = R (1 /
2) = 2 /
3. Moreover, J / ,n ∩ I / = J / ,n ∩ [0 , /
2] = ∅ for every n largeenough, and hence R (1 / (cid:54)∈ I / .Consider now the case of a dyadic rational x = 0 .x . . . x k with k >
1. Let x (cid:48) = 0 .x . . . x k =2 x (mod 1) and assume, by the inductive hypothesis, that K n ∩ ( { x (cid:48) }× [0 , { x (cid:48) }× ( I x (cid:48) ∪ J x (cid:48) ,n )for every n ≥ k −
1, with ( J x (cid:48) ,n ) n ≥ k − a non-increasing sequence of intervals whose intersectionis { R ( x (cid:48) ) } and R ( x (cid:48) ) / ∈ I x (cid:48) = t x ◦ · · · ◦ t x k ([0 , n ≥ k we have K n ∩ ( { x } × [0 , T x ( K n − ∩ ( { x (cid:48) } × [0 , , and hence K n ∩ ( { x } × [0 , { x } × ( t x ( I x (cid:48) ) ∪ t x ( J x (cid:48) ,n − )) . Here, ( J x,n = t x ( J x (cid:48) ,n − )) n ≥ k is a non-increasing sequence of intervals whose intersection is { R ( x ) } = { t x ( R ( x (cid:48) )) } , and R ( x ) / ∈ I x = t x ( I x (cid:48) ).We are left to consider the case when x / ∈ D . For every n ≥
1, let x (cid:112)(cid:112) n = 0 .x . . . x n the n -thtrucantion of the binary expression of x . We want to prove, by induction on n ≥
1, that K n ∩ ( { x } × [0 , { x } × I x (cid:112)(cid:112) n . In fact, for n = 1 we have I x (cid:112)(cid:112) = [0 ,
1] if x = 0 and I x (cid:112)(cid:112) = [0 , /
2] if x = 1. In both cases, K ∩ ( { x } × [0 , { x } × I x (cid:112)(cid:112) and R ( x ) ∈ I x (cid:112)(cid:112) . Moreover, for n > I x n = I x (cid:112)(cid:112) n − if x n = 0 I + x (cid:112)(cid:112) n − if x n = 1 and n is even I − x (cid:112)(cid:112) n − if x n = 1 and n is odd , where I + x (cid:112)(cid:112) n − and I − x (cid:112)(cid:112) n − denote the upper half and the lower half of I x (cid:112)(cid:112) n − , respectively. In allthe cases, the equality K n ∩ ( { x } × [0 , { x } × I x (cid:112)(cid:112) n and the relation R ( x ) ∈ I x (cid:112)(cid:112) n follow byinduction on n , taking into account that0 .x . . . x n − < x < .x . . . x n − x n = 0 , .x . . . x n − < x < .x . . . x n − ∞ if x n = 1 . Then, K ∞ ∩ ( x × [0 , { x } × ( ∩ n ≥ I x (cid:112)(cid:112) n ) = { ( x, R ( x )) } . roof of Proposition 4.2. According to the lemma above, taking the union over x ∈ [0 ,
1) weimmediately get the following equality K ∞ = G ∪ ( ∪ x ∈ D (cid:114) { } ( { x } × I x )) , (29)with G ∩ ( ∪ x ∈ D (cid:114) { } ( { x } × I x )) = { (0 , / } . Equation (29) implies that Cl G ⊂ K ∞ . Therefore, in order to conclude that K ∞ = Cl G , itsuffices to show that the inclusion { x } × I x ⊂ Cl G holds for every x = 0 .x . . . x n ∈ D (cid:114) { } .According to equation (28), for a generic point ( x, y ) ∈ { x } × I x we have y = 0 .x . . . x n v , with v ∈ { , } ∞ . Then, the point ( x, y ) can be arbitrarily approximated by a point ( x (cid:48) , R ( x (cid:48) )) ∈ G with x (cid:48) = 0 .x . . . x n (00) k w , k sufficiently large and w ∈ { , } ∞ any binary word such that ρ ( w ) = v . This concludes the proof of the first part of the proposition.To prove the second part of the proposition, we start by observing that the argument justexposed immediately entails that I x ⊂ (cid:2) lim inf ξ (cid:38) x R ( ξ ) , lim sup ξ (cid:38) x R ( ξ ) (cid:3) The opposite inclusion readily follows by equation (28).
Corollary 4.4.
The map R is not Darboux from the right ( in the sense of [11]) at any dyadicrational x ∈ D (cid:114) { } and its graph G is totally disconnected.Proof. This is an immediate consequence of Proposition 4.2.Using the inclusion G ⊂ K ∞ , we can prove the next proposition. Proposition 4.5. (cid:90) R ( x ) dx = 37 . Proof.
By Remark 4.1, Area K n = Area K n − for every n >
0. Then, taking into account thatArea K = 1, we get Area K n = (cid:16) (cid:17) n (30)for every n ≥
0. Since this vanishes for n → ∞ , we can approximate its integral by the integral A n of the piecewise constant function R n whose graphic is given by the bottom edges of the allrectangles forming K n . Now, R n +1 = R n for every n ≥
1, so we can write (cid:90) R ( x ) dx = lim n →∞ A n . Looking at what happens inside each rectangle of K n − when passing from R n − to R n (compare K and K in Figure 2), we have A n − A n − = 316 Area K n − = 316 · (cid:16) (cid:17) n − Therefore, starting from A = 0 we getlim n →∞ A n = 316 ∞ (cid:88) n = 0 (cid:16) (cid:17) n = 316 ·
167 = 37 . B G and estimate the Hausdorff dimensiondim H G as follows. Proposition 4.6. ≤ dim H G ≤ dim B G = dim B K ∞ = log . Proof.
The two inequalities derive from the projection of G onto [0 ,
1] and from the generalrelation between the Hausdorff dimension and the box-dimension, respectively. The first equalitycomes from the invariance of the box-dimension under closure. The second equality follows fromthe fact that the boxes of the form [ k/ n , ( k + 1) / n ] × [ (cid:96)/ n , ( (cid:96) + 1) / n ] needed to cover K ∞ are exactly the ones contained in K n , which are 3 n , as it is clear by equation (30). Let us start with some global properties of the map R with respect to the Lebesgue measure.We indicate by λ ( A ) the standard Lebesgue measure of a measurable set A ⊂ [0 , N ⊂ [0 ,
1] the subset of 2-normal numbers, that isthe numbers x ∈ [0 ,
1] which admit a binary expansion where all the binary sequences of anygiven length k ≥ / k . In particular, N does not contain any rational number, and hence β ( x ) = β (cid:48) ( x ) for every x ∈ N . We recall that N has full measure λ ( N ) = 1, as proved in [3].For any infinite binary word b ∈ B and any finite binary word w ∈ { , } k , we indicate by f n ( w, b ) the relative frequency of w among all the n subwords of b (cid:112)(cid:112) n + k − of length k , and by f ( w, b ) the asymptotic relative frequency of w in b , if it exists. Namely, we put f n ( w, b ) = |{ i = 1 , . . . , n | w = b i b i +1 . . . b i + k − }| n (31)and f ( w, b ) = lim n →∞ f n ( w, b ) . (32)Then, to say that x = 0 .x x . . . is normal means that f ( w, x x . . . ) = 1 / k for every w ∈ { , } k . This is equivalent to the property that for every k ≥
1, if ( w , w , . . . ) is thedecomposition of x x . . . into contiguous blocks of length k , then the asymptotic relative fre-quency of any w ∈ { , } k in the sequence ( w , w , . . . ) is 1 / k (see [3], [6], [8], [29] and [33]). Proposition 5.1.
The map R is measurable but not bi-measurable. Moreover, the map S definedby equation (22) is a null measure section of R , being S ([0 , ⊂ [0 , (cid:114) N .Proof. The map R is almost everywhere continuous by Proposition 3.1, hence it is measurable.Concerning the section S , for every y ∈ [0 ,
1] the image S ( y ) does not contain any pair ofconsecutive 0’s in its binary expansion β ( S ( y )), which coincides with either σ ( β ( y )) or σ ( β (cid:48) ( y )).Then, we have S ([0 , ⊂ [0 , (cid:114) N , which implies that λ ( S ([0 , A ⊂ [0 ,
1] is the image R ( S ( A )) of the null measure set S ( A ). In particular, this is true forany non-measurable subset of [0 , R is not bi-measurable.We remark that, in the light the uncountability of all the fibers of R discussed in Section 3,part of Proposition 5.1 could be also obtained as a direct consequence of a general theorem byPurves [34] stating that, if f is a bi-measurable map from a standard Borel space X to a Polishspace Y , then at most countably many fibers of f can be uncountable. Proposition 5.2.
