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THE ALGEBRAIC DIFFERENCE OF CENTRAL CANTOR SETS AND

SELF-SIMILAR CANTOR SETS

PIOTR NOWAKOWSKI

Abstract. Let C(a), C(b) ⊂ [0, 1] be the central Cantor sets generated by sequences a, b ∈ (0, 1)N.

The main result in the first part of the paper gives a necessary condition and a sufficient condition

for sequences a and b which inform when C(a)− C(b) is equal to [−1, 1] or is a finite union of closed

intervals. In the second part we investigate some self-similar Cantor sets C(l, r, p), which we call

S-Cantor sets, generated by numbers l, r, p ∈ N, l + r < p. We give a full characterization of the

set C(l1, r1, p) − C(l2, r2, p) which can take one of the form: the interval [−1, 1], a Cantor set, an

L-Cantorval, an R-Cantorval or an M-Cantorval.

1. Introduction

For A,B ⊂ R, we denote by A ± B the set {a± b : a ∈ A, b ∈ B}. The set A − B is called the

algebraic difference of sets A and B. The set A − A is called the difference set of a set A. We will

also write a+A instead {a}+A for a ∈ R. If I ⊂ R is an interval, then by l(I), r(I) we will denote,

respectively, the left and the right endpoint of I.

By a Cantor set we mean a bounded perfect and nowhere dense subset of R.

Given any set C ⊂ R, every bounded component of the set R\C is called a gap of C. A component

of C is called proper if it is not a singleton.

Let us recall the definitions of three types of Cantorvals (compare [13]). A perfect set E ⊂ R is

called an M-Cantorval if it has infinitely many gaps and both endpoints of any gap are accumulated

by gaps and proper components of E. A perfect set E ⊂ R is called an L-Cantorval (respectively,

R-Cantorval) if it has infinitely many gaps, the left (right) endpoint of any gap is accumulated by

gaps and proper components of E, and the right (left) endpoint of any gap is an endpoint of a proper

component of E.

Algebraic differences and sums of Cantor sets were considered by many authors (e.g. [21], [10], [9],

[15], [8], [1], [18], [14]). They appear for example in dynamical systems (see [16]), spectral theory (see

[6],[20]) or number theory (see [8]). One of the known results is the Newhouse gap lemma from [15]

(see Theorem 2.2). This is a condition which implies that the sum of two Cantor sets is an interval.

The most popular types of examined Cantor sets are central Cantor sets and various self-similar

Cantor sets. In our paper we will continue this trend. We will consider two classes of Cantor sets: in

Section 2 we investigate central Cantor sets, and in Section 3 we study the so-called S-Cantor sets,

which are self-similar.

2020 Mathematics Subject Classification. 28A80, 05B10, 11A67.

Key words and phrases. Cantor sets, Cantorvals, algebraic difference of sets, p-adic sets.

Piotr Nowakowski was supported by the GA CR project 20-22230L and RVO: 67985840.

1

http://arxiv.org/abs/2102.11194v1

2 PIOTR NOWAKOWSKI

Every central Cantor subset C of [0, 1] can be uniquely described by a sequence a = (an) ∈ (0, 1)N

(the details are given in Section 2). We then say that C is generated by a and write C = C(a).

The algebraic difference of two central Cantor sets can be either a Cantor set, a finite union of closed

intervals, an L-Cantorval, an R-Cantorval or an M-Cantorval (see [3]). In the paper [11], Kraft proved

that the difference set of a central Cantor set generated by a constant sequence with all terms equal

to α is equal to [−1, 1] if α ≤ 13 , and is a Cantor set if α > 1

3 . Later, it was proved (see [3], [7]) that

the difference set of a central Cantor set C(a), where a ∈ (0, 1)N, is equal to [−1, 1] if and only if

an ≤ 13 for all n ∈ N. In [18], the author proved that if an > 1

3 for all n ∈ N, then the difference set

of C(a) is a Cantor set. In [7] there was given a condition, which implies that the set C(a) − C(a)

is an M-Cantorval. In our paper, we will focus only on the case, when the algebraic difference of

two central Cantor sets is a finite union of closed intervals. In Section 2, we prove a theorem which

concerns the algebraic difference of two different central Cantor sets and is a generalization of the

earlier mentioned theorem concerning the difference set of a central Cantor set. It also gives examples

of central Cantor sets which do not satisfy the assumptions of the Newhouse gap lemma (Theorem

2.2), and still their algebraic difference is equal to [−1, 1].

The algebraic difference (or sum) of self-similar Cantor sets is also widely explored. In [13], the

classes of so-called regular, affine and homogeneous Cantor sets were considered. Every homogeneous

Cantor set is affine, and every affine is regular. There was proved that the algebraic sum of two

homogeneous Cantor sets is, similarly as for central Cantor sets, either a Cantor set, a finite union of

closed intervals, an L-Cantorval, an R-Cantorval or an M-Cantorval. In this paper the authors posed

three open questions:

1. Does the similar result hold generically for affine Cantor sets?

2. What can be said about the topological structure of the sum of two regular Cantor sets?

3. Is it possible to characterize each one of the five possibilities for the algebraic difference of

homogeneous Cantor sets?

In [2], the positive answer to the first question was given. As far as we know, there are still no

answers to the remaining questions.

In Section 3, we present a class of S-Cantor subsets of [0, 1]. They are regular (in the sense given

in [13]), but not affine. For such sets we prove not only that the algebraic sum can be either a Cantor

set, a finite union of closed intervals, an L-Cantorval, an R-Cantorval or an M-Cantorval, but we also

give a full characterization when every class occurs, answering the questions 2. and 3. mentioned

above for this particular class of Cantor sets.

2. The algebraic difference of central Cantor sets

Let us recall the construction of a central Cantor subset of [0, 1] (see e.g. [4]).

An interval I is called concentric with an interval J if they have a common centre.

Let a = (an) be a sequence such that an ∈ (0, 1) for any n ∈ N and I := [0, 1]. In the first step

of the construction we remove from I the open interval P centered at 12 of length a1. Then by I0

and I1 we denote, respectively, the left and the right components of I \P (each of length d1 =1−a12 ).

Generally, assume that for some n ∈ N and t1, t2, . . . , tn ∈ {0, 1} we constructed the interval It1,...,tn

of length dn. Denote by Pt1,...,tn the open interval of length an+1dn, concentric with It1,...,tn . Now, let

THE ALGEBRAIC DIFFERENCE OF CENTRAL CANTOR SETS AND SELF-SIMILAR CANTOR SETS 3

It1,...,tn,0 and It1,...,tn,1 be, respectively, the left and the right components of the set It1,...,tn \ Pt1,...,tn .

By dn+1 denote the common length of both components.

For every n ∈ N, denote

In := {It1,...,tn : (t1, . . . , tn) ∈ {0, 1}n} and Cn(a) :=⋃

In.

Let C(a) :=⋂

n∈NCn(a). Then C(a) is called a central Cantor set.

Define the thickness of a Cantor set C by (see [19])

τ(C) = infG1<G2

max

{

l(G2)− r(G1)

|G1|,l(G2)− r(G1)

|G2|

}

,

where G1, G2 are gaps of the set C, and G1 < G2 means that G1 is on the left of G2.

In the case of central Cantor sets, we have an explicit formula for the thickness.

Lemma 2.1. [9] τ(C((an))) = infn∈N

1−an2an

.

The following theorem is a version for central Cantor sets of a known result about the algebraic

difference of two Cantor sets, which uses the notion of thickness.

Theorem 2.2 (the Newhouse gap lemma; [15]). If a, b ∈ (0, 1)N and τ(C(a))τ(C(b)) ≥ 1, then

C(a)− C(b) = [−1, 1].

Our purpose is to study the algebraic difference C(a) − C(b) of two central Cantor sets. We will

use some ideas from [7].

To denote the concatenation of two sequences t and s, we write tˆs.

Let dn =∏n

i=11−ai2 and gn =

∏ni=1

1−bi2 . By Ias (Ibs , respectively) we will denote the interval Is

from the construction of the set C(a) (C(b), respectively). Let {0, 1}0 = {∅} (the empty sequence),

Ia∅ = Ib∅ = [0, 1] and d0 = g0 = 1. Then we have (see [7, Proposition 1.1 (9)])

Cn (a)− Cn (b) =⋃

p,q∈{0,1}n

(

Iap − Ibq

)

.

