Conditions for the difference set of a central Cantor set to be a Cantorval

Let $C(\lambda )\subset \lbrack 0,1]$ denote the central Cantor set generated by a sequence $ \lambda = \left( \lambda_{n} \right) \in \left( 0,\frac{1}{2} \right) ^{\mathbb{N}}$. By the known trichotomy, the difference set $ C(\lambda )-C(\lambda )$ of $C(\lambda )$ is one of three possible sets: a finite union of closed intervals, a Cantor set, and a Cantorval. Our main result describes effective conditions for $(\lambda_{n})$ which guarantee that $C(\lambda )-C(\lambda )$ is a Cantorval. We show that these conditions can be expressed in several equivalent forms. Under additional assumptions, the measure of the Cantorval $C(\lambda )-C(\lambda )$ is established. We give an application of the proved theorems for the achievement sets of some fast convergent series.


Introduction
By a Cantor set we mean a compact, perfect and nowhere dense subset of the real line. Cantor sets appear in several publications in many different settings, cf. [26] and [29]. They occur in mathematical models involving fractals, iterated functional systems and fractional measures [12]. They play a role in number theory (e.g. in b-ary number representations, and in connection with continued fractions [16]). They are rooted in dynamical systems in the study of homoclinic bifurcations [25], in signal processes and ergodic theory, and also in limit theorems from probability, as it is stated in [26]. Finally, let us mention the applications in spectral theory [11], [30]. Important results describe geometrical properties of Cantor sets via Hausdorff and packing measures and the respective fractal dimensions, and multifractal spectrum [3], [10], [15], [17]. Also, several authors conducted extensive studies on the arithmetic sums [1], [22], [31] and products [29] of two Cantor sets, and their intersections [18], [20].
In our paper we consider central Cantor sets C (λ), which we define using a sequence λ = (λ n ) ∈ 0, 1 2 N . These are Cantor sets such that the ratio of measures of the intervals defined in subsequent steps of the construction is equal to λ n (for the Cantor ternary set, λ n ≡ 1 3 ). We are interested in properties of the difference set C(λ)− C(λ). Several authors examined sums and differences of Cantor sets (see [2], [20] [21], [27], [29], [28]). Considerations about sums and differences of central Cantor sets were often based on the relationship between central Cantor sets and the achievement sets (i.e. sets of all subsums) of convergent series (see Proposition 3.1).
The study of topological properties of the sets of subsums of series has a long history. The first papers on this topic were written over a hundred years ago. In 1914 Kakeya [19] proved that, if x n > ∞ j=n+1 x j for any n, then the set E (x) of all subsums of a series ∞ j=1 x j of positive numbers is a Cantor set. He also showed that E (x) is a finite union of closed intervals if and only if x n ≤ ∞ j=n+1 x j for almost all n's. Kakeya conjectured that the set E (x) is either a Cantor set or a 2 N . If λ n+1 < 1 3 , then the set J s \ (J sˆ0 ∪ J sˆ1 ∪ J sˆ2 ) is a union of two open intervals. We denote them by G 0 s and G 1 s , and call the left and the right gap in J s . If λ n+1 ≥ 1 3 , then J sˆ0 ∩ J sˆ1 = ∅ and J sˆ1 ∩ J sˆ2 = ∅. We denote these intervals by Z 0 s and Z 1 s , and we call them the left and the right overlap in J s . We also assume that d 0 := |I| = 1 and that (1) |J s | = 2d n .
A set is said to be regular closed if it is equal to the closure of its interior. A non-empty regular closed bounded subset C of R is called a Cantorval, if for every connected component I of C with nonempty interior, the both endpoints of I are accumulation points of the union of one-point components of C (see [5], [9]).

Difference sets of central Cantor sets
One of the first results concerning the difference sets of central Cantor sets was proved by Kraft in [21]. He showed that, if a sequence λ = (λ n ) ∈ 0, 1 2 N is constant, that is, λ n = α for any n ∈ N,  (1) a finite union of closed intervals; (2) a Cantor set; (3) a Cantorval.
We will begin our discussion on the difference sets of central Cantor sets by proving that C (λ) − C (λ) is a finite union of intervals if and only if λ n ≥ 1 3 for almost all n's. Similar results, with proofs based on other tools, were shown by Anisca and Ilie (see [2,Theorem 3]).
The following statements hold.
