On the center of distances
aa r X i v : . [ m a t h . GN ] J u l ON THE CENTER OF DISTANCES
WOJCIECH BIELAS, SZYMON PLEWIK, AND MARTA WALCZYŃSKA
Abstract.
In this paper we introduce the notion of the center ofdistances of a metric space, which is required for a generalizationof the theorem by J. von Neumann about permutations of twosequences with the same set of cluster points in a compact metricspace. Also, the introduced notion is used to study sets of subsumsof some sequences of positive reals, as well for some impossibilityproofs. We compute the center of distances of the Cantorval, whichis the set of subsums of the sequence , , , , . . . , n , n , . . . , andalso for some related subsets of the reals. Introduction
The center of distances seems to be an elementary and natural notionwhich, as we think, has not been studied in the literature. It is anintuitive and natural concept which allows us to prove a generalizationof the theorem about permutations of two sequences with the sameset of cluster points in a compact metric space, see Theorem 1. Wehave realized that the computation of the center of distances—even forwell-known examples—is not an easy task because it requires skillfuluse of fractions. We have only found a few algorithms which enablecomputing centers of distances, see Lemmas 3 and 10. So, we presentthe use of this notion for impossibility proofs, i.e., to show that a givenset cannot be the set of subsums, for example Corollary 14.We refer the readers to the paper [8], as it is a good introductionto facts about the set of subsums of a given sequence, cf. also papers[1] and [5] as well as others cited therein. In several papers, the set
Mathematics Subject Classification.
Primary: 54E35; Secondary: 05B10,11B13.
Key words and phrases.
Cantorval, Center of distances.The first and the third author acknowledge support from GAČR project 16-34860L and RVO: 67985840. of all subsums of the sequence , , , , . . . , n , n , . . . , i.e., the set X consisting of all sums P n ∈ A n + P n ∈ B n , where A and B are arbitrary subsets of positive natural numbers, isconsidered. J. A. Guthrie and J. E. Nymann, see [2] and compare[9] and [8, p. 865], have shown that [ , ⊂ X . But, as it can beseen in Corollary 6, we get [ , ⊂ X . For these reasons, we felt thatthe arithmetical properties of X are not well known and described inthe literature. Results concerning these properties are discussed in:Propositions 5, 7 and 8; Corollary 9; Theorems 11, 12, 13, 15 and 16;and also they are presented in Figures 1, 2 and 3.2. A generalization of a theorem by J. von Neumann
Given a metric space X with the distance d , consider the set S ( X ) = { α : ∀ x ∈ X ∃ y ∈ X d ( x, y ) = α } , which will be called the center of distances of X . Suppose that se-quences { x n : n ∈ ω } and { y n : n ∈ ω } have the same set of clusterpoints C . For this, J. von Neumann [7] proved that there exists a per-mutation π : ω → ω such that lim n → + ∞ d ( x n , y π ( n ) ) = 0 . Proofs of theabove statement can be found in [3] and [10]. However, we would liketo present a slight generalization of this result. Theorem 1.
Suppose that sequences { a n } n ∈ ω and { b n } n ∈ ω have thesame set of cluster points C ⊆ X , where ( X, d ) is a compact metricspace. If α ∈ S ( C ) , then there exists a permutation π : ω → ω suchthat lim n → + ∞ d ( a n , b π ( n ) ) = α .Proof. We use the back-and-forth method, which was developed in [4,p. 35–36]. Given α ∈ S ( C ) , we shall renumber { b n } n ∈ ω by establishinga permutation π : ω → ω such that lim n → + ∞ d ( a n , b π ( n ) ) = α. Put π (0) = 0 and assume that values π (0) , π (1) , . . . , π ( m − andinverse values π − (0) , π − (1) , . . . , π − ( m − are already defined. Weproceed step by step as follows. N THE CENTER OF DISTANCES 3 If π ( m ) is not defined, then take points x m , y m ∈ C such that d ( a m , x m ) = d ( a m , C ) and d ( x m , y m ) = α . Then choose b π ( m ) to be thefirst element of { b n } n ∈ ω not already used such that d ( y m , b π ( m ) ) < m .But, if π − ( m ) is not defined, then take points p m , q m ∈ C such that d ( b m , q m ) = d ( b m , C ) and d ( p m , q m ) = α . Choose a π − ( m ) to be the firstelement of { a n } n ∈ ω not already used such that d ( p m , a π − ( m ) ) < m .The set C ⊆ X , as a closed subset of a compact metric space, iscompact. Hence required points x m , y m , p m and q m always exist andalso lim n → + ∞ d ( a n , C ) = 0 = lim n → + ∞ d ( b n , C ) . It follows that α = d ( x m , y m ) = d ( p m , q m ) = lim n → + ∞ d ( a n , b π ( n ) ) . (cid:3) As we have seen, the notion of the center of distances appears in anatural way in the context of metric spaces. Though the computation ofthe center of distances is not an easy task, it can be done for importantexamples giving further information about those objects.3.
