Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-01
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K3 contains P3 as a subgraph but not as an induced subgraph

Statement refuted

Every ordinary subgraph copy of H in G is an induced copy.

Facts & Assumptions

Given: The graphs P3 and K3.

[F1]

Deleting one edge from K3 leaves a spanning subgraph isomorphic to P3 (Subgraphs, induced subgraphs and spanning subgraphs, Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices).

[F2]

An induced embedding must preserve both adjacency and nonadjacency (Induced embeddings and induced copies of a graph).

Counterexample

technique · direct
1.1

By deleting one triangle edge, K3 contains an ordinary subgraph copy of P3.

F1
1.2

Any injection from the three vertices of P3 to the three vertices of K3 is bijective, but the two endpoints of P3 are nonadjacent while their images in K3 are adjacent.

F2
2.1

Thus K3 contains no induced P3, despite containing an ordinary subgraph copy.

step 1.1step 1.2∎

Remarks


\draw[->,line width=.9pt] (2.9,.68)--node[above,font=\scriptsize] {delete $02$} (4.25,.68);

\node[vertex] (p0) at (4.75,0) {$0$}; \node[vertex] (p1) at (5.95,1.35) {$1$}; \node[vertex] (p2) at (7.15,0) {$2$}; \draw[edge] (p0)--(p1)--(p2); \draw[gray!65,dashed,line width=.8pt] (p0)--node[below,font=\scriptsize] {deleted} (p2); \node[caption] at (5.95,-.65) {an ordinary $P_3$ subgraph\\not induced}; \end{tikzpicture} ```

Depends on

Used by

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