Quasi-convex sequences in the circle and the 3-adic integers

Quasi-con v e x sequences in the circle and the 3 -adic inte gers * D. Dikranjan 1 and G ´ abor Luk ´ acs 2 No vember 2, 2018 Abstract In this pape r , we present families of q uasi-con ve x seq uences con ver ging to zero in the circle group T , and the group J 3 of 3 - adic int egers. Thes e seq uences are determin ed by an increasi ng sequen ces of inte gers. For an in creasing se quence a = { a n } ∞ n =0 ⊆ Z , put g n = a n +1 − a n . W e prov e that (a) the set { 0 } ∪ {± 3 − ( a n +1) | n ∈ N } is q uasi-con ve x in T if and on ly if a 0 > 0 and g n > 1 for e very n ∈ N ; (b) the set { 0 } ∪ {± 3 a n | n ∈ N } is quasi -con vex in the group J 3 of 3 -adic integers if and only if g n > 1 for ev ery n ∈ N . Moreo ver , we solv e an open prob lem from [8] by pro vidin g a c omplete characteri zation of the sequen ces a such that { 0 } ∪ {± 2 − ( a n +1) | n ∈ N } is quasi-con ve x in T . Using this result, we also ob tain a charac terizatio n of the sequen ces a such th at the set { 0 } ∪ {± 2 − ( a n +1) | n ∈ N } is quasi -con vex in R . 1. Intr oduction One of t he mains s ources of inspirati on for the theory of topol ogical groups is th e theory of topo- logical v ector spaces, where the notion of con ve xity p lays a prominent role. In this context, the reals R are replaced with the circle group T = R / Z , and linear funct ionals a re replaced by c harac- ters , that is, conti nuous h omomorphi sms to T . By makin g s ubstantial use of characters, V ilenkin introduced the no tion of q uasi-con ve xity for abelian t opological groups as a counterpart of c on vex- ity in top ological vec tor spaces (cf. [13]). The coun terpart of locally con vex spaces are the locally quasi-con ve x groups. This class includes all l ocally compact abelian groups and locally con ve x topological ve ctor spaces (cf. [2]). Every locally quasi-con vex group is maximally almost periodic (briefly , MAP ), which means that its characters separate its points. * 2000 Mathematics Subjec t C lassificatio n : Primary 22A05; Secondary 22B05. K e ywor ds: quasi-co n vex sets, qc-dense sets, torus group, 3 -adic integers 1 The first author acknowledges the fi nan cial aid recei ved from MCYT , MTM 2006- 0203 6 and FEDER fund s. 2 The second a uthor gratefully ack nowledges the generous financial support received from NSERC and the Uni ver - sity of Manitoba , which enabled him to do this research. 2 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s Unlike the “geometrically” transparent property of conv exity , quasi-conv exity remains an ad- mittedly mys terious property . Alt hough locally quasi -con ve x groups have been s tudied by many authors (cf. [1], [2], and [4]), t heir work did not completely rev eal the nature o f the small quasi- con ve x sets. Aus senhofer pro ved t hat the quasi-con vex hull of a finite subset of a MAP group is fi- nite (cf. [1]), and a slight ly more general property was shown in [7] (many examples of finit e quasi- con ve x set s are found i n [3] and [ 10]). Since finite sets are compact, the foll owing st atement—also known as the quasi-con vex c ompactness p r operty —established by Hern ´ andez, and independently , by Bruguera and Mart´ ın-Peinador , can be cons idered a g eneralization of Auss enhofer’ s theorem: A metrizable locally quasi-con vex abeli an group G is complete if and only if t he quasi-conv ex h ull of e very compact subset of G is compact (cf. [9] and [5]). Luk ´ acs extended this result, and prov ed that a metri zable abelian group A is M AP and has the quasi -con ve x compactness prop erty i f and only i f it i s locally qu asi-con ve x and complet e; he also showed that s uch groups are characterized by the property that the ev aluation map α A : A → ˆ ˆ A is a clo sed embedding (cf. [11, I.34]). Interest in the compact quasi-con ve x sets stem s from the theory of Macke y groups (cf. [6] and [10]). The paper [8] is dedicated to the study of countably infinite quasi-con vex sets (in fact, sequences that con ver ge to zero) in some familiar locally compact abelian groups such as R , and the com pact groups T and J 2 ( 2 -adic int egers). In order to formulate the m ain results of [8], we introduce some notations . Let π : R → T denote the canonical projecti on. Since the restricti on π | [0 , 1) : [0 , 1) → T is a bijection, we often identify in the sequel, par abus de language , a number a ∈ [0 , 1) with its image (coset) π ( a ) = a + Z ∈ T . W e pu t T m := π ([ − 1 4 m , 1 4 m ]) for all m ∈ N \{ 0 } . According to st andard n otation in th is area, we use T + to denot e T 1 . For an abelian topologi cal group G , we denote by b G the P ontryagin dual of a G , that is, the group of all characters o f G , endowed with the compact-open topolog y . Definition 1.1. For E ⊆ G and A ⊆ b G , the polars of E and A are defined as E ⊲ = { χ ∈ b G | χ ( E ) ⊆ T + } and A ⊳ = { x ∈ A | ∀ χ ∈ A, χ ( x ) ∈ T + } . (1) The set E is said to be quasi-con vex if E = E ⊲⊳ . Obviously , E ⊆ E ⊲⊳ holds for e very E ⊆ G . Thus, E is quasi-con vex if and o nly if for every x ∈ G \ E there exists χ ∈ E ⊲ such that χ ( x ) 6∈ T + . The set Q G ( E ) := E ⊲⊳ is the smallest quasi- con ve x set of G t hat contains E , and it is called the quasi-conv ex hull of E . In what follows, we rely on the following g eneral property of quasi-con vexity . Pr oposition 1. 