The map R is a singular Borel function, being R ( N ) ⊂ [0 , (cid:114) N . roof. It is enough to prove the last inclusion, since it implies that λ ( R − ([0 , (cid:114) N )) = 1with λ ([0 , (cid:114) N ) = 0, and hence the singularity of R . So, given x = 0 .x x . . . ∈ N and R ( x ) = y = 0 .y y . . . , with y y . . . = ρ ( x x . . . ), we have to prove that y is not normal.Notice that x x . . . = β ( x ) ∈ C and y y . . . ∈ B are both infinite binary words (the lattercoincides with either β ( y ) or β (cid:48) ( y )). Here, we think of the normality of x in terms of asymptoticequi-distribution of contiguous blocks in x x . . . of every given length, as discussed above.By considering the blocks of length 2 in the word x x . . . , and taking into account the substi-tution rule τ defined in (11) and the asymptotic equi-distribution of 01 and 10, guaranteed bythe normality of x , we can immediately conclude that R ( x ) is simply normal, meaning that 0and 1 have the same asymptotic relative frequency 1 / y y . . . .By arguing instead on the blocks of length 8 in the word x x . . . , we will prove thatlim n →∞ ( f n (01 , y y . . . ) + f n (10 , y y . . . )) > lim n →∞ ( f n (00 , y y . . . ) + f n (11 , y y . . . )) , (33)assuming that both the limits exist, otherwise y = R ( x ) cannot be normal and we are done. Thisprevents the word y y . . . to satisfy the asymptotic equi-distribution of the binary subwordsof length 2, and hence implies once again that y = R ( x ) is not normal.To prove (33), let n ≥ f (0 , x x . . . ) = f (1 , x x . . . ) = 1 /
2, forevery δ > m ≥ n of the word ρ ( x x . . . x m ) satisfies thefollowing properties n m = | ρ ( x x . . . x m ) | m = | x x . . . x m | m ∈ (1 / − δ, / δ ) and n > n We want to show that for a suitable choice of m the following inequality holds f n (01 , y y . . . ) + f n (10 , y y . . . ) > f n (00 , y y . . . ) + f n (11 , y y . . . ) + c, (34)where c > n , which gives (33).Let ( w , w , . . . , w m ) be the decomposition of the word x x . . . x m in blocks of length 8, and( v , v , . . . , v m ) be the corresponding decomposition of the word y y . . . y n in blocks of variablelength ≤
8, with v i = ρ ( w i ) for every i = 1 , . . . , k .Based on the normality of x , we can assume m sufficiently large in such a way that the relativefrequency of any binary word w of length 8 in the sequence ( w , w , . . . , w m ) belongs to theinterval (1 / − δ, / + δ ). Then, we consider the pairs of consecutive digits in y y . . . y n ,separating those occurring inside the blocks from those formed by the last digit of a block andthe first digit of the next one (disregarding the empty blocks).The pairs of the latter type are less than m . Concerning the pairs of the former type, directinspection shows that in the images of all binary words of length 8 under ρ , the pairs 01 and10 occur 541 times, while the pairs 00 and 11 occur 228 times. Therefore, inside the blocks( v , v , . . . , v m ) the number of occurrences of the pairs 01 and 10 is at least 541(1 / − δ ) m ,while the number of occurrences of the pairs 00 and 11 at most 228(1 / + δ ) m . Putting alltogether we have f n (01 , y y . . . ) + f n (10 , y y . . . ) ≥ / − δ ) mn ≥ / − δ )8(1 / δ ) , (35) f n (00 , y y . . . ) + f n (11 , y y . . . ) ≤ / + δ ) m + mn ≤ / + δ ) + 18(1 / − δ ) . (36)At this point, it is enough to observe that (35) is greater than (36) for δ sufficiently small.15s a consequence of Proposition 5.2, the map R does not preserve the Lebesgue measure. TheLebesgue measure of the inverse images of dyadic cylinders can be directly computed as follows.According to equation (20), for every y ∈ { , } n and n ≥
1, we have R − ( ξ ([ y ])) (cid:114) { } = (cid:26) ∪ b ∈ B ξ ( (cid:104)(cid:104) σ ( yb ) (cid:105)(cid:105) ) if y = 1 n ∪ b ∈ B ξ ( (cid:104)(cid:104) σ ( yb ) (cid:105)(cid:105) ) ∪ ξ ( (cid:104)(cid:104) σ ( y (cid:48) ) (cid:105)(cid:105) ) otherwise , (37)where y (cid:48) = β ( ξ ( y ) + 1 / n ).Then, by putting | σ ( y ) | = m and recalling that | σ ( y ) | = | y | = n , we get λ ( ∪ b ∈ B ξ ( (cid:104)(cid:104) σ ( yb ) (cid:105)(cid:105) )) = λ ( {(cid:104) a (cid:105) σ ( yb ) | a ∈ A , b ∈ B } ) = λ ( {(cid:104) a (cid:105) σ ( y ) b | a ∈ N n , b ∈ B } )= λ ( ∪ a ∈ N n [ (cid:104) a (cid:105) σ ( y ) ]) = (cid:80) a ∈ N n λ ([ (cid:104) a (cid:105) σ ( y ) ]) = (cid:80) a ∈ N n / |(cid:104) a (cid:105) σ ( y ) | = (cid:80) a ∈ N n / (2 a + ... +2 a n + m ) = (cid:0) (cid:80) a ≥ / a (cid:1) · · · (cid:0) (cid:80) a n ≥ / a n (cid:1) / m = (4 / n / m = 2 n − m / n ≤ (2 / n . Moreover, by taking into account the inclusions ξ ( (cid:104)(cid:104) σ ( y (cid:48) ) (cid:105)(cid:105) ) ⊂ ∪ b ∈ B ξ ( (cid:104)(cid:104) σ ( y (cid:48) (cid:112)(cid:112) k b ) (cid:105)(cid:105) ) , which hold for every k ≥
0, and by applying the above formula for the measure of the right-sideterms, we obtain λ ( ξ ( (cid:104)(cid:104) σ ( y (cid:48) ) (cid:105)(cid:105) )) = 0 . Therefore, based on (37), we have λ ( R − ( ξ ([ y ]))) = 2 n −| σ ( y ) | / n ≤ (2 / n (38)for every y ∈ { , } n and n ≥ R cannot preserve any productmeasure µ p with 0 < p <
1, determined by µ p ( ξ ([ x ])) = p m (1 − p ) n − m where m = | x | , for every x ∈ { , } n and n ≥ R -invariant probability measure can be absolutelycontinuous with respect to Lebesgue measure. Indeed, if there were such a measure ν , then wewould have ν ([0 , (cid:114) N ) = 0, and hence ν ( R − ([0 , (cid:114) N )) = 0. Thus, the density function of ν would vanish almost everywhere on a set containing N .Now we pass to the topological and metric structure of the fibers of R , which we recall to beall uncountable, as a consequence of equation (20) in Section 3. In particular, we will computethe Hausdorff dimension dim H R − ( y ) of the fibers of the rational numbers y ∈ Q and estimatethat of all the other fibers. Let us start with some preliminary results. Proposition 5.3.
For every y ∈ [0 , , the closure Cl R − ( y ) of the fiber R − ( y ) is a nullmeasure Cantor set in [0 , , with a countable remainder Cl R − ( y ) (cid:114) R − ( y ) ⊂ D .Proof. First we prove the inclusion Cl ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) ⊂ ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) ∪ D (39)for every b ∈ C , which implies that Cl R − ( y ) (cid:114) R − ( y ) ⊂ D for every y ∈ [0 ,
1] by (20).Given x ∈ Cl ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ), let ( x n ) n ≥ ⊂ ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) be a sequence such that lim n →∞ x n = x . Then, bythe definition of (cid:104)(cid:104) b (cid:105)(cid:105) in (21), there exists a sequence ( a n ) n ≥ ⊂ A such that x n = ξ ( (cid:104) a n (cid:105) b ), and16ence (cid:104) a n (cid:105) b = β ( x n ), for every n ≥
1. If x ∈ D we are done, otherwise the continuity of β at x (see Section 1) implies that lim n →∞ (cid:104) a n (cid:105) b = β ( x ) ∈ C . If the sequence of the i -th components(( a n ) i ) n ≥ is bounded for every i ≥
1, then by compactness we can replace the sequence ( a n ) n ≥ by a subsequence converging to a ∈ A , and recalling that η is a homeomorphism, we havelim n →∞ (cid:104) a n (cid:105) b = lim n →∞ η (2 a n + η − ( b )) = η ( lim n →∞ a n + η − ( b )) = η (2 a + η − ( b )) = (cid:104) a (cid:105) b , and hence x = ξ ( (cid:104) a (cid:105) b ) ∈ ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ). If instead some sequence (( a n ) i ) n ≥ is unbounded, let m bethe minimum index i ≥ a n ) n ≥ by a suitablesubsequence such that ( a n ) (cid:112)(cid:112) m − = c for some c ∈ { , } m − and every n ≥
1, we have β ( x ) = lim n →∞ (cid:104) a n (cid:105) b = (cid:104) c (cid:105) b (cid:112)(cid:112) m − ∞ , which is absurd, being β ( x ) ∈ C . So this case cannot occur, and we finished the proof of (39).At this point, for any y ∈ (0 , { y } = ∩ n ≥ ξ ([ β ( y ) (cid:112)(cid:112) n ]), and hence R − ( y ) = ∩ n ≥ R − ( ξ ([ β ( y ) (cid:112)(cid:112) n ])), we obtain λ (Cl R − ( y )) = λ ( R − ( y )) = 0 by (38). Similarly,we can obtain λ (Cl R − (0)) = λ ( R − (0)) = 0 starting from { } = ∩ n ≥ ξ ([0 n ]).We are left to prove that Cl R − ( y ) is a Cantor set for every y ∈ [0 , R − ( y ) is 0-dimensional and perfect (see [23, Chapter Four, Section45, Paragraph II]). The 0-dimensionality immediately follows from the fact that Cl R − ( y )cannot contain any interval, having a null measure. For the perfectness, we observe that every x ∈ Cl R − ( y ) (cid:114) R − ( y ) cannot be isolated. On the other hand, if x ∈ R − ( y ) (cid:114) { } , then x = ξ ( η ( c )) for some c ∈ C (see (20)), and we have x = lim n →∞ x n with ( x n = ξ ( η ( c n ))) n ≥ ⊂ R − ( y )defined by ( c n ) i = c i + 2 δ i,n for any i, n ≥
1. Finally, 0 is a limit point for every fiber R − ( y ),and in particular for y = 2 /
3. In fact, 0 = lim n →∞ x n where ( x n ) n ≥ ⊂ R − ( y ) is defined likeabove, starting with any c ∈ C such that ξ ( η ( c )) ∈ R − ( y ) and putting ( c n ) i = c i + 2 nδ i, forany i, n ≥ R , we introduce a partial order relation between infinitebinary sequences b, c ∈ C as follows b (cid:60) c ⇔ there exists m ≥ | c (cid:112)(cid:112) i | ≤ | b (cid:112)(cid:112) i | + m for every i ≥ . (40)We also introduce the induced relation equivalence b ≈ c ⇔ b (cid:60) c and c (cid:60) b . (41)The above relations provide a kind of uniform control on the distribution of the 1’s in the words.In particular, b (cid:60) c and b ≈ c imply analogous relations f (1 , b ) ≥ f (1 , c ) and f (1 , b ) = f (1 , c )respectively, between the asymptotic relative frequencies of 1’s, if they exist. But of of coursethe opposite implications are false.The next two lemmas provide our technical tool for the comparison of fibers. Here, we recallthat S stands for the set of all infinite binary words without any pair of consecutive 0’s. Lemma 5.4.
