Minimum Degree is at Least Connectivity

Theorem
Let $G = \left({V, E}\right)$ be a simple graph.

Then:


 * $\delta \left({G}\right) \geq \kappa \left({G}\right)$

That is, the minimum degree of $G$ is at least its connectivity.

Proof
Pick a vertex $v \in G$ with $\deg_G \left({v}\right) = \delta \left({G}\right)$, that is, a vertex with minimum degree.

Recall that $\Gamma_G \left({v}\right)$ is the neighborhood of $v$ in $G$.

Suppose that $V = \Gamma_G \left({v}\right) \cup \left \{{v}\right\}$.

That is, that $v$ is adjacent to all other vertices of $G$.

Then:
 * $\left\vert{V}\right\vert = \left\vert{\Gamma_G \left({v}\right)}\right\vert + 1 = \delta \left({G}\right) + 1$

By definition, $\kappa \left({G}\right) < \left\vert{V}\right\vert$, so $\kappa \left({G}\right) \leq \delta \left({G}\right)$.

Otherwise, there is at least one vertex of $G$ not adjacent to $v$.

So, deleting $\Gamma_G \left({v}\right)$ from $G$ leaves $v$ isolated.

Thus $\Gamma_G\left({v}\right)$ is a vertex cut of size $\delta \left({G}\right)$.

Since $G$ has a vertex cut of size $\delta \left({G}\right)$, it follows that $\kappa \left({G}\right) \leq \delta \left({G}\right)$.