1 Introduction
Let
$\mathbb {Z}$
be the set of all integers. For
$C,W\subseteq \mathbb {Z}$
and
$c\in \mathbb {Z}$
, define
$$ \begin{align*} C+W:= \{c+w: c\in C,~w\in W\} \quad \text{and} \quad cW:=\{cw: c\in C\}. \end{align*} $$
For the set
$W=\{w\}$
consisting of a single element, we just write
$C+\{w\}$
as
$C+w$
for simplicity. If
$C+W=\mathbb {Z}$
, then C is called an additive complement to W. Furthermore, if no proper subset of C is an additive complement to W, then C is called a minimal additive complement to W. Sometimes, for simplicity, a minimal additive complement is just called a minimal complement throughout this article.
In 2011, Nathanson [Reference Nathanson9, Theorem 8] proved the following interesting theorem.
Theorem 1.1 (Nathanson’s theorem).
Let W be a nonempty, finite set of integers. In
$\mathbb {Z}$
, every complement to W contains a minimal complement to W.
Nathanson [Reference Nathanson9, Problem 11] also posed the following open problem.
Problem 1.2. 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?
In 2012, Chen and Yang [Reference Chen and Yang5, Theorem 1] considered Problem 1.2. For the situation
$\inf W=-\infty $
and
$\sup W=+\infty $
, they made the following progress.
Theorem 1.3 (Chen–Yang).
Let W be a set of integers with
$\inf W=-\infty $
and
$\sup W=+\infty $
. Then, there exists a minimal complement to W.
After Theorem 1.3, it therefore remains to deal with the case
$\inf W>-\infty $
or
$\sup W<+\infty $
. Without loss of generality, we may assume that
$\inf W>-\infty $
since, otherwise, we could consider
$-W$
instead of W. If
$\inf W=w_{0}>-\infty $
, then we can consider the set
$$ \begin{align*} W-(w_{0}-1):=\{w-(w_{0}-1):w\in W\} \end{align*} $$
instead of W. Thus, without loss generality, we may further assume that
$\inf W=1$
. Based on these observations, Chen and Yang [Reference Chen and Yang5, Theorem 2] proved the following theorem.
Theorem 1.4 (Chen–Yang).
Let
$W=\{1=w_1<w_2<\cdots \}$
be a set of integers and
$$ \begin{align*}\overline{W}=\mathbb{Z}^{+}\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots\},\end{align*} $$
where
$\mathbb {Z}^{+}$
is the set of all positive integers.
-
(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.
In 2019, Kiss et al. [Reference Kiss, Sándor and Yang7] pointed out that if
$$ \begin{align*} W=\bigcup\limits_{k=0}^{+\infty} [10^k,2\times 10^k], \end{align*} $$
then it is clear that
$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)=+\infty \quad \text{and} \quad \limsup_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_i})=+\infty. \end{align*} $$
Hence, the criterion
$\lim _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $
in Theorem 1.4(b) cannot be simply improved to
$\limsup _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $
since this will lead to a contradiction between Theorems 1.4(a) and 1.4(b) for this W. Therefore, it is meaningful to think of the following question asked by Yang (private communication).
Problem 1.5. Let
$W=\{1=w_1<w_2<\cdots \}$
be a set of integers and
$$ \begin{align*} \overline{W}=\mathbb{Z}^+\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots\}. \end{align*} $$
Determine the structure of sets W with
$\limsup _{i\rightarrow +\infty }(\overline {w_{i+1}}-\overline {w_i})=+\infty $
which do not have a minimal complement.
Along the same lines, Kiss et al. [Reference Kiss, Sándor and Yang7] investigated certain sets W with
$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)<+\infty \end{align*} $$
which have a minimal complement. Precisely, let X be a set of integers. If there exists a positive integer T such that
$x+T\in X$
for all sufficiently large integers
$x\in X$
, then X is said to be eventually periodic with period T. If W is eventually periodic with
$|\mathbb {N}\setminus W|=+\infty $
, then there is some integer
$T_W$
so that
$w+T_W\in W$
for sufficiently large
$w\in W$
, which means that
$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)\le T_W<+\infty. \end{align*} $$
Kiss, Sándor and Yang focused on eventually periodic sets and provided sufficient conditions and necessary conditions for eventually periodic sets to have a minimal additive complement.