On the center of distances and the set of subsums
Given a metric space X , observe that ∈ S ( X ) and also, if X ⊆ [0 , + ∞ ) and ∈ X , then S ( X ) ⊆ X .If { a n : n ∈ ω } is a sequence of reals, then the set X = { P n ∈ A a n : A ⊆ ω } is called the set of subsums of { a n } . In this case, we have d ( x, y ) = | x − y | . If X is a subset of the reals, then any maximal interval ( α, β ) disjoint from X is called an X - gap . Additionally, when X is a closed set,then any maximal interval [ α, β ] included in X is called an X - interval . Proposition 2. If X is the set of subsums of a sequence { a n } n ∈ ω , then a n ∈ S ( X ) , for all n ∈ ω .Proof. Suppose x = P n ∈ A a n ∈ X . If n ∈ A , then x − a n ∈ X and d ( x, x − a n ) = a n . When n / ∈ A , then x + a n ∈ X and d ( x, x + a n ) = a n . (cid:3) WOJCIECH BIELAS, SZYMON PLEWIK, AND MARTA WALCZYŃSKA
In some cases, the center of distances of the set of subsums of a givensequence can be determined. For example, the unit interval is the setof subsums of { n } n> . So, the center of distances of { n } n> is equalto [0 , ] . Lemma 3.
Assume that ( λ · [0 , b )) ∩ X = λ · X , for a number λ > and a set X ⊆ [0 , b ) . If x ∈ [0 , b ) \ X and n ∈ ω , then λ n x / ∈ X .Proof. Without loss of generality, assume that X ∩ (0 , b ) = ∅ . Thus λ , since otherwise we would get b < λ m t ∈ X , for some t ∈ X and m ∈ ω . Obviously x = λ x ∈ [0 , b ) \ X . Assume that λ n x / ∈ X , so weget λ · [0 , b ) ∋ λ n +1 x = λ · λ n x / ∈ λ · X = ( λ · [0 , b )) ∩ X. Therefore λ n +1 x / ∈ X , which completes the induction step. (cid:3) Theorem 4. If q > and a > , then the center of distances of theset of subsums of a geometric sequence { aq n } n> consists of exactly theterms of the sequence , aq , aq , . . . .Proof. Without loss of generality we can assume a = 1 . The diameter ofthe set X of subsums of the sequence { q n } n > equals q − = P n > q n = X (0) and we always have c n = q n > P i > n +1 1 q i = q n · q − = X ( n ) > c n +1 = q n +1 . So, c = q ∈ X implies that no t > c belongs to S ( X ) . Indeed, c − t < and c + t > q > q − . By Proposition 2, we get q ∈ S ( X ) .The interval ( q ( q − , q ) = ( X (1) , c ) is an X -gap. So, ∈ X impliesthat no t ∈ ( X (1) , c ) belongs to S ( X ) .Now, suppose that t ∈ ( c , X (1)] and t + X (2) < c . Thus X (2) ∈ X implies that t / ∈ S ( X ) . Indeed, X (2) < c < t implies X (2) − t < ,and X (2) + c = X (1) implies that the X -gap ( X (1) , c ) has to contain t + X (2) .If c t + X (2) and t ∈ ( c , X (1)] , then the interval ( X (1) − t, c − t ) is included in the interval [0 , X (2)] . No X -gap of the length q − q ( q − = c − X (1) is contained in the interval [0 , X (2)] . Therefore, there exists x ∈ X ∩ ( X (1) − t, c − t ) —one can even find x being the end of an X -gap, which implies that t / ∈ S ( X ) . Again, by Proposition 2, we get q ∈ S ( X ) . We proceed farther using Lemma 3 with λ = q n . If we take
