2 A lemma

Proof. We divide the proof into three cases.

Case 1: $|U|=\emptyset $ . Choose

$$ \begin{align*}J=\bigg\{a+2i:i=0, 1, \ldots, \bigg\lfloor\frac{b-a}{2}\bigg\rfloor-1\bigg\}\cup \{b\}.\end{align*} $$

Case 2: $|U|=1$ . Write $U=\{u_{1}\}$ . If $u_1-a$ is odd, choose

$$ \begin{align*}J=\bigg\{a+2i:i=0, 1, \ldots, \bigg\lfloor\frac{b-a}{2}\bigg\rfloor-1\bigg\}\cup \{b\}.\end{align*} $$

If $b-a$ and $u_1-a$ are even, choose

$$ \begin{align*}J=\{a\}\cup \bigg\{a+3+2i:i=0, 1, \ldots, \bigg\lfloor\frac{b-a}{2}\bigg\rfloor-3\bigg\}\cup \{b\}.\end{align*} $$

If $b-a$ is odd and $u_1-a$ is even, choose

$$ \begin{align*}J=\{a\}\cup \bigg\{a+3+2i:i=0, 1, \ldots, \bigg\lfloor\frac{b-a}{2}\bigg\rfloor-2\bigg\}\cup \{b\}.\end{align*} $$

Case 3: $|U|\geq 2$ and $u_{i+1}-u_{i}\geq 5$ for any adjacent integers $u_{i}, u_{i+1}\in U$ . If $b-a=8, 9$ or $10$ , then $|U|=2$ . Write $U =\{u_{1}, u_{2}\}$ .

Case 3.1: $b-a=8$ . Then, $\{u_{1}, u_{2}\}=\{a+1, a+6\}, \{a+1, a+7\}\text { or }\{a+2, a+7\}$ . We can choose $J=\{a, a+3, a+5, a+8\}$ .

Case 3.2: $b-a=9$ . Then, $\{u_{1}, u_{2}\}$ is one of $\{a+1, a+6\}, \{a+1, a+7\}, \{a+1, a+8\}$ , $\{a+2, a+7\}, \{a+2, a+8\}$ or $\{a+3, a+8\}$ . We can choose

$$ \begin{align*}J\kern1.5pt{=}\kern1.5pt\{a, a\kern1.6pt{+}\kern1.6pt3, a\kern1.6pt{+}\kern1.6pt5, a\kern1.6pt{+}\kern1.6pt7, a+9\}, \{a, a+2, a+4, a+6, a+9\}\text{ or }\{a, a+3, a+6, a+9\}.\end{align*} $$

Case 3.3: $b-a=10$ . Then, $\{u_{1}, u_{2}\}$ is one of

$$ \begin{align*}\{a+1, a+6\}, \{a+1, a+7\}, \{a+1, a+8\}, \{a+1, a+9\}, \{a+2, a+7\},\end{align*} $$

$$ \begin{align*}\{a+2, a+8\}, \{a+2, a+9\}, \{a+3, a+8\}, \{a+3, a+9\}, \{a+4, a+9\}.\end{align*} $$

We can choose

$$ \begin{align*}J=\{a, a+2, a+5, a+7, a+10\},\quad \{a, a+2, a+4, a+6, a+8, a+10\},\end{align*} $$

$$ \begin{align*}\{a, a+3, a+6, a+8, a+10\}\quad\text{or}\quad\{a, a+3, a+5, a+7, a+10\}.\end{align*} $$

From this point on, we can use induction on $b-a$ . If $b-a=11$ , then $|U|\geq 2$ . If $a+2 \in U$ , then $a+3 \notin U$ . By Case 3.1, we can choose

$$ \begin{align*}\widetilde{J}=\{j_{i}:i\geq1\}\subset [a+3, a+11]\end{align*} $$

with $a+3, a+11\in \widetilde {J}$ and $\widetilde {j}_{i+1}-\widetilde {j_{i}}\in \{2,3\}$ , and we take $J=\{a\}\cup \widetilde {J}.$ If $a+2 \notin U$ , then by Case 3.2, we can choose

$$ \begin{align*}\widetilde{J}=\{j_{i}:i\geq1\}\subset [a+2, a+11]\end{align*} $$

with $a+2, a+11\in \widetilde {J}$ and $\widetilde {j}_{i+1}-\widetilde {j_{i}}\in \{2,3\}$ , and we take $J=\{a\}\cup \widetilde {J}.$

Assume that the result holds whenever $8\leq b-a\leq m$ , where $m\geq 11$ . Now, we consider the case $b-a=m+1$ . If $a+2 \in U$ , then by the hypothesis, we can choose

$$ \begin{align*}\widetilde{J}=\{j_{i}:i\geq1\}\subset [a+3, a+m+1]\end{align*} $$

with $a+3, a+m+1\in \widetilde {J}$ and $\widetilde {j}_{i+1}-\widetilde {j_{i}}\in \{2,3\}$ , and we take $J=\{a\}\cup \widetilde {J}.$ If $a+2 \notin U$ , then by the hypothesis, we can choose

$$ \begin{align*}\widetilde{J}=\{j_{i}:i\geq1\}\subset [a+2, a+m+1]\end{align*} $$

with $a+2, a+m+1\in \widetilde {J}$ and $\widetilde {j}_{i+1}-\widetilde {j_{i}}\in \{2,3\}$ , and we take $J=\{a\}\cup \widetilde {J}.$

By induction, we see that there exists a set $J=\{j_i: i\geq 1\}\subset I$ with $a,b\in J$ and $j_{i+1}-j_{i}\in \{2,3\}$ . This completes the proof of Lemma 2.1.

3 Proof of Theorem 1.6

For $i\geq 2$ , write

$$ \begin{align*}\text{Gap}(V_{i})=\{-c_{i}+b_{j}: j=1, \ldots, i-1\}:= \{h_{j}^{(i)}: j=1, \ldots, i-1\}.\end{align*} $$

Let

$$ \begin{align*}V=\bigcup\limits_{i=1}^{\infty}V_{i}, \quad \text{Gap}(V)=\bigcup\limits_{i=1}^{\infty}\text{Gap}(V_{i}).\end{align*} $$

Then, we have the following three basic facts.

Fact A: $b_{i+1}-b_{i}\leq -5$ for all integers $i\geq 1$ . Clearly, $b_{2}=-6, b_{2}-b_{1}=-5$ . Suppose that $b_{i+1}-b_{i}\leq -5$ for all integers $i<k$ , where $k\geq 2$ . By item (ii), $\{b_{i}\}^{\infty }_{i=1}$ is a decreasing sequence. By item (iv), for any $k\geq 2$ ,

$$ \begin{align*}[-2c_{k-1}, -c_{k}+b_{k-1})\subseteq V_{k}.\end{align*} $$

By item (iii),

$$ \begin{align*}-2c_{k-1}\leq b_{k}-4-c_{k}<b_{k}-c_{k}<-c_{k}+b_{k-1}.\end{align*} $$

Thus,

$$ \begin{align*}b_{k}-c_{k}-i \in [-2c_{k-1}, -c_{k}+b_{k-1}) \quad\text{for } 0\leq i \leq 4,\end{align*} $$

and so

$$ \begin{align*}b_{k}-i=(b_{k}-c_{k}-i)+c_{k}\in c_{k}+V_{k} \quad\text{for } 0\leq i \leq 4.\end{align*} $$

Again, by the definition of $b_{k+1}$ , we have $b_{k+1}-b_{k}\leq -5$ .

Fact B: $C+V=\mathbb {Z}$ . By item (iv), $v_{j+1}-v_{j}\in \{1,2\}$ . Since $b_{k}\rightarrow -\infty $ , by item (ii), we have

$$ \begin{align*}(-\infty,6]\in C+V.\end{align*} $$

For any integer $n\geq 7$ , there exists an i such that $-c_{i-1}\leq n<-c_{i}$ . Hence,

$$ \begin{align*}-c_{i}+b_{1}<-c_{i-1}-c_{i}\leq n-c_{i}<-2c_{i},\end{align*} $$

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

Fact C: if we write $b_{s}=c+v$ for $s\geq 1, c\in C$ and $v\in V$ , then $c=c_{s}$ . For any $k>s$ , by item (iv), $b_{s}-c_{k}\notin V$ . Thus, $c\neq c_{k}$ for all $k>s$ . For any $k\leq s-1$ , noting that $-2c_{s-1}=\min (V\backslash V_{s-1})$ and

