Let $n$ be a natural number and let $S\_n$ be a square $[0,n] \times [0,n]$ in the plane.


We say that a partition $\mathcal{Q} = R\_1 \cup \cdots \cup R\_t$ of $S\_n$ is *simple* if each of the sets $R\_1, \ldots, R\_t$ is connected, has positive area and has diameter at most 1.


**Question.** Does there exist a collection of at most $k$ ($k$ does not depend on $n$) simple partitions $\mathcal{Q}\_1, \ldots, \mathcal{Q}\_k$ of $S\_n$ such that any two points $x,y \in S\_n$ of distance at most 1 are contained in at least one of the sets of the partitions, that is there exist $i \in\{1, \ldots, k\}$ and a set $R \in \mathcal{Q}\_i$ such that $x,y \in R$?



Is it possible to cover all pairs of points at distance at most 1 by constant number of partitions into sets of diameter at most 1?
More generally, replace $S\_n$ by $D$, a union of a collection of closed balls of nonzero radius, and suppose $D$ has diameter greater than 1. Consider partitions of $D$ which are bounded, so each set in a partition has diameter at most 1. Let us further assume that $D$ contains two open balls with centers $x$ and $y$ of distance exactly 1 from one another.


Consider the unit vector $v$ which is the difference of $x$ and $y$. Then the points $tv + x$ and $tv+y$ are also distance 1 apart. Let $R\_t$ be the partition part which contains these two points. Then $R\_t$ is different from $R\_s$ when $t$ and $s$ are distinct real numbers sufficiently small. Thus continuum-many partition parts of diameter 1 are needed to cover just these pairs of points. In particular, more than countably many parts are needed, so finitely many partitions each with a finite number of parts is not enough. Thus we can't find a finite set of simple partitions for the poster even in the case of the unit square.


Gerhard "It's About Cardinality, Not Topology" Paseman, 2015.12.02


