gn.general topology - 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:

$$r(S,k)=sup_{Xsubset mathbb Z^k}frac{Vol(X)}{Vol(X1)}.$$

I claim that for a fixed $S$, $lim_{kto infty}r(S,k)=0$.

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.

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