For other related results about minimal additive complements, one may also refer to [Reference Alon, Kravitz and Larson1–Reference Chen and Fang4, Reference Davenport6–Reference Kwon8, Reference Ruzsa10–Reference Zhou12].
Extending Theorem 1.4(b), we answer Problem 1.5 by the following theorem which can be viewed as a partial solution to Problem 1.2.
Theorem 1.6. Let
$W=\{1=w_1<w_2<\cdots \}$
be a set of integers and
$$ \begin{align*} \overline{W}=\mathbb{Z}^+\setminus W=\{\overline{w_1}<\overline{w_2}<\cdots\}. \end{align*} $$
If there exists a sequence
$i_1<i_2<\cdots $
such that
$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \quad \text{and} \quad \limsup_{t\rightarrow+\infty}|\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|<+\infty, \end{align*} $$
then there does not exist a minimal complement to W.
The following example shows that our new theorem has some advantages compared with Theorem 1.4.
Example 1.7. Let
$$ \begin{align*} W=\{1\}\cup \bigcup\limits_{k=4}^{+\infty}[2^k+9,2^{k+1})=\{1=w_1<w_2<\cdots\}. \end{align*} $$
It is plain that
$$ \begin{align*} \overline{W}=\mathbb{Z}^+\setminus W=\bigcup\limits_{k=1}^{+\infty}[2^k,2^k+8]=\{\overline{w_1}<\overline{w_2}<\cdots\}. \end{align*} $$
Clearly,
$$ \begin{align*} \limsup_{i\rightarrow+\infty}(w_{i+1}-w_i)\le 10 \quad \text{and} \quad \liminf_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_i})\le 1. \end{align*} $$
Thus, both Theorems 1.4(a) and 1.4(b) fail to determine whether such a W has a minimal additive complement. However, it is easy to see that
$$ \begin{align*} \lim_{k\rightarrow\infty}(2^{k+1}-(2^k+8))=+\infty \end{align*} $$
and
$$ \begin{align*} \lim_{k\rightarrow\infty}|\overline{W}\cap(2^k+8,2^{k+1}+8)|=8. \end{align*} $$
Therefore, W does not have a minimal additive complement, thanks to the criterion from Theorem 1.6 with
$\overline {w_{i_{k}}}=2^k+8$
and
$\overline {w_{i_{k+1}}}=2^{k+1}+8$
.
Remark 1.8. We now give a simple explanation to show that the criterion given in Theorem 1.6 is essentially optimal. We first note that
$$ \begin{align*} \limsup_{i\rightarrow+\infty}(\overline{w_{i+1}}-\overline{w_i})\ge \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \end{align*} $$
for those W assumed in Theorem 1.6 which are consistent with the requirement of Problem 1.5. We next show that the condition
$$ \begin{align*} \limsup_{t\rightarrow+\infty}|\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|<+\infty \end{align*} $$
cannot be neglected. To see this, let
$W=\bigcup _{k=0}^{+\infty } [10^k,2\times 10^k]$
. It is clear that for this W, there exists a sequence
$i_1<i_2<\cdots $
with
$\overline {w_{i_t}}=10^{t+1}-1$
such that
$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \quad \text{and} \quad \limsup_{t\rightarrow+\infty} |\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|=+\infty. \end{align*} $$
Then, there exists a minimal additive complement to W by Theorem 1.4(a) since
$\limsup _{i\rightarrow +\infty }(w_{i+1}-w_i)=+\infty $
.
2 Proof of Theorem 1.6
The proof of Theorem 1.6 follows from that of Theorem 1.4(b) with refinements.
