Multigeometric sequences and Cantorvals
aa r X i v : . [ m a t h . C A ] O c t MULTIGEOMETRIC SEQUENCES AND CANTORVALS.
ARTUR BARTOSZEWICZ, MA LGORZATA FILIPCZAK, AND EMILIA SZYMONIK
Abstract.
For a sequence x ∈ l \ c , one can consider the achievementset E ( x ) of all subsums of series P ∞ n =1 x ( n ). It is known that E ( x ) isone of the following types of sets: • finite union of closed intervals, • homeomorphic to the Cantor set, • homeomorphic to the set T of subsums of P ∞ n =1 c ( n ) where c (2 n −
1) = n and c (2 n ) = n (Cantorval).Based on ideas of Jones and Velleman [6] and Guthrie and Nymann[4] we describe families of sequences which contain, according to ourknowledge, all known examples of x ’s with E ( x ) being Cantorvals. Introduction
Suppose that x = ( x (0) , x (1) , x (2) , . . . ) is an absolutely summable se-quence with infinitely many nonzero terms (i.e. x ∈ l \ c ) and let E ( x ) = ( ∞ X n =1 ε n x ( n ) : ε n ∈ { , } ) denote the set of all subsums of the series ∞ P n =1 x ( n ) , called the achievementset of x . It is easily seen that for x = (cid:0) , , , . . . (cid:1) the set E ( x ) is equal tothe Cantor tenary set C, and for x = (cid:0) , , , ... (cid:1) we have E ( x ) = [0 , the example from [6] (due to Velleman and Jones), which will be describedmore precisely in Theorem 2 and Example 5.The following properties of sets E ( x ) were described in 1914 by S. Kakeyain [7]:I. E ( x ) is a compact perfect set.II. If | x ( n ) | > P i>n | x ( i ) | for n sufficiently large, then E ( x ) is homeo-morphic to the Cantor set C .III. If | x ( n ) | P i>n | x ( i ) | for n sufficiently large, then E ( x ) is a finiteunion of closed intervals. Moreover, if | x ( n ) | > | x ( n + 1) | for almostall n and E ( x ) is a finite union of closed intervals, then | x ( n ) | P i>n | x ( i ) | for n sufficiently large.In the same paper Kakeya formulated the hypothesis that, for any x ∈ l \ c , the set E ( x ) is homeomorphic to C or is a finite union of closedintervals. In 1980 it was shown that the Kakeya conjecture is false [14].We recall a number of examples in the literature which demonstrate thefalseness of the conjecture. We use the original notations proposed by theauthors. The notation will be unified later in the paper.A. D. Weinstein and B. E. Shapiro in [14] gave an example of a sequence a with a ( n ) > a ( n + 1) > n , and a ( n ) > P i>n a ( i ) for infinitely many n (hence E ( a ) is not a finite union of intervals), but having the property thatthe set E ( a ) contains an interval. The sequence a is defined by the formulas: a (5 n + 1) = 0 , · − n , a (5 n + 2) = 0 , · − n , a (5 n + 3) = 0 , · − n , a (5 n + 4) = 0 , · − n , a (5 n + 5) = 0 , · − n . So, a = ( 3 · , · , · , · , · , · , . . . ) . However, they did not justify why the interior of E ( a ) is non-empty.Independently, C. Ferens ([3]) constructed a sequence b such that E ( b ) isnot a finite union of intervals but contains an interval, putting b (5 l − m ) = ( m + 3) l − l for m = 0 , , , , l = 1 , , . . . . Therefore b = (7 · , · , · , · , · , · , . . . ) . J. A. Guthrie and J. E. Nymann gave a simpler example of a sequencewhich achievement set is not a finite union of closed intervals and is nothomeomorphic to the Cantor set, defining a sequence by formulas: c (2 n −
1) = n and c (2 n ) = n for n = 1 , , . . . . In a series of papers [4], [11] and [12] J. E. Nymann with J. A. Guthrie andR. A. S´aenz characterized the topological structure of the set of subsums ofinfinite series in the following manner:
Theorem 1.
