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Every hereditary graph class of bounded order has the Erdős–Hajnal property

Example

Let C be a hereditary graph class for which some NN satisfies V(G)N for every GC. Then C has the Erdős–Hajnal property.

Facts & Assumptions

Given: A hereditary class C and a natural number N bounding the order of every member.

[L1]

The definition applies to hereditary classes and asks for one ϵ>0 such that every nonempty G satisfies hom(G)V(G)ϵ (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

[L2]

The homogeneous number is the larger of the clique and stable-set numbers (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}).

[L3]

The logarithm is strictly increasing, satisfies log1=0, and obeys the quotient law (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L4]

For a>0 and xR, ax:=exp(xloga) (Real powers for positive bases, with the zero-base positive-exponent convention).

[L5]

log:(0,)R is the inverse function of exp; in particular exp(logx)=x for x>0 and log(expy)=y for yR (The natural logarithm as the inverse of the exponential function).

Verification

technique · cases
1.1

[assume-case small] If N1, choose ϵ=1; every nonempty member has one vertex and homogeneous number 1 by [L2].

givenL1L2
1.2

[assume-case large] If N2, choose ϵ=1 when N=2, and choose ϵ=log2/logN when N>2. In the latter case logN>log2>log1=0 by [L3], so ϵ>0, and [L4] with [L5] gives Nϵ=exp((log2/logN)logN)=exp(log2)=2.

L3L4L5algebrachoose
2.1

Since log is a strictly increasing bijection onto R with inverse exp, the function exp is strictly increasing as well; so for 1nN the inequality ϵlognϵlogN gives nϵNϵ=2 by [L4].

step 1.2L3L4L5algebra
3.1

In the large case, a graph of order n2 has either an edge, which is a two-vertex clique, or a nonedge, which is a two-vertex stable set; hence hom(G)2nϵ by step 2.1, while for n=1 both sides equal 1. The given hereditary hypothesis places C in the domain of [L1].

givenstep 2.1L1L2algebra
4.1

The cases are exhaustive, and in each [L1] supplies the Erdős–Hajnal property.

step 1.1step 3.1L1cases-exhaustive

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