GENERALIZED CANTOR SETS AND SETS OF SUMS OF CONVERGENT ALTERNATING SERIES

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1 Journal of Applied Analysis Vol. 7, No. 1 (2001), pp GENERALIZED CANTOR SETS AND SETS OF SUMS OF CONVERGENT ALTERNATING SERIES M. DINDOŠ Received September 7, 2000 and, in revised form, February 2, 2001 Abstract. We give a complete characterization of sets of sums of absolutely convergent series with alternating signs. This result finishes wors of various authors 3], 4] 8] who studied this problem. As we will see, the studied sets are exactly sets that can be obtained by generalizing the construction of the Cantor set. Consider the series 1. Introduction ( 1) a n b n, (1) where (a n ) n N is a 0 1 sequence and (b n ) n N is a given (fixed) sequence of real numbers. The series (1) yields different sums for different choices of the sequence (a n ) n N. We would lie to obtain a full characterization of the set of such 1991 Mathematics Subject Classification. 40A21, 26A30. Key words and phrases. Absolutely convergent series, Cantor sets. ISSN c Heldermann Verlag.

2 132 M. DINDOŠ sums, i.e., given a sequence (b n ) n N of real numbers that has limit zero we consider the set S = {r R; r = ( 1) a n b n for some sequence (a n ) n N {0, 1}N }. (2) We call S the set of sums of the sequence (b n ) n N. This problem has already been considered in several papers 3] by K. Menon and 4] 8] by T. Šalát. These authors proved, that if sequence (b n ) n N is real and positive and its sum is finite, then S is a perfect set. They also studied other properties of S. In particular, two important cases have been considered. If b n i=n+1 b i for all n N, then S is just a closed interval. On the other hand, b n > i=n+1 b i for all n N, then S is a Cantor set (we will say it is Cantor-lie ). We thin that the mentioned results omitted the most interesting case when neither of two conditions above hold. We show that in such a case the perfect set S has infinitely many connected components. Essentially, two cases can happen. Either S can be written as a closure of a union of infinitely many disjoint closed intervals (see Example 3.7), or S has empty interior and can be constructed by a process quite similar to the classical construction of the Cantor set. We call such a set a generalized Cantor set. The actual construction is outlined in the Section 2 of this paper. Conversely, we will show that given any generalized Cantor set C, there is a sequence (b n ) n N whose corresponding set of sums S is equal to C. From now on we place certain restrictions on the given sequence (b n ) n N. First of all, the set S depends on the chosen sequence (b n ) n N, but not on the sign of b n ; therefore we can always assume that b n > 0 (terms b n = 0 do not influence S and can be omitted). We also see that if b n = and lim n b n = 0, then S =R. Hence, it suffices to study the case b n <, (3) which means that the series (1) is absolutely convergent. That is, any reordering of the series (1) does not influence on the resulting sum. Thus we can also assume that our sequence (b n ) n N is nonincreasing, i.e., b 1 b 2 b 3... b n > 0. (4) From now on, (3) and (4) will be understood to hold through this paper.

3 GENERALIZED CANTOR SETS Sets of Cantor type Recall first the classical construction of the Cantor set. Starting from a single interval F 1 = 0, 1] at each stage of the process a middle third of each interval is removed yielding a closed set F n which is a union of 2 n 1 disjoint intervals. Then the Cantor set C is the intersection C = n N F n. There is a natural generalization of this process, where the relative length of the removed middle interval may vary. This way we can obtain Cantorlie sets of various Hausdorff dimensions and measures which are all perfect, nowhere dense, c-dense in itself and are also symmetric in certain sense (the Cantor set is also self-similar). In this paper we further generalize the construction above by allowing closed intervals constituting the set F n being non-disjoint. This is a major modification that might seriously influence properties of the resulting set. However as we will see, the construction still preserves many properties of the original Cantor set. Let (r n ) n N, r n (0, 1) for all n N, be any sequence of real numbers and let I be a given closed and bounded interval. Define a decreasing sequence of closed sets F n inductively. First, put F 1 = I. As we are going to see, for each n N the set F n can be written as a union of 2 n 1 disjoint closed intervals of equal length, i.e., we have F n = I n. {0,1,2,...,2 n 1 1} So far, this is clearly the case when n = 1. Now we do the inductive step. Fix {0, 1, 2,..., 2 n 1 1} and define the sets I n+1 2, In as follows. Denote the endpoints of the interval I n by a and b respectively, i.e., In = a, b]. There exists a unique interval J = c, d] such that the intervals I n and J share common middle point and λ(j) λ(i n ) = r n. Here, λ is Lebesgue measure on R. We put Finally, let I n+1 2 = a, c] and I n = d, b]. (5) F n+1 = {0,1,2,...,2 n 1} I n+1. It follows from our construction that F n+1 F n and sets F n are closed.

