Prove that the deletion of edges of a minimum-edge cut of a connected graph $G$ results in a disconnected graph with exactly two components. (Note that a similar result is not true for a minimum vertex cut.)
Suppose that the deletion of edges of a minimum-edge cut of a connected graph $G$ results in a disconnected graph with more than two components. Let $G_1, G_2, G_3,...G_k$ be the connected components obtained by removing minimum-edge cut of a connected graph. so, there exists $k$ edges required to make the graph connected. which contadict our assumption. Am I correct?