N THE CENTER OF DISTANCES 5 advantage of the similarity of X and q n · X we get as a result that no t = 0 different from q n , where n > , belongs to S ( X ) . (cid:3) Note that, when we put a = 2 and q = 3 , then Theorem 4 appliesto the Cantor ternary set. But for a ∈ { , } and q = 4 this theoremapplies to sets C and C which will be defined in Section 4.4. An example of a Cantorval
Following [2, p. 324], consider the set of subsums X = (cid:8)P n> x n n : ∀ n x n ∈ { , , , } (cid:9) . Thus, X = C + C , where C = { P n ∈ A n : 0 / ∈ A ⊆ ω } and C = { P n ∈ B n : 0 / ∈ B ⊆ ω } . Following [6, p. 330], because of its topologicalstructure, one can call this set a
Cantorval (or an M -Cantorval ).Before discussing the affine properties of the Cantorval X we shallintroduce the following useful notions. Every point x in X is deter-mined by the sequence { x n } n> , where x = P n> x n n . Briefly, x n iscalled the n -th digit of x . But the sequence { x n } n> is called a digitalrepresentation of the point x ∈ X . Keeping in mind the formula for thesum of an infinite geometric series, we denote tails of series as follows: C ( n ) = P k = n +1 24 k = · n and C (0) = ; C ( n ) = P k = n +1 34 k = n and C (0) = 1 ; X ( n ) = P k = n +1 54 k = · n and X (0) = .Since X (0) = we get X ⊂ [0 , ] . The involution h : X → X definedby the formula x h ( x ) = − x is the symmetry of X with respect to the point . In order to checkthis, it suffices to note that: P n ∈ A n + P n ∈ B n = x ∈ X ⇒ − x = P n/ ∈ An> n + P n/ ∈ Bn> n ∈ X ; and also that + P n> n = + P n> n +1 = ∈ X . So, we get X = − X and X = h [ X ] . WOJCIECH BIELAS, SZYMON PLEWIK, AND MARTA WALCZYŃSKA Figure 1.
An approximation of the Cantorval X ⊂ [0 , ] .In Figure 1 there are marked gaps ( , ) and ( , ) , both of thelength . Six gaps ( , ) , ( , ) , ( , ) , ( , ) , ( , ) and ( , ) have the length . The rest of gaps are shorter and have lengths notgreater than . To describe intervals which lie in X , we need thefollowing. Let K n = [ , ∩ { P ni =1 x i i : ∀ i x i ∈ { , , , }} . We get K = { } and K = { , , , , } . Keeping in mind C (0) = and C ( n ) < n , we check that f n = n + P n − i =1 24 i > is the smallestreal number in K n . Similarly, using X ( n ) = · n < n and C (0) = 1 ,check that f n | K n | = P ni =1 34 i < is the greatest real number in K n . Infact, we have the following. Proposition 5.
Reals from K n are distributed consecutively at the dis-tance n , from n + P n − i =1 24 i up to P ni =1 34 i , in the interval [ , . There-fore | K n | = (4 n − and | K n +1 | = 4 | K n | + 1 .Proof. Since | K | = 1 and | K | = 5 , the assertions are correct in thesecases. Suppose that K n − = { f n − , f n − , . . . , f n − | K n − | } , where f n − = 34 n − + n − X i =1 i , and f n − j +1 − f n − j = 14 n − for < j < | K n − | − ; in consequence f n − | K n − | = P n − i =1 34 i . Considerthe sum K n − ∪ ( n + K n − ) ∪ ( n + K n − ) ∪ ( n + K n − ) , remove the point n + P n − i =1 34 i > , and then add points f n = n + P n − i =1 24 i and f n = n + P n − i =1 24 i . We obtain the set K n = { f n , f n , . . . , f n (4 n − } , which is what we need. (cid:3) Corollary 6.
