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 of the non-empty closed subsets of X, endowed with the topology induced by (I, p. 101, exerc. 12), now prove that every adhering to 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 is then closed in for the topology (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 of open sets in X. Endow X with a well-ordering; the set P of the which are smallest elements (for this order) of an element of is equipotent to ; if for every , is one of the sets of U having x as smallest element, the set of the for is locally finite and equipotent to , and for and , one has . Using the regularity of X and prop. 4 of I, p. 6, now prove that for every , there exists an open set such that and that, when x runs through P, the are pairwise without common point).
Solution
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