2. ([11, I.3(e)], [8, 2.7]) If f : G → H is a con tinuous homom orphism of abelian topological groups and E ⊆ G , then f ( Q G ( E )) ⊆ Q H ( f ( E )) . Dikranjan and de Leo proved the following theorem. D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 3 Theor em 1.3. Let a = { a n } ∞ n =0 be an increasing sequence of integers, and put g n = a n +1 − a n . (a) ([8, 1.1]) If a 0 > 0 and g n > 1 for every n ∈ N , then K a, 2 = { 0 } ∪ {± 2 − ( a n +1) | n ∈ N } is quasi-con ve x in T . (b) ([8, 1.2]) If g n > 1 for e very n ∈ N , then R a, 2 = { 0 } ∪ {± 2 − ( a n +1) | n ∈ N } is quasi-con vex in R . (c) ([8, 1.4]) If a 0 ≥ 0 and g n > 1 for every n ∈ N , then th e s et L a, 2 := { 0 } ∪ {± 2 a n | n ∈ N } is quasi-con vex in J 2 . The qu estion of whether the conditi ons formulated in T heorem 1 .3(a) are also necessary in order to ass ure that K a, 2 is quasi-conv ex was left open by Dikranjan and de Leo, and in fact, was formu lated as an open prob lem (cf. [8, 5.1(a)]). On e of t he main result s of this paper is the following t heorem, which provides a c omplete solut ion to this open problem. Theor em A. Let a = { a n } ∞ n =0 be an i ncr easing sequence of non-ne gative i nte ger s, a nd put x n = 2 − ( a n +1) . The set K a, 2 = { 0 } ∪ {± x n | n ∈ N } is quasi-con ve x i n T if and only if the following thr ee conditions hold: (i) a 0 > 0 ; (ii) the gaps g n = a n +1 − a n satisfy g n > 1 for all but possibly one inde x n ∈ N ; (iii) if g n = 1 for some n ∈ N , then g n +1 > 2 . Theorem A als o yield s a com plete characterization of the sequences a such that R a, 2 is quasi - con ve x. Theor em B. Let a = { a n } ∞ n =0 be a n incr easing sequence of inte gers, and put r n = 2 − ( a n +1) . The set R a, 2 = { 0 } ∪ {± r n | n ∈ N } is quasi-con vex in R i f and only if th e f ollowing two conditio ns hold: (i) the gaps g n = a n +1 − a n satisfy g n > 1 for all but possibly one inde x n ∈ N ; (ii) if g n = 1 for some n ∈ N , then g n +1 > 2 . It turns out that by replacing the prime 2 with 3 in parts (a) and (c) of Theorem 1.3, one obtains conditions that are not only suffi cient, but also necessary . Theor em C. Let a = { a n } ∞ n =0 be an i ncr easing sequence of non-ne gative i nte ger s, a nd put x n = 3 − ( a n +1) . The set K a , 3 = { 0 } ∪ {± x n | n ∈ N } is quasi-con ve x i n T if and only if the following two conditions hold: (i) a 0 > 0 ; (ii) the gaps g n = a n +1 − a n satisfy g n > 1 for all n ∈ N . Theor em D. Let a = { a n } ∞ n =0 be an i ncr easing sequence of non-ne gative i nte ger s, a nd put y n = 3 a n . The s et L a , 3 = { 0 } ∪ {± y n | n ∈ N } is quas i-con ve x in J 3 if and only if the gaps g n = a n +1 − a n satisfy the condition g n > 1 for all n ∈ N . 4 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s In spit e of t he apparent s imilarity between t he Main Lem ma of [8] and Th eorems 4.2 and 5.2, our techni ques for proving Theorems C and D do differ from th e ones used by Dikranjan and de Leo to prove Theorem 1.3 (a) and (c) (cf. [8]). W e believe that the techniques presented here can also be u sed to prov e app ropriate genera lizations of Theorems C and D for arbitrary primes p > 2 , at least as far as suf ficiency of the conditions is concerne d. The paper is structured as follo ws. In § 2, w e present cond itions that are nece ssary for the quasi- con ve xity of a sequence con verging to zero. These results are used later on, in the proofs of the necessity of the conditions in Theorems A, B, and C. In order to cov er all t hree cases, we consider sequences of rationals of th e form 1 b n in R , wh ere b n | b n +1 for ev ery n ∈ N , as well as t heir images under π in T . § 3 i s dedicated to th e proof of Theorems A and B, the lat ter being an easy consequence of the earlier . In § 4, we p rove Theorem C, and in § 5, t he proo f of Theorem D is presented. 2. Necessary conditions in T and R f or quasi-con vexity Let { b n } ∞ n =0 be an increasing sequence o f natural num bers such th at b n | b n +1 for ev ery n ∈ N , and b 0 > 1 . In this section, we are concerned with finding conditions that are necessary for X = { 0 } ∪  ± 1 b n     n ∈ N  ⊆ T and S = { 0 } ∪  ± 1 b n     n ∈ N  ⊆ R (2) to be quasi-con vex in T and R , respecti vely . W e put q n = b n +1 b n for each n ∈ N . Theor em 2. 1. Suppo se that S is quas i-con ve x in R . Then: (a) |{ n ∈ N | q n = 2 }| ≤ 1 ; (b) if q n = 2 for some n ∈ N , then q n +1 > 4 . For the sake of t ransparency , we break up the proof of Theorem 2.1 in to two lemm as, the first of which holds in ev ery abelian topological group. Lemma 2.2. Let G be an abelian topological gr ou p, and h, h 1 , h 2 ∈ G . Then: (a) h 1 ± h 2 ∈ Q G ( { h 1 , 2 h 1 , h 2 , 2 h 2 } ) ; (b) 4 h ∈ Q G ( { h, 3 h, 6 h } ) ; (c) 5 h ∈ Q G ( { h, 4 h, 8 h } ) . P R O O F . (a) Let χ ∈ { h 1 , 2 h 1 , h 2 , 2 h 2 } ⊲ . Th en χ ( h 1 ) ∈ T + and 2 χ ( h 1 ) = χ (2 h 1 ) ∈ T + , and so χ ( h 1 ) ∈ T 2 . Similarly , χ ( h 2 ) ∈ T 2 . Thus, χ ( h 1 ± h 2 ) = χ ( h 1 ) ± χ ( h 2 ) ∈ T 2 + T 2 = T + . (3) Hence, h 1 ± h 2 ∈ Q G ( { h 1 , 2 h 1 , h 2 , 2 h 2 } ) . (b) Let χ ∈ { h, 3 h, 6 h } ⊲ . Then χ ( h ) ∈ T + , 3 χ ( h ) = χ (3 h ) ∈ T + , and 6 χ ( h ) = χ (6 h ) ∈ T + . Thus, χ ( h ) ∈ { 1 , 3 , 6 } ⊲ = T 6 , and so χ (4 h ) = 4 χ ( h ) ∈ 