Suppose to the contrary that C is a minimal additive complement to W. Then, we clearly have
$\inf C=-\infty $
. Let
$$ \begin{align*}C\cap (-\infty,-\overline{w_{i_1+1}})=\{c_1>c_2>\cdots\}.\end{align*} $$
For any
$s\in \mathbb {Z}^+$
, we assume that
$$ \begin{align} \overline{w_{i_{u_s-1}+1}}\leq -c_s<\overline{w_{i_{u_s}+1}}. \end{align} $$
Since
$\limsup _{t\rightarrow +\infty }|\overline {W}\cap (\overline {w_{i_{t}}},\overline {w_{i_{t+1}}})|<+\infty $
, there is a constant K such that
$$ \begin{align*} \limsup_{t\rightarrow+\infty}|\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|=K, \end{align*} $$
which means that there is some
$t_W$
such that for any
$t\ge t_W$
,
$$ \begin{align} |\overline{W}\cap(\overline{w_{i_{t}}},\overline{w_{i_{t+1}}})|\le K. \end{align} $$
Our proof will be separated into three cases.
Case 1:
$\liminf _{s\rightarrow +\infty } (-c_s-\overline {w_{i_{u_s-1}+1}})<+\infty $
. Let
$$ \begin{align*} h:=\liminf_{s\rightarrow+\infty} (-c_s-\overline{w_{i_{u_s-1}+1}}). \end{align*} $$
Then,
$0\le h<+\infty $
. Define
$$ \begin{align*} H:=\{s:-c_s-\overline{w_{i_{u_s-1}+1}}=h\}. \end{align*} $$
It is clear that H is an infinite set since
$-c_s-\overline {w_{i_{u_s-1}+1}}$
are all integers. Let
$s_0\in H$
be any given integer. Then, there exist integers
$n_{s_0}\in \mathbb {Z}$
and
$w_{s_0}\in W$
such that
$$ \begin{align} n_{s_0}=c_{s_0}+w_{s_0} \quad \text{but}\quad n_{s_0}\neq c+w \end{align} $$
for any
$w\in W$
and
$c\in C$
with
$c\neq c_{s_0}$
, for otherwise,
$C\setminus \{c_{s_0}\}$
will be an additive complement to W, which contradicts the minimal property of C. Recall that
${\inf C=-\infty }$
and
$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty \end{align*} $$
from the condition of our theorem. There exists an integer
$s_0^*> s_0$
such that
$$ \begin{align} n_{s_0}-c_{s}>0 \end{align} $$
for all
$s\in H$
with
$s\ge s_0^*$
and
$$ \begin{align} \overline{w_{i_t+1}}-\overline{w_{i_t}}>|n_{s_0}|+h \end{align} $$
for all
$t\geq u_{s_0^*}-1$
. From (2.3) and (2.4),
$$ \begin{align} n_{s_0}-c_{s}\in \overline{W} \end{align} $$
for
$s\ge s_0^*$
. From (2.5) with
$t=u_s$
,
$$ \begin{align*} \overline{w_{i_{u_s}+1}}-\overline{w_{i_{u_s}}}>|n_{s_0}|+h=|n_{s_0}|-c_s-\overline{w_{i_{u_s-1}+1}}\geq|n_{s_0}|-c_s-\overline{w_{i_{u_s}}} \end{align*} $$
providing that
$s\ge s_0^*$
. In other words,
$$ \begin{align} n_{s_0}-c_s\leq |n_{s_0}|-c_s<\overline{w_{i_{u_s}+1}}. \end{align} $$
However, again from (2.5) with
$t=u_s-1$
,
$$ \begin{align} \overline{w_{i_{u_s-1}+1}}-\overline{w_{i_{u_s-1}}}>|n_{s_0}|+h\geq -n_{s_0}. \end{align} $$
$$ \begin{align} n_{s_0}-c_s\ge n_{s_0}+\overline{w_{i_{u_{s}-1}+1}}>\overline{w_{i_{u_{s}-1}}}, \end{align} $$
provided that
$s\ge s_0^*$
. We conclude from (2.6), (2.7) and (2.9) that
$$ \begin{align} \overline{w_{i_{u_{s}-1}}}<n_{s_0}-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_0}-c_{s}\in \overline{W} \end{align} $$
for all
$s_0,s\in H$
with
$s\ge s_0^*> s_0$
.
