Let T be a tree with sub-trees which each set has a vertex in common - hence T has a vertex in all of its sub-trees?

Question

The question is:

Let T be a tree with sub-trees $T_1,T_2,..,T_n$ such that all trees $T_i,T_j$ have a vertex in common which each set has a vertex in common - show that T has a vertex in all $T_i$.

The difficulty I am having is that if every subtree of T has a vertex in common say $T_4 \cap T_5 \neq \emptyset$ then $T_4 \subset T_5$ or the other way around. Now if this is the case for all subtrees then they are just subsets of larger subtrees that are subsets of T, but since all subtrees have to have a vertex in common they would all have the root of T in common and hence T would have a vertex in all of its subtrees.

This doesn't seem right as we could form subtrees of T not containing the root node if there is more than one level to the tree.

Any thoughts?

Thanks,

Brian

Answer 1

By your own argument the subtrees must form a chain under $\subseteq$; any vertex in the smallest member of the chain is a vertex of $T$ that lies in all $n$ subtrees.