I am attempting to prove this result but I found a case where I can easily disprove this statement.Suppose $G$ is a graph with two nodes u, v and an edge (u,v) and respective self edges, then removing the edge u,v results in two nodes left, therefore leaving the resulting graph disconnected.Can someone show me why I am approaching this question incorrectly?
Can't see how this can be true: any connected undirected graph $G$ contains vertex v such that removing v results in another connected Graph $G'$
graph theory
Suppose to the contrary that removing any vertex of a finite graph $G$ would disconnect it. Then $G$ could not contain any cycles (else we'd have an immediate contradiction) and so $G$ would have to be a tree. However, removing a leaf vertex of $G$ wouldn't disconnect it, contrary again to our assumption.