For p, q ∈ {0, 1}n we define the sequence s ∈ {0, 1, 2, 3}n and the interval Js, putting si := 2pi− qi+1

for i = 1, . . . n, and Js := Iap − Ibq . Then

Cn (a)− Cn (b) =⋃

s∈{0,1,2,3}n

Js

and |Js| = dn + gn for s ∈ {0, 1, 2, 3}n. Observe that

J0 = Ia0 − Ib1 = [−1,−1 + d1 + g1],

J1 = Ia0 − Ib0 = [−g1, d1],

J2 = Ia1 − Ib1 = [−d1, g1],

J3 = Ia1 − Ib0 = [1− d1 − g1, 1].

4 PIOTR NOWAKOWSKI

Moreover, if for some n ∈ N and s ∈ {0, 1, 2, 3}n we have Js = Iap − Ibq , then

Jsˆ0 = Iapˆ0 − Ibqˆ1 = [l(Iap ), l(Iap ) + dn+1]− [r(Ibq)− gn+1, r(I

bq)]

= [l(Js), l(Js) + dn+1 + gn+1],

Jsˆ1 = Iapˆ0 − Ibqˆ0 = [l(Iap ), l(Iap ) + dn+1]− [l(Ibq), l(I

bq ) + gn+1]

= [l(Js) + gn − gn+1, r(Js)− dn + dn+1],

Jsˆ2 = Iapˆ1 − Ibqˆ1 = [r(Iap )− dn+1, r(Iap )]− [r(Ibq)− gn+1, r(I

bq)]

= [l(Js) + dn − dn+1, r(Js)− gn + gn+1],

Jsˆ3 = Iapˆ1 − Ibqˆ0 = [r(Iap )− dn+1, r(Iap )]− [l(Ibq), l(I

bq) + gn+1]

= [r(Js)− dn+1 − gn+1, r(Js)].

Put J∅ = Ia∅ − Ib∅ = [−1, 1] and observe that the above formulas remain true for n = 0 and s = ∅.

Lemma 2.3. For any a ∈ (0, 1)N, n, k ∈ N ∪ {0}, where n > k, we have dn − dn+1 < dk − dk+1.

Proof. It suffices to show that dn − dn+1 < dn−1 − dn, for n > 1, or equivalently 2dn − dn+1 < dn−1.

Dividing both sides of the last inequality by dn−1, we get

1− an −(1− an)(1− an+1)

4< 1,

which holds for all n. �

Lemma 2.4. Assume that a = (an) ∈ (0, 1)N, b = (bn) ∈ (0, 1)N, n ∈ N∪{0} and s ∈ {0, 1, 2, 3}n.

The following equivalences hold:

(1) gndn

≥ an+1 ⇔ l (Jsˆ2) ≤ r (Jsˆ1) ;

(2) dngn

≥ bn+1 ⇔ l (Jsˆ1) ≤ r (Jsˆ2) ;

(3)(

dngn

≥ bn+1 and gndn

≥ an+1

)

⇔ Jsˆ1 ∩ Jsˆ2 6= ∅;

(4) dn+1

gn≥ bn+1 ⇔ Jsˆ0 ∩ Jsˆ1 6= ∅ ⇔ Jsˆ2 ∩ Jsˆ3 6= ∅;

(5) gn+1

dn≥ an+1 ⇔ Jsˆ0 ∩ Jsˆ2 6= ∅ ⇔ Jsˆ1 ∩ Jsˆ3 6= ∅.

Proof. Ad (1) We have

r (Jsˆ1)− l (Jsˆ2) = r(Js)− dn + dn+1 − l(Js)− dn + dn+1 = dn + gn − 2dn + 2dn+1 = gn − dn + 2dn+1.

Hence l (Jsˆ2) ≤ r (Jsˆ1) if and only if gn − dn + 2dn+1 ≥ 0, which is equivalent to gndn

≥ an+1.

Ad (2) The proof is analogous to that of (1).

Ad (3) The assertion follows from (1), (2) and the equivalence

(l (Jsˆ2) ≤ r (Jsˆ1) and l (Jsˆ1) ≤ r (Jsˆ2)) ⇔ Jsˆ1 ∩ Jsˆ2 6= ∅.

Ad (4) Of course, Jsˆ0 ∩ Jsˆ1 6= ∅ if and only if r(Jsˆ0) ≥ l(Jsˆ1). We have

r(Jsˆ0)− l(Jsˆ1) = l(Js) + dn+1 + gn+1 − l(Js)− gn + gn+1 = dn+1 + 2gn+1 − gn.

Thus, Jsˆ0 ∩ Jsˆ1 6= ∅ if and only if dn+1 + 2gn+1 − gn ≥ 0, which is equivalent to dn+1

gn≥ bn+1.

The equivalence Jsˆ0 ∩ Jsˆ1 6= ∅ ⇔ Jsˆ2 ∩ Jsˆ3 6= ∅ follows from the equality

r (Jsˆ0)− l (Jsˆ1) = dn+1 + 2gn+1 − gn = r (Jsˆ2)− l (Jsˆ3) .


Ad (5) The proof is similar to that of (4). �

Lemma 2.5. Assume that a = (an) ∈ (0, 1)N, b = (bn) ∈ (0, 1)N and n ∈ N ∪ {0}. If

dngn

≥ bn+1 andgndn

≥ an+1 and

(

dn+1

gn≥ bn+1 or

gndn

≥ an+1

)

,

then Cn+1 (a)− Cn+1 (b) = Cn (a)− Cn (b).

Proof. Let s ∈ {0, 1, 2, 3}n. By Lemma 2.4 we infer that

Jsˆ1 ∩ Jsˆ2 6= ∅ and Jsˆ0 ∩ Jsˆ1 6= ∅ and Jsˆ2 ∩ Jsˆ3 6= ∅

or

Jsˆ1 ∩ Jsˆ2 6= ∅ and Jsˆ0 ∩ Jsˆ2 6= ∅ and Jsˆ1 ∩ Jsˆ3 6= ∅.

In both cases we have Jsˆ0 ∪ Jsˆ1 ∪ Jsˆ2 ∪ Jsˆ3 = Js. Hence

Cn (a)− Cn (b) =⋃

s∈{0,1,2,3}n

Js =⋃

s∈{0,1,2,3}n

(Jsˆ0 ∪ Jsˆ1 ∪ Jsˆ2 ∪ Jsˆ3)

=⋃

t∈{0,1,2,3}n+1

Jt = Cn+1 (a)− Cn+1 (b) .

�

Now, we can prove the main theorem.

Theorem 2.6. Let a = (an) ∈ (0, 1)N, b = (bn) ∈ (0, 1)N.

(1) If for any n ∈ N∪{0}

(∗)gn+1

dn≥ an+1 or

dn+1

gn≥ bn+1

and

(∗∗)dngn

≥ bn+1 andgndn

≥ an+1,

then C(a)− C(b) = [−1, 1].

(2) If conditions (∗) and (∗∗) hold for sufficiently large n, then C(a) − C(b) is a finite union of

closed intervals.

(3) If C(a)− C(b) = [−1, 1], then condition (∗) holds for all n ∈ N ∪ {0}.

(4) If C(a) − C(b) is a finite union of closed intervals, then condition (∗) holds for sufficiently

large n.

Proof. Ad (1)–(2) Assume that there is n0 ∈ N such that conditions (∗) and (∗∗) hold for all n ≥ n0.

By Lemma 2.5 it follows that

Cn (a)− Cn (b) = Cn0 (a)− Cn0 (b)

for n ≥ n0, and thus

C (a)− C (b) =⋂

n∈N

(Cn (a)−Cn (b)) = Cn0 (a)−Cn0 (b) =⋃

s∈{0,1,2,3}n0

Js,

so C (a) − C (b) is a finite union of closed intervals. If n0 = 0, that is, conditions (∗) and (∗∗) hold

for all n, then C (a)− C (b) = C0 (a)− C0 (b) = [−1, 1] .

6 PIOTR NOWAKOWSKI

Ad (3) Assume on the contrary that condition (∗) does not hold for some n ∈ N. Let s := 0(n−1).

From Lemma 2.4 it follows that Jsˆ0 ∩ Jsˆ1 = ∅ and Jsˆ0 ∩ Jsˆ2 = ∅. Put L := min{l (Jsˆ1) , l (Jsˆ2)}.

Of course, L > r (Jsˆ0) = r (J0(n)) . Let x ∈ (r (J0(n)) , L). Since C(a)− C(b) = [−1, 1], there exists a

sequence u ∈ {0, 1, 2, 3}n such that x ∈ Ju. Then u /∈ {sˆi : i = 0, 1, 2, 3}, so uk > 0 for some k < n.

