Bourbaki

Exercise 16, § 4

General Topology, IX, p. 104

Bourbaki marks this one as harder than the rest.

Let (Xi)i∈I(X_i)_{i \in I} be a non-denumerable family of separated topological spaces, each having at least two distinct points. For every i∈Ii \in I, let ai,bia_i, b_i be two distinct points of XiX_i. Let XX be the product space ∏i∈IXi\prod_{i \in I} X_i, and let YY be the subspace of XX formed by the points (xi)i∈I(x_i)_{i \in I} such that xi=aix_i = a_i except for a denumerable set of indices ii; let bb be the point (bi)(b_i) of XX. In the product space X×YX \times Y, consider the sets A=Y×YA = Y \times Y and B={b}×YB = \{b\} \times Y. Now prove that AA and BB are closed and that there is no pair of open sets U,VU, V in X×YX \times Y such that A⊂U,B⊂VA \subset U, B \subset V and U∩V=∅U \cap V = \varnothing. (Let U,VU, V be two open sets such that A⊂UA \subset U and B⊂VB \subset V; show that there exists an increasing sequence (Hn)(H_n) of denumerable subsets of II, such that, if one denotes by xnx_n the point of YY whose every coordinate of index i∈Hni \in H_n is equal to bib_i, every coordinate of index i∉Hni \notin H_n equal to aia_i, the point (xn+1,xn)(x_{n+1}, x_n) belongs to VV for every integer nn; proving that the sequence (xn+1,xn)(x_{n+1}, x_n) tends to a point of AA, and conclude that U∩V≠∅U \cap V \neq \varnothing). Deduce from this result an example of a connected and non-normal topological ring. Also deduce that the product of a non-denumerable family of separated topological spaces each of which has at least two distinct points cannot be completely normal (use the result above when all the XiX_i are identical to a discrete space formed of two points.)

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.