Let P be a subset of R and V1 (resp. Γ) the vector subspace (resp. the subgroup) of V generated by P. The following conditions are equivalent:
(i) P is closed and symmetric; (ii) P is closed, and P is a root system in V1; (iii) Γ∩R=P.
Assume that these conditions are satisfied. For any α∈P, let α1~ be the restriction of α~ to V1. Then the map α↦α1~ is the canonical bijection from the root system P to P~.
(iii) ⟹ (i): clear.
(i) ⟹ (ii): assume that P is closed and symmetric. First, P satisfies (RS_I) in V1. We show that, if α,β∈P, then sα(β)∈P. This is clear if α and β are proportional. Otherwise, sα(β)=β−n(β,α)α and β−pα∈R for all rational integers p between 0 and n(β,α) (Prop. 9, no. 3), so
β−n(β,α)α∈P since
P is closed and symmetric. Thus,
sα,α1(P)=P, and
P satisfies (RS_{II}). It is clear that
P satisfies (RS_{III}). Thus,
P satisfies (ii), and we have proved the last assertion of the proposition at the same time.
(ii) ⟹ (iii): we show that, if condition (ii) is satisfied, then Γ∩R=P. It is clear that P⊆Γ∩R. Let β∈Γ∩R. Since β∈Γ and P=−P, β=α1+α2+⋯+αk with α1,…,αk∈P. We shall prove that β∈P. This is clear if k=1. We argue by induction on k. We have
0<(β∣β)=i=1∑k(β∣αi), so
(β∣αi)>0 for some index
i. If
β=αi, then
β∈P. Otherwise,
β−αi∈R (Cor. of Th. 1, no. 3), so
β−αi∈P by the induction hypothesis, hence
β∈P since
P is closed.
The conditions of Prop. 23 can be realised with V1=V and yet P=R. For example, this is the case when R is a system of type G2 and P a system of type A2; cf. Plate X.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.