In consequence, x ≥ l (J0(k−1)ˆ1) or x ≥ l (J0(k−1)ˆ2). Using Lemma 2.3, in the first case we get

L > x ≥ l (J0(k−1)ˆ1) = −1 + gk−1 − gk > −1 + gn−1 − gn = l (Jsˆ1) ≥ L,

a contradiction. In the second case, a contradiction is obtained similarly.

Ad (4) Assume to the contrary that (∗) does not hold for infinitely many n ∈ N, and C(a)−C(b) is

a finite union of closed intervals. Then there exists w > 0 such that [−1,−1+w] ⊂ C(a)−C(b). Let

n ∈ N be such that condition (∗) does not hold and w > dn+gn = r (J0(n))+1. Let s := 0(n−1). From

Lemma 2.4 it follows that Jsˆ0∩Jsˆ1 = ∅ and Jsˆ0∩Jsˆ2 = ∅. Put L := min{l (Jsˆ1) , l (Jsˆ2) ,−1+w}.

Of course, L > r (Jsˆ0) = r (J0(n)) . Let x ∈ (r (J0(n)) , L). Since (r (J0(n)) , L) ⊂ [−1,−1 + w] ⊂

C(a)−C(b), there exists a sequence u ∈ {0, 1, 2, 3}n such that x ∈ Ju. Then u /∈ {sˆi : i = 0, 1, 2, 3},

so uk > 0 for some k < n. Consequently, x ≥ l (J0(k−1)ˆ1) or x ≥ l (J0(k−1)ˆ2). Using Lemma 2.3, in

the first case we have

L > x ≥ l (J0(k−1)ˆ1) = −1 + gk−1 − gk > −1 + gn−1 − gn = l (Jsˆ1) ≥ L,

a contradiction. In the second case, we obtain a contradiction in the similar way.

�

The next example shows that the above theorem gives examples of Cantor sets whose algebraic

difference is the interval [−1, 1], despite not satisfying assumptions of the Newhouse gap lemma.

Example 1. Let a = (12 ,14 ,

12 ,

14 , . . . ), b = (14 ,

12 ,

14 ,

12 , . . . ). Then for n ∈ N ∪ {0} we have

d2ng2n

=(14 ·

38)

n

(38 ·14)

n= 1,

d2n+1

g2n+1=

(14 · 38 )

n · 14

(38 · 14 )

n · 38

=2

3.

Hence for all n ∈ Ng2n

d2n−1=

g2n−1

d2n−1·1− b2n

2=

3

2·1

4=

3

8> a2n,

d2n−1

g2n−2=

d2n−2

g2n−2·1− a2n−1

2= 1 ·

1

4= b2n−1,

bn <2

3≤

dn−1

gn−1,

an < 1 ≤gn−1

dn−1.

Therefore, conditions (∗) and (∗∗) hold for every n ∈ N, so C(a)−C(b) = [−1, 1]. Moreover, observe

that τ(C(a)) = τ(C(b)) = min{12 ,

32} = 1

2 , and thus τ(C(a)) · τ(C(b)) = 14 < 1, so the sufficient

condition from the Newhouse gap lemma does not hold.

The characterization of the cases when the set C(a)−C(a) is the interval [−1, 1] or a finite union

of closed intervals has been already proved with use of various methods (see [3], [7]). However, this

result also easily follows from Theorem 2.6.


Corollary 2.7. Let a = (an) ∈ (0, 1)N. Then C(a)− C(a) is equal to:

(1) the interval [−1, 1] if and only if an ≤ 13 for all n ∈ N;

(2) a finite union of closed intervals if and only if the set {n ∈ N : an > 13} is finite.

Proof. For any n we have dngn

= dndn

= 1 and an < 1, so condition (∗∗) holds for all n ∈ N. Sincedn

dn−1= 1−an

2 , the inequality dndn−1

≥ an is equivalent to an ≤ 13 . From Theorem 2.6 we obtain the

assertion. �

Example 2. Condition (∗∗) is not necessary for C(a)−C(b) = [−1, 1]. Let a1 =14 , a2 =

140 , a3 =

140 ,

b1 =320 , b2 =

140 , b3 =

1011 . Then we have

d1 =3

8, d2 =

3

8·39

80=

117

640, d3 = d2

39

80=

4563

51200,

g1 =17

40, g2 =

17

40·39

80=

663

3200, g3 = g2

1

22=

663

70400.

Thus, d2g2

= 38 ·

4017 = 15

17 , b3 =1011 > 15

17 = d2g2. Therefore, condition (∗∗) is not satisfied.

After tedious calculations we can check that C3(a) − C3(b) = [−1, 1]. Choosing next terms of

sequences a and b in such a way that

bn+1 ≤dngn

and an+1 ≤gn+1

dn

for n ≥ 3, we receive sequences which satisfy conditions (∗) and (∗∗), so C(a)−C(b) = C3(a)−C3(b) =

[−1, 1].

3. The algebraic difference of S-Cantor sets

In this section we consider another type of Cantor sets which was examined for example in [17],

[22], [12]. We call them p-Cantor sets. In the class of p-Cantor sets we will distinguish the subclass of

S-Cantor sets. But first, we need some additional notation. Let A ⊂ Z be a finite set, p ∈ N, p ≥ 2.

Set

Ap :=

{

∞∑

i=1

xipi

: xi ∈ A

}

.

This notation comes from [17]. For example, we have Z(p)p = [0, 1], {−p+1,−p+2, . . . , p−2, p−1}p =

[−1, 1], {0}p = {0}, {p − 1}p = {1}, {0, 2}3 = C, where C is the classical Cantor ternary set, and

Z(p) := {0, 1, . . . , p − 1}. Note that, if A ⊂ Z(p), the set Ap consists of all real numbers in [0, 1]

having p-adic expansions with all digits in A.

We will now give some easy but useful properties of the sets Ap.

Proposition 3.1. Let A and B be finite subsets of Z, p ∈ N, p ≥ 2, k ∈ Z. Then:

(i) Ap +Bp = (A+B)p;

(ii) Ap −Bp = (A−B)p;

(iii) kAp = (kA)p.

Proof. (i) We have

Ap +Bp =

{

∞∑

i=1

xipi

: xi ∈ A

}

+

{

∞∑

i=1

yipi

: yi ∈ B

}

=

{

∞∑

i=1

xipi

+∞∑

i=1

yipi

: xi ∈ A, yi ∈ B

}

8 PIOTR NOWAKOWSKI

=

{

∞∑

i=1

xi + yipi

: xi ∈ A, yi ∈ B

}

=

{

∞∑

i=1

zipi

: zi ∈ A+B

}

= (A+B)p.

The proof of (ii) and (iii) is analogous. �

The next proposition is a mathematical folklore and its proof is easy, so we omit it.

Proposition 3.2. Let A be a finite subset of Z and p ∈ N, p ≥ 2. Then:

(i) Ap is a closed set.

(ii) If A has more than 1 element, then Ap is a perfect set.

(iii) If A ( Z(p), then Ap is nowhere dense.

(iv) If A ( Z(p) and A has more than 1 element, then Ap is a Cantor set.

For i, j ∈ Z, we set

〈i, j〉 :=

[i, j] ∩ Z if i < j

{i} if i = j

∅ if i > j.

If (xi) ∈ 〈−p, p〉n, then we will write xn =∑n

i=1xi

pi, for n ∈ N.

In the sequel we will consider sets Ap, where p ∈ N, p > 2, A ⊂ 〈−p+1, p−1〉 and −p+1, 0, p−1 ∈ A.

Note that Ap ⊂ [−1, 1]. Let n ∈ N, x ∈ [−1, 1]. We say that x is (n)-bi-obtainable if for some

k ∈ 〈−pn, pn − 1〉 we have x ∈ [ kpn, k+1

pn] and there exist sequences (yi), (zi) ∈ An such that yn = k

pn

and zn = kpn

+ 1pn.

Remark 3. Observe that the condition stating that there exist sequences (yi), (zi) ∈ An such that

yn = kpn

and zn = kpn

+ 1pn

implies that kpn, kpn

+ 1pn

∈ Ap. Indeed,

k

pn=

n∑

i=1

yipi

+

∞∑

i=n+1

0

pi∈ Ap.

Similarly,

k

pn+

1

pn=

n∑

i=1

zipi

+∞∑

i=n+1

0

pi∈ Ap.

We will often use the following easy observations.

Observation 3.3. (i) For any (xn) ∈ 〈−p, p〉N and any n ∈ N we have

xn =k

pnfor some k ∈ Z.