$$ \begin{align*}b_{s}-c_{k}\leq b_{s}-c_{s-1}<-2c_{s-1},\end{align*} $$

we have $b_{s}-c_{k}\notin V\backslash V_{s-1}$ . Moreover, by item (ii),

$$ \begin{align*}b_{s}\notin V_{s-1}+\{c_{1},\ldots,c_{s-1}\}.\end{align*} $$

Thus, $c\neq c_{k}$ for all $k\leq s-1$ .

It follows that C is a minimal complement to V. Write

$$ \begin{align*}d_{i}=b_{i}-c_{i} \quad\text{for } i=1,2,\ldots.\end{align*} $$

By Fact C, $d_i\in V$ for all $i\geq 1$ . We prove that $d_{i+1}-d_{i}\geq 8$ for all $i\geq 1$ . Since ${c_{i+1}<b_{i+1}+2c_{i}-3}$ and $b_{i+1}<0,c_{i+1}<0$ , we have

$$ \begin{align*}-d_{i+1}<2c_{i}\leq c_{i}+c_{1}=c_{i}-6<c_{i}<c_{i}-b_{1}\leq c_{i}-b_{i}=-d_{i}.\end{align*} $$

Hence,

$$ \begin{align*}d_{i+1}-d_{i}\geq -d_{i}-(c_{i}-6)+1= -b_{i}+7 \geq8.\end{align*} $$

Fix $i\geq 2$ . By Fact A, for $1\leq j\leq i-1$ ,

(3.1) $$ \begin{align} h_{j}^{(i)}-h_{j+1}^{(i)}=-c_{i}+b_{j}-(-c_{i}+b_{j+1})=b_{j}-b_{j+1}\geq 5.\end{align} $$

Moreover, by item (iii),

(3.2) $$ \begin{align} h_{i}^{(i+1)}-h_{1}^{(i)}=-c_{i+1}+b_{i}-(-c_{i}+b_{1}) &=c_{i}-c_{i+1}+b_{i}+1\nonumber\\ &>c_{i}-b_{i+1}-2c_{i}+3+b_{i}+1 \nonumber\\ &=-c_{i}+(b_{i}-b_{i+1})+4\geq15. \end{align} $$

By (3.1) and (3.2), the distance of all adjacent integers of $\text {Gap}(V)$ is no less than 5. For $i\geq 1$ , write

$$ \begin{align*}I_{i}=[d_{i}, d_{i+1}]\cap V.\end{align*} $$

By Lemma 2.1, there exists a set

$$ \begin{align*}W^{(i)}=\{w_{j}^{(i)}: j\geq 1\}\subset I_i\end{align*} $$

such that $w_{j+1}^{(i)}-w_{j}^{(i)}\in \{2,3\}$ and $d_{i}, d_{i+1}\in W^{(i)}$ . Put

$$ \begin{align*}W=\bigcup\limits_{i=1}^{\infty}W^{(i)}.\end{align*} $$

Then, $W=\{w_{i}\}_{i=1}^{\infty }$ satisfies ${w_{i+1}}-{w_{i}}\in \{2,3\}$ for all i.

Furthermore, by item (iii), we can choose suitable $c_{i}$ such that W is infinite and not eventually periodic. Since $|\mathbb {Z}\setminus (C+W)|<+\infty $ , we may assume that

$$ \begin{align*}\mathbb{Z}\setminus(C+W)=\{n_{i}: i=1, \ldots, k\}.\end{align*} $$

Since $C+V=\mathbb {Z}$ , for $i=1,\ldots ,k$ , there exist $c_{s(n_i)}\in C$ and $v_{s(n_i)}\in V$ such that

$$ \begin{align*}n_i=c_{s(n_i)}+v_{s(n_i)}.\end{align*} $$

Hence, $v_{s(n_i)}\notin W$ for all $i=1,\ldots ,k$ . By the construction of W, $v_{s(n_i)}+1\in W$ or ${v_{s(n_i)}-1\in W}$ . Put

$$ \begin{align*} l_i= \begin{cases} c_{s(n_i)}-1 &\text{if}\ v_{s(n_i)}+1\in W,\\ c_{s(n_i)}+1 &\text{otherwise}. \end{cases} \end{align*} $$

