On minimal additive complements of integers

TL;DR

This paper investigates the existence of minimal additive complements in the integers, focusing on sets that are eventually periodic or not, thereby addressing a problem posed by Nathanson.

Contribution

It provides new insights into when minimal additive complements exist for eventually periodic and non-periodic sets, partially resolving Nathanson's problem.

Findings

Minimal complements exist for certain eventually periodic sets.

Non-periodic sets may lack minimal complements.

Results extend understanding of additive complement structures.

Abstract

Let $C,W\subseteq \mathbb{Z}$. If $C+W=\mathbb{Z}$, then the set $C$ is called an additive complement to $W$ in $\mathbb{Z}$. If no proper subset of $C$ is an additive complement to $W$, then $C$ is called a minimal additive complement. Let $X\subseteq \mathbb{N}$. If there exists a positive integer $T$ such that $x+T\in X$ for all sufficiently large integers $x\in X$, then we call $X$ eventually periodic. In this paper, we study the existence of a minimal complement to $W$ when $W$ is eventually periodic or not. This partially answers a problem of Nathanson.

Equations

$$\overline{W}=(\mathbb{Z}\cap (0,+\infty ))\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots \}.$$

$$S\mathrm{mod}m=\{r:r\in \{0,1,\dots ,m-1\},r\equiv s(\text{mod}m)\text{ for some }s\in S\}.$$

$$W=(m\mathbb{N}+X_m)\cup Y^{(0)}\cup Y^{(1)},$$

$$W=\{-3,-1,2,3,4,7,8,12,13,17,18,22,23,\dots \}.$$

$$C=\{j:0\le j\le m-1\text{and}|D_j\cap {\mathbb{Z}}^{-}|=+\infty \}.$$

$$C^{\prime }=\{d\in \mathbb{Z}:d\equiv c(\text{mod}m)\text{for some}c\in C\},$$

$$C_1^{\prime }=\{d\in \mathbb{Z}:d\equiv c(\text{mod}m)\text{for some}c\in C_1\},$$

$$C_2^{\prime }=\{d\in \mathbb{Z}:d\equiv c(\text{mod}m)\text{for some}c\in C_2\}.$$

$$C^{\prime }+(W\setminus Y^{(1)})\text{mod}m=C+X_m\text{mod}m=C_1,$$

$$(D^{\prime }\setminus \{d\})+Y^{(1)}⊉C_2^{\prime }.$$

$$D^{\prime }+Y^{(1)}⊉\{d\in \mathbb{Z}:d\equiv j+y(\text{mod}m)\}.$$

$$n=d+(n-d)\in D_c^{\prime }+(m\mathbb{N}+X_m)\subseteq D^{\prime }+W.$$

$$\Big{(}(D^{\prime}\setminus\{d\})+(W\setminus Y^{(1)})~{}\text{mod}~{}m\Big{)}\subseteq C+X_{m}~{}\text{mod}~{}m=C_{1}.$$

$$\Big{(}(D^{\prime}\setminus\{d\})+(W\setminus Y^{(1)})\Big{)}\cap C_{2}^{\prime}=\emptyset,$$

$$E_i^{-}=\{e:e<0,e\in E,e\equiv i(\text{mod}m)\}.$$

$$y_{+}=\text{max}\{y:y\in Y^{(0)}\cup Y^{(1)}\},$$

$$y_{-}=\text{min}\{y:y\in Y^{(0)}\cup Y^{(1)}\},$$

$${\chi }_E(k)=\{\begin{array}{r}1\text{, if }k\in E;\\ 0\text{, if }k\notin E.\end{array}$$

$${\text{v}}_A$$

$${\text{v}}_{A-m}$$

$$\text{makebox}[0.0pt]⋮$$

$${\text{v}}_{A-im}$$

$$\text{makebox}[0.0pt]⋮$$

$$C=\{l:K\le l<L,l\in E\}\text{mod}T.$$

$$L\le N_0+\text{min}\{0,y_{-}\},$$

$$L-K\ge y_0,$$

$$m\mid L-K,$$

$${\chi }_E(K+i)={\chi }_E(L+i)for-y_0\le i\le +y_0,$$

$$[K,L[\cap E_{i_j}\ne \varnothing \text{for all}{i}_j.$$

$$K-y_0\le K-y_{+}\le e<K<K+y_0,$$

$$L-y_0\le L\le e<L-y_{-}\le L+y_0,$$

$$K-y_0\le K+y_{-}-y_{+}\le e^{{}^{\prime }}<K+y_{+}-y_{-}\le K+y_0.$$

Full text

1. Institute of Mathematics, Budapest University of Technology and Economics, H-1529 B.O. Box, Hungary

2. School of Mathematics and Statistics, Nanjing University of Information

Science and Technology, Nanjing 210044, China

2010 Mathematics Subject Classifications: Primary 11B13, 11B34.

Key words and phrases: Additive complements, minimal complements, periodic sets.