(ii) For every n ∈ N, k ∈ 〈−pn, pn〉 there exists a sequence (wi) ∈ 〈−p, p〉n such that wn = kpn.

(iii) For every n ∈ N, k ∈ 〈−pn + 1, pn − 1〉 there exists a sequence (wi) ∈ 〈−p + 1, p − 1〉n such

that wn = kpn.

(iv) If x ∈ [0, 1pn], then there is a sequence (wi) ∈ 〈0, p − 1〉N such that

∑∞i=n+1

wi

pi= x.

(v) If x ∈ [− 1pn, 0], then there is a sequence (wi) ∈ 〈−p+ 1, 0〉N such that

∑∞i=n+1

wi

pi= x.

We will also need the following technical lemma.

Lemma 3.4. Let p ∈ N, p > 2, A ⊂ 〈−p+ 1, p − 1〉 and −p+ 1, 0, p − 1 ∈ A.


(1) If x ∈ Ap, (xi) ∈ AN is such that x =∑∞

i=1xi

piand, for some n ∈ N and (yi) ∈ 〈−p, p − 1〉n,

x ∈ (yn, yn + 1pn), then xn = yn or xn = yn + 1

pn. Moreover,

(1.1) if xn = yn and yn > 0, then xn = yn or xn = −p+ yn;

(1.2) if xn = yn and yn < 0, then xn = yn or xn = p+ yn;

(1.3) if xn = yn + 1pn

and yn > 0, then xn = yn + 1 or xn = −p+ 1 + yn;

(1.4) if xn = yn + 1pn

and yn < 0, then xn = yn + 1 or xn = p+ yn + 1.

(1.5) if xn = yn and yn = 0, then xn = 0.

(2) If (x, y) is a gap in Ap, (xi), (yi) ∈ AN are such that x =∑∞

i=1xi

piand y =

∑∞i=1

yipi, then there

are n,m ∈ N such that xn + 1, ym − 1 /∈ A and xi = p− 1, yj = −p+ 1 for i > n, j > m.

(2’) If w ∈ [−1, 1] is such that w = vk for some k ∈ N and (vi) ∈ Ak, then w is not an endpoint

of a gap in Ap.

(3) Let x ∈ [−1, 1]. If there exists n ∈ N such that x is (i)-bi-obtainable for i ≥ n, then x ∈ Ap.

(4) Let x ∈ [−1, 1]. If for any k ∈ 〈0, p− 1〉 we have k ∈ A or k − p ∈ A and there is n ∈ N such

that x is (n)-bi-obtainable, then x is (i)-bi-obtainable for i ≥ n.

Proof. Ad (1) By the assumption, we have 0 < x− yn < 1pn. Observe that

|x− xn| =

∣

∣

∣

∣

∣

∞∑

i=n+1

xipi

∣

∣

∣

∣

∣

≤p− 1

pn+1(1− 1p)=

1

pn,

and so − 1pn

≤ xn − x ≤ 1pn. Adding the obtained inequalities we receive − 1

pn< xn − yn < 2

pn. By

Observation 3.3, we have xn − yn = 0 or xn − yn = 1pn.

Ad (1.1) Since yn > 0, we have yn > yn−1. Moreover,

yn +1

pn= yn−1 +

yn + 1

pn≤ yn−1 +

p

pn= yn−1 +

1

pn−1.

Hence x ∈ (yn−1, yn−1+1

pn−1 ). By (1), xn−1 = yn−1 or xn−1 = yn−1+1

pn−1 . In the first case, we have

xn = yn, and in the second case, xn

pn= yn

pn− 1

pn−1 , so xn = yn − p.

Ad (1.2) Since yn < 0, we have yn < yn−1, and so yn + 1pn

≤ yn−1. Moreover,

yn = yn−1 +ynpn

≥ yn−1 −p

pn= yn−1 −

1

pn−1.

Hence x ∈ (yn−1 −1

pn−1 , yn−1). By Observation 3.3, there is a sequence (y′i) ∈ 〈−p, p − 1〉n−1 such

that y′n−1 = yn−1 −1

pn−1 . By (1),

xn−1 = y′n−1 +1

pn−1= yn−1

or

xn−1 = y′n−1 = yn−1 −1

pn−1.

In the first case, we have xn = yn, and in the second case, xn

pn= yn

pn+ 1

pn−1 , so xn = yn + p.

The proofs of (1.3) and (1.4) are similar.

Ad (1.5) We will show that xn−1 = yn−1. Indeed, if |xn−1 − yn−1| > 0, then, by Observation 3.3,

|xn−1 − yn−1| ≥1

pn−1 , and hence

|xn − yn| =

∣

∣

∣

∣

xn−1 − yn−1 +xnpn

−ynpn

∣

∣

∣

∣

≥∣

∣xn−1 − yn−1

∣

∣−

∣

∣

∣

∣

xnpn

−ynpn

∣

∣

∣

∣

≥1

pn−1−

p− 1

pn=

1

pn> 0.

10 PIOTR NOWAKOWSKI

So, xn−1 = yn−1, and thus xn = yn.

Ad (2) First, we will show that there is n ∈ N such that xi = p− 1 for any i > n. On the contrary,

assume that for any k ∈ N there is j > k such that xj k be such that xj < p − 1. Let zi = xi for i 6= j, and zj = p − 1. Since p − 1 ∈ A and

x ∈ Ap, we have z =∑∞

i=1zipi

∈ Ap. Moreover,

z = x+p− 1− xj

pj≤ x+

2(p − 1)

pj< x+

2

pj−1≤ x+

2

pk< y.

So, z ∈ Ap ∩ (x, y), a contradiction. Of course, x < 1, so there is n ∈ N such that xn n.

Now, we will show that xn + 1 /∈ A. On the contrary, assume that xn + 1 ∈ A. We have

x =

∞∑

i=1

xipi

= xn +

∞∑

i=n+1

p− 1

pi= xn +

1

pn.

Let k > n be such that 1pk−1 < y − x. Let wi = xi for i < n, wn = xn + 1, wk = p− 1 and wi = 0 for

the remaining i. Then w =∑∞

i=1wi

pi∈ Ap and

w = xn +1

pn+

p− 1

pk< x+

1

pk−1< y,

a contradiction. Thus, xn + 1 /∈ A. The proof for the sequence (yi) is analogous.

Ad (2’) Immediately follows from (2).

Ad (3) Using the definition of (i)-bi-obtainability and Remark 3, we can find ki ∈ 〈−pi, pi − 1〉

such that x ∈ [kipi, kipi

+ 1pi] and ki

pi∈ Ap for any i ≥ n. Then |x− ki

pi| ≤ 1

pi, so lim

i→∞

kipi

= x. Since Ap is

closed, then x ∈ Ap.

Ad (4) It suffices to show that if x is (n)-bi-obtainable, then it is (n+ 1)-bi-obtainable.

First, suppose that x = kpn

for some k ∈ 〈−pn + 1, pn − 1〉 (of course, 1 and −1 cannot be (n)-bi-

obtainable). Then, from the definition of (n)-bi-obtainability it follows that kpn

∈ Ap and at least one

of kpn

+ 1pn

or kpn

− 1pn

belongs to Ap. Assume that kpn

+ 1pn

∈ Ap (the proof, when kpn

− 1pn

∈ Ap is

similar). Let (yi), (zi) ∈ An be such that yn = kpn

and zn = kpn

+ 1pn. We have x ∈ [ kp

pn+1 ,kp

pn+1 +1

pn+1 ]

and kppn+1 = k

pn∈ Ap. We need to find a sequence (wi) ∈ 〈−p, p − 1〉n+1 such that wn+1 =

kpn

+ 1pn+1 .

Consider the cases.

1o 1 ∈ A. Then put wi = yi for i ≤ n and wn+1 = 1. Then (wi) ∈ An+1 and wn+1 =kpn

+ 1pn+1 .

2o 1 /∈ A. By the assumption we have 1 − p ∈ A. Put wi = zi for i ≤ n and wn+1 = 1− p. Then

(wi) ∈ An+1 and

wn+1 =k

pn+

1

pn+

1− p

pn+1=

k

pn+

1

pn+1.

Therefore, x is (n+ 1)-bi-obtainable.

