Bourbaki

Corollary 3

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

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

Let G be a finite-dimensional Lie group and (Hi)i∈I(H_i)_{i \in I} a family of Lie subgroups of G. Then H=⋂i∈IHiH = \bigcap_{i \in I} H_i is a Lie subgroup of G and

L(H)=⋂i∈IL(Hi). L(H) = \bigcap_{i \in I} L(H_i).
There exists a finite subset J of I such that ⋂i∈JL(Hi)\bigcap_{i \in J} L(H_i) is equal to the intersection M of all the L(H_i). We know that H^* = \bigcap_{i \in J} H_i is a Lie subgroup such that L(H^*) = M (§ 3, no. 8, Corollary 2 to Proposition 29). Let H_0 be the identity component of H^*. It is a Lie subgroup of G and L(H_0) = M. By Corollary 2, H_0 \subset H_i for all i and hence H_0 \subset H \subset H^*, whence the corollary.

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

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Transcribed from the printed volume. Not translated.