A vertex of a minimum vertex cut has a neighbor in every component I'm trying to understand the sol

Avah Knapp 2022-05-22 Answered
A vertex of a minimum vertex cut has a neighbor in every component
I'm trying to understand the solution for the following problem: Prove that κ ( G ) = κ ( G ) when G is a simple graph with Δ ( G ) 3.
The solution goes like this: Let S be the minimum vertex cut, | S | = κ ( G ). Since κ ( G ) κ ( G ) always, we need only provide an edge cut of size |S|. Let H 1 and H 2 be two components of G-S. Since S is a minimum vertex cut, each v S has a neighbor in H 1 and a neighbor in H 2 . The solution conntinues from here...I really have no clue why the part "Since S is a minimum vertex cut, each v S has a neighbor in H 1 and a neighbor in H 2 ." is true.
I tried to show it by contradiction but haven't gotten very far: Suppose otherwise; then there exists a vertex v S which doesn't have a neighbor in H 1 . We can assume that G is connected, hence there is a path joining v with u V ( H 1 )... and I'm stuck. How do I proceed from here? Or should I try to prove this in a completely different way?
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Answers (1)

sag2y8s
Answered 2022-05-23 Author has 10 answers
Step 1
If a vertex v S is not adjacent to any vertices in H 1 , then deleting S { v } is already enough to separate H 1 from the rest of the graph.
Step 2
Therefore S { v } is a smaller vertex cut, which contradicts our assumption that S was a smallest vertex cut.
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