A path Turan problem for infinite graphs

preprint2015arXivOpen access

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 1≤i1<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 k≥2k \geq 2, let p(k)p(k) be the supremum of lim inf⁡n→∞e(Gn)n2\liminf_{ n \rightarrow \infty} \frac{ e(G_n) }{n^2} over all IkI_k-free graphs GG. In 1962, Czipszer, Erdős, and Hajnal proved that p(k)=14(1−1k)p(k) = \frac{1}{4} (1 - \frac{1}{k}) for k∈{2,3}k \in \{2,3 \}. Erdős conjectured that this holds for all k≥4 k \geq 4. This was disproved for certain values of kk by Dudek and Rödl who showed that p(16)>14(1−116)p(16) > \frac{1}{4} (1 - \frac{1}{16}) and p(k)>14+1200p(k) > \frac{1}{4} + \frac{1}{200} for all k≥162k \geq 162. Given that the conjecture of Erdős 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(1−1k)p(k) > \frac{1}{4} ( 1 - \frac{1}{k} ). In particular, the question of whether or not p(4)=14(1−14)p(4) = \frac{1}{4} ( 1 - \frac{1}{4} ) was mentioned by Dudek and Rödl as an open problem. We solve this problem by proving that p(4)≥14(1−14)+1584064p(4) \geq \frac{1}{4} (1 - \frac{1}{4} ) + \frac{1}{584064} and p(k)>14(1−1k)p(k) > \frac{1}{4} (1 - \frac{1}{k}) for 4≤k≤154 \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}

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