Let K be a field, E a right vector space over K, F a left vector space over K, (Mα)α∈A a family of vector subspaces of E and (Nβ)β∈B a family of vector subspaces of F. Then
(α∈A⋂Mα)⊗Kβ∈B⋂Nβ=(α,β)∈A×B⋂(Mα⊗KNβ).
It is obviously sufficient to prove the particular case
(α∈A⋂Mα)⊗KF=α∈A⋂(Mα⊗KF).
Clearly the left hand side of (25) is contained in the right hand side. To prove the converse, we consider a basis (fλ)λ∈L of F. Every element of E⊗KF can then be expressed uniquely in the form ∑λ∈Lxλ⊗fλ, where xλ∈E (§ 3, no. 7, Corollary 1 to Proposition 7); if E′ is a vector subspace of E, the relation ∑λ∈Lxλ⊗fλ∈E′⊗KF is equivalent, by Proposition 14, to xλ∈E′ for all λ∈L. To say that ∑λ∈Lxλ⊗fλ belongs to each of the Mα⊗KF thus means that for all λ∈L and all α∈A, xλ∈Mα, that is xλ∈⋂α∈AMα for all λ∈L, which proves that the right hand side of (25) is contained in the left hand side.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.