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The strong Erdős–Hajnal property implies the Erdős–Hajnal property

Statement

Every hereditary class of finite graphs with the strong Erdős–Hajnal property also has the Erdős–Hajnal property.

Facts & Assumptions

Given: A hereditary class C of finite graphs with the strong Erdős–Hajnal property.

[L1]

The strong Erdős–Hajnal property means that some η>0 makes every graph GC with V(G)2 contain disjoint sets A,B with A,BηV(G) and (A,B) pure (The strong Erdős–Hajnal property for a hereditary graph class).

[L2]

A hereditary class is closed under induced subgraphs (Hereditary graph classes).

[L3]

Every P4-free graph on m vertices has a clique or stable set of size at least m (Every P4-free graph has a clique or stable set of size at least the square root of its order).

[L4]

The Erdős–Hajnal property asks for a positive exponent c such that every nonempty graph in the class has a clique or stable set of size at least V(G)c (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

Proof

technique · direct
1.1

If every nonempty graph in C has one vertex, then 1 is already an Erdős–Hajnal constant, so there is nothing to prove. We therefore assume that C contains some graph with at least two vertices. Let η>0 be the strong Erdős–Hajnal constant from [L1]. Applying [L1] to one nontrivial graph in C gives disjoint sets A and B with A,BηV(G), hence 2η1. Choose c>0 so that 2ηc=1.

L1givenchoose
2.1

We claim that every nonempty graph GC induces a P4-free subgraph on at least V(G)c vertices. We prove this by induction on V(G), and the case V(G)=1 is immediate.

step 1.1given
3.1

Let GC with n:=V(G)2, and assume the claim for smaller orders. By [L1], choose disjoint sets A,BV(G) with A,Bηn and (A,B) pure. By [L2], the induced subgraphs G[A] and G[B] both lie in C, so the induction hypothesis gives induced P4-free subgraphs HAG[A] and HBG[B] with V(HA)Ac and V(HB)Bc.

step 2.1L1L2choose
4.1

Let H:=G[V(HA)V(HB)]. The pair (V(HA),V(HB)) is still pure because it sits inside the pure pair (A,B). No induced P4 can lie entirely in one side, since HA and HB are each P4-free. A 1+3 split is impossible because the lone vertex would be complete or anticomplete to the three opposite vertices, while a vertex of P4 is neither complete nor anticomplete to the other three. A 2+2 split is impossible as well: a complete cross-pair gives four cross edges, and an anticomplete cross-pair gives none, whereas P4 has exactly three edges and is connected. Hence H is P4-free.

step 3.1given
5.1

By [L5], V(H)=V(HA)+V(HB)Ac+Bc2(ηn)c=2ηcnc=nc. This closes the induction and proves the claim from step 2.1.

step 3.1step 4.1L5algebra
6.1

Let G be any nonempty graph in C. By step 5.1, G contains an induced P4-free subgraph H with V(H)V(G)c. Then [L3] gives a clique or stable set in H of size at least V(H)1/2V(G)c/2, and the same set is a clique or stable set in G. By [L4], the exponent c/2 is an Erdős–Hajnal constant for C.

step 5.1L3L4algebra

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