Exercise 25, § 8

a) On a set X, the set of regular topologies, and the set of regular topologies without isolated point, are inductive for the relation « $\mathfrak{T}$ is less fine than $\mathfrak{T}$’ ».

b) A topology $\mathfrak{T}$ on a set X is said to be ultraregular if it is maximal in the set of topologies on X which are regular and without isolated point; for every regular topology without isolated point on X, there therefore exists a finer ultraregular topology. A space whose topology is ultraregular is said to be ultraregular. For a regular topology without isolated point to be ultraregular, it is necessary and sufficient that if A and B are two complementary subsets of X without isolated point, A and B are open for $\mathfrak{T}$. In particular, an ultraregular space is not solvable (I, p. 92, exerc. 11). (To now prove that the condition is sufficient, remark that if $\mathfrak{T}$’ is a regular topology without isolated point, finer than $\mathfrak{T}$, U a nonempty open subset for $\mathfrak{T}$’, $\overline{U}$ its closure for $\mathfrak{T}$’, $\overline{U}$ and X — $\overline{U}$ have no isolated point for $\mathfrak{T}$).

c) In an ultraregular space, the closure of every set without isolated point is an open set, and the interior of a everywhere dense set is everywhere dense; the intersection of two everywhere dense sets is therefore everywhere dense. Conclude that if $\mathfrak{T}_0$ is an ultraregular topology, among all the topologies $\mathfrak{T}$ such that $\mathfrak{T}^* = \mathfrak{T}_0$ (I, p. 103, exerc. 20), there is one which is finer than all the others (necessarily submaximal).