For every b, c ∈ C , the natural bijection φ b,c : (cid:104)(cid:104) b (cid:105)(cid:105) → (cid:104)(cid:104) c (cid:105)(cid:105) defined by φ b,c ( (cid:104) a (cid:105) b ) = (cid:104) a (cid:105) c with a ∈ A , is a homeomorphism. Moreover, φ b,c is a Lipschitz map if b (cid:60) c and c ∈ S , andhence φ b,c is a bi-Lipschitz map if b ≈ c and b, c ∈ S . roof. Given any b ∈ C , we consider the natural bijection φ b : A → (cid:104)(cid:104) b (cid:105)(cid:105) defined by φ b ( a ) = (cid:104) a (cid:105) b = η (2 a + η − ( b )). This is a 1-Lipschitz homeomorphism, being the composition of theisometric map A → A which sends a to 2 a + η − b with the 1-Lipschitz homeomorphism η : A → C . Moreover, for every a, a (cid:48) ∈ A such that a (cid:112)(cid:112) n = a (cid:48) (cid:112)(cid:112) n and a n +1 (cid:54) = a (cid:48) n +1 we put h ( a, a (cid:48) ) = a + . . . + a n + min { a n +1 , a (cid:48) n +1 } , and observe that the maximum index i for which (cid:104) a (cid:105) b (cid:112)(cid:112) i = (cid:104) a (cid:48) (cid:105) b (cid:112)(cid:112) i is given by 2 h ( a, a (cid:48) ) + k bn , where k bn denotes the index of the n -th occurrence of 1 in b .Then, for every b, c ∈ C the identity φ b,c = φ c ◦ φ − b implies that φ b,c is a homeomorphism.Furthermore, if b (cid:60) c and c ∈ S we can see that φ b,c is a Lipschitz map as follows.Let m ≥ | c (cid:112)(cid:112) i | ≤ | b (cid:112)(cid:112) i | + m for every i ≥
1. In particular, we have | c (cid:112)(cid:112) k bi | ≤ | b (cid:112)(cid:112) k bi | + m = i + m, and therefore k bi + 1 ≤ k ci + m +1 ≤ k ci + 2 m + 2 (42)for every i ≥
1, where k bi denotes the index of the i -th occurrence of 1 in b as above, and thelast inequality derives from the fact that c ∈ S does not contain any pair of consecutive 0’s.Given any two different sequences a, a (cid:48) ∈ A , let n be the maximum index i such that a (cid:112)(cid:112) i = a (cid:48) (cid:112)(cid:112) i .Then, by the above observation, we have d ( (cid:104) a (cid:105) b , (cid:104) a (cid:48) (cid:105) b ) = 1 / h ( a,a (cid:48) )+ k bn +1 and d ( (cid:104) a (cid:105) c , (cid:104) a (cid:48) (cid:105) c ) = 1 / h ( a,a (cid:48) )+ k cn +1 , from which by (42) we obtain d ( φ b,c ( (cid:104) a (cid:105) b ) , φ b,c ( (cid:104) a (cid:48) (cid:105) b )) d ( (cid:104) a (cid:105) b , (cid:104) a (cid:48) (cid:105) b ) ≤ m +1 . So, we can conclude that φ b,c is a Lipschitz map with Lipschitz constant 2 m +1 . Lemma 5.5.
For every b, c ∈ C , the bijection ψ b,c = ξ ◦ φ b,c ◦ β | ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) : ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) → ξ ( (cid:104)(cid:104) c (cid:105)(cid:105) ) given by ψ b,c ( ξ ( (cid:104) a (cid:105) b ) = ξ ( (cid:104) a (cid:48) (cid:105) b ) with a ∈ A , is a Lipschitz map if b (cid:60) c and b ∈ S , and hence ψ b,c is a bi-Lipschitz homeo-morphism if b ≈ c and b, c ∈ S .Proof. In the light of the previous lemma, recalling that ξ is 2-Lipschitz, it suffices to show that β | ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) is a Lipschitz map for every b ∈ S .Take any two different real numbers x = ξ ( (cid:104) a (cid:105) b ) and x (cid:48) = ξ ( (cid:104) a (cid:48) (cid:105) b ) in ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ), with a (cid:112)(cid:112) n = a (cid:48) (cid:112)(cid:112) n and a n +1 (cid:54) = a (cid:48) n +1 . Without loss of generality, we assume that a n +1 < a (cid:48) n +1 . Like in the proofof the previous lemma, we have (cid:104) a (cid:105) b (cid:112)(cid:112) h ( a,a (cid:48) )+ k bn = (cid:104) a (cid:48) (cid:105) b (cid:112)(cid:112) h ( a,a (cid:48) )+ k bn . Denoting by w this word, wecan write (cid:104) a (cid:105) b = w c and (cid:104) a (cid:48) (cid:105) b = w c (cid:48) , and hence x = 0 .w c and x (cid:48) = 0 .w c (cid:48) , for suitable c, c (cid:48) ∈ C . Then, we get | x − x (cid:48) | = 0 .x c − .x c (cid:48) ≥ .x − .x
01 = 0 . | w | +1 / | w | +2 while d ( β ( x ) , β ( x (cid:48) )) = d ( (cid:104) a (cid:105) b , (cid:104) a (cid:48) (cid:105) b ) = 1 / | w | +1 . These inequalities immediately imply that β | ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) is a 2-Lipschitz map.Now we focus on the “rational fibers”, that is the fibers R − ( y ) with y ∈ Q .18 roposition 5.6. If y ∈ Q then ξ ( σ ( β ( y ))) , ξ ( σ ( β (cid:48) ( y ))) ∈ Q when defined, and so S ( y ) ∈ Q .Proof. Since any binary expansion of a rational number is periodic, both β ( y ) and β (cid:48) ( y ) havethe form wp ∞ , where w and p are finite binary of length n ≥ (cid:96) ≥
1, respectively. Then,recalling that σ can be interpreted as a true substitution rule applied to the first differences, σ ( β ( y )) and σ ( β (cid:48) ( y )) have both the corresponding form vq ∞ , with v = (cid:110) σ ( w ) if w n = p (cid:96) σ ( wp ) if w n (cid:54) = p (cid:96) and q = (cid:110) σ ( p ) if p (cid:96) = 1 σ (cid:48) ( p ) if p (cid:96) = 0 , (43)where we assume w n = 1 for n = 0 and indicate by σ (cid:48) ( p ) the word σ ( p ) with the first digitcomplemented. In all cases, we have a periodic binary sequence with a period q of length n ,and by applying ξ we get rational number in Q .The next proposition expresses the Hausdorff dimension dim H R − ( y ) for y ∈ Q in terms of thedensity d of the 1’s in the period q of σ ( β ( y )) or σ ( β (cid:48) ( y )) as given by (43), that is d = | q | / | q | . We recall that β ( y ) and β (cid:48) ( y ) coincide for y ∈ [0 , (cid:114) D , while only one of them is defined for y = 0 ,
1. On the other hand, for y ∈ D (cid:114) { , } the period of β ( y ) is 1 and the period of β (cid:48) ( y )is 0, and hence in both cases q = 01 or q = 10 and then d = 1 / d does not depend on the specific choice for the period q in σ ( β ( y )) or σ ( β (cid:48) ( y )), in that d does not change if we choose a different starting digit for theperiod of the word or consider as a period any multiple of a minimal one. In fact, d coincideswith the asymptotic relative frequency f (1 , σ ( β ( y ))) or f (1 , σ ( β (cid:48) ( y ))). Proposition 5.7.
For every y ∈ Q we have dim H R − ( y ) = − log t , with t the unique realnumber in (0 , verifying the equation t + t /d = 1 , where d = | q | / | q | ∈ (0 , is the density of the ’s in the period q of σ ( β ( y )) or σ ( β (cid:48) ( y )) .Proof. According to (20) and (43), and recalling that dim H ( A ∪ B ) = max { dim H A, dim H B } forevery A, B ⊂ [0 , H ξ ( (cid:104)(cid:104) vq ∞ (cid:105)(cid:105) ) with vq ∞ ∈ S . Moreover, thanksto Lemma 5.5 and the bi-Lipschitz invariance of the Hausdorff dimension, we can limit ourselvesto consider the case when v = (cid:15) , since vq ∞ ≈ q ∞ and both vq ∞ and q ∞ belong to S . In addition,by applying once again Lemma 5.5, we can assume that the last digit of q is 1 without changingthe ≈ class. Indeed, q must contain some 1 and thus it can be written as q = q (cid:48) q (cid:48)(cid:48) for some(possibly empty) binary words q (cid:48) and q (cid:48)(cid:48) , whence q ∞ = ( q (cid:48) q (cid:48)(cid:48) ) ∞ = q (cid:48) q (cid:48)(cid:48) q (cid:48) ∞ ≈ ( q (cid:48)(cid:48) q (cid:48) ∞ .Given q = q q . . . q n ∈ { , } n with q n = 1, we put m = | q | and consider the family ofcontractive similarities T = { T a : R → R } a ∈ N m defined by T a ( x ) = x/ a + ... + a m )+ n + x a , where x a = 0 . (cid:104) a (cid:105) q , for every a ∈ N m . Such family T is clearly relatively compact in the space ofall the contractive similarities of R with the topology of the uniform convergence over boundedsets.In terms of binary expressions, since 2( a + . . . + a m ) + n = |(cid:104) a (cid:105) q | , we have T a (0 .x x . . . ) = 0 . (cid:104) a (cid:105) q x x . . . . (44)19his allows us to derive in a straightforward way that ξ ( (cid:104)(cid:104) q ∞ (cid:105)(cid:105) ) = ∪ a ∈ N m T a ( ξ ( (cid:104)(cid:104) q ∞ (cid:105)(cid:105) )) . Indeed, the equality (cid:104) a (cid:105) q q ∞ = (cid:104) a ∞ (cid:105) q ∞ gives the inclusion T a ( ξ ( (cid:104)(cid:104) q ∞ (cid:105)(cid:105) )) ⊂ ξ ( (cid:104)(cid:104) q ∞ (cid:105)(cid:105) ) for every a ∈ N m . On the other hand, for every x = ξ ( (cid:104) a (cid:105) q ∞ ) ∈ ξ ( (cid:104)(cid:104) q ∞ (cid:105)(cid:105) ) with a ∈ A , we have x = ξ ( (cid:104) a a . . . a m (cid:105) q (cid:104) a m +1 a m +2 . . . (cid:105) q ∞ ) ∈ T a (cid:112)(cid:112) m ( ξ ( (cid:104)(cid:104) q ∞ (cid:105)(cid:105) )).From (44), we also see that T a ((0 , ⊂ (0 ,
1) for every a ∈ N m , and T a ((0 , ∩ T a (cid:48) ((0 , ∅ for every a, a (cid:48) ∈ N m with a (cid:54) = a (cid:48) , that is the family T satisfies the so called “open set condition”.Then, we are in position to apply Theorem 2.2 of [28] (see also [17, Theorem 3.11]), in orderto conclude that dim H ( ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) )) = inf { s > | (cid:80) a ∈ N m r sa ≤ } , (45)where r a = 1 / a + ... + a m )+ n denotes the similarity ratio of T a , for every a ∈ N m .Now, recalling that d = m/n , we have (cid:80) a ∈ N m r sa = (cid:80) a ∈ N m / (2( a + ... + a m )+ n ) s = (cid:0) (cid:80) a ≥ / a s (cid:1) · · · (cid:0) (cid:80) a m ≥ / a m s (cid:1) / ns = 2 − ns / (1 − / s ) m = (2 s/d (1 − − s )) − m . Therefore, from (45) we deduce that dim H ( ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) )) is given by the unique s > f ( s ) = 2 s/d (1 − − s ) assumes the value 1. Finally, the equation f ( s ) = 1becomes t + t /d = 1 as in the statement, by putting s = − log t .We remark that actually the open set condition as formulated in the proof above also impliesthat the rational fibers of R have positive Hausdorff measure in the relative Hausdorff dimension(see [28, Theorem 2.2]).To conclude this section, we want to provide a common lower and upper bound for the Hausdorffdimension of all the fiber of R . To this end, we first apply Proposition 5.7 to compute theHausdorff dimension of the two “extremal” rational fibers R − (1) and R − (1 / β (1) = 1 ∞ ⇒ σ ( β (1)) = (01) ∞ ⇒ d = 1 / ⇒ dim H R − (1) = 1 / ,β (1 /
3) = (01) ∞ ⇒ σ ( β (1 / ∞ ⇒ d = 1 ⇒ dim H R − (1 /
3) = log ϕ , where ϕ = (1 + √ / Proposition 5.8.