4 134 M. DINDOŠ Definition 2.1. For a given closed bounded interval I and a sequence (r n ) n N of numbers from (0, 1) consider the sets F n constructed above. We say then that the set C = n N F n is Cantor-lie. Remar 2.2. Clearly, the construction above give us for I = 0, 1] and (r n ) n N = (1/3, 1/3, 1/3,... ) the standard Cantor set. Also apparently, for any interval I and any sequence (r n ) n N the resulting set C is closed, nowhere dense, has cardinality c (the size of continuum) and each point of C is a point of c-density of this set. Now we introduce the indicated modification of the outlined construction. We allow some the elements of the sequence (r n ) n N to be negative, namely we will assume that r n ( 1, 1) for all n N. The only requirement we place is: there is an increasing sequence of integers (n ) N such that r n > 0. (6) As before, we give the construction of the intervals I n+1 2 and I n from I n = a, b] where n = 1, 2,... and = 0, 1,..., 2n 1 1. We find again the unique interval J = c, d] such that the intervals I n and J have common middle point and λ(j) λ(i n) = r n. In particular, if r n > 0 we do not change our construction and define the sets I n+1 2 and I n by (5). If r n = 0, clearly c = d is the middle point I n and we use again (5) to define I n+1 2 and I n+1. Notice however, that these 2+1 two sets are no longer disjoint. Also F n+1 = F n. Finally, if r n < 0 we put I n+1 2 = a, d] and I n = c, b]. (7) Condition (7) means that the sets I n+1 2 and I n have common intersection: the interval J. Also, with F n+1 defined as the union of all I n+1, we have that F n+1 = F n. Definition 2.3. For a given closed bounded interval I and a sequence (r n ) n N of numbers from ( 1, 1) satisfying (6) consider the sets F n F n = {0,1,2,...,2 n 1 1} where I n are constructed by the inductive process defined above. I n

5 GENERALIZED CANTOR SETS 135 We say that the set C = n N F n (8) is a generalized Cantor set, provided C is nowhere dense. When we want to emphasize the dependence of the set C on the initial interval I we write C I instead. Remar 2.4. Since we allow some r n to be negative, it might not be true that the resulting set C has nonempty interior. The following theorem outlines two distinct case that might occur. Proposition 2.5. Let (r n ) n N be a sequence satisfying assumptions of Definition 2.3. Then one of the following is true: (a) C is a generalized Cantor set, i.e., it is nowhere dense. The set C is also perfect, and each point of C is a point of c-density in C. (b) C can be written as a closure of a countable union of disjoint closed intervals. Proof. In follows from (8) that the set C is closed since each F n is a finite union of closed intervals. First, assume that the set C has nonempty interior. It follows that the set C is of the second Baire category. Consider the sets I n for n = 1, 2,... and = 0, 1,..., 2n 1 1. These sets were defined above. For each of intervals I n we can repeat the construction above with the sequence (r n, r n+1, r n+2,... ) yielding a set C I n. Apparently, for each n we have C = C I n. (9) =0,1,...,2 n 1 1 Also, since for fixed n and = 0, 1, 2,..., 2 n 1 1 the intervals I n have the same length we get that the sets C I n and C I n, for two different,, are equal modulo a shift along the real line. In other words, the exists a real number t such that C I n = t + C I n = {t + z; z C I n }. In particular, the set C I n is nowhere dense if and only if the set C I n is. As we now, C is of the second Baire category. The sets on the right hand side of (9) are closed and therefore at least one of them cannot be nowhere dense. From the remar above it follows that all of them are not nowhere dense, i.e., all contain a nonempty interval. This has the following consequence: Given any x C and x J (J an open interval), there exists a closed subinterval J J such that J C.

