Prove that if $G$ has a Hamiltonian path then $k(G-S)≤|S|+1$ for every non-empty proper subset $S$ of $V(G)

Question

State and prove a result analogous to theorem 3.16 that give a necessary condition for a graph to contain a Hamiltonian path.

Theorem 3.16: If $G$ is Hamiltonian then $k(G-S) \leq |S|$ for every nonempty proper subset $S$ of $V(G)$

I'm not sure if I state the analogus correctly, if it's wrong, then no woder I can't prove it. Here it is

Analogus to theorem 3.16:if $G$ has a Hamiltonian path then $k(G-S)≤|S|+1$ for every non-empty proper subset $S$ of $V(G)$.

I state like this because I have found a counter example that show $G$ has Hamiltonian path and $k(G-S) =|S|+1$

Assuming I got the theorem correctly. Since $G$ has a Hamiltonian path, there is a path $P=\{v_1,v_2,...,v_n\}$, let $v_1$ and $v_n$ be the 2 end vertices , I can still begin the path at the cut vertex, can't I?

Answer 1

Assume that $G$ has a Hamiltonian path, then $G$ is connected. Let $k(G-S)=m$, then there are $m$ components $G_1,G_2, ... , G_m$ in which between every pair $G_i , G_{i+1}$ there is a cut vertex $v_i$, so $|S|\geq m-1$, thus $k(G-S)=m \leq |S|+1$