TITLE: covers of $Z^\infty$
QUESTION [11 upvotes]: 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: http://front.math.ucdavis.edu/1008.3868 .

REPLY [5 votes]: In fact there is also a negative answer (again with 3 instead of 4) in the case of the countably infinite sum of Z_2.  View the direct sum with the $\ell_1$-metric as being the collection of all finite subsets of N with the metric which counts the symmetric difference.  Let $[N]^k$ denote all of the k-element subsets of N.  Observe that it suffices to demonstrate that, for each k, if $[N]^k$ is partitioned into finitely many pieces, then there is a piece of the partition containing a sequence $a_i$ $(i < k)$ such that $a_0$ is disjoint from $a_{k-1}$ (thus $a_0$ and $a_{k-1}$ are 2k units apart) and 
$|a_i \Delta a_{i+1}| = 2$ of all $i < k-1$.  But this follows easily from Ramsey's theorem: given any such partition, there is a subset b of N of cardinality 2k, all of whose k element sets are in one piece of the partition (in fact we can find an infinite b if we like).  From this one can now easily construct the sequence of $a_i$ $(i < k)$:
if $b = \{m_i : i < 2k\}$, then set $a_i = \{m_j : i \leq j < i+k\}$.