(2) C (λ) − C (λ) is a finite union of intervals if and only if the set n ∈ N : λ n < 1 3 is finite. Proof. We will only prove (2) (the proof of (1) is similar).
there is s ∈ {0, 1, 2} n such that x ∈ J s . The sequence s is not an extension of 0 (n−1) because x ∈ G 0 0 (n−1) . Hence s k > 0 for some k < n, and consequently Our goal is to find sufficient conditions for the set C (λ) − C (λ) to be a Cantorval. From Theorem 2.2 it follows that we need to assume that λ n < 1 3 for infinitely many terms. From Theorem 2.1 we infer that, for a sequence λ satisfying the above condition, it suffices to prove that C (λ) − C (λ) has nonempty interior. To show this, we will take an interval J t and define the family of gaps G t such that J t \ G t ⊂ C (λ) − C (λ) and int (J t \ G t ) = ∅. The proof of the second condition requires the assumption that λ n ≥ 1 3 for infinitely many terms (Theorem 2.4). To prove the first condition, we will need far stronger assumptions (Theorem 2.5). We now introduce additional notation and definitions used in the next theorems. We consider sequences such that λ n < 1 3 for infinitely many terms. For a fixed sequence λ (and a fixed number k 0 ) we need the sequence (k n ) consisting of all indices greater than k 0 , for which λ kn < 1 3 , and the families G t (n) of some particular gaps in intervals J s for s ∈ {0, 1, 2} kn−1 , t ≺ s.
Let λ = (λ j ) j∈N ∈ 0, 1 2 N be such that λ j < 1 3 for infinitely many terms and λ k 0 +1 > 1 3 for some k 0 ∈ N ∪ {0}. By (k n ) n∈N we denote an increasing sequence such that (λ kn ) n∈N is a subsequence of the sequence (λ j ) j>k 0 consisting of all terms which are less than 1 3 .
for n ≥ m. Then put G t (m) := G t (m). Assume that we have defined families G t (l) for m ≤ l ≤ n. Then define G t (n + 1) as Geometrically, the family G t (n) consists of the leftmost gap and the rightmost gap in the interval J t which appear in C kn (λ) − C kn (λ). The family G t (n) consists of the gaps from G t (n) and the gaps that appear in C kn (λ) − C kn (λ) which are nearest the gaps from n−1 l=m G t (l).
Moreover, let G t := n≥m G t (n) and N is a sequence such that: λ n < 1 3 for infinitely many terms, λ n ≥ 1 3 for infinitely many terms, and there is The following statements hold.
and for any n ≥ m, we have (2) If n ≥ m and G ∈ G t (n), then Proof. Ad (1). From Proposition 1.1(4) we have and therefore r G 0 Ad (2). We prove it inductively. Observe that p (m) = tˆ0 (km−k−1) , q (m) = tˆ2 (km−k−1) , and . Thus, condition (2) holds for n = m. Assume that n ≥ m and for any . Let us consider the cases.
By the inductive hypothesis and (1) In the first case, we have and in the second case, . The proof is analogous to the proof in the previous case.
We will prove inductively that 1}. Let us consider the cases.
Assume now that n ≥ M and H (l) ⊂ G w (l) for l ∈ {M, . . . , n}. Let G ∈ H (n + 1), that is, 1}. We will prove that G ∈ G w (n + 1). Let us consider the cases.
wˆ0 (k n+1 −r−1) ∈ G w (n + 1). If k l > r, then w ≺ v, and therefore G i v ∈ H (l). From the inductive hypothesis it follows that G i v ∈ G w (l), and . Similarly as in the previous case, we show that G ∈ G w (n + 1), which finishes the proof of (3).
1 o There is l < n such that s j = v j for j = 1, . . . , k l − 1 and s k l = v k l .
Writing p := s| (k l − 1), we get G i s ⊂ J s ⊂ J s|k l = J pˆs k l and J v ⊂ J v|k l = J pˆv k l . Since λ k l < 1 3 , the intervals J pˆs k l and J pˆv k l are disjoint. Hence G i s ∩ J v = ∅. 2 o There is r ≤ k n − 1 such that s j = v j for j = 1, . . . , r − 1 and |s r − v r | = 2.
