Bourbaki

Proposition 2

Tag 0AOZ, permanent. Cite it as [Bourbaki, Tag 0AOZ].

Lie Groups and Lie Algebras, chapter IV, Appendix 0. Graphs

Let Γ=(A,S)\Gamma = (A, S) be a forest having only a finite number of vertices.

(i) If Γ\Gamma has at least one vertex, it has a terminal vertex. (ii) If Γ\Gamma has at least two vertices, there is a partition (S′,S′′)(S', S'') of its set of vertices into two non-empty subsets such that two distinct vertices that both belong to S′S' or both belong to S′′S'' are never joined.

Suppose that Γ\Gamma has at least one vertex and let (x0,…,xn)(x_0, \ldots, x_n) be an injective path of maximal length in Γ\Gamma. The vertex x0x_0 cannot be joined to a vertex yy distinct from x0,x1,…,xnx_0, x_1, \ldots, x_n, since otherwise there would exist an injective path in Γ\Gamma of length n+1n + 1, namely (y,x0,…,xn)(y, x_0, \ldots, x_n). The vertex x0x_0 is not joined to any vertex xix_i with 2≤i≤n2 \leq i \leq n, since otherwise (x0,x1,…,xi)(x_0, x_1, \ldots, x_i) would be a circuit in the forest Γ\Gamma. Thus, x0x_0 is terminal.

We shall prove (ii) by induction of the number mm of vertices of Γ\Gamma, the case m=2m = 2 being trivial. Suppose then that m≥3m \geq 3 and that assertion (ii) is proved for graphs with m−1m - 1 vertices. Let aa be a terminal vertex of Γ\Gamma (cf. (i)). We apply the induction hypothesis to the full subgraph of Γ\Gamma whose vertices are the vertices x≠ax \neq a of Γ\Gamma. Thus, there exist two non-empty disjoint subsets S1′S'_1 and S1′′S''_1 of SS with S1′∪S1′′=S−{a}S'_1 \cup S''_1 = S - \{a\}, and such that two distinct vertices in S1′S'_1 (resp. S1′′S''_1) are never joined. Since aa is joined to at most one vertex of Γ\Gamma, we can suppose for example that it is not joined to any vertex in S1′S'_1. The partition (S1′,S1′′∪{a})(S'_1, S''_1 \cup \{a\}) then has the required property. Q.E.D.

For any integer n≥1n \geq 1, denote by AnA_n the graph whose vertices are the integers 1,2,…,n1, 2, \ldots, n and whose edges are the pairs {i,j}\{i, j\} with i−j=±1i - j = \pm 1:

123n−1n∘∘∘⋯∘ \begin{array}{cccccc} 1 & 2 & 3 & n-1 & n \\ \circ & \circ & \circ & \cdots & \circ \end{array}

A graph Γ\Gamma is said to be a chain of length m≥0m \geq 0 if it is isomorphic to Am+1A_{m+1}. This is equivalent to the existence in Γ\Gamma of an injective path (x0,…,xm)(x_0, \ldots, x_m) containing all the vertices, the vertices xix_i and xjx_j not being joined if ∣i−j∣>1|i - j| > 1.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.