Exercise 11, § 2

It is said that a topological space $X$ is solvable if there exist in $X$ two complementary subsets which are both everywhere dense.

a) If a topological space $X$ is solvable, $X$ has no isolated point and every open subspace of $X$ is solvable. Conversely, if $\mathfrak{B}$ is a set of nonempty open subsets of $X$ such that every open subspace $U \in \mathfrak{B}$ is solvable and every nonempty open subset of $X$ contains an element of $\mathfrak{B}$, then $X$ is solvable (consider a $maximale$ family of open subsets belonging to $\mathfrak{B}$, pairwise disjoint).

b) If a Kolmogoroff space $X$ is solvable, every nonempty open subset of $X$ is infinite (use exerc 2 d) of I, p. 89).

c) Let $X$ be a topological space such that: $1^\circ$ the smallest cardinal $\mathfrak{d}$ of the nonempty open subsets of $X$ is infinite; $2^\circ$ there exists a basis $\mathfrak{B}$ of the topology of $X$ such that $\mathrm{Card}(\mathfrak{B}) \leqslant \mathfrak{d}$. Now prove that $X$ is solvable. (Construct by transfinite induction two everywhere dense subsets of $X$, with no common point, using the method of E, III, p. 91, exerc. 24 a).) In particular the rational right is solvable.

d) A topological space $X$ without isolated point is said to be isodyne if for every nonempty open subset $U$ of $X$, one has $\mathrm{Card}(U) = \mathrm{Card}(X)$. In a space $X$ without isolated point, every nonempty open set in $X$ contains a nonempty isodyne open subspace. Deduce that if every isodyne open subspace of $X$ is solvable, $X$ is solvable.