6 136 M. DINDOŠ Indeed, pic any such x and J. We can find a sequence of intervals (K n ) n N (not necessarily unique) such that x K n and K n = I n, for some {0, 1, 2,..., 2n 1 1}. By (6), the length of intervals K n tends to zero, hence there exists n N such that K n J. Since C Kn K n and the set C Kn contains an interval (call it J ) we get that J J C. This establishes our claim. It follows, that the set C can be written as a closure of a union of closed disjoint intervals. Indeed, consider all intervals a i, b i ] with rational endpoints that are contained in C. From the property of C established above it follows that C = i N a i, b i ]. The intervals a i, b i ] are not necessarily disjoint, but each of them is contained in a unique maximal interval c i, d i ] with property a i, b i ] c i, d i ] C. The maximallity of intervals J i = c i, d i ] implies that J i, J j are either disjoint or equal for any pair (i, j) of positive integers. By taing each interval only once we can order them into a disjoint countable (possibly finite) sequence (denoted again by (J i )) for which C = i J i. This proves (b). Next, we consider part (a) of our proposition. Assume that the set C is nowhere dense. First we prove that C is perfect. Tae any x C. As before there is a sequence of intervals (K n ) n N (not necessarily unique) such that x K n and K n = I n, for some {0, 1, 2,..., 2n 1 1}. (10) The endpoints of the interval K n are also endpoints of some other intervals I n+1 that create the set F n+1. Proceeding inductively, we see that the endpoints of the set K n belong to F m for all m, hence to C as well. From this we see that the sequence of all left (or right) endpoints of sets K n converge to x. Therefore C is perfect. Finally, to show that each point of C is a point of c-density in C we just have to chec that the cardinality of C is c. Indeed, once we have this, for any x C we again tae the sets K n as in (10). Each set C Kn C constructed from the interval K n and sequence (r n, r n+1, r n+2,... ) has cardinality c. Since the length of K n goes to zero, the desired c-density follows. The proof that cardinality of the set C is c goes as follows. The ey is to index the intervals I n in a special way. First, we find the smallest integer n for which condition (6) holds, i.e., r n > 0. We loo at the interval I0 n. The sets I0 n+1, I1 n+1, which we denote by J0 1, J 1 1 respectively, were constructed from the interval I0 n by removing the middle part of In 0 using the number r n > 0. In particular, J0 1, J 1 1 are disjoint. Inductively, assume now that we have defined the intervals J n α 1,α 2,...,α n for n = 1, 2,...,, and α i {0, 1} for i = 1, 2,..., n,

7 GENERALIZED CANTOR SETS 137 such that each of J n α 1,α 2,...,α n is just I m j for some m and j. Also, for a fixed n, all intervals J n α 1,α 2,...,α n are disjoint; have the same length and moreover: J n+1 α 1,α 2,...,α n,0 J n α 1,α 2,...,α n, J n+1 α 1,α 2,...,α n,1 J n α 1,α 2,...,α n, for n = 1, 2,..., 1. We give here how to find the intervals Jα n 1,α 2,...,α n for n = + 1. Consider the interval Jα 1,α 2,...,α for some choice of α i s. As assumed, there is a pair (m, j) such that Jα 1,α 2,...,α = Ij m. Using (6) we find the smallest integer s m for which r s > 0. Consider the interval I2 s s m j. It follows that I2 s s m j Im j. We put J +1 α 1,α 2,...,α,0 = Is+1 2 s m+1 j, J +1 α 1,α 2,...,α,1 = Is+1 2 s m+1 j+1. It is obvious that Jα +1 1,α 2,...,α +1 construction we get that has all required properties. Also from the λ(j +1 α 1,α 2,...,α +1 ) 1 2 λ(j α 1,α 2,...,α ). This means that the length of these intervals goes to zero with increasing. Finally, we have that F n for all n. α i {0,1},,2,...,n J n α 1,α 2,...,α n Therefore each number a α1,α 2,α 3,... defined by {a α1,α 2,α 3,...} = n N J n α 1,α 2,...,α n belongs to C. All these numbers are different and their cardinality is c since they are indexed by zero-one sequences. This concludes our proof. Proposition 2.5 gives us that the generalized Cantor set from in Definition 2.3 has all properties which the classical Cantor set has. Let us also point out, that our construction is indeed very general. In particular, it includes many generalizations of Definition 2.1 described in literature. For example, one such generalization requires that at each step of the construction instead of removing the middle interval, we remove two or three (or n) intervals. By removing two intervals at each stage three new equidistantly spaced intervals are created. It is easy to chec that the sequence ( 1/5, 1/3, 1/5, 1/3,... ) does the same job. Another natural problem is whether two different sequences of (r n ) n N can generate same resulting set (starting from the same initial interval I). This cannot happen for sequences from Definition 2.1, i.e., when all r n > 0.

