Having trouble understanding this statement: If G is k-critical, then the minimum degree of its vertices is k-1 I want to prove that if a graph $G$ is $k$-critical, then $deg(v) \ge k-1,$ $ \forall v \in V_G$ , but I have no idea where to begin
A $k-$critical graph is a graph such that the chromatic number of $G$ is $k$ and for any edge in the graph, $G \backslash e$  has a proper colouring of $k-1$ colours

As seen in Bondy and Murty book:
By contradiction:


*

*Let G be a k-critical graph with $\delta$ < k-1

*Let v be a vertex with degree  $\delta$.

*Since G  k-critical , G-v is (k-1) colorible.

*Let (V1,V2,.....Vk-1) be a coloring of G-v.

*By definition v is adjacent to $\delta$ < k-1 vértices.

*Then, there is a Vj partition that doesn't contain any vertex
adjacent to v.(This is because there is k-1 partitions and  $\delta$
< k-1.)

*So we have that :


V1,V2,.. , ( Vj U {v} ) ,...Vk-1            is a (k-1)-coloring of G which is a contradiction.
