A family of Cantorvals

John Ferdinands; Timothy Ferdinands

doi:10.1515/math-2019-0109

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A family of Cantorvals

John Ferdinands and Timothy Ferdinands

Published/Copyright: December 10, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

The set of subsums of the series ∑n=1∞ x_n is known to be one of three types: a finite union of intervals, homeomorphic to the Cantor set, or of the type known as a Cantorval. Bartoszewicz, Filipczak and Szymonik have described a family of series which contained all known examples of subsum sets which are Cantorvals. We construct another family of series which produces new examples of subsum sets which are Cantorvals.

Keywords: subsum sets; Cantor set; Cantorval; multigeometric series

MSC 2010: 40A05

1 Introduction

1.1 Notation

We consider an infinite sequence {x} = {x₂, x, x₃, …} of positive terms.

It is well known that rearranging the terms of an absolutely convergent series does not change the sum of the series. We use ẋ = ∑n=1∞ x_n to denote the series arising from {x}. Throughout this paper such a series ẋ will be assumed to be a convergent series of positive terms.

Definition 1.1

Let ẋ be a convergent series of positive terms.

A subsum of ẋ is a number x ∈ ℝ such that x = ∑n=1∞ c_nx_n where c_n ∈ {0, 1} for all n ≥ 1.
The set E(ẋ) ⊂ ℝ is the set of all subsums of ẋ.
X_n = ∑i=n+1∞ x_i is the n-th tail of ẋ.

We now define the well-known Cantor set 𝓒.

Definition 1.2

We recursively define the following subsets of the interval [0, 1]:

C1=13,23;
Cn=Cn−13∪23+Cn−13 for n ≥ 2.

The ternary Cantor set is 𝓒 = [0, 1] ∖ ( ⋃n=1∞ C_n).

Remark 1

Each C_n is a union of disjoint 2^n–1 disjoint open intervals, each of length 13n .

1.2 Kakeya’s results

The following theorem is found in [1], and refers to results from [2, 3, 4, 5].

Theorem 1.1

For any convergent series of positive terms ẋ, E(ẋ) is exactly one of the following:

a finite union of closed and bounded intervals.
homeomorphic to the Cantor set 𝓒.
homeomorphic to the set 𝓒 ⋃ ( ⋃n=1∞ C_2n–1).

Definition 1.3

A Cantorval is a subset of ℝ that is homeomorphic to 𝓒 ⋃ ( ⋃n=1∞ C_2n–1).

Note that the term Cantorval is used in a more general sense. (See [6], for instance). Our definition above applies to what is often known as an M-Cantorval.

Remark 2

When discussing Cantorvals, the following equality of sets may be useful:

C⋃(⋃n=1∞C2n−1)=[0,1]∖(⋃n=1∞C2n).

The possibilities (i) and (ii) in Theorem 1.1 were first stated by Kakeya in [2] as early as 1914. Kakeya’s results are stated below.

Theorem 1.2

(Kakeya’s Results).

E(ẋ) is a finite union of closed and bounded intervals if x_n ≤ X_n for all but finitely many n.
Furthermore, if {x} is a non-increasing sequence and E(ẋ) is a finite union of closed and bounded intervals, then x_n ≤ X_n for all but finitely many n.
E(ẋ) is homeomorphic to 𝓒 if x_n > X_n for all but finitely many n.

From Theorem 1.1 and Kakeya’s Results, we can deduce the following corollary.

Corollary 1.2.1

If E(ẋ) is a Cantorval, then x_n ≤ X_n for infinitely many n and x_n > X_n for infinitely many n.

It should be noted that these conditions do not guarantee that E(ẋ) is a Cantorval. For instance, let ẋ be the series such that x_2n–1 = 1011n and x_2n = 111n for n ≥ 1. It is the case that x_n > X_n for all odd n, and x_n ≤ X_n for all even n, but yet the set E(ẋ) is homeomorphic to the Cantor set 𝓒. This follows from a result in a paper by Z. Nitecki. (See Remark 16 in [7].)

For a convergent series of positive terms, conditions which guarantee that its subsum set is a Cantorval are not known. Bartoszewicz, Filipczak and Szymonik in [1] describe families of series which contain all known examples of series for which the set of subsums is a Cantorval. In particular, they consider multigeometric series, and they construct a family of such series whose subsum sets are Cantorvals. In this paper we extend their result by constructing a different family of multigeometric series whose subsum sets are new examples of Cantorvals.