8 138 M. DINDOŠ But in a more general case r n ( 1, 1), the sequences (1/7, 1/3, 1/7, 1/3,... ), ( 3/7, 1/5, 1/3, 3/7, 1/5, 1/3,... ) indeed generate the same set. Finally, is it true in the case (b) in Proposition 2.5 (when C = i J i, (11) for some J i, i = 1, 2,..., disjoint closed intervals) that the number of intervals J i can be infinite? As we will see, the answer is positive. 3. The set of sums S In this section we study the set S defined in the introduction. We first deal with a special case considered already in 3] and 7] when additional condition b n > b i for n = 1, 2, 3,... (12) i=n+1 holds. We reformulate the result from the mentioned papers using our notation. The converse part in the following theorem is new. Theorem 3.1. Let (b n ) n N be a nonincreasing and positive sequence satisfying (12). Then the corresponding set of sums S S = {r R; r = ( 1) an b n for some sequence (a n ) n N {0, 1}N } is Cantor-lie, i.e., it can be constructed by the process described in Definition 2.1. More precisely, the starting interval I of the construction is ] I = b n, b n, and the positive sequence (r n ) n N is defined by b n r n = i=n+1 b i n = 1, 2, 3,.... (13) b i i=n Conversely, consider any symmetric interval I = a, a], (a > 0) and a sequence (r n ) n N, r n (0, 1), and denote by C the corresponding Cantorlie set. Then there exists a unique decreasing positive sequence (b n ) n N

9 satisfying (12) given by b n = 1 + r n 2 GENERALIZED CANTOR SETS 139 ( ) n 1 a b i, n = 1, 2, 3,..., (14) whose set of sums is C. That means: r C r = ( 1) a n b n for some sequence (a n ) n N {0, 1}N. Moreover, the zero-one sequence (a n ) n N corresponding to r C is unique. Proof. Notice, that the condition (12) guarantees that any r n defined by (13) belongs to (0, 1). As usual by C we denote the Cantor-lie set determined by the interval I and the sequence (r n ) n N. Let (b n ) n N be a sequence satisfying all required assumptions. Define sets S n, n = 1, 2,..., as follows: S 1 = {0}, (15) n S n+1 = {r R; r = ( 1) a i b i for some a i {0, 1}, i = 1, 2,..., n} n 1. Each S n+1 is the set of all possible partial sums of the first n terms of the series (1). We claim that the each set S n is also the set of all middle points of intervals I n, = 0, 1,..., 2n 1 1, defined in the previous section of this paper. Moreover, the length of each interval I n is exactly 2 i=n b i. The proof of these two claims goes inductively. For n = 1 when I0 1 = I there is nothing to prove. Assume now that these two claims have been established for some n 1. We will prove them for n+1. Consider one particular middle point of the interval I n, call it m. By the inductive assumption, there is a (n 1)-tuple (a 1, a 2,..., a n 1 ) such that m = n 1 ( 1)a i b i. It follows that the numbers m b n and m + b n belong to S n+1. It remains to be seen that they are middle points of the intervals I n+1 2, In+1 2+1, respectively. We do the proof just for the interval I n+1 2. Clearly, the length of is given by the number (1 r n )/2λ(I n ). If we plug-in (13) we get that ) = 2 i=n+1 b i. The middle point of I n+1 2 is given by I n+1 2 λ(i n+1 2 m 1 2 λ(in ) λ(in+1 2 ), which immediately gives m b n. This concludes the inductive argument. Now, we want to show that C = S. We first establish C S. Indeed, pic any r C. There exists a unique sequence ( n ) n N such that r I n n for all n = 1, 2, 3,....

10 140 M. DINDOŠ By m n we denote the middle point of the interval I n n. It follows that m n r as n since λ(i n n ) 0. Finally, from the definition of the sets S n it follows that s S lim dist(s, S n) = 0. n This guarantees that r S because m n S n. Conversely, to prove that S C consider any s S, say s = ( 1)a n b n. By m n we denote the sum n 1 ( 1)a i b i. Apparently, m n s and m n for all n and some n {0, 1, 2,..., 2 n 1 1}. Moreover, clearly I n n I n+1 n+1 I n n n = 1, 2,.... This means the unique point r C that belongs to the intersection n N In n is in fact s, since m n r and m n s. Hence s C. This establishes the first part of our theorem. Now we loo at the second (converse) part of this theorem. Tae any symmetric interval I = a, a] and a sequence (r n ) n N (0, 1)N. Define the sequence (b n ) n N by (14). It follows that 1 2 In particular, this implies ( n 1 a 0 < a ) n 1 b i < b n < a b i. n b i < b n. (16) Hence, using induction we get that b n > 0 and (a n b i) > 0 for all n. Also n b n+1 < a b i < b n, and therefore (b n ) n N is decreasing. Denote by b the limit of this sequence. If b > 0 then eventually the difference a n b i must become negative for large n which by (16) cannot happen. So b = 0. Thus also by (16) we get that n a b i 0 as n, or equivalently a = b i. (17) This means that instead of (16) we can write n a b i = b i < b n. i=n+1

