Ore Graph is Connected

Theorem
Let $G = \left({V, E}\right)$ be an Ore graph of order $n \ge 3$. Then, $G$ is connected.

Proof
Proof by contrapositive.

Suppose $G$ is not connected. Then it has at least two components $C_1$ and $C_2$.

Let $m_1$ and $m_2$ be the number of vertices in $C_1$ and $C_2$, respectively. It is clear that $m_1 + m_2 \le n$.

Since $n \ge 3$, there exist non-adjacent vertices $u$ and $v$ such that $u$ is in $C_1$ and $v$ is in $C_2$.

As $G$ is simple, it follows that $\deg_G(u) \le m_1 - 1$ and $\deg_G(v) \le m_2 - 1$.

Thus,

$\deg_G(u) + \deg_G(v) \le m_1 - 1 + m_2 - 1 = m_1 + m_2 - 2 < m_1 + m_2 = n$

Therefore, $G$ is not an Ore Graph.

Also see

 * Ore's Theorem