In 1933 Karol Borsuk conjectured the following
Can every bounded subset $E$ of $\mathbb{R}^d$ be partitioned into $(d+1)$ sets, each of which has a smaller diameter than $E$?
Whilst new to this field of geometry I still have some open/unclear questions on this topic.
Early in game (to my knowledge even before any counter-example were known) Larman suggested investigating the problem when $E$ is a two-distance set. As a result of this suggestion many counter-examples emerged. Most recently the counter-example of Bondarenko in Dimension 64 using strongly regular graphs which form a two distance set on a sphere.
My question to this is:
How did Larmann come up with two-distance set? Why is it reasonable to think that those kind of sets contain many difficulties intrinsic to the general problem? What are those difficulties?
Another question emerges from a result by Hao Chen on Ball packings with high chromatic numbers from strongly regular graphs. Most likely it seems that this publication emerged from another MathOverflow question by Cantwell. Namely: Chromatic number of graphs of tangent closed balls .
Just as Bondarenko, Chen uses strongly regular graphs to form a spherical two-distance set to construct a ball-packing whose tangency graph is highly chromatic.
My question regards the part where Chen wants to link Borsuk's conjecture to the ball packing problem. As written:
The finite version of the Borsuk conjecture can be formulated as follows: the chromatic number of the unit-distance graph for a set of points with maximum distance $1$ is at most $d+1$. So the chromatic number problem for unit ball packings is the “opposite” of the Borsuk conjecture.
Why can Borsuk's conjecture be regarded as Chen names it. Somehow I don't get the connection between the "classic" conjecture by Borsuk and Chens version.
How is the coloring of tangency graphs of ball-packings connected to Borsuk's conjecture? Why can the chromatic number problem for unit packings be regarded as the "opposite" of Borsuk conjecture? What opposite?
I will be thankful for every kind of advice or tip to my questions.
