False statementConstruction: AI-adaptedVerification: AI-generatedprecheck 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 $K$ 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 $K$ .

1.2

Deleting one edge gives an ordinary subgraph $P_{3}$ , which is not complete and hence does not belong to $K$ .

2.1

Therefore the hereditary class $K$ 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 · two levels

9 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

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