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Leaf-reducible wonderful generalized nice finite families have the Erdős-Hajnal property

Statement

Let $F$ be a generalized nice, leaf-reducible, wonderful finite family. Then $F$ has the Erdős-Hajnal property.

Facts & Assumptions

Given: A generalized nice, leaf-reducible, wonderful finite family $F$ .

[L1]

There exist constants $c_{1} \geq 1$ and $c_{2} > 0$ such that every $\overline{F}$ -free graph satisfies the three-outcome lemma from Rödl initialization upgrades generalized niceness to a restricted set, a complete or anticomplete blockade, or a polynomial clique or stable set.

[L2]

For those constants, every induced subgraph $F$ of an $\overline{F}$ -free graph $G$ with $∣ F ∣ \geq ϵ^{2 c_{1}} ∣ G ∣$ has a complete or anticomplete $(k, ∣ F ∣ / k^{c_{1}})$ -blockade for some $k \in [2, ϵ^{- 1}]$ whenever $G$ has no clique or stable set of size $∣ G ∣^{c}$ , where $c : = \min {(42 c_{1}^{2})^{- 1}, c_{2} / 2} .$ (Large induced subgraphs without a polynomial clique or stable set force complete or anticomplete blockades)

[L3]

If every induced subgraph $F$ of $G$ with $∣ F ∣ \geq ϵ^{2 a} ∣ G ∣$ has a complete or anticomplete $(k, ∣ F ∣ / k^{a})$ -blockade for some $k \in [2, ϵ^{- 1}]$ , then $G$ has an $ϵ$ -restricted induced subgraph with at least $ϵ^{3 a} ∣ G ∣$ vertices (Complete or anticomplete blockade hypotheses force an $ϵ$ -restricted induced subgraph).

[L4]

A nonempty $ϵ$ -sparse graph $H$ satisfies $χ (H) \leq ϵ ∣ H ∣ + 1, ∣ H ∣ \leq χ (H) α (H),$ and the same statement with cliques instead of stable sets holds after taking complements (The greedy colouring bound $χ (G) \leq Δ (G) + 1$ for every nonnull finite graph, The bounds $ω (G) \leq χ (G)$ and $∣ V (G) ∣ \leq χ (G) α (G)$ , A set is $c$ -sparse in $G$ exactly when it is $c$ -dense in $\overline{G}$ , so $c$ -restrictedness is complement-invariant, Complementation swaps cliques with stable sets, so $ω (\overline{G}) = α (G)$ ).

[L5]

If the complement class of a hereditary family has the Erdős-Hajnal property, then so does the family itself (A hereditary class has the Erdős–Hajnal property exactly when its complementary class does, with the same constants, The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

Proof

technique · prove an Erdős-Hajnal exponent for the $\overline{\mathcal F}$-free class and then use complement invariance

1.1

Let $c_{1} \geq 1$ and $c_{2} > 0$ be the constants from [L1], put $q : = 42 c_{1}^{2}, m : = 2^{q}, c : = \min {q^{- 1}, c_{2} / 2},$ and fix a nonempty $\overline{F}$ -free graph $G$ . We will prove that $G$ has a clique or stable set of size at least $∣ G ∣^{c}$ .

L1choosegiven

2.1

If $G$ already has such a clique or stable set, there is nothing to prove. So assume for contradiction that $G$ has no clique or stable set of size $∣ G ∣^{c}$ . Because $c \leq q^{- 1}$ , one has $m^{c} = (2^{q})^{c} \leq 2$ . Hence every nonempty graph with at most $m$ vertices already has a clique or stable set of size at least $∣ G ∣^{c}$ , so this forces $∣ G ∣ > m$ .

step 1.1assume-contraalgebra

3.1

Define $x : = ∣ G ∣^{- 1 / (3 c_{1})}, ϵ : = x^{1 / (7 c_{1})} .$ Then $x \leq m^{- 1 / (3 c_{1})} = 2^{- 14 c_{1}}, ϵ \leq 2^{- 14 c_{1} / (7 c_{1})} = \frac{1}{4},$ so in particular $x < ϵ \leq \frac{1}{4}$ .

step 2.1algebra

4.1

By [L2], every induced subgraph $F$ of $G$ with $∣ F ∣ \geq ϵ^{2 c_{1}} ∣ G ∣$ has a complete or anticomplete $(k, ∣ F ∣ / k^{c_{1}})$ -blockade for some $k \in [2, ϵ^{- 1}]$ . Therefore [L3] applies with $a : = c_{1}$ and gives an $ϵ$ -restricted induced subgraph $S \subseteq G$ with $∣ S ∣ \geq ϵ^{3 c_{1}} ∣ G ∣$ .

step 2.1step 3.1L2L3

5.1

Since $ϵ = x^{1 / (7 c_{1})} = ∣ G ∣^{- 1 / (21 c_{1}^{2})}$ and $c_{1} \geq 1$ , one has $∣ S ∣ \geq ϵ^{3 c_{1}} ∣ G ∣ = ϵ^{- 21 c_{1}^{2} + 3 c_{1}} \geq ϵ^{- 1} .$ After taking complements if necessary, [L4] lets us assume that $S$ is $ϵ$ -sparse.

step 3.1step 4.1L4algebra

6.1

Applying [L4] to the graph $G [S]$ yields $α (G [S]) \geq \frac{∣ S ∣}{ϵ ∣ S ∣ + 1} = \frac{1}{ϵ + ∣ S ∣^{- 1}} \geq \frac{1}{2 ϵ} \geq ϵ^{- 1 / 2} = ∣ G ∣^{1 / (42 c_{1}^{2})} \geq ∣ G ∣^{c},$ because $c \leq (42 c_{1}^{2})^{- 1}$ . This contradicts step 2.1.

step 5.1L4algebra

7.1

Therefore every nonempty $\overline{F}$ -free graph has a clique or stable set of size at least $∣ G ∣^{c}$ , so the complement class of $F$ has the Erdős-Hajnal property. By [L5], $F$ itself has the Erdős-Hajnal property.

step 1.1step 6.1L5discharge-contradiction∎

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Sources

Shenwei Huang, Yiao Ju, and Yidong Zhou, Erdős-Hajnal beyond the five-vertex path, Lemma 3.5 and Lemma 1.12 (standard reference, not scraped)
Tung Nguyen, Alex Scott, and Paul Seymour, Induced subgraph density. VII. The five-vertex path, Theorem 7.4 (standard reference, not scraped)