1 Introduction

Let $\mathbb{N}$ denote the set of nonnegative integers and $\mathbb{Z}$ be the set of integers. For $A,B\subseteq\mathbb{Z}$ and $k\in\mathbb{Z}$, let $A+B=\{a+b:~{}a\in A,~{}b\in B\}$ and $kA=\{ka:a\in A\}$. If $A+B=\mathbb{Z}$, then $A$ is called an additive complement to $B$ in $\mathbb{Z}$. If no proper subset of $A$ is a complement to $B$, then $A$ is called a minimal complement to $B$ in $\mathbb{Z}$.

It is easy to see that if $A\subseteq\mathbb{Z}$ is a (minimal) complement to $B\subseteq\mathbb{Z}$, then $A$ is also a (minimal) complement to $B+d$, $d\in\mathbb{Z}$, where $B+d=\{b+d:b\in B\}$.

In 2011, Nathanson [4] proved the following theorem.

Nathanson’s theorem (See [4, Theorem 8]). Let $W$ be a nonempty, finite set of integers. In $\mathbb{Z}$, every complement to $W$ contains a minimal complement to $W$.

In the same paper, Nathanson also posed the following problem.

Problem (See [4, Problem 11]). Let $W$ be an infinite set of integers. Does there exist a minimal complement to $W$? Does there exist a complement to $W$ that does not contain a minimal complement?

For the second part of the above problem, in 2012, Chen and Yang [2] gave two infinite sets $W_{1}$ and $W_{2}$ of integers such that there exists a complement to $W_{1}$ that does not contain a minimal complement and every complement to $W_{2}$ contains a minimal complement.

For the first part of the above problem, in 2012, Chen and Yang [2] proved the following results.

Theorem A (See [2, Theorem 1]). Let $W$ be a set of integers with $\inf W=-\infty$ and $\sup W=+\infty$. Then there exists a minimal complement to $W$.

By Theorem A, now we only need to consider the cases $\inf W>-\infty$ or $\sup W<+\infty$. Without loss of generality, we may assume that $\inf W>-\infty$.

Theorem B (See [2, Theorem 2]). Let $W=\{1=w_{1}<w_{2}<\cdots\}$ be a set of integers and

[TABLE]

(a) If $\limsup_{i\rightarrow+\infty}(w_{i+1}-w_{i})=+\infty$, then there exists a minimal complement to $W$.

(b) If $\lim_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_{i}})=+\infty$, then there does not exist a minimal complement to $W$.

Let $W=\cup_{k=0}^{\infty}[10^{k},2\times 10^{k}]$. Then it is clear that both $\limsup_{i\rightarrow+\infty}(w_{i+1}-w_{i})=+\infty$ and $\limsup_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_{i}})=+\infty$ hold. Hence $\lim_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_{i}})=+\infty$ in Theorem B (b) cannot be changed to $\limsup_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_{i}})=+\infty$.

In this paper, we will give further results on Nathanson’s problem and deal with some sets $W$ do not satisfy the conditions of Theorem B.

First we give some definitions. Let $S\subseteq\mathbb{N}$. Denote by $S\mod m$ the set of residues of $S$ modulo $m$, i.e.,

[TABLE]

Let $X\subseteq\mathbb{N}$. If there exists a positive integer $T$ such that $x+T\in X$ for all $x\in X$, then we call $X$ periodic with period $T$. If $X\cup C$ is a periodic set for some finite set $C\subseteq\mathbb{N}$, then we call $X$ quasiperiodic. If there exists a positive integer $T$ such that $x+T\in X$ for all sufficiently large integers $x\in X$, then we call $X$ eventually periodic with period $T$. Clearly, a periodic set must be quasiperiodic and a quasiperiodic set must be eventually periodic. If $W$ is eventually periodic with $|\mathbb{N}\setminus W|=+\infty$, then both $\lim_{i\rightarrow+\infty}(w_{i+1}-w_{i})<+\infty$ and $\lim_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_{i}})<+\infty$ hold. Hence $W$ does not satisfy the conditions of Theorem B.

