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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 of finite graphs with the strong Erdős–Hajnal property.
The strong Erdős–Hajnal property means that some makes every graph with contain disjoint sets with and pure (The strong Erdős–Hajnal property for a hereditary graph class).
A hereditary class is closed under induced subgraphs (Hereditary graph classes).
Every -free graph on vertices has a clique or stable set of size at least (Every -free graph has a clique or stable set of size at least the square root of its order).
The Erdős–Hajnal property asks for a positive exponent such that every nonempty graph in the class has a clique or stable set of size at least (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).
Proof
If every nonempty graph in has one vertex, then is already an Erdős–Hajnal constant, so there is nothing to prove. We therefore assume that contains some graph with at least two vertices. Let be the strong Erdős–Hajnal constant from [L1]. Applying [L1] to one nontrivial graph in gives disjoint sets and with , hence . Choose so that .
We claim that every nonempty graph induces a -free subgraph on at least vertices. We prove this by induction on , and the case is immediate.
Let with , and assume the claim for smaller orders. By [L1], choose disjoint sets with and pure. By [L2], the induced subgraphs and both lie in , so the induction hypothesis gives induced -free subgraphs and with and .
Let . The pair is still pure because it sits inside the pure pair . No induced can lie entirely in one side, since and are each -free. A split is impossible because the lone vertex would be complete or anticomplete to the three opposite vertices, while a vertex of is neither complete nor anticomplete to the other three. A split is impossible as well: a complete cross-pair gives four cross edges, and an anticomplete cross-pair gives none, whereas has exactly three edges and is connected. Hence is -free.
By [L5], . This closes the induction and proves the claim from step 2.1.
Let be any nonempty graph in . By step 5.1, contains an induced -free subgraph with . Then [L3] gives a clique or stable set in of size at least , and the same set is a clique or stable set in . By [L4], the exponent is an Erdős–Hajnal constant for .
Depends on
- The strong Erdős–Hajnal property for a hereditary graph class
- Every $P_4$-free graph has a clique or stable set of size at least the square root of its order
- The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class
- Hereditary graph classes
- Edges between disjoint vertex sets; complete, anticomplete, pure and mixed pairs
- The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents
Used by
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Sources
- Nicolas Bousquet, Aurélie Lagoutte, and Stéphan Thomassé, The Erdős-Hajnal Conjecture for Paths and Antipaths, Theorem 2 (standard reference, not scraped)