4 T 6 ⊆ T + . Hence, 4 h ∈ Q G ( { h, 3 h, 6 h } ) . D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 5 (c) Let χ ∈ { h, 4 h, 8 h } ⊲ . Then χ ( h ) ∈ T + , 4 χ ( h ) = χ (4 h ) ∈ T + , and 8 χ ( h ) = χ (8 h ) ∈ T + . Thus, χ ( h ) ∈ { 1 , 4 , 8 } ⊲ ∈ T 8 ∪ ( 15 64 + T 16 ) ∪ ( − 15 64 + T 16 ) . (4) Therefore, χ (5 h ) = 5 χ ( h ) ∈ 5 T 8 ∪ ( 11 64 + 5 T 16 ) ∪ ( − 11 64 + 5 T 16 ) ⊆ T + . (5) Hence, 5 h ∈ Q G ( { h, 4 h, 8 h } ) , as desired. Lemma 2.3. If 1 b n 1 + 1 b n 2 ∈ S , then n 1 = n 2 . P R O O F . Suppose that 1 b n 1 + 1 b n 2 ∈ S , and so there is n 0 ∈ N such that 1 b n 1 + 1 b n 2 = 1 b n 0 . (6) Since the s equence { b n } ∞ n =0 is increasing, the sequ ence { 1 b n } ∞ n =0 is decreasing. Th us, n 0 < n 1 , n 2 , because 1 b n 0 > 1 b n 1 , 1 b n 2 . Consequently , b n 0 | b n 1 , b n 2 , and for a = b n 1 b n 0 and b = b n 2 b n 0 , equation (6) can be re written as 1 a + 1 b = 1 , (7) where a, b ∈ N . Therefore, a = b = 2 , which imp lies that b n 1 = b n 2 . Hence, n 1 = n 2 . P R O O F O F T H E O R E M 2 . 1 (a) Suppose t hat q n 1 = q n 2 = 2 . P ut h 1 = 1 b n 1 +1 and h 2 = 1 b n 2 +1 . Then 2 h 1 = 1 b n 1 and 2 h 2 = 1 b n 2 . Thus, { h 1 , 2 h 1 , h 2 , 2 h 2 } ⊆ S , and so by Lemm a 2.2(a), h 1 + h 2 ∈ Q T ( { h 1 , 2 h 1 , h 2 , 2 h 2 } ) ⊆ Q T ( S ) . (8) Consequently , 1 b n 1 +1 + 1 b n 2 +1 = h 1 + h 2 ∈ S , because S is quasi-con vex. Th erefore, by Lemma 2.3, n 1 + 1 = n 2 + 1 . Hence, n 1 = n 2 , as desired. (b) Suppos e that q n = 2 . For h = 1 b n +2 , on e has q n +1 h = 1 b n +1 and 2 q n +1 h = 1 b n . Thus, { h, q n +1 h, 2 q n +1 h } ⊆ S , and si nce S is quasi-conv ex, Q T ( { h, q n +1 h, 2 q n +1 h } ) ⊆ S . If q n +1 = 3 , then by Lemma 2.2(b), ( q n +1 + 1) h = 4 h ∈ Q T ( { h, 3 h, 6 h } ) = Q T ( { h, q n +1 h, 2 q n +1 h } ) ⊆ S. (9) If q n +1 = 4 , then by Lemma 2.2(c), ( q n +1 + 1) h = 5 h ∈ Q T ( { h, 4 h, 8 h } ) = Q T ( { h, q n +1 h, 2 q n +1 h } ) ⊆ S. (10) In either case, 1 b n +1 + 1 b n +2 = q n +1 h + h = ( q n +1 + 1) h ∈ S. (11) By Lemma 2.3, t his implies th at n + 1 = n + 2 . T his contradiction shows that q n +1 6 = 3 and q n +1 6 = 4 . Finally , we note that by (a), q n +1 6 = 2 . Hence, q n +1 > 4 , as desired. 6 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s In order to prove the analogue of Theorem 2.1 for qu asi-con ve xity of X in T , we first establish a general result relating quasi-con ve xity of su bsets of R to quasi-con vexity of their projections in T . Remark 2.4. T he map π | ( − 1 2 , 1 2 ) : ( − 1 2 , 1 2 ) → T \{ 1 2 } i s a homeomorphism. Indeed, it is continuou s and open because π i s s o, and it is clearly surjecti ve, s o it remains to be seen t hat i t is injective. Suppose that π ( y 1 ) = π ( y 2 ) , wh ere y 1 , y 2 ∈ ( − 1 2 , 1 2 ) . Then t here is n ∈ Z such that y 1 = y 2 + n , and so y 1 − y 2 = n . Consequently , n ∈ ( − 1 2 , 1 2 ) + ( − 1 2 , 1 2 ) = ( − 1 , 1 ) . Therefore, n ∈ Z ∩ ( − 1 , 1 ) = { 0 } . Hence, y 1 = y 2 , as required. Theor em 2.5. Let Y ⊆ ( − 1 2 , 1 2 ) be a compact subset of R . If the pr ojection π ( Y ) is qu asi-con ve x in T , then Y i s quasi-con ve x in R . P R O O F . Since π ( Y ) i s quasi-con ve x i n T and π is a continuous group homomorphism, by Propo- sition 1.2, Q R ( Y ) ⊆ π − 1 ( π ( Y )) . On the other hand , there is 0 < M < 1 2 such that Y ⊆ [ − M , M ] , because Y is compact. Thus, Q R ( Y ) ⊆ [ − M , M ] , as [ − M , M ] is quasi-con ve x (cf. [4, p. 79]). Therefore, by Remark 2.4, π − 1 ( π ( Y )) ∩ ( − 1 2 , 1 2 ) = Y , and so Q R ( Y ) ⊆ π − 1 ( π ( Y )) ∩ [ − M , M ] ⊆ π − 1 ( π ( Y )) ∩ ( − 1 2 , 1 2 ) = Y . (12) Hence, Y is quasi-con vex in R , as desired. Corollary 2.6. Let Y ⊆ R be a compact sub set. If ther e is α 6 = 0 s uch th at αY ⊆ ( − 1 2 , 1 2 ) and π ( α Y ) is quasi-con ve x in T , then Y is quasi-con vex in R . P R O O F . By Theorem 2.5, the set αY is quasi-con vex in R . Th e m ap f : R → R defined by f ( y ) = α y is a topological automorphism of R . Consequently , Y = f − 1 ( αY ) is quasi-con vex in R , because quasi-con vexity is preserved by topological isomorphism s. Theor em 2. 7. Suppo se that X is quas i-con ve x in T . Then: (a) b 0 ≥ 4 ; (b) |{ n ∈ N | q n = 2 }| ≤ 1 ; (c) if q n = 2 for some n ∈ N , then q n +1 > 4 . P R O O F . First, we show that b 0 ≥ 4 . Ass ume that b 0 = 2 o r b 0 = 3 . (The case b 0 = 1 i s not possi- ble, because b 0 > 1 . ) Th en h 1 b 0 i = { 0 , ± 1 b 0 } ⊆ X , and t hus (using the fact that H ⊲ coincides wi th the annihilator H ⊥ for e very subgroup H ) one obtains that X ⊲ ⊆ h 1 b 0 i ⊲ = h 1 b 0 i ⊥ = b 0 Z . (13) Consequently , 1 b 0 + X ⊆ X ⊲⊳ = Q T ( X ) . S ince the onl y accumulation point of X is 0 , and the only accumulation point of 1 b 0 + X is 1 b 0 , it follows t hat Q T ( X ) 6⊆ X , contrary to our assum ption t hat X is quasi-con vex. This contradiction s hows that b 0 ≥ 4 , and hence (a) hold s. Since b 0 ≥ 4 , the set S sati sfies that S ⊆ [ − 1 4 , 1 4 ] ⊆ ( − 1 2 , 1 2 ) , and S is compact (because it is consists of zero and a sequence that con ver ges to 0 ). Clearly , π ( S ) = X , and by our ass umption, it is quasi-conv ex. So, by Theorem 2.5, S is quasi-con vex in R . Thus, by Theorem 2.1, (b) and (c) hold, as desired. D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 7 Theor em 2.8. If X is q uasi-con ve x i n T , then for every