Suppose that $W$ is an eventually periodic set and $m$ is a positive period. By shifting a number, we may assume that $W$ has the following structure:

[TABLE]

where $X_{m}\subseteq\{0,1,\dots,m-1\},Y^{(0)}\subseteq\mathbb{Z}^{-},Y^{(1)}$ are finite sets with $Y^{(0)}~{}\text{mod}~{}m\subseteq X_{m}$ and $(Y^{(1)}~{}\text{mod}~{}m)\cap X_{m}=\emptyset$.

For example, if $W=\{2,4,7,8,9,12,13,17,18,22,23,27,28,\ldots\}$, then by shifting a number $5$, we may assume that

[TABLE]

Hence $m=5$, $X_{m}=\{2,3\}$, $Y^{(0)}=\{-3\}$, $Y^{(1)}=\{-1,4\}$.

In this paper, we study that what conditions are needed to ensure the existence of a minimal complement to $W$. First we prove a sufficient condition.

Theorem 1**.**

Let $W$ be defined in (1). If there exists a minimal complement to $W$, then there exists $C\subseteq\{0,1,\dots,m-1\}$ such that the following two conditions hold:

(a) $C+(X_{m}\cup Y^{(1)})\mod m=\{0,1,\dots,m-1\}$;

(b) For any $c\in C$, there exists $y\in Y^{(1)}$ such that $c+y\not\equiv c^{\prime}+x~{}(\text{mod}~{}m)$ for any $c^{\prime}\in C$ and $x\in X_{m}$.

Remark 1**.**

By the proof of Theorem 1, we know that Theorem 1 also holds when $Y^{(1)}$ is an infinite set with $|Y^{(1)}\cap\mathbb{Z}^{-}|<+\infty$.

Let $m=3$, $X_{m}=\{0\}$, $Y^{(1)}\subseteq 3\mathbb{N}+1$. By Theorem 1, we have the following corollary.

Corollary 1**.**

Let $Y\subseteq 3\mathbb{N}+1$ and $W=3\mathbb{N}\cup Y$. Then there does not exist a minimal complement to $W$.

Remark 2**.**

We can choose an infinite set $Y$ in Corollary 1 such that $W$ is not eventually periodic. Hence, there exists an infinite, not eventually periodic set $W\subseteq\mathbb{N}$ such that $w_{i+1}-w_{i}\in\{1,2,3\}$ for all $i$, and there does not exist a minimal complement to $W$.

Remark 3**.**

If $W\subseteq\mathbb{N}$ is a quasiperiodic set, then $Y^{(1)}=\emptyset$ and the condition (b) in Theorem 1 does not hold. Hence there does not exist a minimal complement to $W$.

In the next step we prove a necessary condition.

Theorem 2**.**

Let $W$ be defined in (1). Suppose that there exists $C\subseteq\{0,1,\dots,m-1\}$ such that the following two conditions hold:

(a) $C+(X_{m}\cup Y^{(1)})\mod m=\{0,1,\dots,m-1\}$;

(b) For any $c\in C$, there exists $y\in Y^{(1)}$ such that $c+y\not\equiv c^{\prime}+x~{}(\text{mod}~{}m)$ for any $c^{\prime}\in C\setminus\{c\}$ and $x\in X_{m}\cup Y^{(1)}$.

Then there exists a minimal complement to $W$.

By Theorems 1 and 2, we have the following corollary.

Corollary 2**.**

Let $W=(m\mathbb{N}+X_{m})\cup Y^{(0)}\cup\{a\},$ where $X_{m}\subseteq\{0,1,\dots,m-1\},~{}(Y^{(0)}\mod m)\subseteq X_{m}$ and $a\not\equiv x~{}(\text{mod}~{}m)$ if $x\in X_{m}$. Then there exists a minimal complement to $W$ if and only if there exists a subset $C\subseteq\{0,1,\dots,m-1\}$ such that:

(a) $C+(X_{m}\cup\{a\})\mod m=\{0,1,\dots,m-1\}$;

(b) For any $c\in C$, $c+a\not\equiv c^{\prime}+x~{}(\text{mod}~{}m)$, where $c^{\prime}\in C\setminus\{c\}$ and $x\in X_{m}$.