Taking p := s| (r − 1), we obtain G i s ⊂ J s ⊂ J s|r = J pˆsr and J v ⊂ J v|r = J pˆvr . Since λ r < 1 2 , the intervals J pˆsr and J pˆvr are disjoint. Therefore, G i s ∩ J v = ∅. 3 o There are l, r such that l ≤ n, k l−1 < r < k l , s j = v j for j = 1, . . . , r − 1 and v r = s r + 1.
Put p := s| (r − 1) and and therefore r G i s ≤ r G 0 p . On the other hand, from (1) it follows that Hence r G i s < l (J v ), and so G i s ∩ J v = ∅. 4 o There are l, r such that l ≤ n, k l−1 < r < k l , s j = v j for j = 1, . . . , r − 1 and v r = s r − 1.
The reasoning is analogous to that in the previous case. This finishes the proof of the first condition in (4).
Assume now that G j u ∈ G t , where u ∈ {0, 1, 2} l , j ∈ {0, 1}, and (u, j) = (s, i). If s = u and i = j, then the condition G i s ∩ G j u = ∅ is obvious. Suppose that s = u. Without loss of generality we can assume that n < l.
The next theorem yields information about relationships between the sets C (λ)−C (λ) and G t . It states that, under the same assumptions as in Lemma 2.3, int (J t \ G t ) = ∅. Moreover, this theorem shows that for t = ∅ the inclusion C (λ) − C (λ) ⊂ [−1, 1] \ G t holds, and it gives the formula for a measure of the set G t . Using these conditions, in Theorem 2.5 we will prove (under additional assumptions) that int (C (λ) − C (λ)) = ∅ and, in consequence, C (λ) − C (λ) is a Cantorval. We will also get the formula for the measure of the set C (λ) − C (λ).
N is a sequence such that: λ n < 1 3 for infinitely many terms, λ n ≥ 1 3 for infinitely many terms, and there exists where the sequence (k n ) consists of all indices greater than k 0 , for which λ kn < 1 3 . Let k ≥ k 0 , t ∈ {0, 1, 2} k and m ∈ N be such that k m−1 ≤ k < k m . The following statements hold.
(1) The set J t \ G t has nonempty interior.
. From Lemma 2.3 it follows that J ⊂ J t , |J| > 0, and for any interval G ∈ G t we have r (G) ≤ lim n→∞ r G 0 p(n) = l (J) or l (G) ≥ lim n→∞ l G 1 q(n) = r (J). Hence G ∩ J = ∅ and, in consequence, J ∩ G t = ∅, which completes the proof of (1).
Write G := G ∅ . On the contrary, suppose that the set (C (λ) − C (λ)) ∩ G is nonempty, that is Hence we obtain G i s ∩ (C kn (λ) − C kn (λ)) = ∅, and consequently there is a sequence v ∈ {0, 1, 2} kn such that Let a n denote the number of elements of the set G (n). From Lemma 2.3 it follows that G i s = G j u for any G i s , G j u ∈ G such that (s, i) = (u, j). Hence a 1 = 2 and a n+1 = 2 + 2 n i=1 a i . It is easy to check that the sequence a n = 2 · 3 n−1 is a solution of this recurrence equation. Since the intervals in G are pairwise disjoint and the length of each interval from G (n) is equal to d kn−1 − 3d kn , we have The next theorem is the main result of our paper. Theorem 2.5. Assume that λ = (λ j ) j∈N ∈ 0, 1 2 N is a sequence such that: λ n < 1 3 for infintely many terms, λ n ≥ 1 3 for infintely many terms, and there is k 0 ∈ N ∪ {0} such that λ k 0 +1 > 1 3 . Let k ≥ k 0 , t ∈ {0, 1, 2} k and m ∈ N be such that k m−1 ≤ k < k m , where the sequence (k n ) consists of all indices greater than k 0 , for which λ kn < 1 3 . Moreover, assume that there exists a sequence (δ n ) n∈N such that for any n ∈ N, the following conditions hold Then we have: (2) The set C (λ) − C (λ) is a Cantorval.
(3) If k = k 0 = 0 and t = ∅, then C (λ) − C (λ) = J t \ G t and Proof. If n ∈ N, s ∈ {0, 1, 2} kn−1 , i ∈ {0, 1}, u ∈ {0, 1, 2} kn , and c G i s = c (J u ), then we write G i s ⊂ * J u . Of course, G i s ⊂ * J u implies G i s ⊂ J u . From Proposition 1.1 and (2.1) it follows that Thus, the condition G i s ⊂ * J u is equivalent to the condition We first prove (1). The basic idea of the proof is to show (under additional assumptions) that any gap G i s , where s ∈ {0, 1, 2} kn−1 , t ≺ s and i ∈ {0, 1}, is covered by an interval J u , where u ∈ {0, 1, 2} kn and t ≺ u, that is, the following condition holds The proof is divided into a few steps. Let m ∈ N be such that k m−1 ≤ k < k m . Fix n ≥ m.
