PropositionStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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A minimal counterexample to a kappa-bound is tau-critical

Statement

Let $H$ be a family of finite graphs, let $τ > 0$ , and let $G$ be an $H$ -free graph with $κ (G) < ∣ V (G) ∣^{τ}$ . If $G$ has the fewest vertices among all $H$ -free graphs with that strict inequality, then $G$ is $τ$ -critical.

Facts & Assumptions

Given: A family $H$ of finite graphs, a real $τ > 0$ , and an $H$ -free graph $G$ that is minimal by order among those satisfying $κ (G) < ∣ V (G) ∣^{τ}$ .

[L1]

A graph is $τ$ -critical exactly when it satisfies the strict inequality $κ (G) < ∣ V (G) ∣^{τ}$ and every proper induced subgraph $H$ satisfies $κ (H) \geq ∣ V (H) ∣^{τ}$ (A tau-critical graph).

[L2]

Every induced subgraph of an $H$ -free graph is again $H$ -free ( $H$ -free and $F$ -free graphs under the induced-subgraph convention, Subgraphs, induced subgraphs and spanning subgraphs).

Proof

technique · direct

1.1

The first clause of [L1] already holds for $G$ by the hypothesis $κ (G) < ∣ V (G) ∣^{τ}$ .

L1given

1.2

Let $H$ be a proper induced subgraph of $G$ . Then [L2] makes $H$ $H$ -free, and $∣ V (H) ∣ < ∣ V (G) ∣$ . By the minimality of $G$ , the graph $H$ cannot satisfy $κ (H) < ∣ V (H) ∣^{τ}$ . Hence $κ (H) \geq ∣ V (H) ∣^{τ}$ .

L2given

2.1

Steps 1.1 and 1.2 are exactly the two clauses in [L1], so $G$ is $τ$ -critical.

step 1.1step 1.2L1∎

Depends on

Used by

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Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Maria Chudnovsky, Alex Scott, Paul Seymour, and Sophie Spirkl, Erdos-Hajnal for graphs with no 5-hole, before Theorem 3.1 (standard reference, not scraped)