An example of a hexagon cut into three pieces of smaller diameter.

The Borsuk problem in geometry, for historical reasons^{[note 1]} incorrectly called Borsuk's conjecture, is a question in discrete geometry. It is named after Karol Borsuk.

Problem

In 1932, Karol Borsuk showed^[2] that an ordinary 3-dimensional ball in Euclidean space can be easily dissected into 4 solids, each of which has a smaller diameter than the ball, and generally n-dimensional ball can be covered with n + 1 compact sets of diameters smaller than the ball. At the same time he proved that n subsets are not enough in general. The proof is based on the Borsuk–Ulam theorem. That led Borsuk to a general question:^[2]

Die folgende Frage bleibt offen: Lässt sich jede beschränkte Teilmenge E des Raumes ${\displaystyle \mathbb {R} ^{n}}$ in (n + 1) Mengen zerlegen, von denen jede einen kleineren Durchmesser als E hat?

The following question remains open: Can every bounded subset E of the space ${\displaystyle \mathbb {R} ^{n}}$ be partitioned into (n + 1) sets, each of which has a smaller diameter than E?

— Drei Sätze über die n-dimensionale euklidische Sphäre

The question was answered in the positive in the following cases:

• n = 2 — which is the original result by Karol Borsuk (1932).
• n = 3 — shown by Julian Perkal (1947),^[3] and independently, 8 years later, by H. G. Eggleston (1955).^[4] A simple proof was found later by Branko Grünbaum and Aladár Heppes.
• For all n for smooth convex bodies — shown by Hugo Hadwiger (1946).^[5][6]
• For all n for centrally-symmetric bodies — shown by A.S. Riesling (1971).^[7]
• For all n for bodies of revolution — shown by Boris Dekster (1995).^[8]

The problem was finally solved in 1993 by Jeff Kahn and Gil Kalai, who showed that the general answer to Borsuk's question is no.^[9] They claim that their construction shows that n + 1 pieces do not suffice for n = 1325 and for each n > 2014. However, as pointed out by Bernulf Weißbach,^[10] the first part of this claim is in fact false. But after improving a suboptimal conclusion within the corresponding derivation, one can indeed verify one of the constructed point sets as a counterexample for n = 1325 (as well as all higher dimensions up to 1560).^[11]

Their result was improved in 2003 by Hinrichs and Richter, who constructed finite sets for n ≥ 298, which cannot be partitioned into n + 11 parts of smaller diameter.^[1]

In 2013, Andriy V. Bondarenko had shown that Borsuk's conjecture is false for all n ≥ 65.^[12] Shortly after, Thomas Jenrich derived a 64-dimensional counterexample from Bondarenko's construction, giving the best bound up to now.^[13][14]

Apart from finding the minimum number n of dimensions such that the number of pieces α(n) > n + 1, mathematicians are interested in finding the general behavior of the function α(n). Kahn and Kalai show that in general (that is, for n sufficiently large), one needs ${\textstyle \alpha (n)\geq (1.2)^{\sqrt {n}}}$ many pieces. They also quote the upper bound by Oded Schramm, who showed that for every ε, if n is sufficiently large, ${\textstyle \alpha (n)\leq \left({\sqrt {3/2}}+\varepsilon \right)^{n}}$.^[15] The correct order of magnitude of α(n) is still unknown.^[16] However, it is conjectured that there is a constant c > 1 such that α(n) > cⁿ for all n ≥ 1.

See also

• Hadwiger's conjecture on covering convex bodies with smaller copies of themselves
• Kahn–Kalai conjecture

Note

References

Further reading

• Oleg Pikhurko, Algebraic Methods in Combinatorics, course notes.
• Andrei M. Raigorodskii, The Borsuk partition problem: the seventieth anniversary, Mathematical Intelligencer 26 (2004), no. 3, 4–12.
• Raigorodskii, Andreii M. (2008). "Three lectures on the Borsuk partition problem". In Young, Nicholas; Choi, Yemon (eds.). Surveys in contemporary mathematics. London Mathematical Society Lecture Note Series. Vol. 347. Cambridge University Press. pp. 202–247. ISBN 978-0-521-70564-6. Zbl 1144.52005.

External links

• Weisstein, Eric W. "Borsuk's Conjecture". MathWorld.