Bourbaki

Exercise 29, § 8

General Topology, I

Bourbaki marks this one as harder than the rest.

a) Let R be a separated equivalence relation in a topological space X. In the space F(X)\mathfrak{F}(X) of the non-empty closed subsets of X, endowed with the topology induced by TΘ\mathfrak{T}_\Theta (I, p. 101, exerc. 12), now prove that every A∈F(X)A \in \mathfrak{F}(X) adhering to X/RX/R is contained in an equivalence class according to R (reason by contradiction).

b) Suppose in addition that X is regular and the relation R is open. Now prove that X/RX/R is then closed in F(X)\mathfrak{F}(X) for the topology TΘ\mathfrak{T}_\Theta (use prop. 11 of I, p. 51).

¶(30) Let X be a regular space. Now prove that, in order that X satisfy the condition (D_v) of I, p. 90, exerc. 7, it is necessary and sufficient that every locally finite family of non-empty open sets in X be countable. (To prove that the condition is necessary, consider a locally finite set U\mathfrak{U} of open sets in X. Endow X with a well-ordering; the set P of the x∈Xx \in X which are smallest elements (for this order) of an element of U\mathfrak{U} is equipotent to U\mathfrak{U}; if for every x∈Px \in P, V(x)V(x) is one of the sets of U having x as smallest element, the set V\mathscr{V} of the V(x)V(x) for x∈Px \in P is locally finite and equipotent to U\mathfrak{U}, and for y∈Py \in P and y>xy > x, one has x∉V(y)x \notin V(y). Using the regularity of X and prop. 4 of I, p. 6, now prove that for every x∈Px \in P, there exists an open set W(x)W(x) such that x∈W(x)⊂V(x)x \in W(x) \subset V(x) and that, when x runs through P, the W(x)‾\overline{W(x)} are pairwise without common point).

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.