For a simple undirected connected Graph $G(V,E)$ I say the edge $xy \in E(G)$ disconnects G if the resulting graph G' does not have a path from every vertex to every other vertex.
Now suppose I have such an edge xy, such that $G-(xy)$ is disconnected. I am wondering whether I can always get a disconnected Graph $G'$ similarly by removing the vertex x or y (just one of them) and all corresponding edges in this situation ?
I need to be careful obviously, as removing the egde $(xy)$ might have created a component consisting of just x or y alone, so if I tried to get a disconnected G' by removing that vertex I might have failed ! However if this problem occurs with both vertices then after removing one I end up with a trivial Graph of 1 vertex, which I define to be disconnected for my purposes, and if it only occurs with 1 vertex, then I just remove the other one, which should give me a disconnected Graph as desired again right ?
My Question is, whether this analysis of the situation is correct; i.e. given that an edge xy will give me disconnected graph $G-{(xy)}$ then I will also be able to obtain a disconnected graph $G-(x)$ or $G-(y)$.
Best Answer
So it turns out the analysis given is correct, since we can always remove a vertex with degree greater then 1 as long as G is connected with $|V(G)| \ge 3 $, and the special case where the size of G is smaller is covered by my special case definition.