We see that Theorems 1 and 2 transfer Nathanson’s problem into a finite modulo version when $W$ is an eventually periodic set. In the next theorem, we give a sufficient and necessary condition, but we cannot bound the module.

Theorem 3**.**

Let W be defined in (1). There exists a minimal complement to $W$ if and only if there exists $T\in\mathbb{Z}^{+},m\mid T$, and $C\subseteq\{0,1,\dots,T-1\}$ such that

(a) $C+(X_{T}\cup Y^{(1)})\mod T=\{0,1,\dots,T-1\}$, where $X_{T}=\cup_{i=0}^{\frac{T}{m}-1}\{im+X_{m}\}$;

(b) for any $c\in C$, there exists $y\in Y^{(1)}$ such that $c+y\not\equiv c^{\prime}+x~{}(\text{mod}~{}T)$ for any $c^{\prime}\in C\setminus\{c\}$ and $x\in X_{T}\cup Y^{(1)}$.

Finally, as a complement to Remark 2, we give the following theorem.

Theorem 4**.**

There exists an infinite, not eventually periodic set $W\subseteq\mathbb{N}$ such that $w_{i+1}-w_{i}\in\{1,2\}$ for all $i$ and there exists a minimal complement to $W$.

Now we pose two problems for further research.

Problem 1**.**

We know that Theorem 1 also holds when $Y^{(1)}$ is infinite. Is Theorem 2 also true when $Y^{(1)}$ is infinite?

Problem 2**.**

Does there exist an explicit formula for the upper bound of $T$ in Theorem 3 using $m,Y^{(0)}$ and $Y^{(1)}$?

2 Proofs

Proof of Theorem 1.

Suppose that $D$ is a minimal complement to $W$. For $i\in\{0,1,\ldots,m-1\}$, let $D_{i}=\{d\in D:~{}d\equiv i~{}(\text{mod}~{}m)\}$ and

[TABLE]

For any $t\in\{0,1,\ldots,m-1\}\setminus C$, the set $\{d\in D:~{}d\equiv t~{}(\text{mod}~{}m)\}+W$ does not contain any sufficiently small negative integers. It follows from $D+W=\mathbb{Z}$ that $C+W~{}\text{mod}~{}m=\{0,1,\dots,m-1\}$. That is, $C+(X_{m}\cup Y^{(1)})~{}\text{mod}~{}m=\{0,1,\dots,m-1\}$.

Next we shall prove (b). Suppose that there exists $c\in C$ such that for any $y\in Y^{(1)}$ there exist $c^{\prime}\in C$ and $x\in X_{m}$ with $c+y\equiv c^{\prime}+x~{}(\text{mod}~{}m)$. We take an integer $d\in D$ such that $d\equiv c~{}(\text{mod}~{}m)$ and we shall prove that $D\setminus\{d\}$ is also a complement to $W$. For any integer $n$, write $n=d^{\prime}+w$, where $d^{\prime}\in D$ and $w\in W$.

Case 1. $d^{\prime}\not=d$. Then $n=d^{\prime}+w\in(D\setminus\{d\})+W$.

Case 2. $d^{\prime}=d$.

Subcase 2.1. $(\{w\}~{}\text{mod}~{}m)\subseteq X_{m}$. In this case, there exists a positive integer $k_{0}$ such that $w+km\in W$ for all integers $k\geq k_{0}$. Since $|D_{c}\cap\mathbb{Z}^{-}|=+\infty$, it follows that there exists an integer $k\geq k_{0}$ such that $d-km\in D$. Hence $n=(d-km)+(w+km)$, where $d-km\in D\setminus\{d\}$ and $w+km\in W$. That is, $n\in(D\setminus\{d\})+W$.

Subcase 2.2. $w\in Y^{(1)}$. Since $\{c+y\}~{}\text{mod}~{}m\subseteq C+X_{m}~{}\text{mod}~{}m$ for any $y\in Y^{(1)}$ and $d\equiv c~{}(\text{mod}~{}m),~{}w\in Y^{(1)}$, it follows that  $\{n\}~{}\text{mod}~{}m=\{d+w\}~{}\text{mod}~{}m\subseteq C+X_{m}~{}\text{mod}~{}m$. Hence there exist a $c^{\prime}\in D$ with $c^{\prime}~{}(\text{mod}~{}m)\in C$ and $x\in W$ with $x~{}\text{mod}~{}m\in X_{m}$ such that $n\equiv~{}c^{\prime}+x~{}(\text{mod}~{}m)$. We choose a sufficiently large integer $k$ such that $c^{\prime}-km\in D$, $c^{\prime}-km\not=d$ and $x+km\in W$. Hence $n=(c^{\prime}-km)+(x+km)$, where $c^{\prime}-km\in D\setminus\{d\}$ and $x+km\in W$.