In Section 2 of this paper we generalize a result from [1] by replacing a hypothesis in which the subsum set of a multigeometric series contains a set of consecutive integers by one in which it contains an arithmetic progression. In Section 3 we prove a result which describes a family of series satisfying the latter hypothesis, and whose subsum sets are Cantorvals. Finally in Section 4 we describe a very simple algorithm for generating infinite families of series whose subsum sets are Cantorvals, and we use it to construct two examples.

A referee of the first draft of this paper pointed the authors to the paper by Banakh, Bartoszewicz, Filipczak and Szymonik [8] which gives much more general sufficient conditions for the subsum set of a multigeometric series to be a Cantorval. In Section 4 we will state some of these conditions and apply them to our two examples. Although [8] does give more general conditions than this paper, these conditions do not completely overlap our results. Furthermore our algorithm for producing Cantorvals is new.

2 The main result in [1] and a generalization

Let k₁, k₂, … k_m and q be constants with 0 < q < 1. Then the sequence

(k1,k2,…km,k1q,k2q,…kmq,k1q2,k2q2,…kmq2,…)

is called a multigeometric sequence, and is denoted by (k₁, k₂, … k_m; q), and its set of subsums by E(k₁, k₂, …, k_m; q). (See [1]). Here is the main result by Bartoszewicz, Filipczak and Szymonik in [1].

Theorem 2.1

Let k₁ ≥ k₂ ≥ ⋯ k_m be positive integers and let ∑i=1m k_i.

Suppose that the set ∑i=1mciki:ci=0 or ci=1 contains the numbers n₀, n₀ + 1, n₀ + 2, … n₀ + n for some positive integers n₀ and n. Then the following are true.

If q ≥ 1n+1 , then E(k₁, k₂, …, k_m; q) contains an interval.
If q < kmK+km , then E(k₁, k₂, …, k_m; q) is not a finite union of intervals.

It follows that if 1n+1 ≤ q < kmK+km , then E(k₁, k₂, …, k_m; q) is a Cantorval. The following theorem generalizes this result.

Theorem 2.2

Let k₁ ≥ k₂ ≥ ⋯ k_m be positive integers, and let K = ∑i=1∞ k_i. Suppose that the set ∑i=1mciki:ci=0 or ci=1 contains the numbers a, a + d, a + 2d, …, a + nd for some positive integers a, d and n. Then the following are true:

If q ≥ 1n+1 , then E(k₁, k₂, …, k_m; q) contains an interval.
If q < kmK+km , then E(k₁, k₂, … k_m; q) is not a finite union of intervals.
If 1n+1 ≤ q < kmK+km , then E(k₁, k₂, … k_m; q) is a Cantorval.

Our proofs are very similar to the proofs of the result by Bartoszewicz, Filipczak and Szymonik in [1].

Proof of (i)

Consider the multigeometric sequence (d, d, …, d; q) with d repeated n times. Let x_r be the r-th term, and let X_r = ∑i=r+1∞ x_i. We show that x_r ≤ X_r for all r.

For any r which is not a multiple of n, x_r = x_r+1, and hence x_r ≤ X_r. Suppose that r = kn for some positive integer k. Then x_kn = dq^k–1, and X_kn = ∑i=0∞ ndq^k+i = ndqk1−q . Hence x_kn ≤ X_kn if and only if dq^k–1 ≤ ndqk1−q , if and only if 1n+1 ≤ q, which we have assumed to be true. Therefore x_r ≤ X_r for all r. It follows from (i) of Kakeya’s results that E(d, d, …, d; q) is a finite union of intervals.

Next we show that ∑n=0∞ aqⁿ + E(d, d, …, d; q) is contained in E(k₁, k₂, …, k_m; q).

Let x ∈ ∑n=0∞ aqⁿ + E(d, d, …, d; q). Then x = (a + aq + aq² + …) + d(p₀ + p₁q + p₂q² +…) for some p_i ∈ {0, 1, 2, …, n}, that is, x = (a + p₀d) + (a + p₁d)q + (a + p₂d)q² +⋯. By hypothesis, each (a + p_jd) has the form ∑i=1m c_ik_i where c_i = 0 or c_i = 1. Therefore we have x ∈ E(k₁, k₂, … k_m; q).

