Minimum Degree is at Least Connectivity

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

If $V = \Gamma_G\left({v}\right) \cup \left\{{v}\right\}$—that is, if $v$ 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 $v$.

Then $\Gamma_G\left({v}\right)$ is a vertex cut of size $\delta\left({G}\right)$, since deleting $\Gamma_G\left({v}\right)$ from $G$ leaves $v$ isolated.

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