Now, we take
$K+2$
different integers
$s_{0,1},s_{0,2},\ldots ,s_{0,K+2}\in H$
. Then, there are
$K+2$
integers
$s_{0,1}^*,s_{0,2}^*,\ldots ,s_{0,K+2}^*$
such that for any j with
$1\le j\le K+2$
,
$$ \begin{align} \overline{w_{i_{u_{s}-1}}}<n_{s_{0,j}}-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_{0,j}}-c_{s}\in \overline{W} \quad (\mbox{for all} ~s\ge s_{0,j}^*) \end{align} $$
from (2.10). Now, let s be a sufficiently large integer so that
$$ \begin{align*}s>\max\{s_{0,1}^*,s_{0,2}^*,\ldots,s_{0,K+2}^*\}\end{align*} $$
and
$u_s-1\ge t_W$
. Then, for such s,
$$ \begin{align*} n_{s_{0,j}}-c_s\in \overline{W}\cap(\overline{w_{i_{u_s-1}}},\overline{w_{i_{u_s}+1}}) \quad (\mbox{for} ~1\le j\le K+2) \end{align*} $$
from (2.11). Moreover, from (2.2),
$$ \begin{align*} |\overline{W}\cap(\overline{w_{i_{u_s-1}}},\overline{w_{i_{u_s}+1}})| = |\overline{W}\cap(\overline{w_{i_{u_s-1}}},\overline{w_{i_{u_s}}})|+|\overline{W}\cap[\overline{w_{i_{u_s}}},\overline{w_{i_{u_{s}}+1}})| \le K+1, \end{align*} $$
from which it follows that there are at least two integers
$j_1, j_2$
with
$1\le j_1<j_2\le K+2$
such that
$$ \begin{align*} n_{s_{0,{j_1}}}-c_s=n_{s_{0,{j_2}}}-c_s \quad \text{that is}\quad n_{s_{0,{j_1}}}=n_{s_{0,{j_2}}}. \end{align*} $$
By (2.3), we clearly have
$$ \begin{align*} n_{s_{0,{j_1}}}=c_{s_{0,{j_1}}}+w_{s_{0,{j_1}}} \quad \text{and} \quad n_{s_{0,{j_2}}}=c_{s_{0,{j_2}}}+w_{s_{0,{j_2}}}, \end{align*} $$
which is a contradiction with (2.3) since
$n_{s_{0,{j_1}}}=n_{s_{0,{j_2}}}$
whereas
$c_{s_{0,{j_1}}}\neq c_{s_{0,{j_2}}}$
.
Case 2:
$\liminf _{s\rightarrow +\infty } (c_s+\overline {w_{i_{u_s}+1}})<+\infty $
. The proof is similar to Case 1. Let
$$ \begin{align*} \ell:=\liminf_{s\rightarrow+\infty} (c_s+\overline{w_{i_{u_s}+1}}). \end{align*} $$
Then,
$0\le \ell <+\infty $
. Define
$$ \begin{align*} L:=\{s:c_s+\overline{w_{i_{u_s}+1}}=\ell\}. \end{align*} $$
It is clear that L is an infinite set since
$c_s+\overline {w_{i_{u_s}+1}}$
are all integers. Let
$s_0\in H$
be any given integer. Then, there exist integers
$n_{s_0}\in \mathbb {Z}$
and
$w_{s_0}\in W$
such that
$$ \begin{align} n_{s_0}=c_{s_0}+w_{s_0} \quad \text{but}\quad n_{s_0}\neq c+w \end{align} $$
for any
$w\in W$
and
$c\in C$
with
$c\neq c_{s_0}$
, for otherwise,
$C\setminus \{c_{s_0}\}$
will be an additive complement to W, which contradicts with the minimal property of C. Since
${\inf C=-\infty }$
and
$$ \begin{align*} \lim_{t\rightarrow+\infty}(\overline{w_{i_t+1}}-\overline{w_{i_t}})=+\infty, \end{align*} $$
from the condition of our theorem, there exists an integer
