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The hereditary class of all finite graphs does not have the Erdős–Hajnal property

Statement

The hereditary class of all finite graphs does not have the Erdős–Hajnal property.

Facts & Assumptions

Given: The class $G$ of all finite graphs.

[L1]

A hereditary class has the Erdős–Hajnal property exactly when some $ϵ > 0$ satisfies $hom (G) \geq ∣ V (G) ∣^{ϵ}$ for every nonempty graph in the class (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

[L2]

For every $n \geq 16$ , some $n$ -vertex graph $G$ satisfies $hom (G) < 3 \log_{2} n$ (For every $n \geq 16$ there is an $n$ -vertex graph with $hom (G) < 3 \log_{2} n$ ).

[L3]

For every $ϵ > 0$ , $\log x / x^{ϵ} \to 0$ as $x \to + \infty$ (The logarithm grows more slowly than every positive real power).

[L4]

For $x > 0$ , $\log_{2} x = \log x / \log 2$ (Change of base and inversion of the positive-base real exponential).

Proof

technique · contradiction

1.1assume-contraL1

Suppose, for contradiction, that $G$ has an Erdős–Hajnal constant $ϵ > 0$ .

1.2L3L4choose

By [L3] and [L4], choose an integer $n \geq 16$ so large that $3 \log_{2} n < n^{ϵ}$ .

2.1step 1.1step 1.2L1L2discharge-contradiction∎

Choose from [L2] an $n$ -vertex graph $G$ with $hom (G) < 3 \log_{2} n < n^{ϵ}$ , contradicting [L1] and step 1.1. Therefore $G$ does not have the Erdős–Hajnal property.

Depends on

Used by

The empty forbidden family is not Erdős–Hajnal Counterexample
Every hereditary graph class has the Erdős–Hajnal property False statement

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

A. Chernikov, MATH 223M notes, sec. 3.1 (standard reference, not scraped)