n ∈ N , 4 6 | b n +1 implies th at q n 6 = 3 . In particular , i f 4 6 | b n for every n ∈ N , then q n 6 = 3 for every n ∈ N . In order to prov e Theorem 2.8, we rely on the following proposi tion, whi ch hol ds i n every abe- lian topol ogical group. For x ∈ G , put T r x ( G ) = { χ ( x ) | χ ∈ b G } . Clearly , T r x ( G ) is a subgroup of T , and T r 2 x ( G ) = 2 T r x ( G ) . Pr oposition 2.9. Let G be an abelian top ological gr oup, and x ∈ G . Then the f ollowing are equi v- alent: (i) 2 x ∈ Q G ( { x, 3 x } ) ; (ii) ± 1 4 6∈ T r x ( G ) (i .e., χ ( x ) 6 = ± 1 4 for every χ ∈ b G ); (iii) 1 2 6∈ T r 2 x ( G ) ; (iv) the subgroup T r 2 x ( G ) o f T ha s no non-zer o 2-torsion eleme nts. P R O O F . (i) ⇒ (ii): Assume that ± 1 4 ∈ T r x ( G ) . Then there is χ ∈ b G such that χ ( x ) = 1 4 . Thus, χ ( x ) ∈ T + , and χ (3 x ) = 3 4 ∈ T + , but χ (2 x ) = 1 2 6∈ T + . Th erefore, 2 x 6∈ Q G ( { x, 3 x } ) , contrary to (i). This contradiction shows that ± 1 4 6∈ T r x ( G ) . (ii) ⇒ (iii): If 1 2 ∈ T r 2 x ( G ) = 2 T r x ( G ) , then there i s χ ∈ b G su ch th at 2 χ ( x ) = 1 2 , and thus χ ( x ) = ± 1 4 , contrary to (ii). Hence, 1 2 6∈ T r 2 x ( G ) . (iii) ⇒ (iv): Since the Pr ¨ ufer group Z (2 ∞ ) is the 2 -torsi on subgroup of T , if T r 2 x ( G ) contains a non-zero 2 -torsion element t of order 2 l , then 2 l − 1 t = 1 2 ∈ T r 2 x ( G ) , contrary t o (iii). Hence, T r 2 x ( G ) ∩ Z (2 ∞ ) = { 0 } . (iv) ⇒ (i i): Since T r 2 x ( G ) = 2 T r x ( G ) , this implication is obvious. (ii) ⇒ (i): Supp ose that χ ∈ { x, 3 x } ⊲ . Then χ ( x ) ∈ T + and 3 χ ( x ) = χ (3 x ) ∈ T + . Consequently , χ ( x ) ∈ {± 1 4 } ∪ T 3 . By (b), th is implies that χ ( x ) ∈ T 3 ⊆ T 2 . Therefore, 2 χ ( x ) = χ (2 x ) ∈ T + . Hence, 2 x ∈ Q G ( { x, 3 x } ) , as desired. Corollary 2.10. Let G be an abelian topol ogical gr oup, and x ∈ G . If (a) the or der of 2 x is finite and odd; or (b) b G is torsion, and the or der of ev ery χ ∈ b G su ch that χ ( x ) 6 = 0 is not divisi ble by 4 ; then 2 x ∈ Q G ( { x, 3 x } ) . P R O O F . (a) Since eac h χ ∈ b G is a hom omorphism , the order of χ (2 x ) divides the order of 2 x . So, if χ (2 x ) = 1 2 , then 2 divides the order of 2 x , contrary to ou r ass umption. Therefore, χ (2 x ) 6 = ± 1 2 for all χ ∈ b G , and the statement follows by Propos ition 2.9(iii). (b) The map ˆ x : b G → T defined by ˆ x ( χ ) = χ ( x ) is a hom omorphism , and so t he order of χ ( x ) divides the order of χ . Thus, if χ ( x ) = ± 1 4 , then 4 divides t he order of χ , con trary to our assumption . Hence, χ ( x ) 6 = ± 1 4 for all χ ∈ b G , and th e statement foll ows by Propos ition 2.9(ii). P R O O F O F T H E O R E M 2 . 8 . Suppos e that q n = 3 , and put x = 1 b n +1 . Then 3 x = 1 b n , and thus { x, 3 x } ⊆ X . Since b n +1 is not divisible by 4 , the order of 2 x is odd, and so by Coroll ary 2.10(a), 2 x ∈ Q T ( { x, 3 x } ) . Consequently , 2 x ∈ Q T ( { x, 3 x } ) ⊆ Q T ( X ) = X , (14) 8 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s because X is quasi-con vex. B y Theorem 2.7, b 1 > b 0 ≥ 4 , and so X = π ( S ) , where S ⊆ [ − 1 4 , 1 4 ] . Therefore, by Remark 2.4, (14) implies th at 2 b n +1 ∈ S . This, howe ver , is imp ossible, because 1 b n +1 < 2 b n +1 < 1 b n . This contradiction shows t hat q n 6 = 3 , as desired. Corollary 2.10 also has a useful application for the group J p of p -adic integers. Corollary 2.11. Let p > 2 b e a prime, and x ∈ J p . Then 2 x ∈ Q J p ( { x, 3 x } ) . P R O O F . Recall that b J p = Z ( p ∞ ) , and s o eve ry element in b J p is of order p k for some k ∈ N . Thus, 4 do es not divide the order of an y element in b J p . Therefore, the statement follows by Corol- lary 2.10(b). 3. Sequence s o f the form 2 − ( a n +1) in T and R In this section, we prove Theorem A and B. The case of Theorem A where g n > 1 for e very n ∈ N follows from Theorem 1.3(a). Thus, the main thrust of the ar g ument in the proof of Theorem A is to sh ow t hat the conclusion remains true if one p ermits g n 0 = 1 for a single index n 0 ∈ N . This is carried out by indu ction on n 0 . W e s tart off with proving sufficienc y of t he condi tions of Theorem A in a special case, where n 0 = 0 and a 0 = 1 . Pr oposition 3.1. Supp ose that a 0 = 1 , a 1 = 2 , 2 < g 1 , and 1 < g n for all n ≥ 2 . Then K a , 2 is quasi-con ve x in T . Notation 3.2. For g ∈ Z , we denote by a + g the sequence whose n th term is a n + g . P R O O F O F P R O P O S I T I O N 3 . 1 . Let a ′ = { a ′ n } ∞ n =0 denote the sequence defined by a ′ n = a n +2 . Then for ev ery n ∈ N , a ′ n = a n +2 ≥ a 2 = a 1 + g 1 ≥ 5 , and (15) g ′ n = a ′ n +1 − a ′ n = a n +3 − a n +2 = g n +2 > 1 . (16) Observe that K a , 2 = {± 1 4 , ± 1 8 } ∪ K a ′ , 2 . (17) Let f : T → T denote the continu ous homomorphism defined by f ( x ) = 8 x . By (17), f ( K a , 2 ) = 8 K a ′ , 2 = K a ′ − 3 , 2 . (18) Consequently , by Proposition 1.2, Q T ( K a, 2 ) ⊆ f − 1 ( Q T ( K a ′ − 3 , 2 )) . (19) By (15), one has a ′ 0 − 3 ≥ 2 , and by (16), ( a ′ n +1 − 3) − ( a ′ n − 3) = g ′ n > 1 for e very n ∈ N . Thus, according to Theorem 1.3(a), K a ′ − 3 , 2 is quasi-con ve x in T , and so by (19), Q T ( K a , 2 ) ⊆ f − 1 ( K a ′ − 3 , 2 ) = 4 [ i = − 3 ( i 8 + K a ′ , 2 ) . (20) D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 9 Since a ′ 0 ≥ 5 by (15), one has K a ′ , 2 ⊆ T 16 . So, for i = − 3 , 3 , 4 , ( i 8 + K a ′ , 2 ) ∩ T + ⊆ ( i 8 + T 16 ) ∩ T + = ∅ . (21) As K a , 2 ⊆ T + and T + = { 1 } ⊳ is quasi-con vex, Q T ( K a, 2 ) ⊆ T + . Therefore, by (21), Q T ( K a, 2 ) ⊆ 2 [ i = − 2 ( i 8 + K a ′ , 2 ) . (22) Recall that K a ′ , 2 ⊆ T 16 . Thus, 7 K a ′ , 2 ⊆ 7 T 16 ⊆ 8 T 16 ⊆ T 2 ⊆ T + , (23) and 7 ∈ K ⊲ a ′ , 2 . Since 7 ∈ {± 1 4 , ± 1 8 } ⊲ , it follows from (17) t hat 7 ∈ {± 1 4 , ± 1 8 } ⊲ ∩ K ⊲ a ′ , 2 = ( { ± 1 4 , ± 1 8 } ∪ K a ′ , 2 ) ⊲ = K ⊲ a , 2 . (24) Consequently , Q T ( K a , 2 ) = K ⊲⊳ a, 2 ⊆ { 1 , 7 } ⊳ = {± 1 4 } ∪ T 7 ∪ ( 1 7 + T 7 ) ∪ ( − 1 7 + T 7 ) . (25) Therefore, Q T ( K a, 2 ) ∩ ( 1 4 + K a ′ , 2 ) ⊆ Q T ( K a , 2 ) ∩ ( 1 4 + T 16 ) ⊆ ( 1 7 + T 7 ) ∩ ( 1 4 + T 16 ) = { 1 4 } . (26) Similarly , Q T ( K a , 2 ) ∩ ( − 1 4 + K a ′ , 2 ) ⊆ {− 1 4 } . (27) Hence, by (22), Q T ( K a , 2 ) ⊆ {± 1 4 } ∪ K a ′ , 2 ∪ ( 1 8 + K a ′ , 2 ) ∪ ( − 1 8 + K a ′ , 2 ) . (28) In light of (17) and (28), we conclude by showing th at Q T ( K a , 2 ) ∩ ( 1 8 + K a ′ , 2 ) ⊆ { 1 8 } . (29 ) It will follow by s ymmetry that Q T ( K a , 2 ) ∩ ( − 1 8 + K a ′ , 2 ) ⊆ {− 1 8 } . (30) T o that end, we prove that for every x ∈ 1 8 + K a ′ , 2 , if x 6 = 1 8 , then there is k ∈ K ⊲ a, 2 such t hat k x 6∈ T + . Let x = 1 8 ± x n ∈ ( 1 8 + K a ′ , 2 ) . Then n ≥ 2 , so x n ∈ T 16 , and (2 a n − 1 − 1)( 1 8 − x n ) = (2 a n − 1 − 1 ) 1 8 − ( 2 a n − 1 − 1 ) x n (31) ≡ 1 − 1 8 − 1 4 + x n = − 3 8 + x n 6∈ T + , (32) (2 a n − 1 − 7 )( 1 8 + x n ) = (2 a n − 1 − 7 ) 1 8 + (2 a n − 1 − 7 ) x n (33) ≡ 1 − 7 8 + 1 4 − 7 x n ≡ 1 3 8 − 7 x n 6∈ T + . (34) 10 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s It remain s to show that 2 a n − 1 − 1 and 2 a n − 1 − 7 are indeed in K ⊲ a, 2 . By (15), a n ≥ 5 for ev ery n ≥ 2 . Th us, 2 a n − 1 − 1 , 2 a n − 1 − 7 ∈ {± 1 4 , ± 1 8 } ⊲ . If 2 ≤ m < n , t hen 2 a n − 1 x m = 2 a n − a m − 2 ≡ 1 0 (because g n > 1 for e very n ≥ 2 ). Consequently , (2 a n − 1 − 1 ) x m = − x m ∈ T + and (2 a n − 1 − 7) x m = − 7 x m ∈ 7 T 16 ⊆ T + . (35) On the other hand, if n ≤ m , then x m ∈ T 2 a m − 1 . Therefore, (2 a n − 1 − 1) x m ⊆ (2 a n − 1 − 1 ) T 2 a m − 1 ⊆ 2 a n − 1 T 2 a m − 1 ⊆ 2 a m − 1 T 2 a m − 1 = T + , (36) (2 a n − 1 − 7) x m ⊆ (2 a n − 1 − 7 ) T 2 a m − 1 ⊆ 2 a n − 1 T 2 a m − 1 ⊆ 2 a m − 1 T 2 a m − 1 = T + . (37) Hence, by (17), 2 a n − 1 − 1 , 2 a n − 1 − 7 ∈ {± 1 4 , ± 1 8 } ⊲ ∩ K ⊲ a ′ , 2 = ( {± 1 4 , ± 1 8 } ∪ K a ′ , 2 ) ⊲ = K ⊲ a , 2 , (38) as desired. Lemma 3.3. Let Y ⊆ T be symmetric, m ∈ N \{ 0 } , and 0 < k ≤ 2 m . (a) If Y ⊆ T m and k Y is quasi -con vex in T , then Y is quasi-con ve x too. (b) If Y ⊆ T 4 m and 4 mY is quasi -con ve x in T , then Y ′ = {± 1 4 m } ∪ Y is quas i-con ve x too. P R O O F . For x ∈ T , let k x k d enote the distance of x from 0 , that is, with som e ab use of notations, k x k = inf n ∈ Z | x + n | . (a) Let f : T → T deno te t he cont inuous homo morphism defined by f ( x ) = k x . Since k Y i s quasi-con ve x in T , by Proposition 1.2, Q T ( Y ) ⊆ f − 1 ( Q T ( f ( Y ))) = f − 1 ( Q T ( k Y )) = f − 1 ( k Y ) = k − 1 [ i =0 ( i k + Y ) . (39) On th e other hand, Q T ( Y ) ⊆ T m , because Y ⊆ T m , and T m = { 1 , . . . , m } ⊳ is quasi-con vex. Thus, Q T ( Y ) ⊆ k − 1 [ i =0 ( i k + Y ) ∩ T m . (40) If x = i k + y ∈ i k + Y , where y ∈ Y and i 6 = 0 , then 1 4 m = 1 2 m − 1 4 m ≤ 1 k − 1 4 m ≤ k i k k − k y k ≤ k x k , (41) with equality if and only if k = 2 m , and (i) y = − 1 4 m and i = 1 ; or (ii) y = 1 4 m and i = k − 1 . D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 11 In either case, x = − y ∈ Y , because Y i s symmetric. In all ot her cases (e.g., when k 6 = 2 m ), o ne has x 6∈ T m . Therefore, Q T ( Y ) ⊆ k − 1 [ i =0 ( i k + Y ) ∩ T m = Y , (42) as desired. (b) First, suppose that m = 1 . Then Y ⊆ T 4 , and so Y ′ ⊆ { ± 1 4 } ∪ T 4 . Th e s et {± 1 4 } ∪ T 4 is quasi-con ve x, because it is equal to { 1 , 3 , 4 } ⊳ . Thus, Q T ( Y ′ ) ⊆ {± 1 4 } ∪ T 4 . (43) Let f : T → T deno te t he cont inuous homomorphi sm defined by f ( x ) = 4 x . Since 4 Y ′ = 4 Y is quasi-con ve x in T , by Proposition 1.2, Q T ( Y ′ ) ⊆ f − 1 ( Q T ( f ( Y ′ ))) = f − 1 ( Q T (4 Y ′ )) = f − 1 (4 Y ′ ) (44) = Y ′ ∪ ( 1 4 + Y ′ ) ∪ ( 1 2 + Y ′ ) ∪ ( − 1 4 + Y ′ ) . (45) As Y ′ ⊆ T + , ( 1 2 + Y ′ ) ∩ ( {± 1 4 } ∪ T 4 ) = {± 1 4 } ⊆ Y ′ . (46) Since 1 4 + Y ′ = ( 1 4 + Y ) ∪ { 0 , 1 2 } ⊆ ( 1 4 + T 4 ) ∪ { 0 , 1 2 } , (47) one has ( 1 4 + Y ′ ) ∩ ( {± 1 4 } ∪ T 4 ) = { 0 , 1 4 } ⊆ Y ′ . (48) Similarly , ( − 1 4 + Y ′ ) ∩ ( {± 1 4 } ∪ T 4 ) = { 0 , 1 4 } ⊆ Y ′ . (49) Hence, by (43), Q T ( Y ′ ) ⊆ Y ′ , as desired. If m > 1 , then put Z = mY and Z ′ = mY ′ = {± 1 4 } ∪ Z . Since Y ⊆ T 4 m , one has Z ⊆ T 4 . By our assumption, 4 Z = 4 mY is quasi-con vex in T . Thus, by what we have shown so f ar , Z ′ = mY ′ is quasi-con ve x. By (a), this implies that Y ′ is quasi-con ve x, because Y ′ ⊆ T m and m ≤ 2 m . Using Lemma 3.3, the assu mption that a 0 = 1 in Proposit ion 3.1 can be sufficiently re laxed to cove r the base of the inductive proof of Theorem A. Corollary 3.4. Suppose that 0 < a 0 , g 0 = 1 , 2 < g 1 , and 1 < g n for all n ≥ 2 . Then K a , 2 is quasi-con ve x. P R O O F . Since the sequence a is increasing, one has K a, 2 ⊆ T 2 a 0 − 1 . The sequence a − a 0 + 1 satisfies the cond itions of Proposi tion 3. 