English

A path Turan problem for infinite graphs

Combinatorics 2015-12-22 v1

Abstract

Let GG be an infinite graph whose vertex set is the set of positive integers, and let GnG_n be the subgraph of GG induced by the vertices {1,2,,n}\{1,2, \dots , n \}. An increasing path of length kk in GG, denoted IkI_k, is a sequence of k+1k+1 vertices 1i1<i2<<ik+11 \leq i_1 < i_2 < \dots < i_{k+1} such that i1,i2,,ik+1i_1, i_2, \ldots, i_{k+1} is a path in GG. For k2k \geq 2, let p(k)p(k) be the supremum of lim infne(Gn)n2\liminf_{ n \rightarrow \infty} \frac{ e(G_n) }{n^2} over all IkI_k-free graphs GG. In 1962, Czipszer, Erd\H{o}s, and Hajnal proved that p(k)=14(11k)p(k) = \frac{1}{4} (1 - \frac{1}{k}) for k{2,3}k \in \{2,3 \}. Erd\H{o}s conjectured that this holds for all k4 k \geq 4. This was disproved for certain values of kk by Dudek and R\"{o}dl who showed that p(16)>14(1116)p(16) > \frac{1}{4} (1 - \frac{1}{16}) and p(k)>14+1200p(k) > \frac{1}{4} + \frac{1}{200} for all k162k \geq 162. Given that the conjecture of Erd\H{o}s is true for k{2,3}k \in \{2,3 \} but false for large kk, it is natural to ask for the smallest value of kk for which p(k)>14(11k)p(k) > \frac{1}{4} ( 1 - \frac{1}{k} ). In particular, the question of whether or not p(4)=14(114)p(4) = \frac{1}{4} ( 1 - \frac{1}{4} ) was mentioned by Dudek and R\"{o}dl as an open problem. We solve this problem by proving that p(4)14(114)+1584064p(4) \geq \frac{1}{4} (1 - \frac{1}{4} ) + \frac{1}{584064} and p(k)>14(11k)p(k) > \frac{1}{4} (1 - \frac{1}{k}) for 4k154 \leq k \leq 15. We also show that p(4)14p(4) \leq \frac{1}{4} which improves upon the previously best known upper bound on p(4)p(4). Therefore, p(4)p(4) must lie somewhere between 316+1584064\frac{3}{16} + \frac{1}{584064} and 14\frac{1}{4}

Keywords

Cite

@article{arxiv.1512.06371,
  title  = {A path Turan problem for infinite graphs},
  author = {Xing Peng and Craig Timmons},
  journal= {arXiv preprint arXiv:1512.06371},
  year   = {2015}
}

Comments

16 pages; Comments are welcome

R2 v1 2026-06-22T12:14:20.552Z