We have shown that E(d, d, …, d; q) is a finite union of intervals. Since ∑n=0∞ aqⁿ + E(d, d, …,; q) is a translation of E(d, d, … d; q), it is a finite union of intervals. Therefore E(k₁, k₂, … k_m; q) contains a finite union of intervals, thus proving (i).□

Proof of (ii)

Now suppose that q < kmK+km . We will show that the sequence (k₁, k₂, … k_m; q) is non-increasing and that x_sm > X_sm for all positive integers m.

Recall that k₁ ≥ k₂ ≥ ⋯ k_m. Hence, to show that the sequence is non-increasing, it is sufficient to show that k_m ≥ k₁q. This is true if and only if q ≤ kmk1 . But q <kmK+km=km(k1+k2+⋯km)+km<kmk1. Therefore the sequence is non-increasing.

Observe that x_sm = k_mq^s–1 and that X_sm = k_mq^s–1 and that X_sm = Kqs1−q . So x_sm > X_sm if and only if k_mq^s–1 > Kqs1−q if and only if q < kmK+km , which we have supposed to be true. From (ii) of Kakeya’s results we conclude that E(k₁, k₂, …, k_m; q) is not a finite union of intervals, thus proving (ii).□

Proof of (iii)

Now suppose that 1n+1 ≤ q < kmK+km Then as previously shown, E(k₁, k₂, …, k_m) contains an interval but is not a finite union of intervals. By Theorem 1.1, E(k₁, k₂, …, k_m; q) is a Cantorval.□

3 A family of Cantorvals

The statement of Theorem 2 describes series whose subsum sets are Cantorvals, but it does not provide simple examples of such series. The next theorem describes such a family of series.

Theorem 3.1

Let (a + 2nd, a + (2n – 2)d, … a + 2d, a, d; q) be a multigeometric sequence with 2nd < a < (2n + 2)d and n ≥ 4. If 12n+2≤q<minda,a−d(n+2)a+(n2+n)d, then E(a + 2nd, a+(2n–2)d, …, a + 2d, a, d; q) is a Cantorval.

The proof of the theorem is contained in the following three lemmas.

Lemma 3.2

If 12n+2 ≤ q, then E(a + 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) contains a finite union of intervals.

Proof

Observe that for the series (a + 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) the set

S=∑i=1mciki:ci=0 or ci=1

contains the arithmetic progression (a, a + d, a + 2d, …, a + 2nd, a + (2n + 1)d). It follows from Theorem 2.2 that if 12n+2 ≤ q, then E(a + 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) contains a finite union of intervals.□

Next we want to show that E(a+ 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) is not equal to a finite union of intervals. We will do so by using (ii) of Kakeya’s results, which implies that if the sequence is non-increasing and x_n > X_n for infinitely many n, E(a + 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) is not a finite union of intervals.

But the terms of the sequence (a+ 2nd, a + (2n – 2)d, …, a + 2d, a_d, d; q) may not be non-increasing. Therefore, in order to apply Kakeya’s result, we must first rearrange the terms so that they are non-increasing.

Remark 3

It is important to note that a convergent series ∑n=1∞ x_n of positive terms and any rearrangement of it will have the same subsum sets. To see this, observe that if x = ∑n=1∞ c_nx_n is a subsum of ∑n=1∞ x_n, then by rearranging the series we will get a subsum of the rearrangement, and since we have absolute convergence, the rearranged sum is also equal to x. By the same reasoning, a subsum of the rearranged series will be a subsum of ∑n=1∞ x_n. Hence any conclusions about the subsum set of the rearranged series will be true for the original series as well.

Lemma 3.3

If we rearrange the terms of the sequence so that dq^n–1 comes between (a + 2d)qⁿ and aqⁿ, then the sequence is non-increasing.