$s_0^*> s_0$
such that
$$ \begin{align} n_{s_0}-c_{s}>0 \end{align} $$
for all
$s\in L$
with
$s\ge s_0^*$
and
$$ \begin{align} \overline{w_{i_t+1}}-\overline{w_{i_t}}>|n_{s_0}|+\ell \end{align} $$
for all
$t\geq u_{s_0^*}$
. From (2.12) and (2.13),
$$ \begin{align} n_{s_0}-c_{s}\in \overline{W} \end{align} $$
for
$s\ge s_0^*$
. By (2.14) with
$t=u_s+1$
,
$$ \begin{align} \overline{w_{i_{u_s+1}+1}}-\overline{w_{i_{{u_s}+1}}}>|n_{s_0}|+\ell\geq n_{s_0}, \end{align} $$
providing that
$s\ge s_0^*$
. From (2.1) and (2.16) for
$s\ge s_0^*$
,
$$ \begin{align} n_{s_0}-c_s<n_{s_0}+\overline{w_{i_{u_s}+1}}\leq n_{s_0}+\overline{w_{i_{u_s+1}}}<\overline{w_{i_{u_s+1}+1}}. \end{align} $$
However, using (2.16) again,
$$ \begin{align*} \overline{w_{i_{u_s}+1}}-\overline{w_{i_{u_s}}}>|n_{s_0}|+\ell= |n_{s_0}|+c_s+\overline{w_{i_{u_s}+1}}, \end{align*} $$
by the definition of
$\ell $
, from which it follows that
$$ \begin{align} n_{s_0}-c_s \geq -|n_{s_0}|-c_s>\overline{w_{i_{u_{s}}}}, \end{align} $$
provided that
$s\ge s_0^*$
. We conclude from (2.15), (2.17) and (2.18) that
$$ \begin{align*} \overline{w_{i_{u_{s}}}}<n_{s_0}-c_s<\overline{w_{i_{u_s+1}+1}} \quad \text{and} \quad n_{s_0}-c_{s}\in \overline{W} \end{align*} $$
for all
$s_0,s\in L$
with
$s\ge s_0^*> s_0$
.
The remaining arguments are the same as in the last paragraph of Case 1, so we omit them.
Case 3:
$ \liminf _{s\rightarrow +\infty } (-c_s-\overline {w_{i_{u_s-1}+1}})= \liminf _{s\rightarrow +\infty } (c_s+\overline {w_{i_{u_s}+1}})=+\infty $
. Since C is a minimal complement to W, for any given integer
$s_0\geq 1$
, there exist integers
$n_{s_0}\in \mathbb {Z}$
and
$w_{s_0}\in W$
such that
$$ \begin{align} n_{s_0}=c_{s_0}+w_{s_0} \quad \text{but}\quad n_{s_0}\neq c+w \end{align} $$
for any
$w\in W$
and
$c\in C$
with
$c\neq c_{s_0}$
. By the conditions of Case 3, there exist infinitely many integers
$s>s_0$
such that
$$ \begin{align*} -c_{s}-\overline{w_{i_{u_s-1}+1}}>|n_{s_0}| \quad \text{and} \quad c_s+\overline{w_{i_{u_s}+1}}>|n_{s_0}|, \end{align*} $$
which implies that
$$ \begin{align} n_{s_0}-c_s\leq |n_{s_0}|-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_0}-c_s\geq -|n_{s_0}|-c_s>\overline{w_{i_{u_s-1}+1}}. \end{align} $$
We conclude from (2.19) and (2.20) that
$$ \begin{align*} \overline{w_{i_{u_{s}-1}+1}}<n_{s_0}-c_s<\overline{w_{i_{u_s}+1}} \quad \text{and} \quad n_{s_0}-c_{s}\in \overline{W} \end{align*} $$
for any
$s>s_0$
.
Using similar arguments to those in Case 1, the validity of the last case clearly follows.
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