A path Turan problem for infinite graphs

TL;DR

This paper investigates the maximum edge density in infinite graphs avoiding increasing paths of a certain length, providing new bounds for the case when the path length is four, and advancing understanding of a longstanding Turan problem.

Contribution

The authors establish new lower and upper bounds for the Turan density in infinite graphs avoiding increasing paths of length four, resolving an open problem and refining previous bounds.

Findings

Proved that p(4) is at least 3/16 + 1/584064

Showed that p(k) > 1/4 (1 - 1/k) for 4 ≤ k ≤ 15

Improved the upper bound for p(4) to 1/4

Abstract

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