An analogue of direct sum decomposition of a cofinite subset of integers

Question

Let $S \subseteq \mathbb Z$ be such that $\mathbb Z \setminus S$ is finite , then is it true that there exists infinite $ S_1,S_2 \subseteq \mathbb Z$ such that $S_1+S_2=S$ and for every $s \in S $ , there exists unique $s_1\in S_1 , s_2 \in S_2$ such that $s=s_1+s_2$ ?

Answer 1

Yes, but for a trivial reason. Just enumerate $S=\{c_1,c_2,\dots\}$ and construct the sets $S_1=\{a_1,a_2,\dots\}$ and $S_2=\{b_1,b_2,\dots$ inductively as follows. Suppose that $a_j,b_j$ are already defined for $j<m$ Choose the least $k$ such that $c_k$ is not of the kind $a_i+b_j$ and choose $a_m$ very large negative and $b_m$ very large positive so that $a_m+b_m=c_k$. Then the addition of $a_m$ to $S_1$ and $b_m$ to $S_2$ won't destroy the properties $S_1+S_2\subset S$ (Here is the only place where the cofiniteness of $S$ is used: we need to know that all numbers large enough in absolute values are in $S$, so sums like $a_m+b_j$, $j<m$ are there. In fact, the (upper) density $1$ will suffice too) and the representation uniqueness if they were there before, but now $c_k$ will be also representable as a sum.