False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

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Every hereditary graph class is closed under taking arbitrary subgraphs

False Statement

Every hereditary graph class is closed under taking arbitrary, not necessarily induced, subgraphs.

Facts & Assumptions

Given: The hereditary class $\mathcal K$ of complete graphs.

[L1]

Complete graphs form a hereditary class (Complete graphs form the hereditary class with forbidden basis $\{\overline K_2\}$ ).

[L2]

$K_3$ contains $P_3$ as an ordinary subgraph but not as an induced subgraph ( $K_3$ contains $P_3$ as a subgraph but not as an induced subgraph).

[F1]

Heredity requires closure under induced subgraphs, not arbitrary edge-deleted subgraphs (Hereditary graph classes).

Refutation

technique · direct

1.1

The graph $K_3$ belongs to $\mathcal K$ .

1.2

Deleting one edge gives an ordinary subgraph $P_3$ , which is not complete and hence does not belong to $\mathcal K$ .

2.1

Therefore the hereditary class $\mathcal K$ is not closed under arbitrary subgraphs.

step 1.1step 1.2F1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Valerio Boncompagni, On hereditary graph classes defined by forbidding Truemper configurations (PhD thesis, 2018) (standard reference, not scraped)