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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26
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Many good 2t-vertex subsets force many homogeneous k-sets

Statement

Let 1≤k≤t be integers, and let C be a class of finite graphs such that every graph in C has the (t,k)-homogeneous property. Let G be a finite graph on n≥2t vertices. Suppose at least half of the sets X∈[V(G)]2t contain a t-element subset T⊆X with G[T]∈C. Then G has at least

12(n2t)k

homogeneous vertex sets of size k.

Facts & Assumptions

Given: Positive integers 1≤k≤t, a class C of finite graphs, integers n≥2t, an n-vertex graph G, and the hypothesis that at least half of the sets X∈[V(G)]2t contain a t-element subset T with G[T]∈C.

[L1]

If a graph lies in C, then every t-element subset of its vertex set contains a homogeneous k-element subset (The (t,k)-homogeneous property, Homogeneous vertex sets and the homogeneous number hom⁡(G)=max⁡{ω(G),α(G)}).

[L2]

For a subset T⊆V(G), the induced subgraph on T is G[T] (Subgraphs, induced subgraphs and spanning subgraphs).

[L3]

Proof

technique · direct
1.1givenL3

Call a set X∈[V(G)]2t good when it contains a t-element subset T with G[T]∈C; by hypothesis, there are at least 12(n2t) good sets.

1.2L1L2choose

If X is good, choose T⊆X with ∣T∣=t and G[T]∈C; then [L1] gives a homogeneous k-element subset K⊆T, and since G[T] is the induced subgraph on T, that same set K is homogeneous in G.

2.1step 1.1step 1.2L4

Let R be the relation between the homogeneous k-element subsets K of V(G) and the good sets X∈[V(G)]2t defined by K⊆X. Step 1.2 shows that every good X is related to at least one K, so [L4] gives ∣R∣≥12(n2t).

3.1step 2.1L3L4

For the relation R of step 2.1, fix a homogeneous k-element subset K of V(G). The good sets X with K⊆X are among the 2t-element supersets of K, and [L3] counts those as (n−k2t−k). If N is the number of homogeneous k-element subsets of V(G), then [L4] gives ∣R∣≤N(n−k2t−k).

4.1step 2.1step 3.1algebra

Comparing steps 2.1 and 3.1 yields N≥12(n2t)/(n−k2t−k)=12(nk)/(2tk).

5.1step 4.1algebra∎

Since n≥2t, each factor in the ratio formula satisfies (n−j)/(2t−j)≥n/(2t) for 0≤j<k, so (nk)/(2tk)≥(n/(2t))k. Therefore N≥12(n/(2t))k.

Depends on

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