For every y ∈ [0 , , the following holds / ≤ dim H R − ( y ) ≤ log ϕ . Proof.
Thanks to (20) and the formula dim H ( A ∪ B ) = max { dim H A, dim H B } holding forevery A, B ⊂ [0 , / ≤ dim H ξ ( (cid:104)(cid:104) b (cid:105)(cid:105) ) ≤ log ϕ for every b ∈ S . Thisimmediately follows from Lemma 5.5, taking into account the obvious relation 1 ∞ (cid:60) b (cid:60) (01) ∞ and the fact that the Hausdorff dimension cannot increase under Lipschitz maps. For the sake of convenience, we extend the notation ρ v introduced in (9) and (10) by definingthe maps ρ nv : { , } ∞ → { , } ∞ for every finite binary word v and every n ≥ ρ v ( w ) = ρ v ( w ) , (46)with w ∈ { , } ∞ , and then for n > ρ nv = ρ ρ n − ( v ) ◦ ρ n − v . (47)We emphasize that, in general ρ nv does not coincide with ( ρ v ) n , the n -th iteration of the map ρ v , when v (cid:54) = (cid:15) (for v = (cid:15) we have ρ (cid:15) = ρ , and hence ρ n(cid:15) = ρ n = ( ρ (cid:15) ) n for every n ≥ ρ nv = ρ ρ n − ( v ) ◦ ρ ρ n − ( v ) ◦ . . . ◦ ρ ρ ( v ) ◦ ρ v , (48)which is a composition of h maps, each equal to ρ or (cid:101) ρ . Namely, the i -th map in the compositionis ρ or (cid:101) ρ depending on the parity of the length | ρ n − i ( v ) | of the image of v under the ( n − i )-thpower of ρ . Nevertheless, a straightforward induction on n ≥ ρ n ( vw ) = ρ n ( v ) ρ nv ( w ) , (49)where only the first two occurrences of n denote iterations of maps.As an consequence of the surjectivity of ρ , and hence of (cid:101) ρ , also all the maps ρ nv are surjective.In fact, as it happens for ρ and (cid:101) ρ , the inverse image ( ρ nv ) − ( w ) is countably infinite if w is afinite non-empty word, while it is uncountable if w is an infinite word, that is w ∈ B .On the other hand, contrary to what happens for ρ and (cid:101) ρ , the fact that w ∈ C is not enoughto guarantee that ρ nv ( w ) ∈ B when n >
1. For example ρ ((01) ∞ ) = ( ρ ρ (110) ◦ ρ )((01) ∞ ) = ( ρ ◦ ρ )((01) ∞ ) = ρ ( (cid:101) ρ ((01) ∞ )) = ρ (0 ∞ ) = (cid:15) . The next two lemmas constitute fundamental technical tools used for establishing the dynamicalproperties of the map R . Lemma 6.1.
For any sequence ( w n ) n ≥ of non-empty finite binary words and any sequence ( k n ) n ≥ of natural numbers with k = 0 and k n ≥ k n − + n (cid:15) ( w n − ) for every n ≥ , there is anuncountable set B ⊂ [ w ] ∩ C ⊂ B such that ρ k n ( b ) ∈ [ w n ] for every b ∈ B and every n ≥ .Proof. For any sequences ( w n ) n ≥ and ( k n ) n ≥ as above, we construct the generic element of B as an infinite concatenation b = v v . . . ∈ C , where the v n ’s are non-empty finite binary wordsinductively defined as follows. We start with v = w . Then, given v , v , . . . , v n − with n > u n = v v . . . v n − and we choose v n to be any of the infinitely many (non-empty) wordsin ( ρ k n u n ) − ( w n ) which ends with 1. We observe that, ρ k n ( u n ) = (cid:15) for every n ≥
1, whatever thechoice of the v n ’s. In fact, ρ k ( u ) = ρ k ( v ) = (cid:15) since k ≥ n (cid:15) ( v ), while for n >
1, taking intoaccount that k n − k n − ≥ n (cid:15) ( w n − ), we have by induction ρ k n ( u n ) = ρ k n − k n − ( ρ k n − ( u n − v n − ))= ρ k n − k n − ( ρ k n − ( u n − ) ρ k n − u n − ( v n − ))= ρ k n − k n − ( w n − )) = (cid:15) . The condition that each v n ends with 1 prevents different choices for the v n ’s to produce thesame final word b , as it can be easily deduced from the fact that | v n | = | w n | does not dependon the specific choice of v n . In this way, as the result of infinitely many infinite choices, one foreach n >
0, we get an uncountable set B of words b = v v . . . ∈ [ w ] ∩ C such that ρ k n ( b ) = ρ k n ( u n v n v n +1 . . . ) = ρ k n ( u n ) ρ k n u n ( v n ) ρ k n u n v n ( v n +1 . . . ) = w n ρ k n u n v n ( v n +1 . . . ) ∈ [ w n ]for every n ≥
1, where the second equality can be obtained by two applications of (49), whilethe third one immediately follows from ρ k n ( u n ) = (cid:15) and ρ k n u n ( v n ) = w n .21 emma 6.2. For any sequence ( w n ) n ≥ of non-empty finite binary words there is an uncount-able set B ⊂ [ w ] ∩ C ⊂ B with the property that for every b ∈ B and every n, m ≥ , thereexists k ≥ m such that ρ k ( b ) ∈ [ w n ] ⊂ B .Proof. Possibly by replacing ( w n ) n ≥ with the concatenation of all its finite initial subsequences,we can assume that each word w n appears infinitely many times in the sequence. Under thisassumptions, it is enough to prove the existence of an uncountable subset B ⊂ [ w ] ∩ C withthe weaker property that for every b ∈ B and n ≥ k ≥ ρ k ( b ) ∈ [ w n ]. Theexistence of such a set B is guaranteed by the previous lemma.We now proceed discussing some asymptotic properties of the orbits of the map R . First of all,we show that the set of rationals Q = Q ∩ [0 ,
1] is R -invariant and establish the asymptoticbehaviour of the restriction of R | Q : Q → Q . Proposition 6.3.
