Bourbaki

Exercise 27, § 4

General Topology, IX, p. 107

Bourbaki marks this one as harder than the rest.

a) Let X be a normal space, (Uλ)λ∈L(U_\lambda)_{\lambda \in L} a locally finite open covering of X. Now prove that there exists a countable open covering (Vn)n∈N(V_n)_{n \in \mathbf{N}} of X, where, for each n∈Nn \in \mathbf{N}, VnV_n is the union of a family (Wv)v∈Kn(W_v)_{v \in K_n} of pairwise disjoint open sets forming a covering of VnV_n finer than (Uλ)λ∈L(U_\lambda)_{\lambda \in L}. (Let (fλ)λ∈L(f_\lambda)_{\lambda \in L} be a continuous partition of unity subordinate to (Uλ)λ∈L(U_\lambda)_{\lambda \in L}. For every finite subset J of L, let WjW_j be the set of the x∈Xx \in X such that fλ(x)<fμ(x)f_\lambda(x) < f_\mu(x) for all the pairs (λ,μ)(\lambda, \mu) such that λ∈L−J\lambda \in L - J, μ∈J\mu \in J. Take for KnK_n the set of subsets of L having n elements.) b) Let X be a topological space, O\mathscr{O} the set of open sets of X. One says that a subset U\mathscr{U} of O\mathscr{O} is quasi-full if it satisfies the following conditions: α)\alpha) U\mathscr{U} is a covering of X; β)\beta) every open subset of a set belonging to U\mathscr{U} belongs to U\mathscr{U}; γ)\gamma) every finite union of sets of U\mathscr{U} belongs to U\mathscr{U}; δ)\delta) every union of a family of sets belonging to U\mathscr{U} and pairwise disjoint, belongs to U\mathscr{U}. Now prove that in order that the space X be paracompact, it is necessary and sufficient that every quasi-full subset U\mathscr{U} of O\mathscr{O} be equal to O\mathscr{O}. (To see that the condition is necessary, use a) to construct a countable locally finite open covering (Vn)n∈N(V_n)_{n \in \mathbf{N}} of X such that Vn∈UV_n \in \mathscr{U} for every n∈Nn \in \mathbf{N}. Then, with the aid of a continuous partition of unity (fn)(f_n) subordinate to

(V_n) and of the function f(x)=∑n>0nfn(x)f(x) = \sum_{n > 0} n f_n(x), construct an open covering (Wn)(W_n) of XX by open sets of U\mathscr{U} such that W‾n⊂Wn+1\overline{W}_n \subset W_{n+1} for every n≥0n \geq 0. To see that the condition is sufficient, given an open covering R\mathscr{R} of XX, consider the set U\mathscr{U} of the open sets of X such that for every set A∈UA \in \mathscr{U}, the intersections with AA of the sets of R\mathscr{R} constitute a covering of AA possessing a finer and locally finite open covering.)

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.