Exercise 25, § 8

a) On a set $X$, the set of regular topologies and the set of regular topologies with no isolated points are inductive sets with respect to the relation "“$\mathfrak{T}$ is coarser 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 have no isolated points; hence given any regular topology on $X$ with no isolated points, there is a finer ultraregular topology. A space is said to be ultraregular if its topology is ultraregular. Show that a regular topology $\mathfrak{T}$ with no isolated points is ultraregular if and only if, whenever $A$ and $B$ are complementary subsets of $X$ which have no isolated points, then $A$ and $B$ are open sets. In particular, an ultraregular space is not solvable ($§ 2$, Exercise 11). (To prove that the condition is sufficient, observe that if $\mathfrak{T}'$ is a regular topology with no isolated points and is finer than $\mathfrak{T}$, and if $U$ is a non-empty open set with respect to $\mathfrak{T}'$ and $\overline{U}$ is its closure in the topology $\mathfrak{T}'$, then $\overline{U}$ and $X - \overline{U}$ have no isolated points with respect to the topology $\mathfrak{T}'$).

c) In an ultraregular space, the closure of every set which has no isolated points is open, and the interior of a dense set is dense; the intersection of two dense sets is therefore dense. Deduce that if $\mathfrak{T}_0$ is an ultraregular topology, then amongst all the topologies $\mathfrak{T}$ such that $\mathfrak{T}^* = \mathfrak{T}_0$ (Exercise 20) there is one (necessarily submaximal) which is finer than all the others.