1, and thus 2 a 0 − 1 K a, 2 = K a − a 0 +1 , 2 is quasi-con ve x. Hence, by Lemma 3.3(a) with k = m = 2 a 0 − 1 , one obtains that K a , 2 is quasi-con ve x, as desired. 12 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s P R O O F O F T H E O R E M A . Suppose that K a, 2 is q uasi-con ve x in T , and put b n = 2 a n +1 . Usin g the notations of Section 2, q n = b n +1 b n = 2 a n +1 − a n = 2 g n . (50) By Theorem 2 .7(a), b 0 = 2 a 0 +1 ≥ 4 , and t hus a 0 ≥ 1 . By Theorem 2.7(b), there is at mos t o ne index n ∈ N such that q n = 2 . Therefore, g n = 1 for at mo st one i ndex n , and so g n > 1 for all but o ne n ∈ N , because { a n } ∞ n =0 is increasing. B y Theorem 2.7(c), if q n = 2 for so me n ∈ N , then q n +1 > 4 . Hence, if g n = 1 for some n ∈ N , then g n +1 > 2 . Con versely , suppos e t hat a satisfies (i)–(iii ). If g n > 1 for all n ∈ N , th en by Theorem 1. 3(a), K a, 2 is quasi-con vex in T , and there is nothing left to prove. So, we may ass ume that there is n 0 ∈ N s uch that g n 0 = 1 . In this case, by (ii), there is pre cisely one such inde x n 0 . W e proceed by induction on n 0 . If n 0 = 0 , then by Corollary 3.4, K a , 2 is q uasi-con ve x in T . For the inducti ve step, suppose that the t heorem ho lds for all sequences such t hat g n 0 = 1 . Let a be a sequ ence that satisfies (i)–(iii) such that g n 0 +1 = 1 . Let a ′ = { a ′ n } ∞ n =0 denote the sequ ence defined by a ′ n = a n +1 − a 0 − 1 . As g n 0 +1 = 1 , by (ii), g n > 1 for all n 6 = n 0 + 1 . In particular , g 0 > 1 , and so a ′ 0 = a 1 − a 0 − 1 = g 0 − 1 > 0 . (51) This shows that a ′ satisfies (i). Since a satisfies (ii)–(iii) and g ′ n = a ′ n +1 − a ′ n = ( a n +2 − a 0 − 1 ) − ( a n +1 − a 0 − 1 ) = a n +2 − a n +1 = g n +1 , (52) so does a ′ , and g ′ n 0 = 1 . Thus, by the inductive hypot hesis, K a ′ , 2 is quasi -con vex in T . Put m = 2 a 0 − 1 . Clearly , 4 mK a ′ + a 0 +1 , 2 = 4 · 2 a 0 − 1 K a ′ + a 0 +1 , 2 = 2 a 0 +1 K a ′ + a 0 +1 , 2 = K a ′ , 2 . (53) The smallest member of the sequence a ′ + a 0 + 1 is a ′ 0 + a 0 + 1 = a 1 = a 0 + g 0 ≥ a 0 + 2 . (54) Therefore, K a ′ + a 0 +1 , 2 ⊆ T 2 ( a ′ 0 + a 0 +1) − 1 = T 2 a 0 + g 0 − 1 ⊆ T 2 a 0 +1 = T 4 m . (55) Hence, by Lemma 3.3(b) applied to the set Y = K a ′ + a 0 +1 , 2 , {± 1 4 m } ∪ K a ′ + a 0 +1 , 2 = { ± 1 2 a 0 +1 } ∪ K a ′ + a 0 +1 , 2 = {± 2 − ( a 0 +1) } ∪ K a ′ + a 0 +1 , 2 = K a, 2 (56) is quasi-con vex in T . This compl etes the proof. P R O O F O F T H E O R E M B . Suppose that R a, 2 is quasi-con vex in R , and put b n = 2 a n +1 . Using the notations of Section 2, q n = b n +1 b n = 2 a n +1 − a n = 2 g n . (57) D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 13 By Theorem 2.1(a), there is at most o ne i ndex n ∈ N such that q n = 2 . Th us, g n = 1 for at most one index n , and so g n > 1 for all but one n ∈ N , because { a n } ∞ n =0 is increasing. By Theorem 2.1(b), if q n = 2 for some n ∈ N , then q n +1 > 4 . Consequently , if g n = 1 for so me n ∈ N , then g n +1 > 2 . Suppose that g n > 1 for all but possibly one i ndex n ∈ N , and if g n = 1 for some n ∈ N , then g n +1 > 2 . Then a ′ n = a n − a 0 + 1 satisfies the same conditions, because g ′ n = a ′ n +1 − a ′ n = g n , and furthermore, a ′ 0 = 1 > 0 . Thus, by Theorem A, π ( R a ′ , 2 ) = K a ′ , 2 is quasi-con ve x in T . W e note that for α = 2 1 − a 0 , one has αR a , 2 = R a ′ , 2 ⊆ [ − 1 4 , 1 4 ] . Therefore, by Corollary 2 .6, R a , 2 is quasi-con ve x in R . 4. Sequence s o f the f orm 3 − ( a n +1) in T In thi s s ection, we present the proof of Theorem C. Recall that the Pontryagin dual b T of T is Z . For k ∈ N , l et η k : T → T denote the continu ous character defined by η k ( x ) = 3 k · x . For m ∈ N , put J m = { k ∈ N | mη k ∈ K ⊲ a , 3 } and Q m = { mη k | k ∈ J m } ⊳ . Lemma 4.1. J 1 = J 2 = N \ a . P R O O F . One has mη k ( x n ) = m · 3 k − a n − 1 . Clearly , 3 i ∈ T + if and on ly if i 6 = − 1 , and 2 · 3 i ∈ T + if and only if i 6 = − 1 . Thus, for m = 1 , 2 , mη k ( x n ) ∈ T + if and only if k − a n − 1 6 = − 1 , or equiv alently , a n 6 = k . Theor em 4.2. If g n > 1 for every n ∈ N , then Q 1 ∩ Q 2 = { ∞ P n =0 ε n x n | ( ∀ n ∈ N )( ε n ∈ {− 1 , 0 , 1 } ) } . In order to pro ve Theorem 4.2, we rely on an auxiliary statement. W e identify points of T w ith ( − 1 / 2 , 1 / 2] . Recall that e very y ∈ ( − 1 / 2 , 1 / 2] can be written in the form y = ∞ X i =1 c i 3 i = c 1 3 + c 2 3 2 + · · · + c s 3 s + · · · , (58) where c i ∈ {− 1 , 0 , 1 } . Lemma 4.3. Let y ∈ T . If y ∈ T + and 2 y ∈ T + , t hen c 1 = 0 i n every r epr esentation of y in t he form (58) . P R O O F . Since c i ∈ {− 1 , 0 , 1 } , one has      ∞ X i =2 c i 3 i      ≤ ∞ X i =2 1 3 i = 1 9 · 3 2 = 1 6 . (59) Thus, | y | ≥    c 1 3    −      ∞ X i =2 c i 3 i      ≥ | c 1 | 3 − 1 6 . (60) 14 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s Since y , 2 y ∈ T + , one has y ∈ T 2 = [ − 1 8 , 1 8 ] . Therefore, | c 1 | 3 ≤ | y | + 1 6 ≤ 1 8 + 1 6 = 7 24 < 1 3 . (61) Hence, c 1 = 0 , as desired. P R O O F O F T H E O R E M 4 . 2 . ( ⊆ ) Let x = ∞ P i =1 c i 3 i be a representation o f x ∈ T in the form (58). Then for ev ery k ∈ N , one has η k ( x ) = 3 k x = ∞ X i =1 c i 3 i − k ≡ 1 ∞ X i = k +1 c i 3 i − k = ∞ X i =1 c k + i 3 i . (62) If x ∈ Q 1 ∩ Q 2 , then for e very k ∈ J 1 = J 2 , the element y = η k ( x ) sati sfies y ∈ T + and 2 y ∈ T + . Thus, by Lemma 4. 