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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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Alon–Pach–Solymosi: if H1 and H2 have the Erdős–Hajnal property, so does the graph obtained from H1 by substituting H2 for a vertex

Statement

Let H1 and H2 be finite simple graphs with the Erdős–Hajnal property, let v∈V(H1), and suppose the substitution H=H1[v→H2] is defined (Substituting one graph for a vertex of another). Then H has the Erdős–Hajnal property.

Facts & Assumptions

Given: Finite simple graphs H1,H2 with the Erdős–Hajnal property, a vertex v∈V(H1), and the substitution H=H1[v→H2]; write h=∣V(H1)∣, so h≥1.

[F1]

A real ϵ>0 is an Erdős–Hajnal constant for a hereditary class C when every nonempty J∈C satisfies hom⁡(J)≥∣V(J)∣ϵ; a finite graph K has the Erdős–Hajnal property when the class of K-free graphs has such a constant (The Erdős–Hajnal property and an Erdős–Hajnal constant for a hereditary graph class).

[F2]

hom⁡(G)=max⁡{ω(G),α(G)}, where ω(G) and α(G) are the largest cardinalities of a clique and of a stable set of G (Homogeneous vertex sets and the homogeneous number hom⁡(G)=max⁡{ω(G),α(G)}, Cliques, stable sets, the clique number ω(G) and stability number α(G)).

[F3]

G is K-free when G has no induced copy of K, an induced copy being the image of an induced embedding (H-free and F-free graphs under the induced-subgraph convention, Induced embeddings and induced copies of a graph).

[L1]

For every family F of finite graphs, the class of F-free finite graphs is hereditary (Every class defined by forbidden induced subgraphs is hereditary).

[L2]

If ϵ>0 is an Erdős–Hajnal constant for the class of K-free graphs and W⊆V(G) is nonempty with ∣W∣ϵ>hom⁡(G), then G[W] has an induced copy of K (If ϵ is an Erdős–Hajnal constant for H and W is a nonempty vertex set with ∣W∣ϵ>hom⁡(G), then G[W] has an induced copy of H).

[L3]

If h≤m≤n=∣V(G)∣ and every m-element W⊆V(G) has an h-element subset S with G[S]≅H1, then the number of h-element sets S⊆V(G) with G[S]≅H1 is at least (n−h+1)h/mh (If every m-element vertex set contains an induced copy of H, then at least (nh)/(mh) of the h-element vertex sets induce a copy of H).

[L4]

With Ψ the set of induced embeddings of H1−v into G and Xφ the extension set of φ∈Ψ, one has ind⁡H1(G)=∑φ∈Ψ∣Xφ∣ and ∣Ψ∣≤n h−1 (The induced copies of H1 in G are counted by summing, over the induced embeddings of H1−v, the number of vertices that extend them at v).

[L5]

If φ∈Ψ and ψ is an induced embedding of H2 into G whose image lies in Xφ, then G has an induced copy of H=H1[v→H2] (An induced copy of H2 inside the extension set of an induced embedding of H1−v yields an induced copy of H1 with H2 substituted for v).

[L6]

For finite sets X≠∅ and Y and a relation R⊆X×Y with row fibres Rx, there is x+∈X with ∣Rx+∣≥∣R∣/∣X∣ (If X is nonempty, some row fibre is at least the average size and some row fibre is at most the average size).

[F4]

ind⁡K(G) is the number of induced embeddings of K into G (The induced-embedding count ind⁡H(G)).

[F5]

G[W]=(W, E(G)∩[W]2), so two vertices of W are adjacent in G[W] exactly when they are adjacent in G (Subgraphs, induced subgraphs and spanning subgraphs).

[F6]

For a>0 and real x, ax=exp⁡(xlog⁡a) (Real powers for positive bases, with the zero-base positive-exponent convention).

[L7]

For a,b>0 and real r,s: ar+s=aras, (ab)r=arbr, (a/b)r=ar/br and (ar)s=ars (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[L8]

The logarithm is continuous and strictly increasing on (0,∞), is onto R, satisfies log⁡(xy)=log⁡x+log⁡y and log⁡(x/y)=log⁡x−log⁡y, and log⁡1=0 (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L9]

The exponential is continuous and strictly increasing on R (The exponential function is strictly increasing).

[F7]

For b>0 with b≠1 and x>0, log⁡bx=log⁡x/log⁡b (The logarithm to a positive base other than one).

[L10]

Every complete ordered field is Archimedean: for every x there is a natural number k≥1 with x<k (Every complete ordered field is Archimedean).

[L11]

Every nonempty subset of N has a least element (The well-ordering principle).

Proof

technique · direct
1.1givenF1F7L8choosealgebra

Choose Erdős–Hajnal constants ϵ1>0 for the class of H1-free graphs and ϵ2>0 for the class of H2-free graphs, and set β=1/(2(h+1)), ϵ0=βmin⁡{ϵ1,ϵ2}, n0=max⁡{4⋅16h, 2h, h2(h+1)} and ϵ=min⁡{ϵ0, 1/log⁡2n0}. Then 0<β≤1/4, ϵ0>0, n0≥64 and ϵ>0, and β(h+1)=1/2 gives 1−βh=1/2+β.

