Proof. Fix $c\in C$ . Since C is minimal, the proper subset $C\setminus \{c\}$ is not an h-fold additive complement to W. Hence, there exists an integer $n_c$ such that

$$ \begin{align*} n_c\notin h(C\setminus\{c\})+W. \end{align*} $$

However, $hC+W=\mathbb Z$ , so there are $x_1,\ldots ,x_h\in C$ and $w_c\in W$ with

$$ \begin{align*} n_c=x_1+\cdots+x_h+w_c. \end{align*} $$

If none of the $x_i$ were equal to c, then $x_1+\cdots +x_h\in h(C\setminus \{c\})$ , and this would put $n_c$ in $h(C\setminus \{c\})+W$ , contrary to the choice of $n_c$ . Therefore, at least one of the $x_i$ is equal to c. Let $r_c$ be the number of indices i for which $x_i=c$ . Then, $1\leq r_c\leq h$ . Let $B_c$ be the sum of the remaining $h-r_c$ terms. Every term contributing to $B_c$ lies in $C\setminus \{c\}$ , so $B_c\in (h-r_c)(C\setminus \{c\})$ and $ n_c=B_c+r_c c+w_c. $ Put $N_c=n_c-B_c$ . Then, $N_c=r_c c+w_c$ .

Now, let $y\in C\setminus \{c\}$ . Suppose, to obtain a contradiction, that $N_c-r_cy\in W$ . Since $B_c$ is a sum of $h-r_c$ elements of $C\setminus \{c\}$ and $r_cy$ is a sum of $r_c$ elements of $C\setminus \{c\}$ , the integer $B_c+r_cy$ belongs to $h(C\setminus \{c\})$ . Hence,

$$ \begin{align*} n_c=B_c+r_cy+(N_c-r_cy) \end{align*} $$

belongs to $h(C\setminus \{c\})+W$ , which contradicts the defining property of $n_c$ . Thus, ${N_c-r_cy\notin W}$ . If $W\subseteq \mathbb Z_{>0}$ and $N_c-r_cy>0$ , then the positive integer $N_c-r_cy$ lies in $\mathbb Z_{>0}\setminus W$ .

Proof. If each of the h labels occurred at most K times, then the total number of objects would be at most $hK$ , which contradicts the existence of $hK+1$ objects. Thus, some label occurs at least $K+1$ times. The second assertion is the usual pigeonhole principle: an injective map from a set of cardinality $K+1$ to a set of cardinality at most K cannot exist.

Proof of Theorem 1.1.

Assume, for contradiction, that there is a minimal h-fold additive complement C to W. Since $\inf W=1$ , we have $W\subseteq \mathbb Z_{>0}$ . The hypotheses imply $K\geq 3$ , since $a_{t-1},a_t,a_{t+1}$ all belong to $\overline W\cap (a_{t-2},a_{t+2})$ for every $t\geq T$ . We first prove that $\inf C=-\infty $ . If $\inf C=m$ for some integer m, then every element of $hC$ is at least $hm$ and every element of $hC+W$ is at least $hm+1$ . This is impossible because $hC+W=\mathbb Z$ . Therefore, C is unbounded below.

After discarding finitely many elements if necessary, enumerate the elements of ${C\cap (-\infty ,-a_3)}$ in strictly decreasing order, $ c_1>c_2>\cdots. $ For each $s\geq 1$ , apply Lemma 2.1 to $c_s$ . We obtain integers $n_s$ , $r_s\in \{1,\ldots ,h\}$ , $B_s\in (h-r_s)(C\setminus \{c_s\})$ and $w_s\in W$ such that, with $N_s=n_s-B_s$ ,

$$ \begin{align*} N_s=r_s c_s+w_s \end{align*} $$

and

$$ \begin{align*} N_s-r_s y\notin W \quad\text{for every }y\in C\setminus\{c_s\}. \end{align*} $$

Since the set of possible values of $r_s$ is finite, there exists an integer $r\in \{1,\ldots ,h\}$ such that $r_s=r$ for infinitely many s. Let $ e_1>e_2>\cdots $ be the subsequence of the $c_s$ for which $r_s=r$ , still written in decreasing order. For each m, let $n_m$ , $N_m$ , $B_m$ and $w_m$ denote the corresponding private data, so that