11 GENERALIZED CANTOR SETS 141 This proves that (b n ) n N satisfies (12). Denote again by S n the sets defined by (15). Combining (14) and (17) yields b n = 1 + r n b i, n = 1, 2, 3,.... (18) 2 i=n In particular, when we compute r n from (18) we get (13). This means that we can use the claims established in the first part of this proof, namely that S n is the set of all middle points of intervals I n, = 0, 1,..., 2n 1 1, for the given sequence (r n ) n N. From here the proof continues as above, eventually showing that C = S. Now we briefly discuss uniqueness of the sequence (b n ) n N. Assume that there exists a Cantor-lie set C which is the set of sums of two different positive decreasing sequences (b n ) n N, (b n) n N satisfying (12). Denote by (r n ) n N, (r n) n N the corresponding positive sequences defined by (13). It follows that these two sequences are also different. However, according to our assumption they both generate the same Cantor-lie set C. This is clearly a contradiction. Indeed, denote by I n, I n the intervals corresponding to (r n ) n N, (r n) n N, respectively. Find the smallest n for which r n r n. Let for example r n < r n. It follows that for all I n+1 I n+1 and I n+1 I n+1. In particular, the right endpoint of the interval I0 n+1 belongs to the Cantorlie set generated by (r n ) n N, but it does not belong to the Cantor-lie set generated by (r n) n N. We conclude this proof by another uniqueness result. We show that for any r C there exists a unique zero-one sequence (a n ) n N for which r = ( 1) a n b n. Assume contrary, i.e., there are two such sequences, call them (a n ) n N, (a n) n N. Denote n 1 n 1 m n = ( 1) a i b i, m n = ( 1) a i bi, respectively. Find the smallest n such that m n m n. It follows that these two points are the centers of two different intervals I n, In, respectively. Since these two intervals are disjoint and for all l n m l I n and m l In, we have that the limits lim n m n and lim n m n cannot be equal. This is a contradiction.

12 142 M. DINDOŠ The next case has also been considered in 7]. Let (b n ) n N satisfy the following condition b n b i, for n = 1, 2, 3,.... (19) i=n+1 Proposition 3.2. Let (b n ) n N be a positive sequence satisfying (19) and b n 0. Then the corresponding set of sums S = {r R; r = ( 1) a n b n for some sequence (a n ) n N {0, 1}N } is the interval ] I = b n, b n. Proof. Pic any r I. We define (a n ) n N inductively by n 1 0, if r ( 1) a i b i, a n = 1, otherwise. Corollary 3.3. Assume that (b n ) n N is a positive sequence whose limit is 0 and there is N such that b n i=n+1 b i for any n. Then the corresponding set S = {r R; r = ( 1) a n b n for some sequence (a n ) n N {0, 1}N } is a union of finite number of disjoint closed intervals. Finally, we are ready to formulate our result for all remaining cases. Thans to Corollary 3.3 we can exclude the case when (19) holds for all n and assume that there exists an increasing sequence (n ) N of positive integers for which b n > b i, for = 1, 2, 3,.... (20) i=n +1 Lemma 3.4. Let (b n ) n N be a nonincreasing positive sequence satisfying (20). Then the corresponding set S has infinitely many connected components. In other words, there is an infinite collection of open disjoint intervals L i, i = 1, 2,..., whose intersection with S is empty but whose endpoints are in S.