Hence, $(D\setminus\{d\})+W=\mathbb{Z}$ which contradicts the fact that $D$ is a minimal complement. Therefore, (b) holds. ∎

Proof of Theorem 2.

Let $C_{1}=C+X_{m}~{}\text{mod}~{}m$, $C_{2}=\{0,1,\dots,m-1\}\setminus C_{1}$,

[TABLE]

[TABLE]

[TABLE]

By (a), we have $C^{\prime}+W=\mathbb{Z}$. Since $C+X_{m}~{}\text{mod}~{}m=C_{1}$, it follows that

[TABLE]

and so $\left(C^{\prime}+(W\setminus Y^{(1)})\right)\cap C_{2}^{\prime}=\emptyset$. It follows from (b) that $C_{2}^{\prime}\neq\emptyset$. Noting that $C^{\prime}+W=\mathbb{Z}$, we have $C^{\prime}+Y^{(1)}\supseteq C_{2}^{\prime}$. Since $Y^{(1)}$ is a finite set, by the proof of Nathanson’s theorem (See [4, Theorem 4, page 2015]), there exists $D^{\prime}\subseteq C^{\prime}$ such that $D^{\prime}+Y^{(1)}\supseteq C_{2}^{\prime}$ and for any $d\in D^{\prime}$,

[TABLE]

Next we shall prove that $D^{\prime}$ is a minimal complement to $W$.

For $i\in C$, let $D_{i}^{\prime}=\{d\in D^{\prime}:d\equiv i~{}(\text{mod}~{}m)\}$. First we prove that $|D_{i}^{\prime}\cap\mathbb{Z}^{-}|=+\infty$ for all $i\in C$. Suppose that there exists a $j\in C$ such that $|D_{j}^{\prime}\cap\mathbb{Z}^{-}|<+\infty$. By (b), there exists a $y\in Y^{(1)}$ such that $j+y\not\equiv c+x~{}(\text{mod}~{}m)$, where $c\in C\setminus\{j\}$, $x\in X_{m}\cup Y^{(1)}$ and so

[TABLE]

Noting that $(\{j+y\}~{}\text{mod}~{}m)\not\subseteq C+X_{m}~{}\text{mod}~{}m=C_{1}$, we have $(\{j+y\}~{}\text{mod}~{}m)\subseteq C_{2}$. It follows that $D^{\prime}+Y^{(1)}\not\supseteq C_{2}^{\prime}$, a contradiction. Hence, $|D_{i}^{\prime}\cap\mathbb{Z}^{-}|=+\infty$ for all $i\in C$.

Next we prove that $D^{{}^{\prime}}$ is a complement. For any integer $n\in C_{1}^{\prime}$, by $C+X_{m}~{}\text{mod}~{}m=C_{1}$, there exists $c\in C$ and $x\in X_{m}$ such that $n\equiv c+x~{}(\text{mod}~{}m)$. Since $|D_{c}^{\prime}\cap\mathbb{Z}^{-}|=+\infty$, there exists a sufficiently small negative integer $d\in D_{c}^{\prime}$ such that $n-d>0$. The congruences $n\equiv c+x~{}(\text{mod}~{}m)$ and $d\equiv c~{}(\text{mod}~{}m)$ imply that $n-d\equiv x~{}(\text{mod}~{}m)$. Hence, $n-d\in m\mathbb{N}+X_{m}$ and so

[TABLE]

Hence $C_{1}^{\prime}\subseteq D^{\prime}+W.$ On the other hand, $D^{\prime}+W\supseteq D^{\prime}+Y^{(1)}\supseteq C_{2}^{\prime}$. Therefore, $D^{\prime}+W=\mathbb{Z}$.

Finally, we prove that $D^{{}^{\prime}}$ is a minimal complement. For any $d\in D^{\prime}$, we have

[TABLE]

It follows that

[TABLE]

and so $(D^{\prime}\setminus\{d\})+W\not\supseteq C_{2}^{\prime}$. Hence $(D^{\prime}\setminus\{d\})+W\not=\mathbb{Z}$.