The interval [ , is included in the Cantorval X . N THE CENTER OF DISTANCES 7
Proof.
The union S { K n : n > } is dense in the interval [ , . Hence,Proposition 5 implies the assertion. (cid:3) Note that it has observed that [ , ⊂ X , see [2] or compare [8].Since X is centrally symmetric with as a point of inversion, thisyields another proof of the above corollary. However, our proof seemsto be new and it is different than the one included in [2].Put C n = n · X = X ∩ [0 , · n ] , for n ∈ ω . So, each C n is an affinecopy of X . Proposition 7.
The subset X \ [ , ⊂ X is the union of pairwise dis-joint affine copies of X . In particular, this union includes two isometriccopies of C n = n · X , for every n > .Proof. The desired affine copies of X are C and + C = h [ C ] , + C and h [ + C ] , and so on, i.e., P ni =1 24 n + C n +1 and h [ P ni =1 24 n + C n +1 ] . (cid:3) Proposition 8.
The subset X \ (( , ∪ ( , ) ∪ ( , )) ⊂ X is theunion of six pairwise disjoint affine copies of D = [0 , ] ∩ X .Proof. The desired affine copies of D = · ( X ∩ [0 , ]) lie as shown inFigure 2. (cid:3) D + D + Dh [ + D ] h [ + D ] h [ D ]
16 14 512 12 23
76 54 531712 32
Figure 2.
The arrangement of affine copies of D . Corollary 9.
The Cantorval X ⊂ [0 , ] has Lebesgue measure .Proof. There exists a one-to-one correspondence between X -gaps and X -interval as it is shown in Figure 3. In view of Propositions 7 and8, we calculate the sum of lengths of all gaps which lie in [0 , ] \ X asfollows: + 6 · · + · + . . . + · (cid:0) (cid:1) n + . . . = + P n > (cid:0) (cid:1) n = . Since − = 1 we are done. (cid:3) WOJCIECH BIELAS, SZYMON PLEWIK, AND MARTA WALCZYŃSKA
16 14 512 12 23
76 54 531712 32
Figure 3.
The correspondence between X -gaps and X -intervals.If we remove the longest interval from n · D , then we get the unionof three copies of D , each congruent to n +1 · D . This observation—weused it above by default—is sufficient to calculate the sum of lengthsof all X -intervals as follows: + + 6 · · + . . . + · (cid:0) (cid:1) n + . . . = 1 . Therefore the boundary X \ Int X is a null set.5. Computing the center of distances
In case of subsets of the real line we formulate the following lemma.
Lemma 10.
Given a set C ⊆ [0 , + ∞ ) disjoint from an interval ( α, β ) ,assume that x ∈ [0 , α ] ∩ C . Then the center of distances S ( C ) is disjointfrom the interval ( α − x, β − x ) , i.e., S ( C ) ∩ (( α, β ) − x ) = ∅ .Proof. Given x ∈ [0 , α ] ∩ C , consider t ∈ ( α − x, β − x ) . We get x α α − x < t < β − x. Since α < x + t < β , we get x + t / ∈ C , also x < t implies x − t / ∈ C .Therefore, x ∈ C implies t / ∈ S ( C ) . (cid:3) We will apply the above lemma by putting suitable C -gaps in placeof the interval ( α, β ) . In order to obtain t / ∈ S ( C ) , we must find x < t such that x + t ∈ ( α, β ) and x ∈ C . For example, this is possiblewhen α < t < α and the interval [0 , α ] includes no C -gap of the lengthgreater than or equal to β − α . But if such a gap exists, then we choosethe required x more carefully. Theorem 11.