13 GENERALIZED CANTOR SETS 143 Proof. Consider the sets S n defined by (15). In particular, we want to loo at the set S n +1 for any n for which (20) holds. Denote by M the largest element and by N the second largest element of this set. The sequence (b n ) n N is nonincreasing, hence n n 1 M = b i, N = b n + b i. In particular, we have that M N = 2b n. Define L = N + b i, M i=n +1 i=n +1 Condition (20) guarantees that L is well defined and has nonempty interior. Also clearly L S =. Finally, the endpoints of this interval are obviously in S. Also for different, the sets L are clearly different. Theorem 3.5. Let (b n ) n N be a nonincreasing positive sequence satisfying (20). Then the corresponding set S = {r R; r = b i. ( 1) a n b n for some sequence (a n ) n N {0, 1}N } has infinitely many connected components. The set S is either a generalized Cantor set from Definition 2.3 or it is a closure of a union of infinitely many disjoint closed intervals. In both cases S can be constructed by the process described in Definition 2.3. More precisely, the starting interval I of the construction is ] I = b n, b n, and the sequence (r n ) n N ( 1, 1)N is defined by r n = b n i=n+1 b i n = 1, 2, 3,.... (21) b i i=n Conversely, consider a symmetric interval I = a, a], (a > 0) and a sequence (r n ) n N ( 1, 1)N that satisfies (6), and denote by C the corresponding set generated by this sequence and the interval I. Then there exists a sequence (b n ) n N (not necessarily unique) whose set of sums is C. That

14 144 M. DINDOŠ means: r C r = ( 1) a n b n for some sequence (a n ) n N {0, 1}N. (22) The zero-one sequence (a n ) n N corresponding to r is not necessarily unique. One way to construct the sequence (b n ) n N is to tae ( ) b n = 1 + r n 1 n a b i, n = 1, 2, 3,.... (23) 2 If the sequence (r n ) n N satisfies r n r n 1 r n, n = 1, 2, 3,..., (24) then (b n ) n N generated by (23) is also nonincreasing and (20) holds. Proof. If we examine (21), we see that the assumption (20) guarantees that r n ( 1, 1) and moreover there is an increasing sequence of integers (n ) N such that r n > 0. This means that our sequence (r n ) n N satisfies condition (6). If we reexamine the proof of Theorem 3.1, we see that this proof does not use positivity of the sequence (r n ) n N. Hence, the fact that the middle points of intervals I n, = 0, 1, 2..., 2n 1 1, belong to the set S n defined by (15), remains true. Recall also, that the above proof of C = S neither use r n > 0 nor disjointness of intervals I n, = 0, 1, 2..., 2n 1 1. The only necessary fact was that lim n λ(i n ) = 0. By (6) this holds here. This gives that again C = S which establishes the first part of Theorem 2.3. Conversely, tae now a symmetric interval I = a, a], (a > 0) and a sequence (r n ) n N ( 1, 1)N that satisfies (6). Denote by C the corresponding set constructed in Section 2. Define (b n ) n N by (23). The claim is that the set of sums S of this sequence is equal to C. By (23), induction gives us n 1 0 < b n < a b i, n = 1, 2, 3,.... In particular, this means that n 0 < a b i, n = 1, 2, 3,.... (25) Hence we see that b i a < and therefore lim n b n = 0. Up to this point we did not use (6). For any n = n that satisfies (6) we have

15 (16), i.e. GENERALIZED CANTOR SETS 145 n 0 < a b i < b n. (26) Taing limit on the both sides yields a = b i. (27) Combining (26) and (27) gives that our sequence (b n ) n N satisfies (20). Notice however, this sequence might not be nonincreasing. In fact, it can happen that b n < b n+1 for some n N. This is not a serious problem. The assumption that (b n ) n N is nonincreasing is not necessary since the first part of Theorem 3.5 does not require it. Nevertheless, if we want to have a nonincreasing sequence, we can simply reorder the sequence (b n ) n N. As we now, reordering does not change sum if the series is absolutely convergent. In case, the additional condition (24) holds, the resulting sequence (b n ) n N is nonincreasing. In fact (24) is equivalent to: (1 + r n )(3 + r n+1 ) 2(1 + r n+1 ). If we divide both sides by 4 and multiply by a n 1 b i we get: ( r ) ( n+1 b n = r ) ( ) n 1 n rn a b i ( ) 1 + r n 1 n+1 a b i. 2 Finally, this is equivalent to b n 1 + r n+1 2 ( a ) n b i = b n+1. Hence, the monotonicity of (b n ) n N follows. Notice also that the condition (24) does not impose any restriction on r n+1, provided r n > 0. This is of course consistent with Theorem 3.1 where condition (24) did not appear. From here we can continue exactly as in the proof of Theorem 3.1. Condition (27) gives us (18) and (13). Therefore immediately C = S. Remar 3.6. As it is already indicated in the statement of Theorem 3.5, it is possible that two different decreasing sequences satisfying (20), say (b n ) n N and (b n) n N, yield the same set of sums C. This is due to the fact

