I'm having some problem understanding the question below:
Let T = (V,E) be a tree. Show that T has a vertex v such that for all e that exists in E, the component of T-e containing v has at least |v|/2 vertices (where |v| = number of vertices, and |v|/2 is rounded up if |v| is odd). Prove that either v is unique or there are 2 adjacent vertices.
What I had interpreted initially was that there exists a vertex v such that if you take that vertex and all of its adjacent vertices to create a component, it would have at least |v|/2 vertices (upper-ceiling). (e.g., there would be at least one vertex in the Tree with degree of (|v|/2)-1 or greater). That is probably a wrong interpretation since the tree below doesn't have vertex with degree (upper-ceiling) |v|/2-1 or greater.
Could someone try to clarify the initial statement for me and help me get started on the proof? Thanks so much!
I understand it to mean that there is a vertex in the "middle" of the tree, so that no matter which edge you remove (breaking it into 2 sub-trees), the sub-tree that contains that vertex is always at least half the size of the original tree.