Let $T_1, \: \ldots, \: T_k$ subtrees of a tree $T$ such that for all $1\leq i Let $T_1,\: \ldots,\: T_k$ subtrees of a tree $T$ such that for all $1\leq i <j \leq k $ the trees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex that is in all $T_i$.

I found this, it is right?
 A: To try to give a brief indication of how to go about the proof:

*

*In any tree, the intersection of any two subtrees is necessarily itself a subtree (possibly empty).

*Given a (not necessarily finite) tree, it is vital to establish that given a family of three subtrees which intersect pairwise the three have a common vertix. This requires a bit of surgery on paths (the argument resembles somewhat that given to show the uniqueness of the path connecting any two vertices in a tree).

*With the above preliminaries in their proper place, one can prove by induction that given a fixed tree
and any nonempty finite family of subtrees which intersect pairwise there will exist a vertex common to all the subtrees in the family. The proof is carried by induction on the cardinality of the index set, the base case is trivial as it refers to a family with just one component and the inductive step is established by fixing one of the subtrees in a given family $P$ indexed by a set of $n+1$ elements and by considering the family $Q$ of just $n$ subtrees obtained by intersecting the remaining $n$ subtrees in the original family with the fixed one. The observation at 2) ensures that this family of restricted subtrees satisfies the condition of pairwise intersection, so that the inductive hypothesis applies safely to this family. Clearly, the intersection of all members of $Q$ coincides with the intersection of all members of $P$, so the latter is nonempty as long as the former is.

A very prosaic description, but perhaps I will add a proper formal version later on, if I find the time.