16 146 M. DINDOŠ that two different sequences (r n ) n N and (r n) n N can give us the same generalized Cantor set C. A good example of this phenomena are the following two series: ( ( 1) a 2 2n+1 7 n + 1 ) ( 1)a 2n+2 7 n, n=0 n=0 (( 1) a 3n+1 + ( 1) a 3n+2 + ( 1) a 3n+3 ) 1 7 n. (28) The first series has the corresponding sequence (r n ) n N equal to (1/7, 1/3, 1/7, 1/3,... ) while (r n ) n N for the second one is equal to ( 3/7, 1/5, 1/3, 3/7, 1/5, 1/3,... ). Also mentioned in Theorem 3.5 is the fact that for some series the correspondence between 0 1 sequences (a n ) n N and the numbers r = ( 1) a n b n C is no longer bijective. Indeed, two different sequences (0, 1, a 3, a 4,... ) and (1, 0, a 3, a 4,... ) yield the same r for the bottom series in (28). Nevertheless, by Proposition 2.5 the cardinality of C is still c. Our next example explores the case when the generated set S = C is of type (b) from Proposition 2.5, i.e., there is a countable (but infinite) collection of closed disjoint intervals J i for which C = i J i. (29) Example 3.7. There exists a sequence (b n ) n N satisfying (20) whose set S is a closure of infinite number of disjoint closed intervals J i (as in (29)). Indeed, consider the series: ( ( 1) a 9 3n+1 6 n + 6 ( 1)a 3n+2 6 n + 5 ) ( 1)a 3n+3 6 n. (30) n=0 The corresponding interval I is I = 24, 24] and (r n ) n N = ( 1/4, 1/5, 1/9, 1/4, 1/5, 1/9,... ). We will show in two steps that the set S is of type (b) and the collection J i is infinite. The first step is to establish that S contains infinitely many mutually disjoint holes, i.e., open intervals L i not in S with end points in S. In the second step we show that the interval 12, 12] is fully contained in S. Step 1: Let the sets S n be defined as in (15). We loo at S 4. Clearly S 4 = { 20, 10, 8, 2, 2, 8, 10, 20}. (31)

17 Let GENERALIZED CANTOR SETS 147 c n = i=n ( 9 6 i i + 5 ) 6 i = 24 6 n. Since c 1 = 4, by looing at S 4 we can claim that the interval L 1 = (14, 16) is not in S but its end points are. From here we proceed inductively. By looing at S 7, because of c 2 = 2/3, we conclude ( L 2 = , ). 6 Similarly (( n 2 ) ( 20 L n = 6 i + 14 n 2 ) ) 6 n 1, 20 6 i n 1. i=0 i=0 Step 2: Here we show that the interval 12, 12] is contained in S. We proceed by induction. The claim is that for any n = 1, 2,... and any x 12, 12] at least one of the following two statements is true: (a) There is s in S 3n+1 for which s x 2/6 n 1. (b) There are s 1, s 2, both in S 3n+1, such that s 2 = s n 2 and s 1 < x < s 2. By looing at (31) we conclude that the above statement holds for n = 1. Assume now that we have established this claim for some n 1. Our aim is to show it for n + 1. Denote P n+1 = {r R; r = ( 1) a n +( 1)b 6 n +( 1)c for some a, b, c {0, 1}}. 6n Then clearly S 3n+4 = S 3n+1 + P n+1 where the addition A + B of two sets A, B is defined standardly by Also P n+1 = A + B = {a + b; a A and b B}. { 20 6 n, 10 6 n, 8 6 n, 2 6 n, 2 6 n, 8 6 n, 10 6 n, 20 } 6 n. (32) Assume first that we have an x 12, 12] for which (a) is true. This means that s 12/6 n x s + 12/6 n and therefore by looing at s + P n+1 we conclude that (for n + 1) (a) must be true if x belongs to one of the intervals s 12 6 n, s 8 ] 6 n, otherwise (b) holds. s 2 6 n, s + 2 ] 6 n, s n, s + 12 ] 6 n,