The set Q ⊂ [0 , is R -invariant. Moreover, R | Q admits only two periodicorbits, namely C = { , / } and C = { , / } , and Q decomposes as the disjoint union of twodense subsets Q = Q ∪ Q , (50) where Q i consists of all the rationals in Q whose forward orbit contains C i , for i = 1 , .Proof. The fact that C = { , / } or C = { , / } are 2-cycles can be trivially verified.Then, the R -invariance of Q follows once we prove that R ( x ) ∈ Q for every x ∈ Q (cid:114) { , } .Given any x ∈ Q (cid:114) { , } , we can write β ( x ) = wp ∞ ∈ C with w and p finite binary words ofminimal length such that | p | >
0. Then, by applying ρ we obtain ρ ( β ( x )) = ρ ( wp ∞ ) = ρ ( w ) ρ w ( p ) ρ wp ( p ) ρ wp ( p ) . . . If | p | is even, | wp k | has the same parity of | w | , and hence ρ wp k = ρ w for every k ≥
1. If instead | p | is odd, | wp k | has the same parity of | w | + k , and so ρ wp k coincides with ρ w for k even whileit coincides with the complementary map (cid:101) ρ w for k odd. Thus, we can rewrite ρ ( β ( x )) as ρ ( β ( x )) = (cid:26) ρ ( w )( ρ w ( p )) ∞ if | p | is even ρ ( w )( ρ w ( p ) (cid:101) ρ w ( p )) ∞ if | p | is odd . In both cases ρ ( β ( x )) is periodic, which implies that R ( x ) = ξ ( ρ ( β ( x )) ∈ Q . This concludes theproof of the R -invariance of Q .Now, we pass to prove that the forward orbit of any x ∈ Q (cid:114) { , } contains one of C or C .From the last formula we derive that ρ ( β ( x )) ∈ C , that is ρ ( β ( x )) = β ( R ( x )), except when | p | is even and | ρ w ( p ) | = 0, in which case ρ ( β ( x )) = β (cid:48) ( R ( x )) = ρ ( w )0 ∞ . Here, we have twopossibilities, either | ρ ( w ) | = 0 and then R ( x ) = 0 or else | ρ ( w ) | > β ( R ( x )) = v ∞ with n (cid:15) ( v ) ≤ n (cid:15) ( ρ ( w )) (in fact v = ρ ( w ) (cid:112)(cid:112) k −
0, where k the position of the last 1 in ρ ( w )).By iterating the process, we can conclude that R n ( x ) admits a purely periodic binary expansionfor every n ≥ n (cid:15) ( w ).On the other hand, for x (cid:48) ∈ Q (cid:114) { , } such that β ( x (cid:48) ) = q ∞ with q a finite binary word ofminimal length such that 0 < | q | < | q | , we have ρ ( β ( x (cid:48) )) = (cid:26) ( ρ ( q )) ∞ if | q | is even( ρ ( q ) (cid:101) ρ ( q )) ∞ if | q | is odd . The minimality of q , implies that | ρ ( q ) | = | q | < | q | if | q | is even, while | ρ ( q ) (cid:101) ρ ( q ) | = 2 | ρ ( q ) | =2 | q | < | q | if | q | is odd. In this last case, | ρ ( q ) (cid:101) ρ ( q ) | is even, | ρ ( q ) (cid:101) ρ ( q ) | = | ρ ( q ) (cid:101) ρ ( q ) | /
2, and ρ ( β ( x (cid:48) )) = β ( R ( x (cid:48) )). Therefore, either R ( x (cid:48) ) = ξ ( ρ ( β ( x (cid:48) ))) or R ( x (cid:48) ) = ξ ( ρ ( β ( x (cid:48) ))) admits apurely periodic binary expansion whose period length is strictly less than | q | . By iteration, weeventually get R n ( x (cid:48) ) = 0 or R n ( x (cid:48) ) = 1 for a sufficiently large n .22t this point, we are left to show that Q and Q are dense, or equivalently that they meet allthe dyadic intervals ξ ([ w ]) ⊂ [0 ,
1] with w any finite binary word.First, let us argue for Q . We put n = n (cid:15) ( w ), and define a sequence of finite binary words p , p , . . . , p n by induction on i decreasing from n to 0, as follows. We start with p n = 0, andgiven p i with 0 < i ≤ n , we let p i − be any even length element of ( ρ ρ i − ( w ) ) − ( p i ). The existenceof such an element is guaranteed by the surjectivity of ρ ρ i − ( w ) and the possibility of appendinga 0, if needed to make the length even, without changing the image under ρ ρ i − ( w ) .Then, we put x = ξ ( wp ∞ ) ∈ ξ ([ w ]). By induction on i , we get R i ( x ) = ξ ( ρ i ( w ) p ∞ i )for every i = 0 , . . . , n , as follows. The base of the induction is the trivial case of i = 0, whilethe inductive step is given by R i ( x ) = R ( ξ ( ρ i − ( w ) p ∞ i − )) (the inductive hypothesis)= ξ ( ρ ( ρ i − ( w ) p ∞ i − )) (since ρ i − ( w ) p ∞ i − ∈ C )= ξ ( ρ i ( w ) ρ ρ i − ( w ) ( p ∞ i − )) (thanks to equation (47))= ξ ( ρ i ( w )( ρ ρ i − ( w ) p i − ) ∞ ) (since p i − has even length)= ξ ( ρ i ( w ) p ∞ i ) (by definition of p i − ) . In particular, we have R n ( x ) = ξ ( ρ n ( w ) p ∞ n ) = ξ (0 ∞ ) = 0, as desired.This concludes the proof of the density of Q . The same argument, but starting from p n = 1instead of p n = 0, proves that Q is dense.We remark that C and C are the only periodic R -orbits in Q and that Q does not containany dense R -orbit. Thus, in the following, when considering any other periodic R -orbit or anydense R -orbit, we can always assume that they are disjoint from Q , avoiding in this way thetechnicalities due to the double binary expansion of the dyadic rationals. Proposition 6.4.
The set of periodic points of R is dense in [0 , and contains uncountablymany n -periodic points for any given minimal period n ≥ .Proof. In the light of the above observation, apart from the 2-cycles C and C , all the other n -periodic R -orbits are contained in [0 , (cid:114) Q and then they bijectively correspond to those of ρ through the unique binary expansion β ( x ) of every x ∈ [0 , (cid:114) Q .Therefore, the first part of the statement follows once we prove that for any non-empty finitebinary word w there are uncountably many infinite binary ρ -periodic words b in [ w ] ∩ C , whoseperiod is the vanishing order n (cid:15) ( w ). In fact, it is enough to observe that the correspondingpoints ξ ( b ) in the generic dyadic interval ξ ([ w ]) ⊂ [0 ,
1] are R -periodic with the same period.Fixed any non-empty finite binary word w , we put n = n (cid:15) ( w ) ≥ b = w w w . . . w k . . . ∈ C , where w = w and for every integer k ≥ w k be any of the infinitely many (non-empty) words in ( ρ nw w ...w k − ) − ( w k − ) which ends with 1. As in the proof of Lemma 6.1, thislast condition guarantees that there are no duplicates in the resulting words b , which are thusuncountably many. Then, we have ρ n ( b ) = ρ n ( w w w . . . w k . . . )= ρ n ( w ) ρ nw ( w ) ρ nw w ( w ) . . . ρ nw w ...w k − ( w k ) . . . = (cid:15)w w . . . w k − . . . = b , and hence b has period n . 23or the second part we need the additional fact that for every n ≥ w in ρ − n ( (cid:15) ), for example w = σ n − (0), and that such aword w can be chosen so that no ρ k ( w ) is a prefix of w for k < n , for example w = 00 σ n − (0).This last property immediately implies that ρ k ( b ) (cid:54) = b for every as above and k = 1 , . . . , n − n is the minimal period of b .In particular, Proposition 6.4 tells us that the set Fix R of fixed points of R is uncountable,meaning that the graph of R crosses the diagonal uncountably many times. In fact, by taking w = 0 (cid:96) in the previous proof, the resulting word b = w w w . . . ∈ C belongs to Fix ρ , and hence ξ ( b ) ∈ Fix R . In the special case when w = w = 0 and each w k is chosen to have minimal lengthamong the words with the prescribed properties, that is either w k = σ ( w k − ) or w k = σ ( (cid:101) w k − )according to the parity of | w w . . . w k − | , we get as b the word b = 00101 (cid:81) ∞ h =0 (1 · h (01) · h ) = 00101111010101111111010101010101 . . . . (51)The corresponding real number x = ξ ( b ) = 0 . . . . (52)turns out to be the largest element in Fix R , as we will see in a while.Now, we want to make more explicit the procedure given in the proof of Proposition 6.4 inthe specific case of fixed points. To any sequence a = ( a , a , . . . ) ∈ A = N ω , we associates a(distinct) fixed point (cid:104) a (cid:105) F ∈ Fix ρ ⊂ C and a corresponding (distinct) fixed point x a = ξ ( (cid:104) a (cid:105) F ) =0 .b a ∈ Fix R ⊂ [0 , b and x associated to the null sequence.First, we observe that a non-empty finite binary word w can be completed to a fixed point wb ∈ Fix ρ for a suitable b ∈ B if and only if ρ ( w ) is a prefix of it. Moreover, if this is the casethen there is a unique word b ∈ S such that b w = wb ∈ Fix ρ . The word b w can be constructed,by generalizing the construction of b outlined above. Since ρ ( w ) is a prefix of w , we can write w = ρ ( w ) v for a certain word v , which can be easily seen to be non-empty. Then, we define b w as the infinite concatenation b w = wv v . . . ∈ Fix ρ , where the words v k for k ≥ v and putting v k = σ ( v k − ) or v k = σ ( (cid:101) v k − ) if | wv . . . v k − | is even or odd, respectively (compare with the con-struction of b above). Then b = v v . . . ∈ S , since each v k contains no pair of consecutive0’s and ends with 1. Moreover, these properties of v k make it uniquely determined by v k − forevery k ≥
1, which implies the unicity of b in S by induction.Now, given any a = ( a , a , . . . ) ∈ A , we put (cid:104) a (cid:105) F = b and define the sequence ( (cid:104) a (cid:105) F n ) n ≥ ⊂ Fix ρ as follows. For every n ≥
1, let k n be the position of the n -th occurrence of 1 in (cid:104) a (cid:105) F n − , and w an be the word obtained from (cid:104) a (cid:105) F n − (cid:112)(cid:112) k n by inserting the word 0 a n before the last digit 1. Theword w an share with (cid:104) a (cid:105) F n − (cid:112)(cid:112) k n the property that its image ρ ( w an ) = ρ ( (cid:104) a (cid:105) F n − (cid:112)(cid:112) k n ) is a prefix of it,so we can set (cid:104) a (cid:105) F n = b w an . Notice that | w an | = k n + 2 a n ≥ k n > k n − + 2 a n − = | w an − | for every n >
1. Since the sequence ( k n ) n ≥ is increasing and (cid:104) a (cid:105) F n (cid:112)(cid:112) k n − = (cid:104) a (cid:105) F n − (cid:112)(cid:112) k n − for every n ≥ (cid:104) a (cid:105) F = lim n →∞ (cid:104) a (cid:105) F n , (53)which is a fixed point for R , admitting by construction all the prefixes w an for n ≥ R , we need to prove that the previous constructionproduces all the points in Fix ρ . Actually, we prove something more in the following Lemma. Lemma 6.5.
The map φ : A → Fix ρ given by a (cid:55)→ φ ( a ) = (cid:104) a (cid:105) F is a Lipschitz homeomorphism. roof. For any b ∈ Fix ρ , we consider the sequence a b = ( a b , a b . . . ) ∈ A , where a bn is thenumber of (disjoint) pairs of 0 s immediately preceding the n -th occurrence of 1 in b . Then, themap ψ : Fix ρ → A given by b (cid:55)→ a b is the inverse of φ , as straightforward verification shows.Now, for any a, a (cid:48) ∈ A we have that d ( a, a (cid:48) ) = 1 / n +1 if and only if a (cid:112)(cid:112) n = a (cid:48) (cid:112)(cid:112) n and a n +1 (cid:54) = a (cid:48) n +1 ,or equivalently, with the above notations, w an = w a (cid:48) n and w an +1 (cid:54) = w a (cid:48) n +1 . Therefore, (cid:104) a (cid:105) F and (cid:104) a (cid:48) (cid:105) F share the same prefix of length | w an | = k n + 2 a n ≥ k n ≥ n , while they have different prefixesof length | w an +1 | = k n +1 + 2 a n +1 . So, we have 1 / k n +1 +2 a n +1 ≤ d ( φ ( a ) , φ ( a (cid:48) )) ≤ / n +1 , whichmeans that φ is a Lipschitz map (second inequality) and that ψ is continuous at φ ( a ) (firstinequality). Proposition 6.6.
