Delete a vertex,how to show that graph remains connected? How can I show that at each connected graph(there is a path from any point to any other point in the graph),there is a vertex,so that if we delete this vertex and the edges that conclude to it,the graph remains connected?
 A: Since $G$ is connected, it has a spanning tree. Consider a leaf of this tree.
A: Consider the following graph: It is connected, you can get from any vertex to any vertex. But what happens if you take away one vertex? Then it stops being connected.

In Graph theory there is something called the connectivity of a graph. It tells you how many vertices need to be deleted before it becomes disconnected. In this case it is one. Because taking out B makes it disconnected.

Let G be any graph. If G has a vertex with degree 1 take it away. See it is still connected.
Suppose G has no vertex of degree $1$ Then look at a spanning tree of the graph. Then this tree will have a leaf(a vertex with degree 1) take it away and see the tree is still connected. So putting back the edges will not take away the connectivity of the graph.
A: Let $G$ be a connected graph with at least two vertices. Choose vertices $u,v\in V(G)$ so as to maximize the distance $d(u,v)$, the length of a shortest path from $u$ to $v$. If $w\in V(G)-\{v\}$, then a shortest path from $u$ to $w$ does not go through $v$; if it did, we would have $d(u,w)\gt d(u,v)$. Thus every vertex of $G-v$ is connected to $u$ by a path in $G-v$; i.e., the graph $G-v$ is connected.
A: Sounds incorrect to me. If this is true, let $G = (V, E)$ be such graph. Then, there is one vertex $v_1$, removing which from $G$ we get another coherent graph $G_1 = G - v_1$.
Applying the same principle, there exists some $v_2$ such that $G_2 = G_1 -v_2$ is coherent. By induction, it's easy to see that the empty graph is coherent.
That contradicts axiom (CG1) on p.10 in the above-cited definition paper.