Now, assume that x ∈ ( kpn, kpn+1 + 1

pn) for some k ∈ 〈−pn, pn − 1〉. Then there is k′ ∈ 〈0, p − 1〉

such that x ∈ [kp+k′

pn+1 ,kp+k′

pn+1 + 1pn+1 ]. Since x is (n)-bi-obtainable, there exist (yi), (zi) ∈ An such that

yn = kpn

and zn = kpn

+ 1pn. We will find a sequence (vi) ∈ An+1 such that vn+1 = kp+k′

pn+1 . Consider

the cases.

1o k′ ∈ A. Put vi = yi for i ≤ n and vn+1 = k′. Then (vi) ∈ An+1 and vn+1 =kp+k′

pn+1 .


2o k′ /∈ A. Since k′ ≥ 0, by the assumption, we have k′ − p ∈ A. Put vi = zi for i ≤ n and

vn+1 = k′ − p. Then (vi) ∈ An+1, and

vn+1 =k

pn+

1

pn+

k′ − p

pn+1=

kp + k′

pn+1.

Now, we will find a sequence (ui) ∈ An+1 such that un+1 =kp+k′

pn+1 + 1pn+1 . Consider the cases.

1o k′ + 1 ∈ A. Then put ui = yi for i ≤ n and un+1 = k′ + 1. Then (ui) ∈ An+1 and un+1 =kp+k′

pn+1 + 1pn+1 .

2o k′ + 1 /∈ A. If k′ < p − 1, then k′ + 1 ∈ 〈1, p − 1〉 and, by the assumption, k′ + 1 − p ∈ A. If

k′ = p− 1, then k′ + 1− p = 0 ∈ A. In both cases put ui = zi for i ≤ n and un+1 = k′ + 1− p. Then

(ui) ∈ An+1, and

un+1 =k

pn+

1

pn+

k′ − p+ 1

pn+1=

kp+ k′

pn+1+

1

pn+1.

Therefore, x is (n+ 1)-bi-obtainable. �

Let p ∈ N, p ≥ 2 and A ( Z(p) has more than 1 element. In Proposition 3.2 (iv) it was pointed

out that the set Ap is a Cantor set. Every such a set will be called a p-Cantor set. In the class of

p-Cantor sets we will distinguish some special subclass. Let l, r, p ∈ N, p > 2, l + r < p. We will

consider Cantor sets of the form C(l, r, p) := A(l, r, p)p where

A(l, r, p) := 〈0, l − 1〉 ∪ 〈p− r, p − 1〉.

We call such a set a special p-Cantor set or, in short, an S-Cantor set. This will not lead to any

misunderstandings because p will be always established. A set C(l, r, p) has the following construction.

In the first step, we divide [0, 1] into p subintervals with equal lengths and we enumerate them (starting

from 0). Then we remove p − l − r consecutive intervals starting from the one with number l. So,

there remains l intervals on the left and r intervals on the right of the emergent gap. We continue

this procedure for the remaining intervals. Observe that in the construction of a central Cantor set

we divide intervals into three (usually not equal) parts, and then we delete the middle one. So, there

is clearly no inclusion between these two families of Cantor sets. Note that, unlike central Cantor

sets, S-Cantor sets do not have to be symmetric but are always self-similar. However, we will also

examine symmetric S-Cantor sets.

Let us introduce some more notation. If A ⊂ Z, A 6= ∅, then we write

diam(A) := sup{|a− b| : a, b ∈ A},

∆(A) := sup{b− a : a, b ∈ A, a < b, (a, b) ∩A = ∅},

I(A) :=∆(A)

∆(A) + diam(A).

Theorem 3.5. [5] Let A ⊂ Z be a nonempty finite set, p ∈ N, p ≥ 2. Then Ap is an interval if and

only if 1p≥ I(A).

Remark 4. The assertion of the above theorem is equivalent to: Ap is an interval if and only if

p ≤ 1 + diam(A)∆(A) .

Corollary 3.6. Let p ∈ N, p > 2, A,B ⊂ Z(p), 0, p−1 ∈ A and 0, p−1 ∈ B. Then Ap−Bp = [−1, 1]

if and only if ∆(A−B) ≤ 2.

12 PIOTR NOWAKOWSKI

Proof. It is easily seen that diam(A−B) = 2p−2. Hence the inequality p ≤ 1+ diam(A−B)∆(A−B) is equivalent

to p− 1 ≤ 2p−2∆(A−B) , and so to ∆(A−B) ≤ 2. �

Using the above corollary and Lemma 3.4, we can prove the main theorem of this section, which

gives a full characterization of the sets of the form C(l1, r1, p) − C(l2, r2, p). Some inequalities with

parameters p, l1, r1, l2, r2 will play the key role in this theorem. We will assume that these numbers

are natural such that p > 2, l1 + r1 < p, l2 + r2 < p. Consider the following conditions:

(S1) l1 + l2 + r2 ≥ p or l1 + r1 + r2 ≥ p;

(S2) l1 + r1 + l2 ≥ p or r1 + l2 + r2 ≥ p;

(S3) l1 + r1 + l2 + r2 ≤ p;

(S1∗) l1 + l2 + r2 > p or l1 + r1 + r2 > p;

(S2∗) l1 + r1 + l2 > p or r1 + l2 + r2 > p.

Obviously (S1) follows from (S1∗) and (S2) from (S2∗). Moreover, if (S3) holds, then (S1) and (S2)

do not hold (so (S1∗) and (S2∗) do not hold as well).

In the theorem we will consider the following combinations of conditions:

(1) (S1) ∧ (S2)

(2) (S3)

(3) (S1∗) ∧ ¬(S2)

(4) (S2∗) ∧ ¬(S1)

(5) ¬(S1∗) ∧ ¬(S2∗) ∧ ¬(S3) ∧ ¬((S1) ∧ (S2)).

It is easy to see that for any parameters exactly one of the conditions (1)–(5) holds.

Remark 5. Observe that, when we consider the conditions (S1), (S1∗), (S2), (S2∗), (S3) for the set

C(l2, r2, p)− C(l1, r1, p) = −(C(l1, r1, p)− C(l2, r2, p)),

we need to swap the indices ”1” and ”2” in the original inequalities. This means that conditions

(S1), (S1∗), (S2), (S2∗) for the set C(l2, r2, p)−C(l1, r1, p) are the same as conditions (S2), (S2∗), (S1),

and (S1∗), respectively, for the set C(l1, r1, p)−C(l2, r2, p), while condition (S3) does not change. In

particular, (4) holds for C(l1, r1, p)− C(l2, r2, p) if and only if (3) holds for C(l2, r2, p)− C(l1, r1, p).

Theorem 3.7. Let l1, r1, l2, r2, p ∈ N, p > 2, l1 + r1 < p, l2 + r2 < p.

(1) C(l1, r1, p)− C(l2, r2, p) = [−1, 1] if and only if (S1) and (S2) hold.

(2) C(l1, r1, p)− C(l2, r2, p) is a Cantor set if and only if (S3) holds.

(3) C(l1, r1, p)−C(l2, r2, p) is an L-Cantorval if and only if (S1∗) holds, but (S2) does not hold.

(4) C(l1, r1, p)−C(l2, r2, p) is an R-Cantorval if and only if (S2∗) holds, but (S1) does not hold.

(5) C(l1, r1, p)− C(l2, r2, p) is an M-Cantorval if and only if (S1∗), (S2∗), (S3) do not hold and

at least one of (S1) or (S2) also does not hold.

Proof. Since for any parameters exactly one of the conditions given in (1)–(5) holds and a set cannot

have at the same time two of the following forms: the interval [−1, 1], a Cantor set, an L-Cantorval,

an R-Cantorval or an M-Cantorval, we only need to prove the implications ”⇐”.


Let us put

A := A(l1, r1, p) = 〈0, l1 − 1〉 ∪ 〈p− r1, p − 1〉,

B := A(l2, r2, p) = 〈0, l2 − 1〉 ∪ 〈p− r2, p− 1〉.

Then A−B is equal to

〈−p+ 1, l1 + r2 − p− 1〉 ∪ 〈−l2 + 1, l1 − 1〉 ∪ 〈−r1 + 1, r2 − 1〉 ∪ 〈p − r1 − l2 + 1, p− 1〉.

Observe that the second and the third component contain 0. Hence A−B is equal to

〈−p+ 1, l1 + r2 − p− 1〉 ∪ 〈min{−l2,−r1}+ 1, max{l1, r2} − 1〉 ∪ 〈p − r1 − l2 + 1, p− 1〉.

Let

L = 〈l1 + r2 − p,min{−l2,−r1}〉,

R = 〈max{l1, r2}, p − r1 − l2〉.

Then A−B = 〈−p+ 1, p − 1〉 \ (L ∪R).