The fixed point set
Fix R is a uncountable subset of [0 , (cid:114) D and its closure Cl(Fix R ) is a null measure Cantor in [0 , with a countable remainder Cl(Fix R ) (cid:114) Fix R ⊂ D .Proof. The inclusion Fix R ⊂ [0 , (cid:114) D follows by Proposition 6.3. As a consequence, the maps ξ and β restrict to a bijective maps Fix ρ ↔ Fix R , and hence Fix R is uncountable.We now prove that Cl(Fix R ) ⊂ Fix R ∪ D , which trivially gives Cl(Fix R ) (cid:114) Fix R ⊂ D . Theargument is similar to the first part of the proof of Proposition 5.3. Let x ∈ Cl(Fix R ) and( x n ) n ≥ ⊂ Fix R be a sequence such that lim n →∞ x n = x . Then, by Lemma 6.5 there is asequence ( a n ) n ≥ ⊂ A such that x n = ξ ( φ ( a n )) for every n ≥
1. If x ∈ D we are done, otherwisethe continuity of β at x implies that lim n →∞ φ ( a n ) = β ( x ) ∈ C . If the sequence of the i -thcomponents (( a n ) i ) n ≥ is bounded for every i ≥
1, then by compactness we can replace thesequence ( a n ) n ≥ by a subsequence converging to a ∈ A , and by the continuity of φ we have β ( x ) = φ ( a ), that is x = ξ ( φ ( a )) ∈ Fix R . If instead some sequence (( a n ) i ) n ≥ is unbounded, let m be minimum index i ≥ a n ) n ≥ by a suitable subsequence such that the sequence of truncations (( a n ) (cid:112)(cid:112) m − ) n ≥ is constant, andhence φ ( a n ) (cid:112)(cid:112) k m − , where k m − is the index of the ( m − d for every n ≥
1. Then, we get β ( x ) = lim n →∞ φ ( a n ) = d ∞ , which is absurdsince β ( x ) ∈ C . This concludes the proof of the inclusion Cl(Fix R ) ⊂ Fix R ∪ D .Concerning the null measure property, let us consider the map φ ◦ φ − ∞ : (cid:104)(cid:104) ∞ (cid:105)(cid:105) → Fix ρ , where φ ∞ : A → (cid:104)(cid:104) ∞ (cid:105)(cid:105) and φ : A → Fix ρ are the homeomorphisms defined in the proof ofLemma 5.4 and in Lemma 6.5, respectively. This map sends (cid:104) a (cid:105) ∞ ∈ (cid:104)(cid:104) ∞ (cid:105)(cid:105) to (cid:104) a (cid:105) F ∈ Fix ρ forevery a ∈ A , and it can be shown to be a 1-Lipschitz map as follows. Given any a, a (cid:48) ∈ A suchthat a (cid:112)(cid:112) n = a (cid:48) (cid:112)(cid:112) n and a n +1 (cid:54) = a (cid:48) n +1 we put h ( a, a (cid:48) ) = a + . . . + a n + min { a n +1 , a (cid:48) n +1 } , as in the proofof Lemma 5.4. Then, the maximum index i for which (cid:104) a (cid:105) ∞ (cid:112)(cid:112) i = (cid:104) a (cid:105) ∞ (cid:112)(cid:112) i is given by 2 h ( a, a (cid:48) ) + n ,while the maximum index i for which (cid:104) a (cid:105) F (cid:112)(cid:112) i = (cid:104) a (cid:48) (cid:105) F (cid:112)(cid:112) i is given by 2 h ( a, a (cid:48) ) + k n , where k n is theindex of the n -th occurrence of 1 in the word (cid:104) a (cid:105) F n − defined in the construction of (cid:104) a (cid:105) F justbefore Lemma 6.5. Since k n ≥ n , we can conclude that φ ◦ φ − ∞ is a 1-Lipschitz map. From this,arguing as in the proof of Lemma 5.5, we can derive the Lipschitz condition for the map ξ ◦ φ ◦ φ − ∞ ◦ β : ξ ( (cid:104)(cid:104) ∞ (cid:105)(cid:105) ) → Fix
R .
Therefore, thanks to Proposition 5.8 and its proof, we have dim H Fix R ≤ dim H ξ ( (cid:104)(cid:104) ∞ (cid:105)(cid:105) ) =log ϕ <
1, which implies that Fix R has a null Lebesgue measure. Since the remainderCl(Fix R ) (cid:114) Fix R is countable, also Cl(Fix R ) has a null Lebesgue measure.As a consequence, Cl(Fix R ) is topologically 0-dimensional. On the other hand, it is a perfectset, because for every a ∈ A , we have ξ ( (cid:104) a (cid:105) F ) = lim n →∞ ξ ( (cid:104) a n (cid:105) F ) with a n ∈ A defined by( a n ) i = a i + δ i,n , for any i, n ≥
1. So, Cl(Fix R ) is a Cantor set (see [23]).25e observe that the proof of Proposition 6.6 is essentially based on the parametrization of Fix ρ given by Lemma 6.5, which is nothing else than an explicit reformulation in the case of fixedpoints of the general construction provided in the proof of Proposition 6.4 of periodic points ofany period. So, we expect that similar results hold for periodic points of any given period.The idea is that, in order to generate all the periodic orbits of a given period n >
1, onecould start from a certain finite set of “simplest” ones, and than progressively enlarge this setby iterating in a suitable order the following procedure. Take any such n -orbit ( b , b , . . . , b n ),insert some pairs of consecutive 0’s in any position of any b i , and then reconstruct all the orbitaccordingly.For example, starting from the 2-cycle (1 ∞ , ∞ ) of ρ , corresponding to the 2-cycle C of R (see Proposition 6.3), one can obtain in this way the Thue-Morse 2-cycle ( u, (cid:101) u ) of ρ describedin the next remark, and hence a corresponding 2-cycle for R . Remark 6.7.
Consider the well known Thue-Morse substitution rule on B (cid:98) τ : (cid:110) (cid:55)→ (cid:55)→ , and the digit-wise generated map (cid:98) τ : B → B . The map (cid:98) τ has two fixed points, the Thue-Morsesequence starting with u = 01101001100101101001011001101001 . . . and its complementary sequence (cid:101) u , the Thue-Morse sequence starting with , which was used in [42] to found the combinatorics of words. Recalling the definition of τ in (13) and noting that τ (0) = (cid:98) τ (0) = 0110 and τ (1) = (cid:98) τ (1) = 1001 , we can easily verify that ( u, (cid:101) u ) is a 2-cycle of ρ . Then, letting t = ξ ( u ) be the Thue-Morsenumber, we have the corresponding 2-cycle ( t, − t ) of R . Concerning the periodic points of R having higher order n , in analogy with the description ofthe maximal fixed point x , we limit ourselves to consider the periodic points whose binaryexpansion is the “simplest” infinite binary word among those generated by the constructiondescribed in the proof of Proposition 6.4. Namely, the binary word obtained by starting fromthe word w = 0 and choosing each w k to be the shortest word in ( ρ nw w ...w k − ) − ( w k − ), thatis the image of w k − under a suitable composition of σ ’s and (cid:101) σ ’s.The case of n even is not so interesting, since we always obtain 0 . ∞ , while the binary expressionof the “simplest” periodic point of odd order n = 2 (cid:96) + 1 ≥ x (cid:96) = 0 . (01) (cid:96) − (cid:81) ∞ h =0 (cid:0) (2 (cid:96) +1) h + (cid:96) (01) (2 (cid:96) +1) h + (cid:96) − (2 (cid:96) −
1) 2 (2 (cid:96) +1) h + (cid:96) (01) (2 (cid:96) +1) h +2 (cid:96) (2 (cid:96) +1) h +2 (cid:96) (01) (2 (cid:96) −
1) 2 (2 (cid:96) +1) h +2 (cid:96) (cid:1) . (54)In order to see that this point has order n = 2 (cid:96) + 1, we write it as 0 .w (cid:81) ∞ h =0 ( u h v h ), with w = (01) (cid:96) − , u h = 1 (2 (cid:96) +1) h + (cid:96) (01) (2 (cid:96) +1) h + (cid:96) − (2 (cid:96) −
1) 2 (2 (cid:96) +1) h + (cid:96) ,v h = (01) (2 (cid:96) +1) h +2 (cid:96) (2 (cid:96) +1) h +2 (cid:96) (01) (2 (cid:96) −
1) 2 (2 (cid:96) +1) h +2 (cid:96) . (55)Then, we first note that the word w vanish to the empty word (cid:15) in exactly n iterates of ρ .Moreover, by a somewhat tedious but straightforward calculation, repeatedly using the rules ρ : (cid:26) (01) k (cid:55)→ k k (cid:55)→ (01) k , k ≥
1, one checks that ρ nw ( u ) = (cid:101) ρ ◦ ρ (cid:96) − ( u ) = w , while ρ nwu v ...u h ( v h ) = ρ (cid:96) +1 ( v h ) = u h and ρ nwu v ...v h ( u h +1 ) = ρ (cid:96) +1 ( u h +1 ) = v h for every h ≥
0. Putting all those equation together, one can conclude that w (cid:81) ∞ h =0 ( u h v h ) is afixed point for ρ n , and hence that 0 .w (cid:81) ∞ h =0 ( u h v h ) is a fixed point for R n .The rest of this section is devoted to analyse some chaotic properties of the map R . Proposition 6.8.
The set of points with a dense R -orbit is uncountable and dense in [0 , .Proof. To get the existence of uncountably many points with a dense R -orbit, it suffices to takethe sequence ( w n ) n ≥ in Lemma 6.2 with the property that it contains all the non-empty finitebinary words. The density of the points with a dense R -orbit follows immediately observingthat we can take any non-empty finite word as w . Proposition 6.9.
For every interval I ⊂ [0 , , there exists n ≥ such that R n ( I ) = [0 , .In particular, R is topologically ( strongly ) mixing and has sensitive dependence on initial con-ditions.Proof. Let w be any non-empty finite binary word such that ξ ([ w ]) ⊂ I , and set n = n (cid:15) ( w ).Pick y ∈ [0 , ρ nw , there is v ∈ C such that ρ nw ( v ) = β ( y ), and hence ρ n ( wv ) = ρ n ( w ) ρ nw ( v ) = β ( y ) as well. Then, for x = ξ ( wv ) we have x ∈ I and R n ( x ) = y .This implies sensitivity to initial conditions with sensitivity constant 1 /
2. Since every open setis a countable union of open intervals, it also follows that, for every open sets
A, B ⊂ [0 , n ≥ R m ( A ) ∩ B (cid:54) = ∅ for every m > n , that is R is topologically mixing. Corollary 6.10.