For any x ∈ l \ c , the set E ( x ) is one of the following types: (i) a finite union of closed intervals; (ii) homeomorphic to the Cantor set; (iii) homeomorphic to the set E ( c ) (of subsums of the sequence ( , , , , , . . . ) ). Note, that the set E ( c ) is homeomorphic to C ∪ S ∞ n =1 S n − , where S n denotes the union of the 2 n − open middle thirds which are removed from[0 ,
1] at the n -th step in the construction of the Cantor ternary set C . Suchsets are called Cantorvals (to emphasize their similarity to unions of inter-vals and to the Cantor set simultaneously). Formally, a Cantorval (moreprecisely, an M -Cantorval - compare [9]) is a non-empty compact subset S of the real line such that S is the closure of its interior, and both end-points of any component with non-empty interior are accumulation pointsof one-point components of S .Theorem 1 states that the space l can be decomposed into four sets c , C , I and MC , where I consists of sequences x with E ( x ) equal to a finite unionof intervals, C consists of sequences x with E ( x ) homeomorphic to the Cantorset, and MC consists of sequences x with E ( x ) being Cantorvals. Some ARTUR BARTOSZEWICZ, MA LGORZATA FILIPCZAK, AND EMILIA SZYMONIK algebraic properties and topological (Borel) classification of these subsets of l have been recently discussed in [1].Finally, in Jones’ paper [6] there is presented a sequence d = ( 35 , , , , · , · , · , · , · ( 19109 ) , . . . ) . In [6], R. Jones shows a continuum of sequences generating Cantorvals,indexed by a parameter q , by proving that, for any positive number q with15 ∞ X n =1 q n < q < ) the sequence( 35 , , , , q, q, q, q, q , . . . )is not in C nor I , so it belongs to MC . Based on Jones’ idea, we will describeone-parameter families of sequences which contain (in particular) a, b, d andmany others. 2. The main result.
For any q ∈ (0 , ) we will use the symbol ( k , k , . . . , k m ; q ) to denote thesequence ( k , k , . . . , k m , k q, k q, . . . , k m q, k q , k q , . . . , k m q , . . . ). Suchsequences we will call multigeometric. Theorem 2.
Let k > k > · · · > k m be positive integers and K = P mi =1 k i .Assume that there exist positive integers n and n such that each of num-bers n , n + 1 , . . . , n + n can be obtained by summing up the numbers k , k , . . . , k m (i.e. n + j = P mi =1 ε i k i with ε i ∈ { , } , j = 1 , . . . , n ). If q > n +1 then E ( k , . . . , k m ; q ) has a nonempty interior . If q < k m K + k m then E ( k , . . . , k m ; q ) is not a finite union of intervals. Consequently, if n + 1 q < k m K + k m then E ( k , . . . , k m ; q ) is a Cantorval. Proof.
Denote x q = ( k , . . . , k m ; q ). We start with showing that, for q < k m K + k m , E ( x q ) is not a finite union of closed intervals. Observe first, thatthe sequence x q is non-increasing. Indeed, from the inequality q < k m K + k m , itfollows that qK + qk m < k m , and k m > qK − q > qK > qk . Moreover, using the same inequality, we obtain X i>m x q ( i ) = K ∞ X j =1 q j = K q − q < k m . Hence, for any n ∈ N , we have x q ( nm ) > P i>nm x q ( i ) and, according to thesecond sentence of the Kakeya property III, E ( x q ) is not a finite union ofclosed intervals.Suppose now that q > n +1 and consider the sequence y = (1 , . . . , , q, . . . , q, q , . . . , q , . . . )with n repetitions of each term. Note that, for any k ∈ N , the sum X j>nk y ( j ) = q k − nq − q is, by inequality q > n +1 , bigger than or equal to y ( nk ) = q k − . Therefore,for any i ∈ N y ( i ) X j>i y ( j )and again from the property III, we obtain that E ( y ) has non-empty interior.To end the proof, we show that n ∞ X j =0 q j + E ( y ) ⊂ E ( x q ) . If t ∈ n P ∞ j =0 q j + E ( y ), then there exist p i ∈ { , , . . . , n } , i = 0 , , , . . . such that t = ( n + n q + . . . ) + ( p + p q + . . . ) . Therefore t = ( n + p ) + ( n + p ) q + . . . ARTUR BARTOSZEWICZ, MA LGORZATA FILIPCZAK, AND EMILIA SZYMONIK belongs to E ( x q ). (cid:3) Examples.
Using the latter theorem, we can easily check that sequences a, b and d generate Cantorvals, because they belong to appropriate one-parameterfamilies, indexed by q . Example 3.