18 148 M. DINDOŠ If we have x 12, 12], for which (b) holds for n, then the sum {s 1, s 2 }+ P n+1 contains the set { s 16 6 n, s 10 6 n, s 8 6 n, s 2 6 n, s n, s n, s n, s + 16 } 6 n where s = (s 1 + s 2 )/2, i.e. x s 18/6 n, s + 18/6 n ]. Hence, we conclude that (for n + 1) (a) must be true if x is in one of the intervals s 18 6 n, s 16 ] 6 n s n, s + 10 ] 6 n,, s 10 6 n, s 8 ] 6 n s 16 6 n, s 18 ] 6 n,, s 2 6 n, s + 2 ] 6 n, and (b) is true in all other cases. This finishes the inductive argument. Finally, from (a) and (b) we have that dist(x, S 3n+1 ) 3/6 n 1, for any x 12, 12], which means that x S. 4. Size of the set S There are two natural but unrelated ways to measure the size of a set A R. The first one is given in terms of its Baire category. A set is small if it is of the first Baire category. The second one is to consider the measure or the Hausdorff dimension of a given set A. As before, let λ be Lebesgue measure and dim(a) means the Hausdorff dimension of a set A. The results of the previous two sections give us the following: Proposition 4.1. Let S be the set of sums of a nonincreasing positive sequence (b n ) n N whose sum is finite. Denote ] I = b n, b n. Then exactly one of the following is true: (a) The set S is closed and nowhere dense, i.e., of the first Baire category. Also λ(i \ S) > 0. (b) For any open interval J that intersects S the set J S is of the second Baire category. Also always λ(s) > 0. If (19) holds we have I = S, otherwise λ(i \ S) > 0. There are some natural questions that remain unanswered by Proposition 4.1. We would lie to now whether in case (a) the measure of S can be zero. If the answer is positive, what is dim(s)?

19 GENERALIZED CANTOR SETS 149 Example 4.2. Given any m 0, λ(i)) there is a closed nowhere dense set S for which λ(s) = m. The construction is very simple. We give a sequence (r n ) n N that generates a Cantor-lie set of such measure. Then Theorem 3.1 gives the corresponding sequence (b n ) n N. Consider r n = (λ(i) m)/2 n m + (λ(i) m)/2 n 1. Indeed, the measure of the set F n (defined in Section 2) is λ(f n ) = m + λ(i) m 2 n 1. Consequently, the set C = n F n has measure m. See also 8] for a different proof of this result. For the sae of the completeness, let us give here a similar result from 6] about the Hausdorff dimension of the set S. Example 4.3. Given any d 0, 1] there is a closed nowhere dense set S for which dim(s) = d. Again we define a sequence (r n ) n N that generates a Cantor-lie set C of a given dimension. Consider first any d (0, 1). We tae (r n ) n N = (r, r, r,... ) for some r (0, 1). Then the set C generated by this sequence has a property of self similarity in the following sense. There exists two real numbers a, b R such that ( C = a + 1 r ) ( 2 C b + 1 r ) 2 C = C 1 C 2. (33) Here the sets on the right hand side are disjoint. The multiplication of a set A by a number is defined standardly by A = {a; a A}. This has the following consequence: If m d denotes the d-dimensional Hausdorff measure then necessarily m d (A) = d m d (A). From (33) follows that ( ) 1 r d m d (C) = m d (C 1 ) + m d (C 2 ) = 2 m d (C). 2 Clearly, if 2 (1 r)/2] d 1 then m d (C) must be either 0 or. Thus, the Hausdorff dimension of the set C must be exactly the point where ( ) 1 r d 2 = 1. 2 Given any d (0, 1) we see that r should be taen equal to r = /d (0, 1). If d = 1 the previous example gives a set for which λ(c) > 0, i.e., dim(c) = 1. Finally, if d = 0 any increasing sequence (r n ) n N with limit one generates a set C of Hausdorff dimension zero. This follows from a slight modification of the argument we gave above (see 6] for more details).

20 150 M. DINDOŠ References 1] Dindoš, M., On a typical series with alternating signs, Real Anal. Exchange 25(2) (1999/2000), ] Dindoš, M., On sign distribution in relatively convergent series, Acta Math. Hungar. 91(4) (2001), (to appear). 3] Menon, P. K., On a class of a perfect set, Bull. Amer. Math. Soc. 54 (1948), ] Šalát, T., Absolutely convergent series and dyadic expansions (in Slova), Math.- Phys. J. Slova Acad. Sci. 9(1) (1959), ] Šalát, T., A note on absolutely convergent series (in Slova), Math.-Phys. J. Slova Acad. Sci. 7(3) (1957), ] Šalát, T., Absolut onvergente Reihen und das Hausdorffsche Mass, Czechoslova Math. J. 84(9) (1959), ] Šalát, T., On sums of certain convergent series (in Slova), Math.-Phys. J. Slova Acad. Sci. 4(4) (1954), ] Šalát, T., On certain properties of positive series (in Slova), Acta Univ. Comenian. Fac. Rerum Natur. 2 (1957), Martin Dindoš University of North Carolina Phillips Hall CB 3250 Chapel Hill NC USA