Relation between the connectivity and the minimum degree of a graph 
Prove that for every graph $G$, (a) $κ(G) ≤ δ(G)$; (b)   if $δ(G)≥n−2,$ then $κ(G)=δ(G).$

Here $κ(G)$ is the connectivity, which is the minimum size of a cut set of $G$. And  $δ(G)$ is the minimum degree of $G$.
I have part (a) I'm having a hard time with part (b). 
 A: $\kappa(G)$ is presumably the vertex connectivity of $G$. 
Suppose $\delta(G)=n-1$. Then no matter what vertices you take the graph will remain connexted.
Suppose $\delta(G)=n-2$. What does a graph with minimum degree $n-2$ look like? What does the complement look like? it has connected components with $2$ or $1$ vertices. Therefore if you take vertices away from $G$ the complement will always remain disconnected unless you remove $n-2$ vertices. Since the complement of a disconnected graph is connected we conclude we need to remove $n-2$ vertices to make $G$ disconnected.
A: Prove that for any graph G: 
a) k(G)≤δ(G)
Proof: Let G be a graph of order n and let v∈V(G), be a vertex in G such that deg⁡(v)=δ(G). Then we can disconnect v from G by removing δ(G) edges. Hence the connectivity of G must be smaller or equal to δ(G).
b) If δ(G)≥n-2, then k(G)=δ(G)
 
Proof: Let G be a graph of order n such that δ(G)≥n-2. If δ(G)=n-1 then G is complete and by definition k(G)=n-1= δ(G). Note that δ(G) cannot be bigger then n-1 for simple graphs, so we are left with the case for δ(G)=n-2. Let S={v1,u2,...,vn-2 } be a subset of V(G) such that k(G-S)=0. Suppose we can remove one vertex from S, and still have G-S be disconnected. But then |V(G-S)|=n-(n-3)=3, and the lowest degree for a vertex in G-S is δ(G-S)=δ(G)-|S|=n-2-(n-3)=1, so it must be connected (draw a graph to prove this to yourself). Therefore, if δ(G)=n-2 then it must be the case that k(G)=δ(G)=n-2