Let us write N := N (s, i) and consider two cases.
Define a sequence u ∈ {0, 1, 2} kn in the following way: Since k < N , we have t ≺ s| (N − 1) ≺ u. Moreover, G 0 s ⊂ * J u because from (2.1) and Proposition 1.1 it follows that Then t ≺ u and G 1 s ⊂ * J u because This finishes the proof of Claim 1.
Let us consider two cases.
Define the sequence u ∈ {0, 1, 2} kn in the following way: Since k < k m ≤ k l = N , we have that t ≺ v| (N − 1) ≺ u. Moreover, G 0 s ⊂ * J u because from (2.1) and Proposition 1.1 it follows that By assumption, G s N p ⊂ * J v . Let us define the sequence u ∈ {0, 1, 2} kn in the following way: Similarly as in case 1 o , we get t ≺ u and r (J u ) − r G 1 s = δ n+1 , so G 1 s ⊂ * J u . This completes the proof of Claim 2.
Hence x ∈ J sˆ0 ∪ J sˆ1 ∪ J sˆ2 or x ∈ G 0 s ∪ G 1 s . In the first case, condition (2.3) is satisfied for u = sˆj. Let us assume that x ∈ G 0 s ∪ G 1 s . We may assume that x ∈ G 0 s (if x ∈ G 1 s , the proof is analogous). Since x / ∈ G t , G 0 s / ∈ G t (m). Using Claim 3 we deduce that there is a sequence u ∈ {0, 1, 2} kn such that t ≺ u and G i s ⊂ * J u . This completes the proof of (2.3) for n = m. Assume now that n ≥ m and condition (2.3) holds for n, i.e. there is a sequence v ∈ {0, 1, 2} kn such that t ≺ v and x ∈ J v . Since λ kn+1 , . . . , λ k n+1 −1 ≥ 1 3 , Proposition 1.1 (8) shows that there is a sequence s ∈ {0, 1, 2} k n+1 −1 such that t ≺ s and x ∈ J s . From Claim 3 and (α n+1 ) we infer the existence of a sequence u ∈ {0, 1, 2} k n+1 such that t ≺ u and x ∈ J u , which finishes the proof of (2.3) for n + 1.
Remark 1. If a sequence (δ n ) satisfies condition (2.1), then The next proposition shows that the assumptions of Theorem 2.5 can be written as a system of infinitely many equations. In the next section we will use such a version of condition (2.1) to examine achievement sets of some fast convergent series.
3 . Then r = k n and r + 1 = k n+1 for some n ∈ N.
We now show that the second equality holds. If k n+1 − k n = 1, then λ kn , λ kn+1 < 1 3 , and consequently 3d kn+1 = 4d kn − d kn−1 . Hence On the other hand, if k n+1 − k n > 1, then λ kn < 1 3 and λ kn+1 ≥ 1 3 , and consequently 6d kn+1 = 7d kn − d kn−1 . Using the first equality, we get which finishes the proof of the second equality in (2.1). The third equality results from the second one because It may seem unnatural to describe covering the gaps via equalities as in Theorem 2.5. Our initial idea was to describe it by inequalities. However, it turned out that the system of received inequalities is equivalent to condition (2.1). It suggests that sequences satisfying the assumptions of Theorem 2.5 are quite exceptional. In the next section we will argue why the family of such sequences is interesting.

The sets os subsums of series
We will use the results obtained in the previous section to examine the sets of subsums of series.
Let x = (x j ) j∈N be a nonincreasing sequence of positive numbers such that the series ∞ j=1 x j is convergent. The set x j is called the achievement set of x. The sum and the remainders of a series we denote by S := ∞ j=1 x j and r n := ∞ j=n+1 x j . Obviously, r 0 = S. If x n > r n for n ∈ N, then the series is called fast convergent.
We will examine the difference set E (x) − E (x) for some convergent series with positive terms.
We first check which series correspond to sequences satisfying the assumptions of Theorem 2.5. It turns out that they have a very simple form.