The $l_i$ may appear repeatedly. Without loss of generality, we may assume that

$$ \begin{align*}L=\{l_1,\ldots,l_m\}.\end{align*} $$

Then, $m\leq k$ . By item (iii), $l_i\not \in C$ for all $1\leq i\leq m$ . Write

$$ \begin{align*}C_0=C\cup L.\end{align*} $$

Since $\{b_{i}\}^{\infty }_{i=1}$ , $\{c_{i}\}^{\infty }_{i=1}$ are decreasing sequences and $b_1=-1$ , $c_1=b_2=-6$ , there exists an integer $h(n_i)\geq 2$ such that

(3.3) $$ \begin{align}b_{h(n_i)}<c_{s(n_i)}\leq b_{h(n_i)-1}.\end{align} $$

Write

$$ \begin{align*}t(k)=\max{\{h(n_1),\ldots,h(n_k)\}}-1.\end{align*} $$

For $j=1,\ldots , t(k)$ , let

(3.4) $$ \begin{align} C_{j}= \begin{cases} C_{j-1}\setminus \{c_{j}\} &\text{if}\ C_{j-1}\setminus \{c_{j}\}+W=\mathbb{Z},\\ C_{j-1}&\text{otherwise}. \end{cases} \end{align} $$

For $\mu =1,\ldots ,m$ , let

(3.5) $$ \begin{align} C_{t(k)+\mu}= \begin{cases} C_{t(k)+\mu-1}\setminus \{l_{\mu}\} &\text{if}\ C_{t(k)+\mu-1}\setminus \{l_{\mu}\}+W=\mathbb{Z},\\ C_{t(k)+\mu-1}&\text{otherwise}. \end{cases} \end{align} $$

Write

$$ \begin{align*}C^{*}=C_{t(k)+m}.\end{align*} $$

Then, $C^{*}+W=\mathbb {Z}$ . Now we shall prove $C^{*}$ is a minimal complement to W.

If there exists an integer $c_{j}\in C^{*}$ with $1\leq j\leq t(k)$ , then by the construction of $C_{j}$ in (3.4),

$$ \begin{align*}C_{j-1}\setminus \{c_{j}\}+W\neq \mathbb{Z}.\end{align*} $$

Thus, $C^{*}\setminus \{c_{j}\}+W\neq \mathbb {Z}$ . If there exists an integer $l_{\mu }\in C^{*}$ for some $1\leq \mu \leq m$ , then by the construction of $C_{t(k)+\mu }$ in (3.5),

$$ \begin{align*}C_{t(k)+\mu-1}\setminus \{l_{\mu}\}+W\neq \mathbb{Z}.\end{align*} $$

Thus, $C^{*}\setminus \{l_{\mu }\}+W\neq \mathbb {Z}$ .

Next, we shall show that for all $j>t(k)$ ,

$$ \begin{align*}b_j\not\in (C^{*}\cap L)+W.\end{align*} $$

In fact, if $b_{j}=(c_{s(n_i)}-1)+w$ for some $c_{s(n_i)}-1\in C^{*}$ and $w\in W$ , then

$$ \begin{align*}b_{j}-c_{s(n_i)}=w-1\geq 0.\end{align*} $$

By (3.3) and the fact that $\{b_{i}\}^{\infty }_{i=1}$ is a decreasing sequence, we see that the above inequality cannot hold. Similarly,

$$ \begin{align*}b_{j}\neq (c_{s(n_i)}+1)+w'\end{align*} $$

for any $c_{s(n_i)}+1\in C^{*}$ and $w'\in W$ . Since $C^{*}+W=\mathbb {Z}$ , there exist $c\in C, w\in W$ such that

$$ \begin{align*}b_{j}=c+w.\end{align*} $$

By Fact B, $c=c_{j}$ . Thus, $C^{*}\setminus \{c_{j}\}+W\neq \mathbb {Z}$ for all $j>t(k)$ . Therefore, $C^{*}$ is a minimal complement to W. This completes the proof of Theorem 1.6.