The map R is chaotic in the sense of Devaney ( see e.g. [13]) .Proof. It follows by Propositions 6.4, 6.8 and 6.9.Concerning distributional chaos, a concept introduced within the context of topological dynam-ical systems in 1994 (see [37]), the map R satisfies the strongest form of distributional chaos,namely the distributional chaos of type 1, DC1-chaos in short [1]. Actually, we prove that theDC1-chaos exhibited by R is uniform in the sense of [30], as specified in the next proposition. Proposition 6.11.
The map R is uniformly DC1-chaotic, meaning that it admits an uncount-able uniformly DC1-scrambled set, that is an uncountable subset S ⊂ [0 , such that for every δ > and some δ > the following two conditions hold, where denotes the cardinality, lim sup n →∞ n { k = 0 , . . . , n − such that | R k ( x ) − R k ( y ) | < δ } = 1 , (56)lim inf n →∞ n { k = 0 , . . . , n − such that | R k ( x ) − R k ( y ) | < δ } = 0 , (57) for every x (cid:54) = y in S .Proof. We will explicitly construct a set S = { x α } α ∈ A satisfying the properties (56) and (57).As the set of indices A we take the image A = µ ( B ) ⊂ B of the map µ : B → B defined by µ ( b ) n = (cid:26) b k if n = p k with p prime and k ≥
10 otherwise , b ∈ B . We observe that A is an uncountable set, being µ injective, and that all theelements α ∈ A share the same n -th component α n = 0 for infinitely many n ≥
1. Furthermore,since each α = µ ( b ) contains infinite copies of b as subsequences, any two different elements α, α (cid:48) ∈ A cannot be ultimately coinciding, that is for every m ≥ n ≥ m with α n (cid:54) = α (cid:48) n .Now, let ( e n ) n ≥ be the increasing sequence of even natural numbers inductively defined by e = 1 and e n = n (cid:80) n − k =1 e k for n > , (58)and let ( v n ) n ≥ and ( u n ) n ≥ be the two sequences of binary words inductively defined by v = 1 and v n = σ e n ( v n − ) for n > ,u = 001 and u n = 00 σ e n ( u n − ) for n > . (59)Clearly, we have n (cid:15) ( v n ) = n (cid:15) ( u n ) = (cid:80) nk =1 e k for every n ≥
1. Moreover, the equality σ (1 k ) =1 k implies that v n = 1 (cid:80) nk =2 ek , (60)for every n > α ∈ A we associate the sequence of non-empty finite binary words ( w αn ) n ≥ , where w α can be arbitrarily chosen, while for n ≥ w αn = (cid:110) v n if α n = 0 u n if α n = 1 . We set k n = n (cid:15) ( w α ) + . . . + n (cid:15) ( w αn − ) and observe that k n is the same for all α ∈ A .Then, we apply Lemma 6.1 the sequences ( w αn ) n ≥ and ( k n ) n ≥ and pick a single binary word b α ∈ [ w α ] ∩ C , out of the uncountable resulting set B , such that ρ k n ( b α ) ∈ [ w αn ] for every n ≥ S = { x α = ξ ( b α ) } α ∈ A . Since all the words b α and ρ k ( b α ) belong to C , by theirconstruction in the proof of Lemma 6.1, we have that all the points x α are distinct and that R k ( x α ) = ξ ( ρ k ( b α )) for every k ≥ α ∈ A .To see that S is uniformly DC1-scrambled, take any α (cid:54) = α (cid:48) in A . For any δ >
0, let c ( δ ) ≥ / c ( δ ) / < δ , in such a way that the interval ξ ([ ρ h ( w αn )]) has width λ ( ξ ([ ρ h ( w αn )]) < δ for every h = 1 , . . . , n (cid:15) ( w αn ) − c ( δ ) = (cid:80) nk =1 e k − c ( δ ) . (61)By construction, there is a sequence ( n i ) i ≥ such that α n i = α (cid:48) n i for every i ≥
1. By (61), thisimplies that1 k n i +1 { k = 1 , . . . , k n i +1 such that | R k ( x α ) − R k ( x α (cid:48) ) | < δ } ≥ (cid:80) n i k =1 e k − c ( δ ) k n i +1 . (62)Thanks to equation (58), we have (cid:80) n i k =1 e k = (cid:80) n i − k =1 e k + 2 e n i = 2( n i + 1) e n i n i , (63)while k n i +1 = (cid:80) n i k =1 n (cid:15) ( w αk ) = (cid:80) n i k =1 (cid:80) kh =1 e h = (cid:80) n i − k =1 (cid:80) kh =1 e h + (cid:80) n i h =1 e h ≤ n i − (cid:80) n i − k =1 e k + (cid:80) n i k =1 e k = 2( n i − e ni n i + (cid:80) n i k =1 e k = 2( n i − e ni n i + 2( n i + 1) e n i n i . (64)28herefore, the right hand side of (62) is bounded below by n i + 1 n i + n i − c ( δ ) k n i +1 , which tends to 1 as n i diverges. This completes the proof of property (56).To prove (57), we consider a different sequence ( n i ) i ≥ such that α n i (cid:54) = α (cid:48) n i for every i ≥ | R k ni ( x α ) − R k ni ( x α (cid:48) ) | > /
8. So, putting δ = 1 / k n i +1 { k = 1 , . . . , k n i +1 such that | R k ( x α ) − R k ( x α (cid:48) ) | < δ } ≤ k n i +1 − (cid:80) n i k =1 e k k n i +1 . Recalling the equations (63) and (64), we see that the last quantity vanishes as n i diverges,which gives property (57).We emphasize that in the above proof the family of starting words ( w α ) α ∈ A is completelyarbitrary. This means that the DC1-scrambled sets for R form a dense subset in the space of allsubsets of [0 ,
1] with the Hausdorff pseudo-distance. In particular, the DC1-scrambled set S canbe chosen to be a dense subset of any interval I ⊂ [0 , w αn ) n ≥ in the proof, it could be shown that actually R is transitively DC1-chaotic,that is all the points of the uncountable DC1-scrambled set S have dense orbits. Remark 6.12.
Proposition 6.11 is interesting in view of the results of [39] and [40] , in whichthe author proves that there is a residual subset of in the space of bounded functions of Baireclass , and of Baire class as well, whose elements are neither Devaney nor Li-Yorke chaotic ( see [20] and [26]) . It follows that the map R , although generated by a very simple erasingsubstitution rule, is “much more chaotic” than the topologically generic map in its Baire classand also in the higher one. In the final part of this section we address topological entropy.We recall that for a map f : X → X with X a metric space, the topological entropy can bewritten as [14, 4] h ( f ) = lim ε → lim sup n →∞ log r ( n, ε ) n , (65)where r ( n, ε ) is the maximum cardinality of an ε -separated set (that is a set whose points areat least ε -apart from each other) in the metric d n ( x, y ) = max ≤ i ≤ n | f i ( x ) − f i ( y ) | . (66)Although this definition is expressed in metric terms, it turns out to depend only on theunderlying topology — for a modern general reference on topological entropy see [15].For continuous interval maps, positive topological entropy is known to be equivalent distribu-tional chaos, and in particular to DC1-chaos. In our more general context, the positivity of theentropy of the map R should be proved directly. In fact, in the next proposition we prove that h ( R ) = ∞ . This can be thought as a consequence of the fact that the substitution ρ is “efficientenough” in erasing the prefixes, so that all the points “forget” quickly where they came from.Namely, Proposition 2.1 tells us that the vanishing order of any finite binary word w satisfiesthe inequality n (cid:15) ( w ) ≤ (cid:98) log | w |(cid:99) + 2 . (67)29 roposition 6.13. The map R has infinite topological entropy.Proof. For every k, n ≥ w , w , . . . , w n ∈ { , } k , we applyLemma 6.1 to get a single binary word b w ,w ,...,w n ∈ [ w ∩ C such that for every i = 0 , , . . . , nρ i(cid:96) ( b w ,w ,...,w n ) ∈ [ w i , where (cid:96) = 2 (cid:98) log | w |(cid:99) + 2 ≥ n (cid:15) ( w i ) according to Proposition 2.1. Then, we put S k,n = { x w ,w ,...,w n = ξ ( b w ,w ,...,w n ) | w , w , . . . , w n ∈ { , } k } ⊂ [0 , . (68)Since all the words b w ,w ,...,w n and ρ k ( b w ,w ,...,w n ) belong to C , by their construction in the proofof Lemma 6.1, we have that all the points x w ,w ,...,w n are distinct and that R k ( x w ,w ,...,w n ) = ξ ( ρ k ( b w ,w ,...,w n )) for every k ≥ w , w , . . . , w n ∈ { , } k .We note that, for every k, n ≥ S k,n has cardinality S k,n = 2 ( n +1) k and it is ε -sepa-rated in the metric d n if ε < / k +1 . Indeed, given any two different points x = x w ,w ,...,w n and y = x w (cid:48) ,w (cid:48) ,...,w (cid:48) n in S k,n , we have w i (cid:54) = w (cid:48) i for some i = 0 , . . . , n , and hence | R i ( x ) − R i ( y ) | ≥ ε ,as it can be easily deduced from x ∈ ξ ([ w i y ∈ ξ ([ w (cid:48) i k = (cid:100)− log ε (cid:101) − h ( R ) ≥ lim ε → lim sup n →∞ log 2 ( n +1) k n = lim ε → lim n →∞ ( n + 1) log 2 ( (cid:100)− log ε (cid:101)− n = ∞ . Recently some attention has been devoted to the study of finer properties of topological entropy,in particular to the characterization of the points around which the entropy concentrates. Moreprecisely, one can introduce the notion of relative entropy h ( f, K ) for a map f : X → X , with X a metric space, and a subset K ⊂ X , as the limit (65) where r ( n, (cid:15) ) is defined by constrainingthe ε -separated sets to be contained in K . Then, following [44], if f has positive entropy, x ∈ X is called an entropy point (resp. a full entropy point) if h ( f, K ) > h ( f, K ) = h ( f )) forevery closed neighborhood of x in X . Remark 6.14.
It is known that, in case of continuous maps with positive entropy on a com-pact metric space, minimality implies that every point is a full entropy point [44] . The system ([0 , , R ) is not minimal (there are periodic points of every period), nevertheless every point isa full entropy point, as an immediate consequence of Proposition 6.9. We recall that a Liouville number is a real number x such that, for every n , there exists asequence of rationals p i /q i ( p i , q i ∈ Z , q i (cid:54) = 0) verifying (cid:12)(cid:12)(cid:12) x − p i q i (cid:12)(cid:12)(cid:12) < q ni . (69) Proposition 7.1.
