When n = 1, the case is trivial, since the graph itself is a tree and hence it is a
spanning tree.
When n = 2, the graph is isomorphic to the cycle C4 and therefore result follows
from Theorem 2.2.2.
When n > 2, consider any spanning tree T of K2n. In T, the intersection
N(v,) n N(v2) of the open neighborhoods the vertices V! and v2 contains exactly
one vertex. Because, if it has more than one vertex they will create a circuit and if it
has no vertex, then T will be disconnected. Let u; be that vertex and degT (uj) = 2