To prove that \( G - v \) is connected for any vertex \( v \in V(G) \), given that for every vertex \( v \in V(G) \) there is a Hamiltonian path of \( G \) with initial vertex \( v \), we need to show that removing any single vertex \( v \) from \( G \) does not disconnect the graph. Let's break this down step by step.

### Definitions and Properties

1. **Hamiltonian Path**: A path that visits every vertex exactly once.
2. **Connected Graph**: A graph is connected if there is a path between any two vertices.

### Given Condition

For every vertex \( v \in V(G) \), there exists a Hamiltonian path starting at \( v \). This means that we can find a path that starts at \( v \) and visits all other vertices exactly once.

### Proof

#### Step 1: Assume for Contradiction

Assume, for the sake of contradiction, that there exists a vertex \( v \in V(G) \) such that \( G - v \) is not connected. This means that removing \( v \) from \( G \) results in at least two disconnected components, say \( C_1 \) and \( C_2 \).

#### Step 2: Hamiltonian Path Starting at \( v \)

Since there is a Hamiltonian path starting at \( v \), let \( P_v \) denote this path. Let the path be:
\[
P_v = v, v_1, v_2, \ldots, v_{n-1}
\]
where \( v_1, v_2, \ldots, v_{n-1} \) are the vertices of \( G \) visited by the Hamiltonian path after starting at \( v \).

#### Step 3: Effect of Removing \( v \)

Removing \( v \) from \( G \) should leave a path that visits all other vertices exactly once (the remaining part of the Hamiltonian path):
\[
P_{v - v} = v_1, v_2, \ldots, v_{n-1}
\]

Since \( P_v \) was a Hamiltonian path, all vertices \( v_1, v_2, \ldots, v_{n-1} \) are distinct and connected in sequence. This sequence of vertices should span \( G - v \).

#### Step 4: Disconnected Components

If \( G - v \) is disconnected, there must be no path between at least one pair of vertices from different components \( C_1 \) and \( C_2 \) in the subgraph. However, since \( P_{v - v} \) (the path obtained by removing \( v \) from \( P_v \)) includes all vertices of \( G \) except \( v \), it must span all vertices of \( G - v \), connecting them in sequence.

#### Step 5: Contradiction

The existence of a Hamiltonian path in \( G \) that visits all vertices, when removed \( v \), implies that \( G - v \) must be connected. This contradicts our initial assumption that \( G - v \) is disconnected.

Therefore, our assumption that \( G - v \) is disconnected must be false. Thus, \( G - v \) is connected.

### Conclusion

Since the removal of any vertex \( v \in V(G) \) does not disconnect \( G \), we conclude that \( G \) is 2-connected. This means that \( G \) is connected and remains connected upon the removal of any single vertex.

Would you like more details or have any questions?

Here are 5 questions you might want to ask next:
1. How does the concept of 2-connectedness extend to higher connectivity?
2. Can you give an example of a 2-connected graph and a non-2-connected graph?
3. How does Menger's Theorem relate to the concept of 2-connected graphs?
4. What are some practical applications of Hamiltonian paths in real-world problems?
5. How can we determine the connectivity of a given graph?

Tip: When dealing with graph connectivity, leveraging properties of paths and cycles, such as Hamiltonian paths, can provide strong insights into the graph's structure and resilience to vertex removal.

Understand why the graph G − v remains connected upon removing any vertex v. This problem delves into graph theory concepts including Hamiltonian paths and graph connectivity, with a detailed proof using Menger's Theorem. Suitable for advanced undergraduate students and anyone interested in advanced graph theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation