English

Pach's selection theorem does not admit a topological extension

Combinatorics 2018-09-18 v2

Abstract

Let U1,,Ud+1U_1,\dots, U_{d+1} be nn-element sets in RdR^d and let u1,,ud+1\langle u_1,\ldots,u_{d+1}\rangle denote the convex hull of points uiu_i in UiU_i (for all ii) which is a (possibly degenerate) simplex. Pach's selection theorem says that there are sets Z1U1,,Zd+1Ud+1Z_1 \subset U_1,\dots, Z_{d+1} \subset U_{d+1} and a point uu in RdR^d such that each Zi>c1(d)n|Z_i| > c_1(d)n and uu belongs to z1,...,zd+1\langle z_1,...,z_{d+1} \rangle for every choice of z1z_1 in Z1,,zd+1Z_1,\dots,z_{d+1} in Zd+1Z_{d+1}. Here we show that this theorem does not admit a topological extension with linear size sets ZiZ_i. However, there is a topological extension where each Zi|Z_i| is of order (logn)(1/d)(\log n)^(1/d).

Cite

@article{arxiv.1610.05053,
  title  = {Pach's selection theorem does not admit a topological extension},
  author = {Imre Bárány and Roy Meshulam and Eran Nevo and Martin Tancer},
  journal= {arXiv preprint arXiv:1610.05053},
  year   = {2018}
}

Comments

8 pages

R2 v1 2026-06-22T16:22:43.686Z