Week 5: Separation of Graphs

Proof. “⇐” Suppose G has a cut vertex v. Then Gr v has at least two components G1 and G2. Given vertices vi ∈ V (Gi), i = 1, 2. Each path connecting v1 and v2 must contain the vertex v. So any two paths between v1 and v2 must contain the common internal vertex v. This is a contradiction. “⇒” Given two distinct vertices u and v of G, let d(u, v) denote the distance between u and v. We apply induction on d(u, v) to show that there are internally disjoint paths between u and v. For d(u, v) = 1, there exists an edge e = uv in G. Since u and v are not cut vertices, we have both G r u and G r v connected. If e is a cut edge, then there is a third vertex adjacent with u or v, say adjacent with u; it turns out that u is a cut vertex of G, which is a contradiction. So e cannot be a cut edge; consequently, e is contained in a cycle C. We thus have two internally disjoint paths uev and C r e between u and v.