Is it possible to cover $Z^\infty$ (the infinite direct sum of $Z$'s with the $l\_1$-metric) by a finite set of collections of subsets $U^0,...,U^n$ such that each collection $U^i$ consists of uniformly bounded sets $U\_j^i$ that are 4-disjoint (the distance between any two subsets $U\_j^i$, $U\_k^i$ in each $U^i$ is at least 4)? The motivation is here: <https://arxiv.org/abs/1008.3868> .



covers of $Z^\infty$
The answer is NO even if we replace $4$ by $3$.
Let me sketch a proof. This is based upon the following lemma.


Lemma. Fix $S>0$ and for an integer $k$ conisder in $\mathbb Z^k$ sets $X$ of diameter at most $S$. Denote by $Vol(X)$ the number of points in $X$ and denote by $X1$ the set of points of distance at most $1$ from $X$. Now let $\delta(S,k)$ be the supremum over all $X$ of diameter at most $S$ of the ratio: 


$$r(S,k)=sup\_{X\subset \mathbb Z^k}\frac{Vol(X)}{Vol(X1)}.$$


I claim that for a fixed $S$, $lim\_{k\to \infty}r(S,k)=0$.


Let us skip the proof of the lemma and instead deduce the claim. Suppose by contradiction that the answer is positive. Then for every $k$ we will get a solution to the problem in $\mathbb Z^k$ with the fixed number of sets ($U^0,...,U^n$) such that each $U^i\_j$ is of the diameter at most $S$. Now, chose such $k$ that $r(S,k)<\frac{1}{2n}$ and let us deduce the contradiction. 


From Lemma it follows that the supremum of asimptotic density of each set $U^i$ in $\mathbb Z^k$ is less than $\frac{1}{n+1}$. Indeed, since the distance between different components of $U^i$ is $4$, every point of $U^i1$ that does not belong to $U^i$ is on distance one from at most one component of $U^i$. And lemma gives us the inequality (that should be understood as assymptotic in $\mathbb Z^k$)
$$Vol(U^i)<\frac{1}{2n} Vol(U^i1)\le \frac{1}{2n}Vol(\mathbb Z^k)$$
Hence $\mathbb Z^k$ can not be covered by $U^0,...,U^n$.


It is clear where this proof breakes if we conisder $2$-disjoint sets. In this case one point of $U^i1$ can be on distance $1$ to many components of $U^i$ and the above inequality will not hold. But for $3$-disjoint sets this works.


As for the proof of the lemma, I think, it should be rather standard.


