# Covering a set $A \subset \Bbb R$ by two families of disjoint intervals taken from given intervals

Let $A \subset \Bbb R$ be a bounded set. Every element $a \in A$ is the center of some given open interval, let's denote it by $I_a=(a-r_a, a+r_a)$. I'm interested in knowing the following:

Can we always color some of the given intervals $I_a$ in red and some others in blue such that intervals of the same color are disjoint, and the colored intervals together contain all of $A$?

I know that there is some constant number of colors that suffices for all possible $A$ and $I_a$, and this holds even in higher dimensions for different constants; this is Besicovitch's covering theorem. Thus we can rephrase the question as:

what is the optimal constant for Besicovitch's covering theorem in $\Bbb R$?

Note that $A$ is not necessarily closed, so it seems we cannot choose some "extremal" values such as largest intervals to construct a greedy approach to coloring.

A now-deleted answer linked to this post containing a proof. However, I don't understand the proof at all. So it suffices if someone would write it in a way that is easier to comprehend.

• Is $r(a)$ given or can we choose it? – abnry Sep 10 '16 at 21:37
• @abnry It is given. – Emolga Sep 10 '16 at 21:48
• $I_{a}\subseteq A$? – ecrin Jan 13 at 9:23