(3.1) $$ \begin{align} N_m=r e_m+w_m, \end{align} $$

(3.2) $$ \begin{align} N_m-r y\notin W \quad\text{for every }y\in C\setminus\{e_m\}. \end{align} $$

Because $e_m\to -\infty $ , after discarding finitely many initial terms, we may assume that $-r e_m\geq a_2$ for every m. For each m, there is then a unique integer $u_m\geq 2$ such that

(3.3) $$ \begin{align} a_{u_m-1}\leq -r e_m<a_{u_m}. \end{align} $$

The sequence $u_m$ tends to infinity. Indeed, if $u_m$ were bounded along an infinite subsequence, then $-r e_m$ would be bounded above along that subsequence by one of the finitely many corresponding $a_u$ , which contradicts $e_m\to -\infty $ .

Define

(3.4) $$ \begin{align} \alpha_m=-r e_m-a_{u_m-1}, \quad \beta_m=r e_m+a_{u_m}. \end{align} $$

By (3.3), both $\alpha _m$ and $\beta _m$ are nonnegative integers, and $\beta _m>0$ unless $-re_m=a_{u_m}$ , which cannot occur because the right inequality in (3.3) is strict. We split into three exhaustive cases.

First, suppose that

$$ \begin{align*} \rho = \liminf_{m\to\infty}\alpha_m<+\infty. \end{align*} $$

Since the $\alpha _m$ are nonnegative integers, $\rho $ is a nonnegative integer and the set

$$ \begin{align*} H=\{m:\alpha_m=\rho\} \end{align*} $$

is infinite. Fix $m_0\in H$ and write $N=N_{m_0}$ . Since $e_m\to -\infty $ , there exists $M_1>m_0$ such that

(3.5) $$ \begin{align} N-r e_m>0 \end{align} $$

for all $m\in H$ with $m\geq M_1$ . Since $a_{t+1}-a_t\to \infty $ and $u_m\to \infty $ , there exists $M_2\geq M_1$ such that, for all $m\in H$ with $m\geq M_2$ , we have $u_m\geq T$ and

(3.6) $$ \begin{align} a_{u_m-1}-a_{u_m-2}>|N|+\rho, \quad a_{u_m+1}-a_{u_m}>|N|+\rho. \end{align} $$

Let $m\in H$ with $m\geq M_2$ . Because $m>m_0$ , the elements $e_m$ and $e_{m_0}$ are distinct. From (3.2), with $y=e_m$ , and from (3.5), we get

(3.7) $$ \begin{align} N-r e_m\in\overline W. \end{align} $$

Since $m\in H$ , (3.4) gives

(3.8) $$ \begin{align} -r e_m=a_{u_m-1}+\rho. \end{align} $$

Using the first inequality in (3.6),

$$ \begin{align*} a_{u_m-1}-a_{u_m-2}>|N|+\rho\geq -N-\rho. \end{align*} $$

Adding $N+\rho +a_{u_m-2}$ to both sides gives $ N+a_{u_m-1}+\rho>a_{u_m-2}. $ Together with (3.8), this is

(3.9) $$ \begin{align} N-r e_m>a_{u_m-2}. \end{align} $$

Using the second inequality in (3.6),

$$ \begin{align*} a_{u_m+1}=a_{u_m}+(a_{u_m+1}-a_{u_m})>a_{u_m}+|N|+\rho. \end{align*} $$

Since $a_{u_m}>a_{u_m-1}$ and $|N|\geq N$ , this implies

(3.10) $$ \begin{align} a_{u_m+1}>a_{u_m-1}+N+\rho=N-r e_m. \end{align} $$

From (3.7), (3.9) and (3.10), we have proved that, for every fixed $m_0\in H$ , there exists $M(m_0)$ such that

(3.11) $$ \begin{align} N_{m_0}-r e_m\in\overline W\cap(a_{u_m-2},a_{u_m+1}) \subseteq\overline W\cap(a_{u_m-2},a_{u_m+2}) \end{align} $$

whenever $m\in H$ and $m\geq M(m_0)$ .