Proof

Consider the first few terms of the sequence:

a+2nd,a+(2n−2)d,…,a+2d,a,d,(a+2nd)q,…,(a+2d)q,aq,…

We move d so that it is between (a + 2d)q and aq, and in general we move dq^n–1 between (a + 2d)qⁿ and aqⁿ. The first few terms of the rearranged sequence become:

a+2nd,a+(2n−2)d,…,a+2d,a,(a+2nd)q,(a+(2n−2)d)q,…,(a+2d)q,d,aq,(a+2nd)q2,(a+(2n−2)d)q2,…,(a+2d)q2,dq,aq2,…

To show that this rearrangement is non-increasing, it is sufficient to show that:

a ≥ (a + 2nd)q;
(a + 2d)q ≥ d;
d ≥ aq.

We first prove 1. By one of the hypotheses of Theorem 3.1 we have that

q≤a−d(n+2)a+(n2+n)d.

Hence we see that

(a+2nd)q≤(a+2nd)a−d(n+2)a+(n2+n)d.

By another of the hypotheses of Theorem 3.1, 2nd < a, and hence a + 2nd < 2a. Also a – d < a.

Therefore

(a+2nd)q≤2aa(n+2)a+(n2+n)d. (1)

By still another hypothesis of Theorem 3.1, n ≥ 4. It follows that

(n+2)a+(n2+n)d≥6a+20d>2a.

Combining this with (1) gives

(a+2nd)a≤2aa2a=a

thus proving 1.

To prove 2 we see that by the hypotheses of Theorem 3.1, a > 2nd and q ≥ 12n+2 , and so

(a+2d)q>2nd+2d2n+2=d.

To prove 3 we see that by yet another of the hypotheses of Theorem 3.1, q < da , which implies that d > aq.□

Lemma 3.4

The set of subsums of the rearrangement described in Lemma 3.3 is not a finite union of intervals.

Proof

We will show that in the rearranged series there are infinitely many terms which are strictly greater than their tails. First we show that the term a in the rearranged series is strictly greater than its tail.

Let

K=(a+2nd)+(a+(2n−2)d)+⋯+(a+2d)+a+d.

The tail of a is d + ∑n=1∞ Kqⁿ = d + Kq1−q . Hence a is strictly greater than its tail if and only if a > d + Kq1−q , if and only if q < a−da−d+K . Now oberve that

K=(n+1)a+d+2d(1+2+⋯+n)=(n+1)a+d+2dn(n+1)2=(n+1)a+(n2+n+1)d.

Substituting this value for K in the inequality for q, we get that a is strictly greater than its tail if and only if q<a−d(n+2)a+(n2+n)d. But this is true by one of the hypotheses of Theorem 3.1.

For every positive integer n, the tail of aqⁿ is given by

dqn+∑i=1∞Kqn+i=dqn+Kqn+11−q.

We have shown above that a > d + Kq1−q . It follows that aqⁿ > dqⁿ + Kqn+11−q for every positive integer n. Therefore there are infinitely many terms which are strictly greater than their tails. It follows from (ii) of Kakeya’s results that the subsum set of the rearranged series is not a finite union of intervals.□

Remark 4

In Remark 3, we showed that rearranging the terms of a series with positive terms does not change its set of subsums. It follows from Lemma 3.4 that E(a+ 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) is not equal to a finite union of intervals.

We can now give the proof of Theorem 3.1.

Proof of Theorem 3.1

By Lemma 3.2, E(a+ 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) contains a finite union of intervals. By Remark 4, E(a+ 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) is not equal to a finite union of intervals. Hence by Theorem 1.1, E(a+ 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) is a Cantorval.□

4 Two examples

We consider the question of how to construct a family of multigeometric series which satisfies the somewhat complicated hypotheses of Theorem 3.1. We need values of d, a and n which satisfy the following conditions:

2nd < a < (2n + 2)d;
n ≥ 4;
12n+2<minda,a−d(n+2)a+(n2+n)d.

Proposition 1

If (1) and (2) are satisfied, then (3) is also satisfied.

Proof

Suppose that (1) and (2) are true. From (1) we see that a < (2n + 2)d implies that 12n+2 < da . By a straightforward calculation we see that (3) is true if and only n2+3n+2n<ad. Since n ≥ 4 and 2nd < a, we see that

n2+3n+2n=n+3+2n<n+3+1=n+4≤2n<ad.

□

Example 4.1

(17, 15, 13, 11, 9, 1; q) is a Cantorval if 110<q<437

Proof

Let n = 4. Then we need 8 < ad < 10. We could choose a = 9 and d = 1, which gives us the sequence (17, 15, 13, 11, 9, 1; q). Then

a−d(n+2)a+(n2+n)d=9−1(4+2)9+(42+4)1=437.

