I'm going through a set of notes on Graph Theory. The following theorem is proved:
Theorem 1: Let $G$ be a graph of order $n$ ($ n \geq 3$) such that $ \delta(G) \geq \frac{n}{2} $. Then $G$ is Hamiltonian.
I'm happy with the statement and proof of this theorem. The notes then prove the following theorem:
Theorem 2: Let $G$ be a graph of order $n$ ($n \geq 3$) such that $G$ is connected and $ \delta(G) \geq \frac{k}{2} $, where $ k < n $. Then $G$ has a path of length k.
I'm happy with the proof of this. The notes then make the following remark
Remark: We do need $k < n $, e.g. $ G = K_k $, and we do need $G$ to be connected e.g. two disjoint copies of $K_k$.
I don't understand this remark. Clearly $k$ can't be bigger than $n$ (as there obviously can't be a path through more than $n$ vertices on a graph with $n$ vertices). So the remark must mean we can't have $k = n $. But if $k = n$, then isn't this just the same as Theorem 1? I don't see how the example relates, either; if $ G = K_n $, then the theorem still works?
Thanks for your help!