Choose $K+1$ distinct indices $m_{0,1},\ldots ,m_{0,K+1}$ in H. Since H is infinite, we may choose $m\in H$ so large that

$$ \begin{align*} m\geq \max_{1\leq j\leq K+1}M(m_{0,\, j}) \quad\text{and}\quad u_m\geq T. \end{align*} $$

Then, (3.11) gives

$$ \begin{align*} N_{m_{0,\, j}}-r e_m\in \overline W\cap(a_{u_m-2},a_{u_m+2}) \quad(1\leq j\leq K+1). \end{align*} $$

The latter set has cardinality at most K by the hypothesis of the theorem. Hence, by Lemma 2.2, two of these $K+1$ integers are equal. Thus, there exist $1\leq j_1<j_2\leq K+1$ such that $ N_{m_{0,\, j_1}}-r e_m=N_{m_{0,\, j_2}}-r e_m $ and, hence,

(3.12) $$ \begin{align} N_{m_{0,\, j_1}}=N_{m_{0,\, j_2}}. \end{align} $$

By (3.1), $ N_{m_{0,\, j_2}}=r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}}. $ Substituting this into (3.12) yields

(3.13) $$ \begin{align} N_{m_{0,\, j_1}}=r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}}. \end{align} $$

The element $e_{m_{0,\, j_2}}$ is not equal to $e_{m_{0,\, j_1}}$ . Also, $B_{m_{0,\, j_1}}\in (h-r)(C\setminus \{e_{m_{0,\, j_1}}\})$ . Therefore, $B_{m_{0,\, j_1}}+r e_{m_{0,\, j_2}}$ belongs to $h(C\setminus \{e_{m_{0,\, j_1}}\})$ . Since $n_{m_{0,\, j_1}}=B_{m_{0,\, j_1}}+N_{m_{0,\, j_1}}$ , (3.13) gives

$$ \begin{align*} n_{m_{0,\, j_1}} =B_{m_{0,\, j_1}}+r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}} \in h(C\setminus\{e_{m_{0,\, j_1}}\})+W, \end{align*} $$

which contradicts the private choice of $n_{m_{0,\, j_1}}$ . This contradiction disposes of the first case.

We may now assume that

$$ \begin{align*} \liminf_{m\to\infty}\alpha_m=+\infty. \end{align*} $$

Suppose next that

$$ \begin{align*} \ell = \liminf_{m\to\infty}\beta_m<+\infty. \end{align*} $$

Since the $\beta _m$ are positive integers, $\ell $ is a positive integer and the set $ L=\{m:\beta _m=\ell \} $ is infinite. Fix $m_0\in L$ and set $N=N_{m_0}$ . As before, there exists $M_1>m_0$ such that

(3.14) $$ \begin{align} N-r e_m>0 \end{align} $$

for all $m\in L$ with $m\geq M_1$ . Since $u_m\to \infty $ and the gaps $a_{t+1}-a_t$ tend to infinity, there exists $M_2\geq M_1$ such that, for all $m\in L$ with $m\geq M_2$ , we have $u_m\geq T$ and

(3.15) $$ \begin{align} a_{u_m}-a_{u_m-1}>|N|+\ell, \quad a_{u_m+2}-a_{u_m+1}>|N|+\ell. \end{align} $$

Let $m\in L$ with $m\geq M_2$ . Since $m>m_0$ , we may use (3.2) with $y=e_m$ ; together with (3.14), it gives

(3.16) $$ \begin{align} N-r e_m\in\overline W. \end{align} $$

Because $m\in L$ , we have

(3.17) $$ \begin{align} r e_m+a_{u_m}=\ell \quad\text{so}\ -r e_m=a_{u_m}-\ell. \end{align} $$

The first inequality in (3.15) gives

$$ \begin{align*} a_{u_m}-a_{u_m-1}>|N|+\ell\geq -N+\ell. \end{align*} $$

Adding $N-\ell +a_{u_m-1}$ to both sides gives $ N+a_{u_m}-\ell>a_{u_m-1}. $ Using (3.17), this becomes

(3.18) $$ \begin{align} N-r e_m>a_{u_m-1}. \end{align} $$

For the upper bound, observe that

$$ \begin{align*} a_{u_m+2}-a_{u_m} =(a_{u_m+2}-a_{u_m+1})+(a_{u_m+1}-a_{u_m})>|N|+\ell. \end{align*} $$

Since $|N|+\ell \geq N-\ell $ , we obtain

(3.19) $$ \begin{align} a_{u_m+2}>a_{u_m}+N-\ell=N-r e_m. \end{align} $$

From (3.16), (3.18) and (3.19), for every fixed $m_0\in L$ , there exists $M(m_0)$ such that

(3.20) $$ \begin{align} N_{m_0}-r e_m\in\overline W\cap(a_{u_m-1},a_{u_m+2}) \subseteq\overline W\cap(a_{u_m-2},a_{u_m+2}) \end{align} $$

whenever $m\in L$ and $m\geq M(m_0)$ .

