ATRACTOR

Given a set \(F\), we want to find a partition of it such that each part has a diameter smaller than the diameter of \(F\) and the number of parts is as small as possible.

We are used to talking of diameter only for circles and spheres, but we can also talk of the diameter of any bounded and non-empty set if we define it as being the supreme of the distances between any two points in that set. In that case we have, for example, that the diameter of a square is the length of one of its diagonals and the diameter of an equilateral triangle the length of one of its sides:

Diameters of some sets in the plane

Let \(F\) be a set with diameter \(d\). We call Borsuk number of \(F\) and denote by \(a(F)\) the smallest number of subsets of \(F\) with diameter smaller than \(d\) that form a partition of \(F\).

For example, if we divide a circle with diameter \(d\) into two parts, at least one of them has diameter \(d\). However, there exists at least one way of breaking it into three parts in such a way that each has diameter smaller than \(d\). The following figure shows one such partition:

The Borsuk number of a circle is 3

But can other sets in the plane have a Borsuk number different from 3?

Certainly, if you try, you can quickly find a bounded set in the plane whose Borsuk number is 2...

Of course! A rectangle, or an ellipse, or a line segment... All these sets in the plane have Borsuk number equal to 2:

Rectangles (not squares), ellipses

(not circles) and line segments, all have Borsuk number equal to 2.

There are many sets with Borsuk number smaller than 3. But is there any set in the plane with Borsuk number larger than 3?

The answer is NO: whatever bounded set* in the plane we pick, that set can be broken into, at most, three parts with smaller diameter than itself, so its Borsuk number is at most 3.

Suppose we are considering a set \(F\) in the plane with diameter \(d\). Consider a regular hexagon, with distance \(d\) between opposing sides and containing our set \(F\). (You can believe that such hexagon always exists!)

What if we go from the plane to three-dimensional space? What Borsuk numbers have the bounded sets in this space? We can start by looking at a few examples…

A cone whose slant height is larger than the diameter of the base has Borsuk number equal to 2, since it is possible to partition it in two parts of smaller diameter by cutting it with a plane parallel to the base.

Remark: We leave as an exercise to the reader the calculation of the Borsuk number of a cone when the length of the slant height is equal to the diameter of the base of the cone. Is it equal to one of the previous - 2 or 3 - or different from both?

And what is the Borsuk number of a sphere three-dimensional ball?

It is clearly not possible to partition it into two parts of smaller diameter...

... and it is also not possible to partition it into three parts of smaller diameter...

... but it is possible to do it using four parts!

How?

We indicate two answers:

or

Therefore, if \(F\) is a three-dimensional sphere ball, \(a(F)=4\).

We have seen examples of three-dimensional sets with Borsuk numbers 2, 3 and 4. Is there any three-dimensional set with Borsuk number 5?

Once again the answer is NO. All bounded three-dimensional sets have Borsuk number at most 4. The proof of this result is based on a lemma stating that any tridimensional set of diameter \(d\) is contained in an octahedron with opposite faces at distance \(d\) from each other and with three of its vertices cut by planes orthogonal to its diagonals at distance \(\frac{d}{2}\) from the center.

... and that can be partitioned into four parts of diameter smaller than \(d\), thus partitioning \(F\) into four parts of smaller diameter.

We have seen what happens to the Borsuk number for sets in the plane and three-dimensional space. What about Borsuk numbers of higher dimensional spaces?

In an analogous way to what we have seen above for a three-dimensional sphere, one can show that an \(n\)-dimensional sphere has Borsuk number equal to \(n+1\). Is it always the case, like in space or three-dimensional space, that the \(n\)-dimensional sphere \(n>2\) has the largest possible Borsuk number among all \(n\)-dimensional sets?

Some thought that YES, just like in dimensions 2 and 3 the maximal Borsuk number is the disk and the sphere, in (larger)dimension \(n\) would be that of the \(n\)-dimensional sphere

K. Borsuk, the mathematician that initially studied these questions, formulated in 1933 the following conjecture:

Borsuk's conjecture:

This conjecture occupied many mathematicians that for many years tried to verify or falsify it. The examples of dimensions 2 and 3 suggested that it may be true, but the problem remained open until 1992! Only at that time was it published the first counter-example for this conjecture, finally showing that it was false. The authors of this breakthrough were Jeff Kahn and Gil Kalai.

Borsuk’s conjecture was thus not correct, since Kahn and Kalai showed that in high-dimensional spaces it is possible to construct counter-examples, showing in fact that the Borsuk number can grow exponentially as the dimension goes to infinity.

During the time when a proof of Borsuk’s conjecture was being attempted, several upper bounds for the possible Borsuk numbers in spaces of dimension \(n\) were attained, in an attempt to attain at last the conjectured bound.

In 1955, using partitions of \(n\)-dimensional cubes, H. Lenz proved that for any \(n\)-dimensional body \(F\) we have \(a(F) \leq(\sqrt{n}+1)^{n}\). This values is significantly far from the value conjectured by Borsuk since, for instance, in \(n=4\) it only guarantees that any body can be partitioned into 81 parts of smaller diameter.

In 1982, using partitions of \(n\)-dimensional balls, Lassak proved that for any \(n\)-dimensional body \(F\) we have \(a(F)\leq (2^{n-1}+1)\), which, for \(n=4\), already guarantees that any body can be partitioned into 9 parts of smaller diameter.

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