Proposition 2
Tag 0AOZ, permanent. Cite it as [Bourbaki, Tag 0AOZ].
Lie Groups and Lie Algebras, chapter IV, Appendix 0. Graphs
Let be a forest having only a finite number of vertices.
(i) If has at least one vertex, it has a terminal vertex. (ii) If has at least two vertices, there is a partition of its set of vertices into two non-empty subsets such that two distinct vertices that both belong to or both belong to are never joined.
Suppose that has at least one vertex and let be an injective path of maximal length in . The vertex cannot be joined to a vertex distinct from , since otherwise there would exist an injective path in of length , namely . The vertex is not joined to any vertex with , since otherwise would be a circuit in the forest . Thus, is terminal.
We shall prove (ii) by induction of the number of vertices of , the case being trivial. Suppose then that and that assertion (ii) is proved for graphs with vertices. Let be a terminal vertex of (cf. (i)). We apply the induction hypothesis to the full subgraph of whose vertices are the vertices of . Thus, there exist two non-empty disjoint subsets and of with , and such that two distinct vertices in (resp. ) are never joined. Since is joined to at most one vertex of , we can suppose for example that it is not joined to any vertex in . The partition then has the required property. Q.E.D.
For any integer , denote by the graph whose vertices are the integers and whose edges are the pairs with :
A graph is said to be a chain of length if it is isomorphic to . This is equivalent to the existence in of an injective path containing all the vertices, the vertices and not being joined if .
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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