The center of distances of the Cantorval X consists ofexactly the terms of the sequence , , , . . . , n , n , . . . .Proof. We get { n : n > } ∪ { n : n > } ⊆ S ( X ) , by Proposition 2.Also, the diameter of X is and ∈ X , hence no t > belongs to S ( X ) . N THE CENTER OF DISTANCES 9
We use Lemma 10 with respect to the gap ( α, β ) = ( , ) . Keeping inmind the affine description of X , we see that the set X ∩ [0 , ] has agap ( , ) of the length . For t ∈ ( , ) \ { } , we choose x in X such that x ∈ ( − t, − t ) . So, if t ∈ ( , ) \ { } , then t / ∈ S ( X ) .Similarly using Lemma 10 with the gap ( α, β ) = ( , ) , we check thatfor t ∈ ( , ] \ { } there exists x in X such that x ∈ ( − t, − t ) .Hence, if t ∈ ( , ] \ { } , then t / ∈ S ( X ) . Analogously, using Lemma10 with the gap ( α, β ) = ( , ) , we check that if < t < ,then t / ∈ S ( X ) .So far, we have shown that numbers and are the only elementsof S ( X ) ∩ ( , + ∞ ) . For the remaining part of the interval [0 , + ∞ ) theproof proceeds by induction on n , since X and n · X are similar. (cid:3) Denote Z = [0 , ] \ Int X . Thus the closure of a X -gap is the Z -intervaland the interior of a X -interval is the Z -gap. Theorem 12.
The center of distances of the set Z is trivial, i.e., S ( Z ) = { } .Proof. If α > , then { α, − α } ∩ Z = ∅ , hence ∈ Z implies α / ∈ S ( Z ) . If α ∈ { , } , then ∈ ( , ) ⊂ Z implies α / ∈ S ( Z ) .Indeed, the number belongs to the Z -gap ( , ) and the number + belongs to the Z -gap ( , and the number − belongs tothe Z -gap ( , ) . Also ∈ Z implies ( , ∩ S ( Z ) = ∅ , since ( , is a Z -gap. For the same reason ∈ Z implies / ∈ S ( Z ) and ∈ Z implies that no α ∈ ( , ) belongs to S ( Z ) . Since < − < and < + < , then ∈ Z implies / ∈ S ( Z ) . But, if α ∈ ( , ) , then ∈ Z implies α / ∈ S ( Z ) . Indeed, < − α < and < + α < . So far, we have shown S ( Z ) ∩ [ , + ∞ ) = ∅ . In fact, sets [ n +1 , n ] and S ( Z ) are always disjoint, since n · ( Z ∩ [0 , , n ] ∩ Z . Therefore α ∈ [ n +1 , n ] ∩ S ( Z ) implies n · α ∈ [ , ∩ S ( Z ) , a contra-diction. Finally, we get S ( Z ) = { } . (cid:3) Now, denote Y = Z ∩ X = X \ Int X . Thus, each X -gap is also the Y -gap, and the interior of an X -interval is the Y -gap. Theorem 13. S ( Y ) = { } ∪ { n : n ∈ ω } , i.e., the center of dis-tances of the set Y consists of exactly of the terms of the sequence , , , . . . , n , . . . .Proof. Since the numbers , , (= P n> n ) , , and are in Y , weget S { ( n +1 , n ) : n ∈ ω } ∩ S ( Y ) = ∅ , as in the proof of Theorem 12. We see that ∈ S ( Y ) , because ( Y ∩ [0 , ]) + 1 = Y ∩ [1 , ] . Moreover ( Y ∩ [0 , ] + ) ∪ ( Y ∩ [0 , ] + ) ⊂ Y , so ∈ S ( Y ) . Similarly, we check that n ∈ S ( Y ) . (cid:3) Corollary 14.
