Ore Graph is Connected

Theorem
Let $G = \struct {V, E}$ be an Ore graph.

Then $G$ is connected.

Proof
Let $G$ be an Ore graph of order $n$.

$G$ is not connected.

Then it has at least two components.

Call these components $C_1$ and $C_2$.

Thus, there exist non-adjacent vertices $u$ and $v$ such that $u$ is in $C_1$ and $v$ is in $C_2$.

Let $m_1$ and $m_2$ be the number of vertices in $C_1$ and $C_2$ respectively.

It is clear that:
 * $m_1 + m_2 \le n$

By definition of Ore graph, $G$ is simple.

Thus it follows that:
 * $\map {\deg_G} u \le m_1 - 1$ and $\map {\deg_G} v \le m_2 - 1$

Thus:

That is:
 * $\map {\deg_G} u + \map {\deg_G} v < n$

But by definition of Ore graph:
 * $\map {\deg_G} u + \map {\deg_G} v \ge n$

By Proof by Contradiction, it follows that our assumption that $G$ is not connected was false.

Hence the result.

Also see

 * Ore's Theorem