3, the coeffic ient of 1 3 in (62) is zero, that is, c k +1 = 0 . Since J 1 = J 2 = N \ a holds by Lemm a 4.1, we h a ve shown that if c i 6 = 0 , then i − 1 6∈ N \ a , so i − 1 ∈ a , and therefore i = a n + 1 for some n ∈ N . Hence, x has the form x = ∞ X n =0 c a n +1 3 a n +1 = ∞ X n =0 ε n x n , (63) where ε n = c a n +1 ∈ {− 1 , 0 , 1 } . ( ⊇ ) Let k ∈ J 1 = J 2 = N \ a (see Lemma 4.1), and x = ∞ P n =0 ε n x n . Then η k ( x ) = ∞ P n =0 ε n η k ( x n ) . Let n 0 be the first index such that η k ( z n 0 ) 6≡ 1 0 . Then k < a n 0 + 1 , and since k 6∈ a , one has k < a n 0 . Consequently , η k ( z n 0 ) ≤ 1 9 . Thus, η k ( z n 0 + i ) ≤ 1 9 i +1 , bec ause g n > 1 i mplies th at a n 0 + i − a n 0 ≥ 2 i . Therefore,      ∞ X n = n 0 ε n η k ( x n )      ≤ ∞ X n = n 0 | η k ( x n ) | = ∞ X i =0 | η k ( z n 0 + i ) | ≤ ∞ X i =0 1 9 i +1 = 1 9 · 9 8 = 1 8 . (64) So, η k ( x ) = ∞ X n =0 ε n η k ( x n ) ≡ 1 ∞ X n = n 0 ε n η k ( x n ) ∈ T 2 . (65) Hence, η k ( x ) ∈ T + and 2 η k ( x ) ∈ T + , as desired. Lemma 4.4. If g n > 1 f or all n ∈ N , g 0 > 0 , and 0 ≤ k < l in N , t hen a k − 1 ∈ J m for m = 3 a l − a k ± 2 , that is, mη a k − 1 ∈ K ⊲ a , 3 . P R O O F . Let χ = mη a k − 1 and n ∈ N . If n < k , then a n + 2 ≤ a k (as g n > 1 ), and so χ ( x n ) = mη a k − 1 ( x n ) = m · 3 a k − a n − 2 ≡ 1 0 . (66) D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 15 If k ≤ n < l , then a n + 2 ≤ a l (since g n > 1 ), so a n − a k + 2 ≤ a l − a k , and thus χ ( x n ) = m · 3 a k − a n − 2 = m 3 a n − a k +2 = 3 a l − a k ± 2 3 a n − a k +2 ≡ 1 ± 2 3 a n − a k +2 ∈ T + , (67) because a n − a k + 2 ≥ 2 . Finally , if l ≤ n , then χ ( x n ) = 3 a l − a k ± 2 3 a n − a k +2 = 1 3 a n − a l +2 ± 2 3 a n − a k +2 . (68) Since k < n and g n > 1 , on e has a k − a n ≥ 2 , and s o 2 3 a n − a k +2 ∈ [0 , 2 81 ] . Furthermore, n ≥ l implies that a n ≥ a l , and so 1 a n − a l +2 ∈ [0 , 1 9 ] . Therefore, χ ( x n ) ∈ [0 , 11 81 ] ⊆ T + . This show that χ = mη a k − 1 ∈ K ⊲ a , 3 . Remark 4.5. Lemma 4.4 heavily depends o n the ass umption that a 0 > 0 . Indeed, if a 0 = 0 , th en η a 0 − 1 = η − 1 is not defined, and thus Lemma 4.4 makes no sense (and so f ails) for k = 0 . P R O O F O F T H E O R E M C . Suppose that K a , 3 is quasi-conv ex, and put b n = 3 a n +1 . Using th e notations of Section 2, q n = b n +1 b n = 3 a n +1 − a n = 3 g n . (69) By Theorem 2.7(a), b 0 ≥ 4 , and thus a 0 > 0 . Since 4 does not divide b n +1 , by Theorem 2.8, q n 6 = 3 . Consequently , g n 6 = 1 . Hence, g n > 1 for ev ery n ∈ N , because { a n } ∞ n =0 is increasing. Con versely , suppose that a 0 > 0 and g n > 1 for e very n ∈ N , and let x ∈ Q T ( K a , 3 ) \{ 0 } . W e s how that x ∈ K a , 3 . By Theorem 4.2, x can be e xpressed in the form x = N P n =0 ε n x n , where N ∈ N ∪ {∞} , and ε n i ∈ {− 1 , 1 } and n i < n i +1 for ever y i , bec ause Q T ( K a ) ⊆ Q 1 ∩ Q 2 . Assume that N > 0 . By replacing x w ith − x if necessary , we may assume that ε n 0 = 1 . W e put ρ = ε n 1 , so that x = x n 0 + ρx n 1 + x ′ , where x ′ =    N P i =2 ε n i x n i if N > 1 0 if N = 1 . (70) Put k = n 0 , l = n 1 , m = 3 a l − a k + 2 ρ , and χ = mη a k − 1 . By Lemma 4.4, χ ∈ K ⊲ a , 3 . Since k < l and g k > 1 , one has that a l − a k ≥ 2 . Thus, one obtains that χ ( x n 0 ) = (3 a l − a k + 2 ρ ) η a k − 1 ( x k ) = (3 a l − a k + 2 ρ ) 1 9 = 3 a l − a k − 2 + 2 ρ 9 ≡ 1 2 ρ 9 (71) χ ( x n 1 ) = (3 a l − a k + 2 ρ ) η a k − 1 ( x l ) = (3 a l − a k + 2 ρ ) · 3 a k − a l − 2 = 1 9 + 2 ρ 3 a l − a k +2 . (72) Therefore, χ ( x ) = χ ( x n 0 ) + ρχ ( x n 1 ) + χ ( x ′ ) = ρ 3 + 2 3 a l − a k +2 + χ ( x ′ ) . (73) 16 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s W e claim that | χ ( x ′ ) | ≤ 11 8 · 3 4 . (74) If N = 1 , then x ′ = 0 , and (74) holds trivially , so we may assum e that N > 1 . Put s = n 2 . Since g n > 1 for all n ∈ N , one has a s + j − a s ≥ 2 j , and thus η a k − 1 ( x s + j ) = 3 a k − a s + j − 2 ≤ 3 a k − a s − 2 j − 2 (75) for all i ∈ N . Consequentl y , | χ ( x ′ ) | ≤ N X i =2 | χ ( x n i ) | ≤ ∞ X j = s | χ ( x j ) | = ∞ X j =0 | χ ( x j + s ) | = m ∞ X j =0 | η a k − 1 ( x j + s ) | ! (76) ≤ m ∞ X j =0 3 a k − a s − 2 j − 2 ! ≤ m 3 a s − a k +2 ∞ X j =0 1 3 2 j ! = m 3 a s − a k +2 · 9 8 = m 8 · 3 a s − a k , (77) because s = n 2 ≤ n i < n i +1 for all i ∈ N . Since k < l < s , g k > 1 , and g l > 1 , one has that a s − a l ≥ 2 and a s − a k ≥ 4 . Th erefore, | χ ( x ′ ) | ≤ m 8 · 3 a s − a k ≤ 3 a l − a k + 2 8 · 3 a s − a k = 1 8  1 3 a s − a l + 2 3 a s − a k  ≤ 1 8  1 3 2 + 2 3 4  = 11 8 · 3 4 , (78) as required. Hence, using the fact that a l − a k ≥ 2 , we ob tain that 2 3 a l − a k +2 + | χ ( x ′ ) | ≤ 2 3 4 + 11 8 · 3 4 = 27 8 · 3 4 = 1 24 . (79) By (73), this impli es that χ ( x ) 6∈ T + , contrary to t he assumpt ion t hat x ∈ Q T ( K a , 3 ) . Hence, N = 0 , and x = x n 0 ∈ K a, 3 , as desired. 5. Seque nces of the form 3 a n in J 3 In this section, we present the proof of Theorem D. Recall that th e Pontryagin dual b J 3 of J 3 is the Pr ¨ ufer group Z (3 ∞ ) . For k ∈ N , let ζ k : J 3 → T denote the cont inuous character defined by ζ k (1) = 3 − ( k + 1) . For m ∈ N , put J m = { k ∈ N | mζ k ∈ L ⊲ a , 3 } and Q m = { mζ k | k ∈ J m } ⊳ . Lemma 5.1. J 1 = J 2 = N \ a . P R O O F . One has mζ k ( y n ) = m · 3 a n − k − 1 . Clearly , 3 i ∈ T + if and only if i 6 = − 1 , and 2 · 3 i ∈ T + if and only if i 6 = − 1 . Thus, for m = 1 , 2 , mζ k ( y n ) ∈ T + if and only if a n − k − 1 6 = − 1 , or equiv alently , a n 6 = k . Theor em 5.2. If g n > 1 for every n ∈ N , then Q 1 ∩ Q 2 = { ∞ P n =0 ε n y n | ( ∀ n ∈ N )( ε n ∈ {− 1 , 0 , 1 } ) } . D. Dikra njan and G. Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s 17 P R O O F . Reca ll that ev ery element x ∈ J 3 can be writt en i n t he form x = ∞ P i =0 c i · 3 i , wh ere c i ∈ {− 1 , 0 , 1 } . For x represented in this form, one has ζ k ( x ) = ∞ X i =0 c i · 3 i − k − 1 ≡ 1 k X i =0 c i · 3 i − k − 1 = k +1 X i =1 c k − i +1 3 i . (80) ( ⊆ ) If x ∈ Q 1 ∩ Q 2 , then for every k ∈ J 1 = J 2 , the element y = ζ k ( x ) satisfies y ∈ T + and 2 y ∈ T + . Thus, by Lemma 4.3, the coeffi cient o f 1 3 in (80) is zero, that is, c k = 0 . Since J 1 = J 2 = N \ a holds by Lem ma 5.1, we have sh own that i f c i 6 = 0 , then i 6∈ N \ a , so i ∈ a , and therefore i = a n for some n ∈ N . Hence, x has the form x = ∞ X n =0 c a n · 3 a n = ∞ X n =0 ε n y n , (81) where ε n = c a n ∈ {− 1 , 0 , 1 } . ( ⊇ ) Let k ∈ J 1 = J 2 = N \ a (see Lemma 5.1), and x = ∞ P n =0 ε n y n . Let n 0 denote t he largest index such t hat ζ k ( y n 0 ) 6≡ 1 0 . Then a n 0 ≤ k , and s ince k 6∈ a , one has a n 0 < k . Consequ ently , ζ k ( y n 0 ) ≤ 1 9 . Thus, ζ k ( y n 0 − i ) ≤ 1 9 i +1 , because g n > 1 implies that a n 0 − a n 0 − i ≥ 2 i . Therefore,      n 0 X n =0 ε n ζ k ( y n )      ≤ n 0 X n =0 | ζ k ( y n ) | = n 0 X i =0 | ζ k ( y n 0 − i ) | ≤ n 0 X i =0 1 9 i +1 < ∞ X i =0 1 9 i +1 = 1 9 · 9 8 = 1 8 . (82) So, ζ k ( x ) = n 0 X n =0 ε n ζ k ( y n ) ∈ T 2 . (83) Hence, ζ k ( x ) ∈ T + and 2 ζ k ( x ) ∈ T + , as desired. Lemma 5.3. If g n > 1 for a ll n ∈ N and 0 ≤ k < l in N , then a l + 1 ∈ J m for m = 3 a l − a k ± 2 , that is, mζ a l +1 ∈ L ⊲ a, 3 . P R O O F . Let χ = mζ a l +1 and n ∈ N . If l < n , then a l + 2 ≤ a n (as g l > 1 ), and so χ ( y n ) = mζ a l +1 ( y n ) = m · 3 a n − a l − 2 ≡ 1 0 . (84) If k < n ≤ l , then a k + 2 ≤ a n (since g k > 1 ), so a l − a n + 2 ≤ a l − a k , and thus χ ( y n ) = mζ a l +1 ( y n ) = m · 3 a n − a l − 2 = m 3 a l − a n +2 = 3 a l − a k ± 2 3 a l − a n +2 ≡ 1 ± 2 3 a l − a n +2 ∈ T + , (85) because a l − a n + 2 ≥ 2 . Finally , if 0 ≤ n ≤ k , then χ ( y n ) = 3 a l − a k ± 2 3 a l − a n +2 = 1 3 a k − a n +2 ± 2 3 a l − a n +2 . ( 86) Since n < l and g n > 1 , on e has a l − a n ≥ 2 , and so 2 3 a l − a n +2 ∈ [0 , 2 81 ] . Furthermore, k ≥ n implies that a k ≥ a n , and so 1 3 a k − a n +2 ∈ [0 , 1 9 ] . Therefore, χ ( y n ) ∈ [0 , 11 81 ] ⊆ T + . Th is shows that χ = mζ a l +1 ∈ L ⊲ a , 3 . 18 D. Dikra njan and G . Luk ´ acs / Quasi-co n ve x sequen ces in the cir cle and the 3 -adi c inte ger s P R O O F O F T H E O R E M D . Suppose that there i s n 0 ∈ N such that g n 0 = 1 . Then y n 0 +1 = 3 y n 0 , and t hus by Corollary 2.11, 2 y n 0 ∈ Q J 3 ( { y n 0 , 3 y n 0 } ) ⊆ Q J 3 ( L a, 3 ) . Since 2 y n 0 6∈ L a, 3 , t his s hows that L a , 3 is not quasi-con vex. Con versely , su ppose that g n > 1 for eve ry n ∈ N , and let x ∈ Q J 3 ( L a, 3 ) \{ 0 } . W e show that x ∈ L a , 3 . By Theorem 5.2, x can b e expressed in the form x = ∞ P n =0 ε n y n , where ε n ∈ {− 1 , 0 , 1 } , be- cause Q J 3 ( L a , 3 ) ⊆ Q 1 ∩ Q 2 . By discarding all s ummands with ε n = 0 , w e obtain that x = N P i =0 ε n i y n i , where N ∈ N ∪ {∞} , and ε n i ∈ {− 1 , 1 } and n i < n i +1 for ever y i . Ass ume that N > 0 . By replac- ing x with − x if necessary , we may assume that ε n 0 = 1 . W e put ρ = ε n 1 . Since n 2 ≤ n i for e very i ≥ 2 , y n 2 divides y n i for ev ery i ≥ 2 . Thus , x = y n 0 + ρy n 1 + y n 2 z , where z ∈ J 3 . Put k = n 0 , l = n 1 , and χ = ( ρ 3 a l − a k + 2) ζ a l +1 . By Lemma 5.3, χ ∈ L ⊲ a, 3 , because χ = ρ (3 a l − a k + 2 ρ ) ζ a l +1 and ρ = ± 1 . One has ζ a l +1 ( y n 2 ) = 3 a n 2 − a l − 2 ≡ 1 0 , because n 2 > l and g l > 1 , and consequ ently a n 2 − a l ≥ 2 . So, ζ a l +1 ( y n 2 z ) = 0 in T , because ke r ζ a l +1 = 3 a l +2 J 3 is an ideal in th e t opological ring J 3 . Therefore, χ ( y n 2 z ) = 0 , and hence χ ( x ) = χ ( y n 0 ) + ρχ ( y n 1 ) . (87) Since k < l and g k > 1 , one has that a l − a k ≥ 2 . Thus, one obtains that χ ( y n 0 ) = ( ρ 3 a l − a k + 2)3 a k − a l − 2 = ρ 9 + 2 3 a l − a k +2 , and (88) χ ( y n 1 ) = ( ρ 3 a l − a k + 2)3 − 2 = ρ 3 a l − a k − 2 + 2 · 3 − 2 ≡ 1 2 9 . (89) Therefore, χ ( x ) = χ ( y n 0 ) + ρχ ( y n 1 ) = ρ 3 + 2 3 a l − a k +2 . (90) Since a l − a k ≥ 2 , one has a l − a k + 2 ≥ 4 , and consequentl y 2 3 a l − a k +2 ≤ 2 81 . So, χ ( x ) 6∈ T + , contrary to t he assumption that x ∈ Q J 3 ( L a , 3 ) . Hence , N = 0 , and x = y n 0 ∈ L a, 3 , as desired. Acknowledgmen ts W e are grateful to Karen Kipper for her ki nd help i n proof-reading th is paper for grammar and punctuation. Refer ences [1] L. Aussenhofer . Contributions to the duality theory of abe lian top ological group s and to the th eory of nuclea r group s. Dissertationes Math . (Rozprawy Mat.) , 384:11 3, 1999. [2] W . Banaszczyk. 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Department of Mathematics and Computer Science Depar tment of Mathemat ics Univ ersity of Udine Univ ersity of Manitoba V ia delle Scienze, 208 – Loc. Rizzi, 33100 Udine W i nnipeg, Manitoba, R3T 2N2 Italy Canada e-mail: dikranja@dimi.uniud .it e-mail: lukacs@cc.umanitoba.ca