The Weinstein-Shapiro sequence ( [14] ).It is clear that if E ( x ) is a Cantorval, α = 0 and αx = ( αx (1) , αx (2) , . . . ) ,then E ( αx ) is a Cantorval too. To simplify a notation we multiply thesequence a by and consider the family of sequences a q = (8 , , , , q ) for q ∈ (0 , ) . Summing up , , , and , we can get any natural numberbetween n = 4 and n + n = 26 . Therefore, by Theorem 2, for any q satisfying inequalities q < , the sequence a q generates a Cantorval. Obviously, the number used in [14] belongs to [ , ) . It is not difficult to check (using III) that a q ∈ I for q > . Example 4.
The Ferens sequence ( [3] ).For the family of sequences b q = (7 , , , , q ) K is equal to , n = 3 and n = 19 . Hence, for any q ∈ [ , ) , b q gener-ates a Cantorval. In particular, the sequence (7 , , , , ) , obtained fromthe Ferens sequence by multiplication by a constant, generates a Cantorval.Note that b q ∈ I , for q > . Example 5.
The Jones-Velleman sequence ( [6] ). Applying Theorem 2 to the sequence d q = (3 , , , q ) we obtain K = 9 , n = 2 and n = 5 , so for any q ∈ [ , ) , E ( d q ) is aCantorval. Moreover d q ∈ I for q > . We can also consider analogous sequences for more than three ’s. Infact, any sequence x q = (3 , , . . . , | {z } k − times ; q ) with q ∈ [ k , k +5 ) , generates a Cantorval.Note that for k = 1 and k = 2 the argument of Theorem 2 breaks down,because k > k +5 . It means, in particular, that Theorem 2 does not applyto the Guthrie and Nymann example c = (cid:0) , (cid:1) . However, we can apply Theorem 2 to ”shortly defined” sequences. Indeed,for the sequence (4 , , q ), numbers K , n and n are the same as for d q .It is not difficult to check that, to keep the interval [ n +1 , k m K + k m ) non-empty, m should be greater than 2.There is a natural question if Theorem 2 precisely describes the set of q with ( k , . . . , k m ; q ) ∈ MC . The upper bounds, for all mentioned examplesare exact, because ( k , . . . , k m ; q ) ∈ I , for q > k m K + k m . However, this is nottrue for all sequences satisfying the assumptions of Theorem 2. Example 6.
For the sequence h q = (10 , , , , , , q ) , we have K = 47 , n = 5 and n = 37 . Therefore the interval [ n +1 , k m K + k m ) = [ , ) is non-empty.However, for h = (10 , , , , , , ) and any n ∈ N , we have P i> n − h ( i ) =( ) n − (2 + · − ) = 4( ) n − < h (7 n − . It means that h / ∈ I . Since > , we have h / ∈ C and so h ∈ MC .It is not difficult to check, using III again, that h q / ∈ I if and only if q < . ARTUR BARTOSZEWICZ, MA LGORZATA FILIPCZAK, AND EMILIA SZYMONIK
Observe, that E ( k , . . . , k m ; q ) ⊂ P Ki =1 C q , where C q = E ((1; q )) and P Ki =1 C q denotes the algebraic sum. In [2] it is proved that, if q < K +1 then P Ki =1 C q is homeomorphic to the Cantor set. The following theoremimproves this result. Theorem 7.
Let x = ( k , ..., k m ; q ) be a multigeometric sequence and Σ := ( m X i =1 ε i k i : ( ε i ) mi =1 ∈ { , } m ) .If q < /card (Σ) then E ( x ) is a Cantor set.Proof. Clearly, E ( x ) = Σ + qE ( x ). Suppose that q < /card (Σ) and theset E ( x ) has a nonempty interior. Therefore E ( x ) has positive Lebesguemeasure λ ( E ( x )) and λ ( E ( x )) ≤ card (Σ) · q · λ ( E ( x )) < λ ( E ( x ))which gives a contradiction. (cid:3) Using the latter theorem to the Weinstein-Shapiro sequence a q = (8 , , , , q )(compare Example 3) we obtain Σ of cardinality 25. It means that E ( a q ) ∈ C for q ∈ (cid:0) , (cid:1) . We do not know what is the type of E ( a q ) for q ∈ (cid:2) , (cid:1) .Analogously, E ( b q ) ∈ C for q ∈ (cid:0) , (cid:1) (compare Example 4), E ( d q ) ∈ C for q ∈ (cid:0) , (cid:1) (compare Example 5) and E ( h q ) ∈ C for q ∈ (cid:0) , (cid:1) (compareExample 6).4. The generalization of Guthrie-Nymann example.
