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.