I am interested in proving the statement:
Let $S\subseteq\mathbb{N}$ such that for every $r\in\mathbb{N}$ and for every $k_{1}$, $k_{2}$, $\ldots$, $k_{r}\in\mathbb{N}$, the set $\big(S-k_{1}\big) \cap \big(S-k_{2}\big) \cap \ldots \cap \big(S-k_{r}\big)$ has positive density in $\mathbb{N}$. Then the set $S$ must have density $1$. The density I have in mind is not natural density but Banach density.
To prove the statement, imagine that such a set $S$ does not have full density.
If $S = \bigcup_{i=0}^{j} a\mathbb{N}+i$ is the union of $j < a$ arithmetic progressions, we can see that for $r =a$ and taking $k_{1} = 0, k_{2} = 1, \ldots, k_{a} = a$, the intersection is empty — a contradiction. Similarly, we would also reach a contradiction for a set $S$ of the form $\bigcup_{s=1}^{n}\Big(\bigcup_{i=0}^{j_{s}} a_{s}\mathbb{N}+i \Big)$, where all $a_{s}$ are distinct, $j_{s} < a_{s}$ for each $1 \leq s \leq n$. So, this would also be true if the set $S$ were contained in $\bigcup_{s=1}^{n}\Big(\bigcup_{i=0}^{j_{s}} a_{s}\mathbb{N}+i \Big)$ of the above form.
However, for a general $S$, I am neither sure of the statement nor its proof. I would appreciate any help or feedback.
