Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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Every hereditary graph class has a finite forbidden induced-subgraph basis

False Statement

Every hereditary graph class has a finite minimal forbidden induced-subgraph basis.

Facts & Assumptions

Given: The hereditary class B\mathcal B of finite bipartite graphs.

[F1]

For every k1k\ge1, C2k+1C_{2k+1} is an odd cycle, and distinct lengths give nonisomorphic graphs (Empty and complete graphs, complete bipartite graphs, and the convention that PnP_n and CnC_n have nn vertices).

[F2]

A minimal forbidden induced subgraph is outside the class while all proper induced subgraphs are inside (Minimal forbidden induced subgraphs and forbidden bases).

[L2]

The family of all such minimal graphs is the unique minimal basis (Every hereditary graph class is determined by its unique minimal forbidden induced subgraphs).

Refutation

technique · contradiction
1.1

Suppose the minimal forbidden basis of B\mathcal B is finite.

assume-contra
1.2

For every k1k\ge1, the graph C2k+1C_{2k+1} is not bipartite. Every proper induced subgraph of this chordless cycle is a disjoint union of paths, hence is bipartite. Thus C2k+1C_{2k+1} is minimally forbidden.

L1F1F2
2.1

The minimal basis therefore contains the pairwise nonisomorphic graphs C3,C5,C7,C_3,C_5,C_7,\ldots.

step 1.2L2
3.1

This is an infinite family, contradicting step 1.1. Hence a hereditary class need not have a finite minimal forbidden basis.

step 1.1step 2.1discharge-contradiction

Remarks

C3C5C7¢¢¢C3;C5;C7;:::arethepairwisenonisomorphicminimalforbiddengraphs

Depends on

Used by

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Sources