1.2F6L8L9

For a≥1 and s≤t one has as≤at, because log⁡a≥0 and both exp⁡ and log⁡ are increasing; and for 0<a≤b and r>0 one has ar≤br for the same reason.

1.3L1F1

By [L1] the class of H-free graphs is hereditary, so it is a class for which [F1] can supply a constant.

2.1step 1.1step 1.2L10L11F6choose

Let G be a finite simple graph with n=∣V(G)∣≥n0 and hom⁡(G)<nϵ0. The set of natural numbers k with k≥nβ is nonempty by [L10], so it has a least element m by [L11]; since nβ>0 we have m≥1, and m−1<nβ, so nβ≤m<nβ+1≤2nβ, the last step because n≥1 gives nβ≥1.

2.2step 1.1step 1.2F2F6F7L8algebra

Turning to the small orders, let J be any nonempty H-free graph with n′=∣V(J)∣<n0. If n′≥2 then two distinct vertices of J form a clique or a stable set, so hom⁡(J)≥2; and n′ϵ<n0ϵ≤n01/log⁡2n0=exp⁡(log⁡n0⋅log⁡2/log⁡n0)=2, so hom⁡(J)≥n′ϵ. If n′=1 then hom⁡(J)=1=1ϵ.

3.1step 1.1step 2.1step 1.2L7algebra

Since n≥h2(h+1)=h1/β, raising to the power β gives nβ≥h, so m≥h; and β≤1/4 with n≥64 gives m<n1/4+1≤n. Hence h≤m≤n.

4.1step 2.1step 3.1step 1.2L2L7F5F3

Let W⊆V(G) with ∣W∣=m. Then W≠∅ and ∣W∣ϵ1=mϵ1≥(nβ)ϵ1=nβϵ1≥nϵ0>hom⁡(G), using ϵ0≤βϵ1 and n≥1; so by [L2] the graph G[W] has an induced copy of H1, that is, an h-element S⊆W with G[S]≅H1.

5.1step 3.1step 4.1L3F3F4

By [L3] the number g of h-element sets S⊆V(G) with G[S]≅H1 is at least (n−h+1)h/mh, and each such S carries at least one induced embedding of H1 into G, distinct sets carrying distinct embeddings because their images differ; so ind⁡H1(G)≥g≥(n−h+1)h/mh>0.

6.1step 5.1L4L6F4

By [F4] and step 5.1 the set Φ of induced embeddings of H1 into G is nonempty, so the set Ψ of [L4] is nonempty as well, since each member of Φ restricts into it. Applying [L6] to the relation pairing φ∈Ψ with the members of Φ restricting to it, whose row fibres have sizes ∣Xφ∣ and whose total size is ind⁡H1(G) by [L4], gives φ+∈Ψ with ∣Xφ+∣≥ind⁡H1(G)/∣Ψ∣.

7.1step 1.1step 2.1step 5.1step 6.1L4L7algebra

Since ∣Ψ∣≤n h−1 and n≥n0≥2h gives n−h+1>n/2, and m<2nβ by step 2.1, we get ∣Xφ+∣≥(n−h+1)hmh n h−1>(n/2)h2hnβhn h−1=n1−βh4h=4−hn1/2nβ.

8.1step 1.1step 2.1step 7.1step 1.2L7algebra

From n≥4⋅16h we get n1/2≥2⋅4h, so 4−hn1/2≥2 and step 7.1 gives ∣Xφ+∣>2nβ>m≥nβ. In particular Xφ+ is nonempty.

9.1step 2.1step 8.1step 1.2L2L7F5F3

Therefore ∣Xφ+∣ϵ2≥(nβ)ϵ2=nβϵ2≥nϵ0>hom⁡(G), so by [L2] the graph G[Xφ+] has an induced copy of H2; the corresponding induced embedding has image inside Xφ+ and, adjacency in G[Xφ+] agreeing with adjacency in G, it is an induced embedding of H2 into G.

10.1step 2.1step 9.1step 1.2L5F3

By [L5] applied to φ+ and that embedding, the graph G of step 2.1 has an induced copy of H and so is not H-free. Hence an H-free graph G with ∣V(G)∣=n≥n0 cannot satisfy hom⁡(G)<nϵ0, and therefore satisfies hom⁡(G)≥nϵ0≥nϵ, the last inequality by step 1.2 and ϵ≤ϵ0.

11.1step 1.3step 10.1step 2.2F1∎

Steps 10.1 and 2.2 together give hom⁡(G)≥∣V(G)∣ϵ for every nonempty H-free graph G, whatever its order, and the class of H-free graphs is hereditary by step 1.3; that is, ϵ is an Erdős–Hajnal constant for it and H has the Erdős–Hajnal property.

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