Therefore, $D^{\prime}$ is a minimal complement to $W$. ∎

Proof of Theorem 3.

Assume that the set $W$ satisfies the conditions of Theorem 3. Applying Theorem 2 with $m=T$, it follows that $W$ has a minimal complement.

Suppose that $W$ has a minimal complement $E$. We will prove that there exist a positive integer $T$ and a set $C\subseteq\{0,1,\dots,T-1\}$ satisfying the conditions of Theorem 3.

For $0\leq i<m$, let

[TABLE]

Let $0\leq i_{1}<i_{2}<\dots<i_{t}<m$ be the sequence of indices with $|E^{-}_{i_{j}}|=\infty$. It is clear that there exists an integer $N_{0}$ such that, $e\in E$ and $e\leq N_{0}$ imply that $e\in E_{i_{j}}$ for some $i_{j}$. It follows from Theorem 1 that $Y^{(0)}\cup Y^{(1)}\neq\emptyset$. Let

[TABLE]

[TABLE]

and $y_{0}=\max\{y_{+},-y_{-},y_{+}-y_{-}\}$. Let $\chi_{E}(k)$ denote the characteristic function of the set $E$, i.e.,

[TABLE]

Let $A=N_{0}+\text{min}\{0,y_{-}\}$. We consider the following vectors:

[TABLE]

It is clear that there are infinitely many vectors $\textbf{v}_{A},\textbf{v}_{A-m},\dots,\textbf{v}_{A-im},\dots,$ each of them has $2y_{0}+1$ coordinates, which are [math] or $1$. Since there are at most $2^{2y_{0}+1}$ different vectors, by the pigeon hole principle, there exist a vector v and an infinite sequence $0\leq k_{1}<k_{2}<\cdots{}$ such that $\textbf{v}_{A-k_{i}m}=\textbf{v}$ for all positive integer $i$.

Define $L=A-k_{1}m$ and choose a sufficiently large integer $k_{i}$ such that $k_{i}m-k_{1}m\geq y_{0}$ and $[A-k_{i}m,A-k_{1}m[\cap E_{i_{j}}\neq\emptyset$ for every index $i_{j}$. Let $K=A-k_{i}m$, $T=L-K$ and

[TABLE]

Now we shall prove that such an integer $T$ and a set $C$ satisfy the conditions of Theorem 3. By definitions, $K$ and $L$ have the following properties.

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

First we prove that $C+(X_{T}\cup Y^{(1)})~{}\text{mod}~{}T=\{0,1,\ldots,T-1\}$, where $X_{T}=\bigcup_{i=0}^{\frac{T}{m}-1}\{im+X_{m}\}$. For any integer $l$ with $K\leq l<L$, there exist $e\in E$ and $w\in W$ such that $l=e+w$. As $w\geq\text{min}\{0,y_{-}\}$, it follows from (2) that $e=l-w<L-\text{min}\{0,y_{-}\}\leq N_{0}$, and so $e\in E_{i_{j}}^{-}$ for some integer $i_{j}$. Suppose that $w\in Y^{(0)}\cup Y^{(1)}$. Then we have $y_{-}\leq w\leq y_{+}$ and $K-y_{+}\leq e=l-w<L-y_{-}$. We have three cases.

Case 1. $K-y_{+}\leq e<K$. Noting that

[TABLE]

by (5), we have $e+(L-K)\in E$. By $K-y_{+}\leq e<K$ and (3), it follows that $K\leq L-y_{+}\leq e+(L-K)<L$. Let $c=e+(L-K)$. Then $c~{}\text{mod}~{}T\in C$ and $l\equiv c+w~{}(\text{mod}~{}T)$. Hence $l~{}\text{mod}~{}T\in C+(X_{T}\cup Y^{(1)})~{}\text{mod}~{}T$.

Case 2. $K\leq e<L$. It follows that $e~{}\text{mod}~{}T\in C$ and so $l~{}\text{mod}~{}T\in C+(X_{T}\cup Y^{(1)})~{}\text{mod}~{}T$.

