Exercise 20, § 8
General Topology, I, 138
Bourbaki marks this one as harder than the rest.
a) An open subset of a topological space is said to be regular if (§ 1, Exercise 3); equivalently, is regular if is the interior of a closed set. is said to be semi-regular (and its topology is said to be semi-regular) if the regular open subsets of form a base of . Show that if satisfies axiom then is semi-regular.
b) Prove that the intersection of two regular open sets is a regular open set. Deduce that the regular open sets with respect to form a base of a topology on , which is coarser than and semi-regular; is said to be the semi-regular topology associated with . Show that for each subset of which is open in , the closure of with respect to is the same as its closure in , and that if is accessible for the isolated points of with respect to are the same as the isolated points of with respect to . Show that is Hausdorff if and only if is [use Exercise 3 d) of § 1], and finally that if is a regular space then the continuous mappings of into are the same in both topologies and .
c) Let be a semi-regular space and let be its topology. Show that a topology on is such that if and only if there is a set of dense subsets of (in the topology ) such that every finite intersection of sets of belongs to , and such that the topology is generated by the union of and the set of subsets of which are open in the topology . [Consider the dense open subsets (in the topology ) and notice that every open set in is the intersection of a dense open set (in the topology ) and an open set (in the topology ).] Hence construct examples of non-regular Hausdorff topologies which are finer than a regular topology (and even finer than a topology of a compact metrizable space).
Solution
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