Observe that L has at most one element if and only if l1 + r2 − p ≥ −l2 or l1 + r2 − p ≥ −r1 and

R has at most one element if and only if p− r1 − l2 ≤ l1 or p− r1 − l2 ≤ r2. So we have

(3.1)

(S1) ⇔ |L| ≤ 1 ⇔ l1 + r2 − p+ 1 /∈ L

(S∗1) ⇔ L = ∅ ⇔ l1 + r2 − p /∈ L

(S2) ⇔ |R| ≤ 1 ⇔ p− r1 − l2 − 1 /∈ R

(S∗2) ⇔ R = ∅ ⇔ p− r1 − l2 /∈ R

Moreover, since p+min{−l2,−r1} > p− r1 − l2 and

R ∩ (p + L) = 〈max{l1, r2}, p− r1 − l2〉 ∩ 〈l1 + r2, p +min{−l2,−r1}〉,

we have

(3.2) (S3) ⇔ R ∩ (p+ L) 6= ∅ ⇔ p− r1 − l2, l1 + r2 ∈ R ∩ (p+ L).

Note that −p + 1, 0, p − 1 ∈ A − B, so the assumptions of Lemma 3.4 are satisfied. Observe also

that if R∩ (p+L) = ∅, then k ∈ A−B or k− p ∈ A−B for any k ∈ 〈0, p− 1〉. Indeed, if k /∈ A−B,

then k ∈ R, and thus 0 > k − p /∈ L, so k − p ∈ A − B. Hence if (S3) does not hold, then the

assumptions of Lemma 3.4 (4) are satisfied.

By Proposition 3.1 we have

C(l1, r1, p)− C(l2, r2, p) = Ap −Bp = (A−B)p.

Since A−B has more than one element, (A−B)p is perfect, by Proposition 3.2.

Ad (1) From Corollary 3.6 it follows that the equality C(l1, r1, p)−C(l2, r2, p) = [−1, 1] is equivalent

to the inequality ∆(A− B) ≤ 2, which holds if and only if L and R have at most one element. And

this is satisfied if and only if (S1) and (S2) hold.

Ad (2) We have to show that (A − B)p is nowhere dense. Let x ∈ (A − B)p, (xi) ∈ (A − B)N be

such that x =∑∞

i=1xi

piand ε > 0. We will find an interval (y, z) ⊂ (x − ε, x + ε), which is disjoint

with (A−B)p. Let n ∈ N be such that 2pn

< ε. Define y, z ∈ [0, 1] in the following way:

yi = xi = zi for i ≤ n,

14 PIOTR NOWAKOWSKI

yn+1 = zn+1 = p− r1 − l2 − 1,

yn+2 = p− r1 − l2 − 2, zn+2 = yn+2 + 3 = p− r1 − l2 + 1,

yi = p− 1, zi = −p+ 1 for i > n+ 2

and let y =∑∞

i=1yipi, and z =

∑∞i=1

zipi. Then yi, zi ∈ 〈−p+ 1, p− 1〉 for i ∈ N. Hence

z − y =3

pn+2−

∞∑

i=n+3

2p− 2

pi=

1

pn+2> 0.

We have |x − y| ≤∑∞

i=n+12p−2pi

= 2pn

< ε. Similarly, |z − x| < ε, and thus (y, z) ⊂ (x − ε, x + ε).

Let w ∈ (y, z). We will show that w /∈ (A − B)p. On the contary, assume that w ∈ (A − B)p. So,

there is a sequence (wi) ∈ (A−B)N such that w =∑∞

i=1wi

pi. Since

yn+2 = p− r1 − l2 − 2 ≥ p− r1 − l2 − l1 − r2 ≥ 0,

we have

y = yn+1 +yn+2

pn+2+

∞∑

i=n+3

p− 1

pi> yn+1

and

z = yn+1 +p− r1 − l2 + 1

pn+2−

∞∑

i=n+3

p− 1

pi< yn+1 +

p

pn+2< yn+1 +

1

pn+1,

and so w ∈ (yn+1, yn+1 +1

pn+1 ). By Lemma 3.4 (1), wn+1 = yn+1 or wn+1 = yn+1 +1

pn+1 . Consider

the cases.

1o wn+1 = yn+1. We have

w ∈ (y, z) =

(

yn+2 +

∞∑

i=n+3

p− 1

pi, zn+2 −

∞∑

i=n+3

p− 1

pi

)

=

(

yn+2 +1

pn+2, zn+2 −

1

pn+2

)

=

(

yn+2 +1

pn+2, yn+2 +

3

pn+2−

1

pn+2

)

=

(

yn+2 +1

pn+2, yn+2 +

2

pn+2

)

.

Using again Lemma 3.4 (1), we have

wn+2 = yn+2 + 1 = p− r1 − l2 − 1

or

wn+2 = yn+2 + 2 = p− r1 − l2.

Since (S2) does not hold, by (3.1), we obtain |R| > 1, and consequently {p−r1−l2−1, p−r1−l2} ⊂ R.

Thus wn+2 ∈ R, a contradiction.

2o wn+1 = yn+1 +1

pn+1 . Since yn+1 + 1 > 0, by Lemma 3.4 (1.1), we have

wn+1 = yn+1 + 1 = p− r1 − l2 ∈ R

or

wn+1 = −p+ yn+1 + 1 = −p+ p− r1 − l2.

Since p − r1 − l2 ∈ R and wn+1 ∈ A − B, we must have wn+1 = −r1 − l2, but, because (S3) holds,

by (3.2), p− r1 − l2 ∈ (L+ p), and hence −r1 − l2 ∈ L. So, wn+1 /∈ (A−B), a contradiction. Thus,

(y, z) ∈ (x− ε, x+ ε) \ (A−B)p, so (A−B)p is nowhere dense. It is also perfect, so it is a Cantor set.

Ad (3)–(5) First, we will show some additional conditions (C1), (C2) and (C3).


(C1) If conditions (S2) and (S3) do not hold, then every gap in (A−B)p is accumulated on the

left by infinitely many gaps and proper components of (A−B)p.

First, observe that 1 ∈ A − B. Indeed, if 1 /∈ A− B, then 1 ∈ R, and so l1 = r2 = 1. Since (S2)

does not hold, we have r1+ l2 < p−1. Hence r1+ l2+ l1+ r2 < p+1, i.e. (S3) holds, a contradiction.

Thus, 1 ∈ A−B.

Now, let x be the left endpoint of a gap in (A−B)p and (xi) ∈ (A−B)N be such that x =∑∞

i=1xi

pi.

Then, by Lemma 3.4 (2), there is n ∈ N such that

xi = p− 1 for i > n.

Let ε > 0 and m > n be such that 1pm

< ε. We will find intervals [y, z] ⊂ (x− ε, x) ∩ (A − B)p and

(s, t) ⊂ (x− ε, x) \ (A−B)p. Define y, z, s, t in the following way:

yi = zi = si = ti = xi for i ≤ m,

ym+1 = 0, zm+1 = 1, sm+1 = p− r1 − l2 − 2, tm+1 = sm+1 + 3 = p− r1 − l2 + 1,

yi = zi = 0, si = xi = p− 1, ti = −p+ 1 for i > m+ 1

and put y =∑∞

i=1yipi, z =

∑∞i=1

zipi, t =

∑∞i=1

tipi, s =

∑∞i=1

sipi. Observe that z > y, x > t and

t− s =3

pm+1−

∞∑

i=m+2

2p − 2

pi=

1

pm+1> 0.

Since 1 ∈ A−B, we have 1 /∈ R, so every element of R is greater than 1. Because (S2) does not hold,

from (3.1) we obtain p− r1 − l2 − 1 ∈ R. Hence p− r1 − l2 − 1 > 1, and therefore

sm+1 = p− r1 − l2 − 2 ≥ 1 = zm+1.

Consequently, s > z. Thus

0 < x− t < x− s < x− z < x− y =

∞∑

i=m+1

p− 1

pi=

1

pm< ε,

which implies [y, z] ⊂ (x− ε, x) and (s, t) ⊂ (x− ε, x).

Now we will show that [y, z] ⊂ (A−B)p. Let w ∈ [y, z]. Then

w ∈ [y, z] =[

ym+1, zm+1

]

=

[

ym+1, ym+1 +1

pm+1

]

.

Since ym+1 = 0 ∈ (A−B), zm+1 = 1 ∈ (A−B) and yi = zi = xi ∈ (A−B) for i ≤ m, the point w is

(m+1)-bi-obtainable. Since (S3) does not hold, by (3.2), we have R∩(p+L) = ∅. By Lemma 3.4 (4),

w is (i)-bi-obtainable for i > m+1, and by Lemma 3.4 (3), w ∈ (A−B)p, and thus [y, z] ⊂ (A−B)p.