Case 3. $L\leq e<L-y_{-}$. Noting that

[TABLE]

by (5), we have $e-(L-K)\in E$. Since $L\leq e<L-y_{-}$ and (3), it follows that $K\leq e-(L-K)<K-y_{-}<L$. Let $c=e-(L-K)$. Then $c~{}\text{mod}~{}T\in C$ and $l\equiv c+w~{}(\text{mod}~{}T)$. Hence $l~{}\text{mod}~{}T\in C+(X_{T}\cup Y^{(1)})~{}\text{mod}~{}T$.

Suppose that $w\in T\mathbb{N}+X_{T}$. Since $w\geq 0$ and $e=l-w<L\leq N_{0}$, we have $e\in E_{i_{j}}^{-}$ for some integer $i_{j}$, and so $e\equiv i_{j}~{}(\text{mod}~{}m)$. It follows from (6) that there exists an integer $e^{{}^{\prime}}$ such that $e^{{}^{\prime}}\in E_{i_{j}}^{-}$, $K\leq e^{{}^{\prime}}<L$ and $e^{{}^{\prime}}\equiv i_{j}~{}(\text{mod}~{}m)$. Let $w\equiv x~{}(\text{mod}~{}m)$, where $x\in X_{m}$. Obviously, $l\equiv e+w\equiv i_{j}+x\equiv e^{{}^{\prime}}+x~{}(\text{mod}~{}m)$. By (4) there exists an integer $u$ with $0\leq u<\frac{T}{m}$ such that $l\equiv e^{{}^{\prime}}+um+x~{}(\text{mod}~{}T)$. Thus $l~{}\text{mod}~{}T\in C+(X_{T}\cup Y^{(1)})~{}\text{mod}~{}T$.

Therefore, $C+(X_{T}\cup Y^{(1)})~{}\text{mod}~{}T=\{0,1,\ldots,T-1\}$.

In the next step we show that the second condition of Theorem 3 holds. For any integer $e\in E$ with $K\leq e<L$, there exists a $w\in W$ such that $e+w\neq e^{{}^{\prime}}+w^{{}^{\prime}}$ for any $e^{\prime}\neq e$, $e^{\prime}\in E$ and $w^{\prime}\in W$.

If $w\in(m\mathbb{N}+X_{m})\cup Y^{(0)}$, by $e<L\leq N_{0}$ and $e\in E$, then there exists a positive integer $s$ such that $e+w=(e-sm)+(w+sm)$, where $e-sm\in E$ and $w+sm\in W$. This is a contradiction.

Now we may assume that $w\in Y^{(1)}$. It is enough to prove that $e+w\not\equiv e^{{}^{\prime}}+w^{{}^{\prime}}~{}(\text{mod}~{}T)$ for any $e^{{}^{\prime}}(\not=e)\in E$, $K\leq e^{{}^{\prime}}<L$ and $w^{{}^{\prime}}\in X_{T}\cup(Y^{(1)}~{}\text{mod}~{}T)$. Suppose that such $e^{\prime}$ and $w^{\prime}$ exist, i.e., $e+w\equiv e^{\prime}+w^{\prime}~{}(\text{mod}~{}T)$.

If $w^{{}^{\prime}}\in X_{T}$, then $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}+tT$ for some integer $t$. Hence there exists a positive integer $s$ such that $e+w=(e^{{}^{\prime}}-sm)+(w^{{}^{\prime}}+sm+tT)$, where $e^{{}^{\prime}}-sm\in E$ and $w^{{}^{\prime}}+sm+tT\in W$. This is a contradiction.

Now we assume $w^{{}^{\prime}}\in Y^{(1)}$. Then $K+y_{-}\leq e+w,e^{{}^{\prime}}+w^{{}^{\prime}}<L+y_{+}$. By $T=L-K\geq y_{0}\geq y_{+}-y_{-}$, it follows that either $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}$ or $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}+T$ or $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}-T$.

Case 1. $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}$. Then we have $e=e^{{}^{\prime}}$ and $w=w^{{}^{\prime}}$, a contradiction.

Case 2. $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}+T$. It follows that $K+y_{-}\leq e^{{}^{\prime}}+w^{{}^{\prime}}<K+y_{+}$, and so

[TABLE]

By (5), we have $e^{{}^{\prime}}+T\in E$, and then $e+w$ has another representation $(e^{{}^{\prime}}+T)+w^{{}^{\prime}}$. Therefore $w=w^{{}^{\prime}}$ and $e=e^{{}^{\prime}}+T$, which contradicts with $K\leq e,e^{{}^{\prime}}<L$.