The map R admits uncountably many fixed points that are Liouville numbers,hence transcendental real numbers.Proof. For the existence of uncountably many Liouville fixed points, it suffices to observe that,by construction, the fixed point 0 . (cid:104) a (cid:105) F is a Liouville number as soon as the sequence a ∈ A increases fast enough (see definition (53) and Proposition 6.6). By a classic result [25], all thefixed points which are Liouville numbers are transcendental.A similar result could be established analogously for periodic points of any given period. How-ever, this simple argument does not apply to the largest fixed point, for which we have toprovide a different proof of transcendence. 30 roposition 7.2. The largest fixed point x of the map R is transcendental.Proof. Let p ( n, x ) be the cardinality of the set of words of length n appearing in the binaryexpansion of x . Then we have lim sup n →∞ p ( n, x ) n < . (70)Indeed, given any n >
3, let k be the least positive integer such that k ≥ log ( n/ · k ≥ n and 3 · k − < n . Then, we split the binary expansion b of x as the product b = b (cid:48) b (cid:48)(cid:48) , where b (cid:48) = 00101 (cid:81) k − h =0 · h (01) · h and b (cid:48)(cid:48) = (cid:81) ∞ h = k · h (01) · h . The length of b (cid:48) is5 + 9 (cid:80) k − h =0 h = 5 + 9(2 k −
1) = 9 · k − < n − < n . (71)Therefore, there are less than 6 n different subwords of b having length n and starting in b (cid:48) . We are left to consider the different subwords of length n that are contained in b (cid:48)(cid:48) . Dueto the inequality 3 · k ≥ n , each of them consists of a (possibly empty) sequence of 1’sfollowed/preceded by a (possibly empty or truncated on the right/left) sequence of 01’s. So,there are at most 2 n + 3 such subwords. Therefore, p ( n, x ) < n + 3 for every n ≥
1, whichgives (70). Thus, by Theorem 3.1 in [7], x is transcendental.Along the same lines we can prove the following. Proposition 7.3.
The “simplest” periodic point x (cid:96) of period (cid:96) + 1 defined in (54) is transcen-dental for each (cid:96) ≥ .Proof. We know that the general form for the binary expansion of x (cid:96) is b (cid:96) = w (cid:81) ∞ h =0 u h v h where w, u h and v h are the words in (55). Proceeding as above, let k be the least positive integer suchthat 2 (2 (cid:96) +1) k + (cid:96) ≥ n and 2 (2 (cid:96) +1)( k − (cid:96) < n . We then split the binary expansion of b (cid:96) as b (cid:48) (cid:96) b (cid:48)(cid:48) (cid:96) , where b (cid:48) (cid:96) = w (cid:81) k − h =0 u h v h and b (cid:48)(cid:48) (cid:96) = (cid:81) ∞ h = k u h v h . Now, one shows that the length of b (cid:48) (cid:96) is bounded aboveby 12 · (cid:96) +1 n , which also gives a bound for the number of different subwords having length n and starting in b (cid:48) (cid:96) . Moreover, by the same argument used above yields that b (cid:48)(cid:48) (cid:96) contributes withat most 2 n + 3 to p ( n, b (cid:96) ). Many natural questions concerning the map R do not have yet an answer. We limit ourselvesto list a few of them.Concerning basic properties of the map, we propose the following. Question 8.1.
What is the Hausdorff dimension of the graph G of R ? We note that the self-affine structure of G discussed in Section 4 seems not to fit into any ofthe known criteria for the computation of the Hausdorff dimension.From the ergodic point of view, the main question is probably the next one. Question 8.2.
Does R admit an invariant measure which is not purely atomic? We recall that the singularity of R prevents the existence of absolutely continuous (with respectto Lebesgue) invariant measures.Although the main interest in studying the dynamics of a map such as R comes from a naturalgeneralization of topological dynamics questions to a wider class of functions, like Baire class 1functions, it could be of interest also in an ergodic context to model specific “non-equilibrium”situations in which events that are impossible at one time become possible at another time.31oncerning points with a dense orbit, heuristics indicate that there is plenty of them, as wehave uncountability for two different reasons: in the procedure described in the proof of Lemma6.2 there are countably many steps each involving countably many choices; in Proposition 6.8there are uncountably many possible choices for the sequence ( w n ) n ≥ .Therefore, it seems reasonable to ask the following. Question 8.3.
Is the set of points with a dense R -orbit of second category ( or even co-meagre ) ?Does it have positive ( or even full ) Lebesgue measure?
About the arithmetic properties of R , in the light of the results of Section 7, and of the factthat the Thue-Morse numbers is known to be transcendental, one could ask the question below. Question 8.4.
Is every fixed point of R transcendental? Is every periodic point of R not in C or C transcendental? Finally, as said in the Introduction, the map R is the model-case of a more general class of in-terval maps generated by erasing block substitutions, meaning (in the binary case) substitutionrules s : { , } k → { , } ∗ such that at least one word w of length k is mapped to (cid:15) . Lookingat these more general maps poses of course new problems. Question 8.5.
What of the properties we established for the map R are true for interval mapsgenerated by which classes of ( binary ) erasing block substitutions? What new phenomena arisein this extended context? Let us list some speculation about this last question.1. While the map ρ has an “almost-inverse” map σ (up to insertions of the word 00), this isnot true for a general erasing substitution rule s as above. Indeed, in the s -preimages of agiven infinite word v ∈ { , } ω , there could be in principle infinitely many words (cid:81) ∞ i =1 w i with w i ∈ { , } k , such that s ( w i ) (cid:54) = (cid:15) for every i ≥
1. This could make the topologicalstructure of the fibers much more complex than in our present case.2. Even assuming that there is only one word w (cid:15) ∈ { , } k mapped by s to the empty word, if w (cid:15) (cid:54) = 0 k we have to consider separately three objects, that is the set D of dyadic rationalsin [0 ,
1] and the two sets S = { x ∈ [0 , | x = 0 .v, v = s ( w ∞ ) , w ∈ { , } ∗ } ,S = { x ∈ [0 , | x = 0 .ww ∞ (cid:15) , w ∈ { , } nk } . It is not difficult to see that these sets coincide in the case when w (cid:15) = 0 k , but this is nottrue in general. Moreover, if s − ( (cid:15) ) ⊂ { , } k has more than one element, the interval map f s : [0 , → [0 ,
1] generated by the symbolic action of s is in general discontinuous also atsome (actually uncountably many) irrationals.3. For suitable erasing substitution rules s , the map f s can have stronger properties than ρ .For instance, s can be chosen so that f s is a Darboux function.A first study of interval maps generated by general erasing block substitutions is done in [12],but several problems are still open, in particular concerning point 1 above. Acknowledgements
This work is partially supported by the research project PRIN 2017S35EHN 004 “Regular andstochastic behaviour in dynamical systems” of the Italian Ministry of Education and Research.32 ist of symbols N natural numbers (including 0) D ⊂ Q dyadic rationals and all rationals, respectively ϕ the golden ratio (1 + √ / Q = Q ∩ [0 ,
1] rationals in [0 , D = D ∩ [0 ,
1] dyadic rationals in [0 , N ⊂ [0 ,
1] normal binary numbers in [0 , A = N ω infinite sequences of natural numbers B = { , } ω infinite binary sequences/words C ⊂ B infinite binary sequences/words which are not eventually 0 C (cid:48) ⊂ B infinite binary sequences/words which are not eventually 1 S ⊂ C infinite binary sequences/words with no pair of consecutive 0’s η : A → C map ( a , a , . . . ) (cid:55)→ a a . . . (1) ξ : B → [0 ,
1] map ( b , b , . . . ) (cid:55)→ x = 0 .b b . . . (3) β : (0 , → C map x (cid:55)→ ( b , b , . . . ), the unique sequence in C such that x = 0 .b b . . . (4) β (cid:48) : [0 , → C (cid:48) map x (cid:55)→ ( b , b , . . . ), the unique sequence in C (cid:48) such that x = 0 .b b . . . (4) | s | , | w | length of a sequence s or word w , respectively | s | , | w | number of 1’s in a binary sequence or word, respectively (cid:101) s , (cid:101) w binary sequence/word obtained by complementing s/w , respectively s (cid:112)(cid:112) n , w (cid:112)(cid:112) n n -truncation of a sequence s or a word w , respectively (5)[ s ] , [ w ] set of all infinite sequences/words having prefix s / w , respectively (6) { , } ∗ possibly empty finite binary sequences/words { , } ∞ possibly empty finite or infinite binary sequences/words ρ : { , } ∞ → { , } ∞ the main substitution map (7) τ : { , } ∞ → { , } ∞ an equivalent 2-block substitution map(defined on even/infinite length binary sequences/words) (11) τ : { , } ∞ → { , } ∞ the substitution map inverse to τ (13) σ : { , } ∞ → { , } ∞ the “simplest” substitution map inverse to ρ (14) R : [0 , → [0 ,
1] the real map corresponding to ρ (16) S : [0 , → [0 ,
1] the section of R corresponding to σ (22) (cid:15) the empty sequence/word n (cid:15) vanishing order function of binary sequences/words under the action or ρ (12) (cid:101) ρ the substitution map complementary to ρ (8) ρ k ρ or (cid:101) ρ , depending on k being an even or odd natural, respectively (9) ρ v auxiliary substitution map, coinciding with ρ | v | , such that ρ ( vw ) = ρ ( v ) ρ v ( w ) (10) ρ nv auxiliary substitution map such that ρ n ( vw ) = ρ n ( v ) ρ nv ( w ) (48) (cid:104) a (cid:105) b binary sequence/word obtained from b by inserting 0 a k before each k -th 1 (2) (cid:104)(cid:104) b (cid:105)(cid:105) the auxiliary set {(cid:104) a (cid:105) b | a ∈ A } used to construct the fibers of ρ and R (15 , 21) (cid:104) a (cid:105) F the element of Fix ρ generated from the “simples” one b according to a ∈ A (53) b the “simplest” fixed point (cid:104) ∞ (cid:105) F of the map ρ (51) x the largest fixed point ξ ( b ) of the map R (52) x (cid:96) the “simplest” periodic point of the map R having odd order 2 (cid:96) + 1 (54) C , C the two rational R -cycles { , / } and { , / } respectively Q , Q the subsets of Q attracted by C and C respectively (50)33 eferences [1] F. Balibrea, J. Sm´ıtal and M. ˇStef´ankov´a, The three versions of distributional chaos ,Chaos Solitons Fractals 23 (2005), 1581-1583.[2] A. Belshaw and P. Borwein,
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