Neither Z nor Y is the set of subsums of a sequence.Proof. Because S ( Z ) = { } , thus Proposition 2 decides the case with Z . Also, this proposition decides the cases with Y , since ∈ Y and P n ∈ ω n = . (cid:3) Let us add that the set of subsums of the sequence { n n ∈ ω is includedin Y . One can check this, observing that each number P n ∈ A n , wherethe nonempty set A ⊂ ω is finite, is the right end of an X -interval.6. Digital representation of points in the Cantorval X Assume that A = { a n } n> and B = { b n } n> are digital representa-tions of a point x ∈ X , i.e., P n> a n n = P n> b n n = x, where a n , b n ∈ { , , , } . We are going to describe dependencies be-tween a n and b n . Suppose n is the least index such that a n = b n .Without loss of generality, we can assume that a n = 2 < b n = 3 ,bearing in mind that X ( n ) = · n . And then we say that A is chasing B (or B is being caught by A ) in the n -step: in other words, P i>n a i i = P i>n b i i + n and a k = b k for k < n . If it is never the casethat a k = 5 and b k = 0 , then it has to be b k + 3 = a k for all k > n . Insuch a case, we obtain P i>n a i i = P i>n b i i + n , since P i>n i = n . N THE CENTER OF DISTANCES 11
Suppose n is the least index such that a n = 5 and b n = 0 , thus B ischasing A in the n -step. Proceeding this way, we obtain an increasing(finite or infinite) sequence n < n < . . . such that | a n k − b n k | = 5 and | a i − b i | = 3 for n < i / ∈ { n , n , . . . } . Moreover, A starts chasing B inthe n k -step for even k ’s and B starts chasing A in the n k -step for odd k ’s, for the rest of steps changes of chasing do not occur. Theorem 15.
Assume that x ∈ X has more than one digital represen-tation. There exists the finite or infinite sequence of positive naturalnumbers n < n < . . . and exactly two digital representations { a n } n> and { b n } n> of x such that: • a k = b k , as far as < k < n ; • a n = 2 and b n = 3 ; • a n k = 5 and b n k = 0 , for odd k ; • a n k = 0 and b n k = 5 , for even k > ; • a i ∈ { , } and a i − b i = 3 , as far as n k < i < n k +1 ; • a i ∈ { , } and b i − a i = 3 , as far as n k +1 < i < n k +2 .Proof. According to the chasing algorithm described above in step n k the roles of chasing are reversed. But, if the chasing algorithm does notstart, then the considered point has a unique digital representation. (cid:3) The above theorem makes it easy to check the uniqueness of digitalrepresentation. For example, if x ∈ X has a digital representation { x n } n> such that x n = 2 and x n +1 = 3 for infinitely many n , thenthis representation is unique. Indeed, suppose A = { a n } n> and B = { b n } n> are two different digital representations of a point x ∈ X suchthat a k = b k , as far as < k < n and a n = 2 < b n = 3 . ByTheorem 15, never the digit 3 occurs immediately after the digit 2 indigital representations of x for digits greater than n , since it has to be | a k − b k | > for k > n .The map { a n } n> P i> a i i is a continuous function from the Can-tor set (a homeomorphic copy of the Cantor ternary set) onto the Can-torval such that the preimage of a point has at most two points. In fact,the collection of points with two-point preimages and its complementare both of the cardinality continuum. By the algorithm describedabove, each sequence n < n < . . . of positive natural numbers deter-mines exactly two sequences { a n } n> and { b n } n> such that P n> a n n = P n> b n n , and vice versa. In the following theorem, we will use the abbreviation B = { n : n > } ∪ { n : n > } . Theorem 16.
Let A ⊂ B be such that B \ A and A are infinite. Thenthe set of subsums of a sequence consisting of different elements of A is homeomorphic to the Cantor set.Proof. Fix a non-empty open interval I . Assume that { a n } n> is thedigital representation of a point P n> a n n ∈ I . Choose natural numbers m > k such that numbers P kn =1 a n n and m + P kn =1 a n n belong to I .Then choose j > m so that a j ∈ B \ A , where a = 2 or a = 3 . Finallyput b n = a n , for < n k ; b j = 5 ; b j +1 = 2 and b i = 0 for othercases. Since b m = 0 , we get P i> b i i ∈ I . Theorem 15 together withconditions b j = 5 and b j +1 = 2 imply that the point P i> b i i is not inthe set of subsums of A . Thus, this set being dense in itself and closedis homeomorphic to the Cantor set. (cid:3) References [1] A. Bartoszewicz and Sz. Głąb,
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Institute of Mathematics, University of Silesia, Bankowa 14, 40-007Katowice, Poland; Institute of Mathematics of the Czech Academyof Sciences, Žitná 25, 115 67 Praha 1, Czech Republic
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