Bourbaki

Exercise 7, § 3

Topological Vector Spaces, IV, p. 59

Bourbaki marks this one as harder than the rest.

a) Let E be a (DF) space, U a convex, balanced and closed set in E, which absorbs the bounded parts of E, (xn)(x_n) a sequence of points of CU\mathbf{C} U. Prove that there exists a neighbourhood V of 0 in E which contains none of the xnx_n. (Let (An)(A_n) be a countable basis of the canonical bornology of E. Prove that one can define by recurrence a sequence (λn)(\lambda_n) of numbers >0> 0 and a sequence (Vn)(V_n) of neighbourhoods of 0, convex, balanced and closed, such that λnAn⊂⋂j=1∞Vj\lambda_n A_n \subset \bigcap_{j=1}^\infty V_j, λnAn⊂U\lambda_n A_n \subset U and xn∈CVnx_n \in \mathbf{C} V_n for every n. For this, the λj\lambda_j and VjV_j being determined for j≤nj \leq n, first take λn+1\lambda_{n+1} such that λn+1An+1⊂U\lambda_{n+1} A_{n+1} \subset U and λn+1An+1⊂Vj\lambda_{n+1} A_{n+1} \subset V_j for every j≤nj \leq n, then take Vn+1V_{n+1} containing the closure of the convex envelope of the union of the λjAj\lambda_j A_j for j≤n+1j \leq n + 1.) b) Deduce from a) that if M is a subset of E containing a countable everywhere dense set, the topology induced on M by the strong topology of the bidual E′′E'' of E is identical with the topology induced by that of E. In particular, the convergent sequences in E are the same for the topology of E and the topology induced by the strong topology of E′′E''; for every metrizable subset M of E, the topology induced on M by the topology of E is identical with the topology induced by the strong topology of E′′E''. c) Deduce from a) that if there exists a countable everywhere dense set in E, then E is infratonnelé. d) Deduce from b) and exerc. 6 that if every bounded subset of E is metrizable for the topology induced by that of E, then E is infratonnelé.

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.