Math 409 Lecture 6 Minimum Spanning Trees

Proof. This involves proving P ⇔ Q where P is the statement T is a spanning tree of G and Q is the statement T has n − 1 edges. In both statements we are allowed to add on the fact that T is a spanning connected subgraph of G. This addition is implicit in P but is a crucial addition needed for Q if the lemma is to be true. (Q ⇒ P ): Suppose T is a spanning connected subgraph of G with n − 1 edges. We need to show that T is acyclic which we will prove by the method of contradiction. So suppose there is a circuit C in T using the vertices v1, v2, . . . , vk. This circuit contains k edges which are all in T . There are n− k vertices remaining in G and each of them is touched by at least one edge of T since T is a spanning and connected subgraph of G. None of these edges can come from the circuit C since the remaining n− k vertices are not vertices of C. Therefore, T has at least n−k more edges which implies that T has at least k+(n−k) = n edges which contradicts our assumption Q. Therefore, T is acyclic and is a spanning tree of G.