Bourbaki

Exercise 20, § 6

Topological Vector Spaces, II, p. 86

It is set in small type as supplementary.

Let F=R(A),G=RAF = \mathbf{R}^{(A)}, G = \mathbf{R}^A, where A is any infinite set; suppose that F and G are put in separating duality by the bilinear form ⟨x,y⟩=∑α∈Ax(α) y(α)\langle x, y \rangle = \sum_{\alpha \in A} x(\alpha) \, y(\alpha). a) Let N be an additive subgroup of G; we denote by N∗N^* the subgroup of the x∈Fx \in F such that ⟨x,y⟩\langle x, y \rangle is an integer for all y∈Ny \in N and by N∗∗N^{**} the subgroup of the z∈Gz \in G such that ⟨x,z⟩\langle x, z \rangle is an integer for all x∈N∗x \in N^*. If N‾\overline{N} is the closure of N for the topology σ(G,F)\sigma(G, F), show that N∗N^* is closed in F for σ(F,G)\sigma(F, G) and that N∗∗=N‾N^{**} = \overline{N} (to establish this last point, use GT, VII, § 1.3, prop. 6, projecting N on the finite dimensional coordinate varieties of G). b) Suppose that A=NA = N. Let M be a closed subgroup of F for σ(F,G)\sigma(F, G); show that if V is the largest vector subspace contained in M, then M is the topological direct sum of V and of a closed subgroup P that is a free Z\mathbf{Z}-module having an enumerable base. (Consider F as the union of an increasing sequence (Fn)(F_n) of finite dimensional vector subspace and apply GT, VII, § 1.2, th. 2 and § 1, exerc. 7.) P is discrete (for the topology induced by σ(F,G)\sigma(F, G)), if, and only if, P is of finite rank. c) Deduce from a) and b) that when A=NA = N, every closed subgroup of G (when G carries the product topology σ(G,F)\sigma(G, F)) can be transformed, by an automorphism of the topological group G, in a product RI×ZJ\mathbf{R}^I \times \mathbf{Z}^J, where I and J are two sets of N without common elements. d) In the space E=RNE = \mathbf{R}^N, carrying the topology σ(E,E∗)\sigma(E, E^*), show that the subgroup ZN\mathbf{Z}^N is closed and does not contain any line, even though it is not a free Z\mathbf{Z}-module (A, VII, p. 59, exerc. 8); the results of b) do not therefore extend when A is not enumerable.

Solution

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