We have just mentioned that Theorem 2 does not work for sequences(3 , q ) and (3 , , q ). However, Guthrie and Nymann have proved that c = (3 , ) ∈ MC . Following their method we will find q < n +1 such that(3 , , . . . , | {z } k − times ; q ) ∈ MC . Theorem 8.
For any sequence of the form x k = (3 , , . . . , | {z } k − times ; 12 k + 2 ) , the set E ( x k ) is a Cantorval.Proof. We know that x k / ∈ I , because k +2 < k +5 (compare with Example5). It remains to prove that E ( x k ) contains an interval.For a sake of clarity, we will prove a thesis for k = 2 , i.e. we will showthat E ( x ) ⊃ [3 , t = 3 + ∞ X i =1 ε i i with ε i = { , . . . , } belongs to E ( x ).Since E ( x ) is closed and the set { P ni =1 ε i i : ε i ∈ { , . . . , } , i n, n =0 , , . . . } is dense in [3 , n X i =1 ε i i ∈ E ( x )for any n = 0 , , . . . , ε i = 0 , . . . , n = 0, we have 3 ∈ E ( x ).Suppose that any number of the form(1) t ′ = 3 + n − X i =1 ε i i belongs to E ( x ). It means that there exist a i , b i , c i ∈ { , } such that t ′ = 3 + n − X i =1 a i + 2 b i + 2 c i i . Let t = 3 + n X i =1 ε i i . If ε n = 0 , , , t = t ′ + 3 a n + 2 b n + 2 c n n for some t ′ and suitable a n , b n and c n .Suppose that ε n = 1. Hence t = t ′ + 16 n = t ′ − n − + 3 + 2 + 26 n = t ′′ + 3 + 2 + 26 n . If t ′ > t ′′ satisfies (1) and the proof is complete. If t ′ = 3 then t = 3+ 16 n = 2+(1 − n − )+ 76 n = 2+( 56 + 56 + · · · + 56 n − )+ 3 + 2 + 26 n ∈ E ( x ) . To show that for a fixed k >
2, any n = 0 , , ... and ε i = 0 , ..., k + 13 + n X i =1 ε i (2 k + 2) i ∈ E ( x k )and hence [3 , ∈ E ( x k ), one can repeat the previous considerations, usingthe equality3 + 1(2 k + 2) n = 2 + (cid:18) − k + 2) n − (cid:19) + 3 + 2 k (2 k + 2) n . (cid:3) Note that, even for special sequences considered in this paper, it is veryhard to distinguish sequences belonging to C from sequences belonging to MC . In particular, for any sequence of the form x q = (3 , , . . . , q ) , where 2’s repeats itself k -times, x q ∈ I if and only if q > k +5 , and, byTheorem 7, x q ∈ C for q < card (Σ) = k +2 . ✲ C k +2 MC ? k MC k +5 I r r r r r We have no idea what are the types of sets E ( x q ) for q ∈ ( k +2 , k ). Finally, go back to the Guthrie and Nymann sequence c = (cid:0) , (cid:1) . Z.Nitecki, in [10], proved that for q < the sequence c q = (3 , q )belongs to C . The same conclusion follows easily from Theorem 7. It is notdifficult to check that x q ∈ I if and only if q > . ✲ C ? MC I r r r r We do not know what is the type of E ( x q ) for q ∈ ( , ).At last, let us consider one more example from [10] (due to Kenyon). Example 9.
The achievement set E ( f ) of the sequence f = (6 , ) (in ournotation) is M -Cantorval. To prove it, Nitecki observes that is equal to mod and each element of Z can be obtained by summing up the numbers2 and 1 (compare the proof of Theorem 8). Then he makes use of the Bairecategory theorem. By our mind, this fact can be explained in a much simplerway. Indeed, f = 12 (12 ,
2; 14 ) = 12 (12 , , , · , · , · , . . . ) . Hence E ( f ) = 12 E (12 , , , · , · , . . . ) = 12 E ( c ) ∪
12 ( E ( c ) + 12) and E ( f ) is of the same form as E ( c ) . In general, it is easy to observe (in thesame way as above) that the sequences ( k , k , . . . , k m ; q ) and ( q n k , q n k , . . . ,q n m k m ; q ) for integers n , n , . . . , n m are in the same set among of C , I or MC . Observe, for instance, that (2 , ) ∈ I and (3 , ) ∈ MC . However,each element of Z can be obtained by summing up and , but can notbe obtained by summing up and . Acknowledgement
The authors are greatly indebted to the referees for several helpful com-ments improving the paper. In particular they wish to express their thanksfor the idea of Theorem 7.
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