By Theorem 3.1, E(17, 15, 13, 11, 9, 1; q) is a Cantorval if 110<q<min19,437, or 110<q<437 .□

Example 4.2

E(41, 37, 33, 29, 25, 21, 2; q) is a Cantorval if 112<q<19207 .

Proof

Suppose that n = 5, so that 10 < ad < 12. If we choose d = 2 and a = 21, we then get the sequence (41, 37, 29, 25, 21, 2; q). Using Theorem 3.1 we see that E(41, 37, 33, 29, 25, 21, 2; q) is a Cantorval if 112<q<min221,19207, or 112<q<19207 .□

It should be noted that the sequence (17, 15, 13, 11, 9, 1) satisfies the hypotheses of the Bartoszewicz, Filipczak and Szymonik result, since the set ∑i=1mciki:ci=0 or ci=1 contains the numbers 9, 10, 11, …, 17, 18, but the result cannot be used to show that E(17, 15, 13, 11, 9, 1; q) is a Cantorval. With their notation, n = 10, k_m = 1 and K = 66, so that the interval 1n+1,kmK+km=111,167 is empty.

As promised in Section 1, we shall now state the more general results found in [8]. We begin with some definitions.

Definition 4.1

Let A ⊂ ℝ be a compact set containing more than one point.

diamA = sup{|a – b| : a, b ∈ A} is the diameter of A.
Δ(A) = sup{|a – b| : a, b ∈ A, (a, b) ∩ A = ∅}. Note that Δ(A) gives the largest gap in A.
I(A)=Δ(A)Δ(A)+diam(A).
i(A) = inf{I(B) : B ⊂ A, |B| ≥ 2}.

Let k₁ ≥ k₂ ≥ ⋯ ≥ k_m be positive real numbers and let S = ∑i=1mciki:ci∈{0,1} . Also let q ∈ (0, 1).

Theorem 4.1

[8]

E(k₁, k₂, …, k_m; q) is an interval if and only if q ≥ I(S).
E(k₁, k₂, … k_m; q) contains an interval if q ≥ i(S).
E(k₁, k₂, …, k_m; q) is a Cantor set of zero Lebesgue measure if q < 1|S| .

For Example 4.1, the sequence (17, 15, 13, 11, 9, 1; q), we find that

S={0,1}∪{9,10,11,…,18}∪{20,21,22,…,46}∪{48,49,50,…,57}∪{65,66}.

It follows that diamS = 66, Δ(S) = 8, I(S) = 437 , and i(S) = 127 . (Note that the set B = {20, 21, 22, …, 46} ⊂ S and I(B) = 127 .) Finally |S| = 51. By Theorem 4.1, we see that E(17, 15, 13, 11, 9, 1; q) is a Cantor set of zero measure if q < 151 , contains an interval if 127 ≤ q, and is an interval if q ≥ 437 . The proof of Lemma 3.4 in Section 3, together with Remark 4 which follows it, implies that the subsum set E(a + 2nd, a + (2n – 2)d, …, a + 2d, a, d; q) is not a finite union of intervals if q<minda,a−d(n+2)+(n2+n)d. Since a = 9, d = 1 and n = 4, it follows that E(17, 15, 13, 11, 9, 1; q) is not a finite union of intervals if q < 437 . Therefore E(17, 15, 13, 11, 9, 1; q) is a Cantor set of zero measure if q < 151 , is a Cantorval if 110 ≤ q < 437 , and is an interval if q ≥ 437 .

For Example 4.2, the sequence (41, 37, 33, 29, 25, 21, 2; q), we have diam(S) = 188, Δ(S) = 19, I(S) = 19207 , i(S) = 3149 , and |S| = 84. Hence by the same reasoning, E(41, 37, 33, 29, 25, 21, 2; q) is a Cantor set of zero measure if q < 184 , is a Cantorval if 3149 ≤ q < 19207 and is an interval if q ≥ 19207 .

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Received: 2019-04-05

Accepted: 2019-09-27

Published Online: 2019-12-10

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Articles in the same Issue

https://doi.org/10.1515/math-2019-0109

Keywords for this article

subsum sets; Cantor set; Cantorval; multigeometric series

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