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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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K3K_3 has ν(K3)=1<2=τ(K3)\nu(K_3)=1<2=\tau(K_3), so König's equality needs bipartiteness

Counterexample

In the triangle K3K_3, every matching has at most one edge, while every vertex cover has at least two vertices. Hence ν(K3)=1<2=τ(K3)\nu(K_3)=1<2=\tau(K_3).

Facts & Assumptions

Given: The complete graph K3K_3 on three vertices.

[L1]

König's equality applies to finite bipartite graphs (König's theorem: ν(G)=τ(G)\nu(G)=\tau(G) for every finite bipartite graph).

Verification

Verification technique: direct.

1.1

Any two edges of K3K_3 share a vertex, so a matching has at most one edge; any one edge gives ν(K3)=1\nu(K_3)=1.

given
1.2

Deleting one vertex leaves an edge, so one vertex is not a cover; two vertices cover all edges, giving τ(K3)=2\tau(K_3)=2.

1.3

Thus ν(K3)<τ(K3)\nu(K_3)<\tau(K_3), and [L1] shows precisely why this does not contradict König's theorem: K3K_3 is not bipartite.

L1
2.1

This triangle is a finite witness that bipartiteness is a necessary hypothesis for the equality.

step 1.1step 1.2step 1.3∎

Depends on

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