Now, we will show that (s, t) ∩ (A − B)p = ∅. Let v ∈ (s, t). Assume on the contrary that there

exists a sequence (vi) ∈ (A−B)N such that v =∑∞

i=1vipi. Since sm+1 ≥ 1 > 0 and si = p− 1 > 0 for

i > m+ 1, we have sm < s. Thus sm < s < t < x, and so

v ∈ (s, t) ⊂

(

sm, sm +1

pm

)

=

(

sm, sm +∞∑

i=m+1

p− 1

pi

)

=

(

sm, xm +∞∑

i=m+1

xipi

)

= (sm, x) .

16 PIOTR NOWAKOWSKI

Using Lemma 3.4 (1), we have vm = sm or vm = x. If vm = x, by Lemma 3.4 (2’), x is not an

endpoint of a gap, a contradiction. So, vm = sm. We have

v ∈ (s, t) =

(

sm+1 +

∞∑

i=m+2

p− 1

pi, tm+1 −

∞∑

i=m+2

p− 1

pi

)

=

(

sm+1 +1

pm+1, tm+1 −

1

pm+1

)

=

(

sm+1 +1

pm+1, sm+1 +

3

pm+1−

1

pm+1

)

=

(

sm+1 +1

pm+1, sm+1 +

2

pm+1

)

.

Using again Lemma 3.4 (1), we have

vm+1 = sm+1 + 1 = p− r1 − l2 − 1

or

vm+1 = sm+1 + 2 = p− r1 − l2.

Since (S2) does not hold, by (3.1), we obtain |R| > 1, and consequently {p−r1−l2−1, p−r1−l2} ⊂ R.

Thus, vm+1 ∈ R, a contradiction.

Thus, (s, t) ⊂ (x− ε, x) \ (A−B)p, which finishes the proof of (C1).

The following (symmetric to (C1)) condition follows directly from (C1) and Remark 5.

(C2) If conditions (S1) and (S3) do not hold, then every gap in (A−B)p is accumulated on the

right by infinitely many gaps and proper components of (A−B)p.

Now, we will prove the last condition (C3).

(C3) If (S1∗) holds, but condition (S2) does not hold, then every gap in (A−B)p has an adjacent

interval (contained in (A−B)p) on the right.

Let x be the right endpoint of a gap in (A− B)p and (xi) ∈ (A− B)N be such that x =∑∞

i=1xi

pi.

By Lemma 3.4 (2), there is n ∈ N such that xi = −p+ 1 for i > n. Let

yi = xi for i ≤ n,

yi = 0 for i > n.

Put y =∑∞

i=1yipi. We will show that [x, y] ⊂ (A−B)p. Let w ∈ [x, y]. Since

w ∈ [x, y] =

[

y −∞∑

i=n+1

p− 1

pi, y

]

=

[

y −1

pn, y

]

,

there is q ∈ [− 1pn, 0] such that w = y + q. By Observation 3.3 (v), there is a sequence (wi)

∞i=n+1 ∈

〈−p+1, 0〉N such that∑∞

i=n+1wi

pi= q, and so y+

∑∞i=n+1

wi

pi= w. Since (S1∗) holds, by (3.1), we have

L = ∅. Therefore, 〈−p+1, 0〉 ⊂ A−B, and so wi ∈ A−B for i > n. For i ≤ n put wi = yi ∈ A−B.

We found a sequence (wi) ∈ (A − B)N such that w =∑∞

i=1wi

pi. Thus, w ∈ (A − B)p, which finishes

the proof of (C3).

Now, we can prove (3). Assume that (S1∗) holds, but (S2) does not hold. Then (S3) also does

not hold and (A−B)p is not an interval, by (1), so it has a gap. Hence the assumptions of (C1) and

(C3) are satisfied. Therefore, by (C1), every gap in (A−B)p is accumulated on the left by infinitely

many gaps and proper components of (A−B)p, and by (C3), every gap has an adjacent interval on

the right. Thus, (A−B)p is an L-Cantorval, which finishes the proof of (3).

Since minus L-Cantorval is an R-Cantorval, by Remark 5, (4) follows from (3).


Now, we prove (5). Assume that (S1∗), (S2∗), (S3) do not hold and at least one of (S2) or (S1)

also does not hold. If (S2) does not hold, then, by (C1), every gap in (A−B)p is accumulated on the

left by infinitely many gaps and proper intervals contained in (A − B)p. If (S1) also does not hold,

then, from (C2) we infer that every gap in (A− B)p is accumulated on the right by infinitely many

gaps and intervals, and thus we get the assertion.

Assume that (S1) holds, but (S2) does not hold. The other case again easily follows from Remark

5 and the fact that minus M-Cantorval is an M-Cantorval. First, we will show that −1 ∈ A − B or

−p + 2 ∈ A− B. If −1 /∈ A− B, then −1 ∈ L, and hence r1 = l2 = 1. Since (S2) does not hold, we

have l1 + r1 + l2 < p, and thus l1 3. From the fact that (S1)

holds, but (S1∗) does not, it follows that l1 + r2 = p− 1. We have

l1 + r2 − p− 1 = −2 ≥ −p+ 2.

The fact that (S1∗) does not hold and −1 ∈ L together with (3.1) gives us that L = {−1}. So,

−p+ 2 ∈ A−B. Thus, we proved that −1 ∈ A−B or p− 2 ∈ A−B.

We will assume that −1 ∈ A−B. If p− 2 ∈ A−B, then the proof is similar, with small difference,

which we will point out.

Let x be the right endpoint of a gap in (A− B)p and (xi) ∈ (A− B)N be such that x =∑∞

i=1xi

pi.

Then, by Lemma 3.4 (2), there is n ∈ N such that xi = −p + 1 for i > n. Let ε > 0 and m > n be

such that 1pm

< ε. We will find intervals [y, z] ⊂ (x, x+ε)∩ (A−B)p and (s, t) ⊂ (x, x+ε)\ (A−B)p.

Define y, z, s, t in the following way:

yi = zi = si = ti = xi for i ≤ m,

ym+1 = −1, zm+1 = 0, sm+1 = tm+1 = l1 + r2 − p− 1,

ym+2 = zm+2 = 0, sm+2 = p− r1 − l2 − 2, tm+2 = sm+2 + 3 = p− r1 − l2 + 1,

yi = zi = 0, si = p− 1, ti = −p+ 1 for i > m+ 1

and put y =∑∞

i=1yipi, z =

∑∞i=1

zipi, t =

∑∞i=1

tipi, s =

∑∞i=1

sipi. We will show that y, z, s, t ∈ (x, x+ ε).

Of course, y, z, s, t are greater than x. We have

y < z = x+

∞∑

i=m+1

p− 1

pi= x+

1

pm< x+ ε

and also

t− s =3

pm+2−

∞∑

i=m+3

2p − 2

pi=

1

pm+2> 0,

so s < t. Moreover, since (S1∗) does not hold, by (3.1), l1 + r2 − p ∈ L, and in consequence

l1 + r2 − p ≤ 0,

and thus

t = zm+l1 + r2 − p− 1

pm+1+p− r1 − l2 + 1

pm+2−

∞∑

i=m+3

p− 1

pi≤ z+

−1

pm+1+p− r1 − l2

pm+2< z+

−1

pm+1+

p

pm+2= z.

Hence s < t < z < x+ ε, which ends the proof that y, z, s, t ∈ (x, x+ ε).

18 PIOTR NOWAKOWSKI

We will show that [y, z] ⊂ (A−B)p. Let w ∈ [y, z]. Then

w ∈ [y, z] =

[

zm −1

pm+1, zm

]

= [ym+1, zm+1].

Since ym+1 = −1 ∈ A − B, zm+1 = 0 ∈ A − B and yi = zi = xi ∈ A − B for i ≤ m, the point w

is (m + 1)-bi-obtainable (if −1 /∈ A − B it suffices to change ym+1 = p − 1, zm+1 = p − 2). Since

(S3) does not hold, by (3.2), we have R ∩ (p + L) = ∅. By Lemma 3.4 (4), w is (i)-bi-obtainable for

i > m+ 1, so, by Lemma 3.4 (3), w ∈ (A−B)p, and thus [y, z] ⊂ (A−B)p ∩ (x, x+ ε).

