Bourbaki

Exercise 21, § 2

Topological Vector Spaces, II, p. 68

Bourbaki marks this one as harder than the rest.

a) Let AiA_i (1≤i≤r1 \leq i \leq r) r>n+1r > n + 1, be a family of convex subsets of Rn\mathbf{R}^n such that any r−1r - 1 of the AiA_i have a non-empty intersection; show that the r sets AiA_i have a non-empty intersection (Helly's theorem). (Let xix_i be a point of the intersection of the AjA_j with indexes j≠ij \neq i; there exist r numbers λi\lambda_i which are not all zero and are such that ∑i=1rλi=0\sum_{i=1}^r \lambda_i = 0 and ∑i=1rλixi=0\sum_{i=1}^r \lambda_i x_i = 0; in this last equation take to one side those terms with λi≥0\lambda_i \geq 0 and to the other those with λi<0\lambda_i < 0.) b) Given a family of compact convex sets in Rn\mathbf{R}^n, show that the intersection of all the sets of the family is non-empty if the intersection of any selection of n+1n + 1 sets of the family is non-empty. c) In Rn\mathbf{R}^n, let K be a convex set and (Ai)1≤i≤r(A_i)_{1 \leq i \leq r} be a family of r>n+1r > n + 1 convex sets. Suppose that for every selection of n+1n + 1 indices (ik)(i_k) each less than or equal to r there exists a∈Rna \in \mathbf{R}^n such that a+Ka + K contains each of the AikA_{i_k}. Show that then there exists b∈Rnb \in \mathbf{R}^n such that b+Kb + K contains all the AiA_i. Show that similar results hold if « contains » is replaced by « is contained in » or by « meets in a non-empty set ». (For each index i, consider the set CiC_i of the x∈Rnx \in \mathbf{R}^n for which x+K⊃Aix + K \supset A_i (or x+K⊂Aix + K \subset A_i, or (x+K)∩Ai≠∅(x + K) \cap A_i \neq \varnothing)). Generalize to any family of compact convex sets of Rn\mathbf{R}^n.

EXERCISES

Solution

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Transcribed from the printed volume. Not translated.