Choose $K+1$ distinct indices $m_{0,1},\ldots ,m_{0,K+1}$ in L, and choose $m\in L$ so large that

$$ \begin{align*} m\geq\max_{1\leq j\leq K+1}M(m_{0,\, j}) \quad\text{and}\quad u_m\geq T. \end{align*} $$

By (3.20), the $K+1$ integers $N_{m_{0,\, j}}-r e_m$ all lie in $\overline W\cap (a_{u_m-2},a_{u_m+2})$ , which has cardinality at most K. Hence, by Lemma 2.2, there exist indices $1\leq j_1<j_2\leq K+1$ such that

$$ \begin{align*} N_{m_{0,\, j_1}}-r e_m=N_{m_{0,\, j_2}}-r e_m. \end{align*} $$

Cancelling the common integer $r e_m$ gives $ N_{m_{0,\, j_1}}=N_{m_{0,\, j_2}}. $ Using (3.1) for $m_{0,\, j_2}$ , we obtain

$$ \begin{align*} N_{m_{0,\, j_1}}=r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}}. \end{align*} $$

Since $j_1\neq j_2$ , the elements $e_{m_{0,\, j_1}}$ and $e_{m_{0,\, j_2}}$ are distinct. The element $B_{m_{0,\, j_1}}$ is a sum of $h-r$ elements of $C\setminus \{e_{m_{0,\, j_1}}\!\}$ , and $r e_{m_{0,\, j_2}}$ is a sum of r elements of $C\setminus \{e_{m_{0,\, j_1}}\!\}$ . Thus, $B_{m_{0,\, j_1}}+r e_{m_{0,\, j_2}}\in h(C\setminus \{e_{m_{0,\, j_1}}\!\})$ and, therefore,

$$ \begin{align*} n_{m_{0,\, j_1}} =B_{m_{0,\, j_1}}+r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}} \in h(C\setminus\{e_{m_{0,\, j_1}}\!\})+W. \end{align*} $$

This contradicts the private choice of $n_{m_{0,\, j_1}}$ . Thus, the second case is impossible.

It remains to consider the case

(3.21) $$ \begin{align} \liminf_{m\to\infty}\alpha_m=+\infty \quad\text{and}\quad \liminf_{m\to\infty}\beta_m=+\infty. \end{align} $$

Choose arbitrary distinct indices $m_{0,1},\ldots ,m_{0,K+1}$ and put

$$ \begin{align*} Q=\max_{1\leq j\leq K+1}|N_{m_{0,\, j}}|. \end{align*} $$

By (3.21) and $u_m\to \infty $ , there exists m larger than all the $m_{0,\, j}$ such that $u_m\geq T$ and

$$ \begin{align*} \alpha_m>Q, \quad \beta_m>Q. \end{align*} $$

For every j, since $m>m_{0,\, j}$ , (3.2) with $y=e_m$ applies. We first show that the relevant differences are positive and lie in a bounded window. From $\alpha _m=-r e_m-a_{u_m-1}>Q\geq |N_{m_{0,\, j}}|$ , we obtain $ -r e_m>a_{u_m-1}+|N_{m_{0,\, j}}|. $ Adding $N_{m_{0,\, j}}$ gives

(3.22) $$ \begin{align} N_{m_{0,\, j}}-r e_m>a_{u_m-1}+N_{m_{0,\, j}}+|N_{m_{0,\, j}}| \geq a_{u_m-1}. \end{align} $$

From $\beta _m=r e_m+a_{u_m}>Q\geq |N_{m_{0,\, j}}|$ , we obtain $ -r e_m<a_{u_m}-|N_{m_{0,\, j}}|. $ Adding $N_{m_{0,\, j}}$ gives

(3.23) $$ \begin{align} N_{m_{0,\, j}}-r e_m<a_{u_m}+N_{m_{0,\, j}}-|N_{m_{0,\, j}}| \leq a_{u_m}. \end{align} $$