Now, we will show that (s, t) ∩ (A − B)p = ∅. Let v ∈ (s, t). Assume on the contrary that there

exists a sequence (vi) ∈ (A−B)N such that v =∑∞

i=1vipi. Since (S2) does not hold, by (3.1), we have

p− r1 − l2 − 1 ∈ R, and so p− r1 − l2 − 1 ≥ 0. Therefore,

s = sm +l1 + r2 − p− 1

pm+1+

p− r1 − l2 − 2

pm+2+

∞∑

i=m+3

p− 1

pi

≥ sm +l1 + r2 − p− 1

pm+1+

−1

pm+2+

1

pm+2> sm −

p

pm+1= sm −

1

pm

and

t = sm +l1 + r2 − p− 1

pm+1+

p− r1 − l2 + 1

pm+2−

∞∑

i=m+3

p− 1

pi

< sm +l1 + r2 − p− 1

pm+1+

p

pm+2= sm +

l1 + r2 − p

pm+1≤ sm,

because l1 + r2 − p ≤ 0. Hence

v ∈

(

sm −1

pm, sm

)

=

(

xm −∞∑

i=m+1

p− 1

pi, sm

)

=

(

xm +∞∑

i=m+1

xipi, sm

)

= (x, sm).

Thus, using Lemma 3.4 (1), we obtain vm = sm or vm = x, but the second case is impossible (by

Lemma 3.4 (2’)). Hence vm = sm. We also have

s = sm+1 +p− r1 − l2 − 2

pm+2+

∞∑

i=m+3

p− 1

pi≥ sm+1 +

−1

pm+2+

1

pm+2= sm+1

and

t = sm+1 +p− r1 − l2 + 1

pm+2−

∞∑

i=m+3

p− 1

pi< sm+1 +

p

pm+2= sm+1 +

1

pm+1,

and therefore

v ∈

(

sm+1, sm+1 +1

pm+1

)

.

Thus, by Lemma 3.4 (1), vm+1 = sm+1 or vm+1 = sm+1+1 and since vm = sm, we have vm+1 = sm+1

or vm+1 = sm+1 + 1. However,

sm+1 + 1 = l1 + r2 − p /∈ A−B,

so vm+1 = sm+1. Since

v ∈ (s, t) =

(

sm+2 +

∞∑

i=m+3

p− 1

pi, sm+1 +

sm+2 + 3

pm+2−

∞∑

i=m+3

p− 1

pi

)

=

(

sm+2 +1

pm+2, sm+2 +

2

pm+2

)

,


we have, by Lemma 3.4 (1),

vm+2 = sm+2 + 1 = p− r1 − l2 − 1

or

vm+2 = sm+2 + 2 = p− r1 − l2,

but since (S2) does not hold, we have {p − r1 − l2 − 1, p − r1 − l2} ∩ (A − B) = ∅, a contradiction.

Thus, (s, t) ⊂ (x, x+ ε) \ (A−B)p. Finally, we obtain (5). �

Corollary 3.8. Let l, r, p ∈ N, p > 2, l + r < p. Then

(1) C(l, r, p) −C(l, r, p) = [−1, 1] if and only if

2l + r ≥ p or l + 2r ≥ p;

(2) C(l, r, p) −C(l, r, p) is a Cantor set if and only if

2l + 2r ≤ p;

(3) C(l, r, p) −C(l, r, p) is an M-Cantorval if and only if

2l + r p.

Proof. Putting l in the place of l1 and l2, and r in the place of r1 and r2, in the conditions from

Theorem 3.7, we receive the assertion. �

Example 6. Let r1 = 1, l1 = 1, r2 = 1, l2 = 2, p = 4. We will show that C(l1, r1, p)− C(l2, r2, p) =

[−1, 1]. Of course, l1 + r1 < p, l2 + r2 < p. Moreover, l1 + l2 + r2 ≥ p and l1 + l2 + r1 ≥ p, and

consequently conditions (S1) and (S2) are satisfied. So, C(l1, r1, p) − C(l2, r2, p) = [−1, 1]. At the

same time, 2l1 + 2r1 ≤ p and 2l2 + r2 ≥ p, thus, by Corollary 3.8, C(l2, r2, p)− C(l2, r2, p) = [−1, 1],

but C(l1, r1, p)− C(l1, r1, p) is a Cantor set.

Example 7. Let r1 = 2, l1 = 3, r2 = 3, l2 = 1, p = 7. We will show that C(l1, r1, p) − C(l2, r2, p)

is an L-Cantorval and, by symmetry, C(l2, r2, p) − C(l1, r1, p) is an R-Cantorval. We have l1 + r1 <

p, l2+ r2 < p. Moreover, l1+ r1+ r2 > p, so (S1∗) holds and l1+ r1+ l2 < p, r1+ l2+ r2 < p, therefore

(S2) does not hold. In consequence,

C(l1, r1, p)− C(l2, r2, p) = {−6,−5,−4,−3,−2,−1, 0, 1, 2, 5, 6}7

is an L-Cantorval, and

C(l2, r2, p)− C(l1, r1, p) = {−6,−5,−2,−1, 0, 1, 2, 3, 4, 5, 6}7

is an R-Cantorval. At the same time, 2l1 + r1 ≥ p and 2r2 + l2 ≥ p, thus, by Corollary 3.8,

C(l1, r1, p)− C(l1, r1, p) = C(l2, r2, p)− C(l2, r2, p) = [−1, 1].

Let us introduce the notion of a symmetric S-Cantor set. Observe that C(l, r, p) is symmetric with

respect to 12 if and only if l = r. Then the set C(l, p) := C(l, l, p) is called a symmetric S-Cantor set.

Corollary 3.9. Let l1, l2, p ∈ N, p > 2, 2l1 < p, 2l2 < p. Then

(1) C(l1, p)− C(l2, p) = [−1, 1] if and only if

2l1 + l2 ≥ p or l1 + 2l2 ≥ p;

20 PIOTR NOWAKOWSKI

(2) C(l1p)− C(l2, p) is a Cantor set if and only if

2l1 + 2l2 ≤ p;

(3) C(l1, p)− C(l2, p) is an M-Cantorval if and only if

2l1 + l2 p.

Proof. Putting l1 in the place of r1, and l2 in the place of r2 in Theorem 3.7, we obtain the assertion.

�

Using the above corollary, we can receive a result similar to that obtained by Kraft in [11] (although

with additional possibility).

Corollary 3.10. Let l, p ∈ N, p > 2, 2l < p. Then

(1) C(l, p)− C(l, p) = [−1, 1] if and only if

l

p≥

1

3;

(2) C(l, p)− C(l, p) is a Cantor set if and only if

l

p≤

1

4;

(3) C(l, p)− C(l, p) is an M-Cantorval if and only if

l

p∈

(

1

4,1

3

)

.

Example 8. Let l1 = 2, l2 = 1, p = 5. Then 2l1 + l2 ≥ p, so C(l1, p)− C(l2, p) = [−1, 1]. Moreover,l1p≥ 1

3 and l2p< 1

4 , therefore C(l1, p)− C(l1, p) = [−1, 1], but C(l2, p)− C(l2, p) is a Cantor set.

Example 9. Let l = 1, p = 3. Observe that the set C = C(l, p) is the classical Cantor ternary set.

By Corollary 3.10 we get the known result C − C = [−1, 1].

Example 10. Let l = 2, p = 7. Then lp∈ (14 ,

13), and therefore

C(l, p)− C(l, p) = {−6,−5,−4,−1, 0, 1, 4, 5, 6}7

is an M-Cantorval.

Remark 11. The classical Cantor set is often defined as {0, 2}3, that is, the set of all numbers in

[0, 1] that can be written in the ternary system using only zeros and twos. Above examples show

that Theorem 3.7 gives us new examples of Cantorvals and Cantor sets that can be described in the

similar manner. For example, the set {−6,−5,−4,−1, 0, 1, 4, 5, 6}7 from Example 10, which is an

M-Cantorval, can be described as the set of all numbers in [−1, 1] which can be written in septenary

system without using −3,−2, 2 and 3.

Acknowledgement. The author would like to thank Tomasz Filipczak for many fruitful conversa-

tions during the preparation of the paper.


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Institute of Mathematics, Lodz University of Technology, ul. Wolczanska 215, 93-005 Lodz, Poland

Institute of Mathematics, Czech Academy of Sciences, Zitna 25, 115 67 Prague 1, Czech Republic

Email address: [email protected]

http://arxiv.org/abs/2003.04214

arXiv:2102.11194v1 [math.CA] 22 Feb 2021

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