Case 3. $e+w=e^{{}^{\prime}}+w^{{}^{\prime}}-T$. Then we have $L+y_{-}\leq e^{{}^{\prime}}+w^{{}^{\prime}}<L+y_{+}$. Thus $L-y_{0}\leq L+y_{-}-y_{+}\leq e^{{}^{\prime}}<L+y_{+}-y_{-}\leq L+y_{0}$. It follows from (5) that $e^{{}^{\prime}}-T\in E$ which implies that $e+w=(e^{{}^{\prime}}-T)+w^{{}^{\prime}}$. Therefore $w=w^{{}^{\prime}}$ and $e=e^{{}^{\prime}}-T$, which is a contradiction because $K\leq e,e^{{}^{\prime}}<L$.

The proof of Theorem 3 is completed. ∎

Proof of Theorem 4.

By induction we can construct $\{d_{i}\}_{i=1}^{\infty},~{}\{W_{i}\}_{i=1}^{\infty}$ and  $\{c_{i}\}_{i=1}^{\infty}$ such that

(i) $d_{1}=-1,~{}W_{1}=\{1,2,\ldots,12\},~{}c_{1}=-3$;

(ii) $d_{i}$ is the largest negative integer $\not\in W_{i-1}+\{c_{1},c_{2},\ldots,c_{i-1}\}$ for $i\geq 2$;

(iii) $c_{i}<d_{i}+2c_{i-1}$ for all $i\geq 2$;

(iv) for $i\geq 2$, let $W_{i}=W_{i-1}\cup\left([-2c_{i-1},-2c_{i}-1]\setminus\cup_{j=1}^{i-1}\{-c_{i}+d_{j}\}\right).$

Let $W=\cup_{i=1}^{\infty}W_{i}$ and $C=\{c_{i}\}_{i=1}^{\infty}$.

Now we prove that $C$ is a minimal complement to $W$.

First we prove $d_{i+1}-d_{i}\leq-2$ for all integers $i\geq 1$. Clearly $d_{2}=-3,~{}d_{2}-d_{1}=-2$. Suppose that $d_{i+1}-d_{i}\leq-2$ for all integers $i<k~{}(k\geq 2)$. Since

[TABLE]

[TABLE]

it follows that $d_{k}-c_{k},~{}d_{k}-1-c_{k}\in W_{k}$ and then  $d_{k},d_{k}-1\in W_{k}+\{c_{k}\}$. Hence $d_{k+1}\leq d_{k}-2$. By (iv), we have $w_{j+1}-w_{j}\in\{1,2\}$. Since $d_{k}\rightarrow-\infty$, by (ii) we have $(-\infty,9]\subseteq W+C$. For any integer $n\geq 10$, there exists an $i$ such that $-c_{i-1}\leq n<-c_{i}$. Hence

[TABLE]

and so $n-c_{i}\in W_{i}$, that is, $n\in W_{i}+\{c_{i}\}$. Therefore, $W+C=\mathbb{Z}$.

Next, we prove that the complement $C$ is minimal. For any positive integer $i$, we write $d_{i}=c+w$ with $c\in C$ and $w\in W$. Now we shall prove that $c=c_{i}$. By (iv), we have $d_{i}-c_{j}\not\in W$ for all integers $j>i$. Hence $c\not=c_{j}$ for all integers $j>i$. Since $-2c_{i-1}$ is the minimal value of $W\setminus W_{i-1}$ and for any positive integers $j\leq i-1$, $d_{i}-c_{j}\leq d_{i}-c_{i-1}<-2c_{i-1}$, it follows that $d_{i}-c_{j}\not\in W\setminus W_{i-1}$ for all positive integer $j\leq i-1$. Noting that $d_{i}\not\in W_{i-1}+\{c_{1},\ldots,c_{i-1}\}$, we have $d_{i}\not\in W+\{c_{1},c_{2},\ldots,c_{i-1}\}$. Hence $c=c_{i}$.

Therefore, $C$ is a minimal complement to $W$. Furthermore, by (iii), we can choose suitable $c_{i}$ such that $W$ is infinite and not eventually periodic. ∎

3 Acknowledgement

This work was done during the third author visiting to Budapest University of Technology and Economics. He would like to thank Dr. Sándor Kiss and Dr. Csaba Sándor for their warm hospitality. We would like to thank the anonymous referee very much for the detailed comments.