Equations (3.22) and (3.23) show in particular that $N_{m_{0,\, j}}-r e_m>0$ . Hence, (3.2) gives

$$ \begin{align*} N_{m_{0,\, j}}-r e_m\in \overline W\cap(a_{u_m-1},a_{u_m}) \subseteq \overline W\cap(a_{u_m-2},a_{u_m+2}) \end{align*} $$

for all $1\leq j\leq K+1$ . The final set has cardinality at most K, so by Lemma 2.2, there exist indices $1\leq j_1<j_2\leq K+1$ such that

$$ \begin{align*} N_{m_{0,\, j_1}}-r e_m=N_{m_{0,\, j_2}}-r e_m. \end{align*} $$

Cancelling $r e_m$ gives $N_{m_{0,\, j_1}}=N_{m_{0,\, j_2}}$ . Equation (3.1), applied to $m_{0,\, j_2}$ , gives

$$ \begin{align*} N_{m_{0,\, j_1}}=r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}}. \end{align*} $$

The indices $j_1$ and $j_2$ are distinct, so $e_{m_{0,\, j_2}}\neq e_{m_{0,\, j_1}}$ . Consequently, $B_{m_{0,\, j_1}}+r e_{m_{0,\, j_2}}$ is an element of $h(C\setminus \{e_{m_{0,\, j_1}}\!\})$ . Since $n_{m_{0,\, j_1}}=B_{m_{0,\, j_1}}+N_{m_{0,\, j_1}}$ , we have

$$ \begin{align*} n_{m_{0,\, j_1}} =B_{m_{0,\, j_1}}+r e_{m_{0,\, j_2}}+w_{m_{0,\, j_2}} \in h(C\setminus\{e_{m_{0,\, j_1}}\!\})+W, \end{align*} $$

which contradicts the private choice of $n_{m_{0,\, j_1}}$ .

The cases exhaust the possibilities and all lead to contradictions. Therefore, no minimal h-fold additive complement to W exists.

Proof. Choose an integer $K_0$ and an integer $T_0$ such that

$$ \begin{align*} |\overline W\cap(a_t,a_{t+1})|\leq K_0 \quad(t\geq T_0). \end{align*} $$

For $t\geq T_0+2$ , the interval $(a_{t-2},a_{t+2})$ is the disjoint union of the four open intervals

$$ \begin{align*} (a_{t-2},a_{t-1}),\quad (a_{t-1},a_t),\quad (a_t,a_{t+1}),\quad (a_{t+1},a_{t+2}), \end{align*} $$

together with the three selected points $a_{t-1},a_t,a_{t+1}$ . Each open interval contains at most $K_0$ elements of $\overline W$ , and the three selected points belong to $\overline W$ . Therefore,

$$ \begin{align*} |\overline W\cap(a_{t-2},a_{t+2})|\leq 4K_0+3 \quad(t\geq T_0+2). \end{align*} $$

Theorem 1.1 applies with $K=4K_0+3$ .

Proof. This statement is a direct reformulation of Theorem 1.1 in the explicit language of diagonal multi-term sumsets. To see the necessity of stating this independently of the classical case $h=1$ , observe that minimality here is defined with respect to the source set C, not the iterated sumset $hC$ . Suppose, for contradiction, that such a coordinate-minimal set C exists. Even if the sumset $D=hC$ is viewed as an ordinary additive complement to W satisfying $D+W=\mathbb Z$ , a proper sub-complement of D is generally not of the restricted diagonal form $hC'$ with $C'\subsetneq C$ . Therefore, the nonexistence of coordinate-minimal diagonal complements is a structurally distinct phenomenon that cannot be formally deduced from the classical one-fold obstruction.

Question 4.4. Can the four-neighbourhood bound in Theorem 1.1 be replaced by the sharper one-window count appearing in Chen and Ding’s proof [Reference Chen and Ding4] while retaining an explicit finite-fold threshold depending only on h and that single-window bound?

Question 4.5. Is there a nontrivial asymptotic growth threshold for the local density $|\overline W\cap (a_t,a_{t+1})|$ , depending only on W and the selected gap sequence, which still prevents the existence of minimal finite-fold additive complements?

Question 4.6. For ordered multi-term additions $C_1+\cdots +C_h+W=\mathbb Z$ , what structural conditions on W prevent simultaneous coordinate minimality, in the sense that no coordinate set $C_j$ can be properly thinned while all other coordinates remain fixed?