LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Every induced subgraph of a cograph is a cograph

Statement

If $G$ is a cograph and $X \subseteq V (G)$ , then the induced subgraph $G [X]$ is also a cograph.

Facts & Assumptions

Given: A cograph $G$ and a subset $X \subseteq V (G)$ .

[L1]

A cograph is either null or is built from one-vertex graphs by finitely many disjoint unions and complete connections, and every cograph with at least two vertices has a last step of one of those two kinds (Cographs by the singleton, disjoint-union, and complete-connection recursion).

[L2]

If $G_{1} = (V_{1}, E_{1})$ and $G_{2} = (V_{2}, E_{2})$ are vertex-disjoint, then for every $X \subseteq V_{1} \cup V_{2}$ the induced subgraph of their disjoint union on $X$ is the disjoint union of $G_{1} [X \cap V_{1}]$ and $G_{2} [X \cap V_{2}]$ , while the induced subgraph of $G_{1} \nabla G_{2}$ on $X$ is $G_{1} [X \cap V_{1}] \nabla G_{2} [X \cap V_{2}]$ (The complete connection of two disjoint graphs, Subgraphs, induced subgraphs and spanning subgraphs).

Proof

technique · induction on a cograph construction of $G$

1.1givenL1base

We argue by induction on the recursive definition of cographs. If $G$ is the null graph or a one-vertex graph, then $G [X]$ is again the null graph or a one-vertex graph, so it is a cograph by [L1].

1.2L1ih

Assume now that $G$ is nontrivial, and that the claim is already known for the two smaller cographs in the last construction step of $G$ . By [L1], there exist vertex-disjoint smaller cographs $G_{1}$ and $G_{2}$ such that $G$ is either their disjoint union or their complete connection. Put $X_{i} : = X \cap V (G_{i})$ for $i = 1, 2$ . The induction hypothesis gives that $G_{i} [X_{i}]$ is a cograph for $i = 1, 2$ .

2.1step 1.2L2

If $G$ is the disjoint union of $G_{1}$ and $G_{2}$ , then [L2] gives $G [X] = G_{1} [X_{1}] ⊔ G_{2} [X_{2}]$ , a disjoint union of cographs. If $G$ is $G_{1} \nabla G_{2}$ , then [L2] gives $G [X] = G_{1} [X_{1}] \nabla G_{2} [X_{2}]$ , a complete connection of cographs. In either case the recursive definition shows that $G [X]$ is a cograph.

3.1step 1.1step 2.1discharge-induction∎

Steps 1.1 and 2.1 complete the induction, so every induced subgraph of a cograph is a cograph.

Depends on

Used by

Dependency tree · two levels

4 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

Tero Harju, Lecture Notes on Combinatorial Structures in Graph Theory, Section 5.3 (standard reference, not scraped)