Exercise 21, § 9

One says that a topological space X is completely separated (and that its topology is completely separated) if, for every pair of distinct points $x, y$ of X, there exists a closed neighbourhood of $x$ and a closed neighbourhood of $y$ having no common point.

a) Every regular space is completely separated. Every subspace of a completely separated space is completely separated. Every product of completely separated spaces is completely separated. Every topology finer than a completely separated topology is a completely separated topology.

b) Every completely separated minimal space X (exerc. 20) is compact. (Reason by contradiction, by considering an ultrafilter $\mathfrak{U}$ on X and the filter basis formed by the open sets belonging to $\mathfrak{U}$.)

c) Let $\theta$ be an irrational number > 0. For every point $(x, y) \in \mathbf{Q} \times \mathbf{Q}^+$, and every $\alpha > 0$, let $B_\alpha(x, y)$ be the set formed by $(x, y)$ and the points $(z, 0)$ of $\mathbf{Q}^2$ such that $|z - (x + \theta y)| < \alpha$ or $|z - (x - \theta y)| < \alpha$. Let $\mathfrak{V}(x, y)$ be the set of the $B_\alpha(x, y)$ where $\alpha$ runs through the set of real numbers > 0. Prove that there exists a separated topology $\mathfrak{T}$ on $\mathbf{Q} \times \mathbf{Q}^+$ such that, for every $(x, y)$, $\mathfrak{V}(x, y)$ is a fundamental system of neighbourhoods of $(x, y)$ for $\mathfrak{T}$. Prove that whatever the points $a, b$ of $\mathbf{Q} \times \mathbf{Q}^+$ may be, every closed neighbourhood of $a$ meets every closed neighbourhood of $b$ for the topology $\mathfrak{T}$. Prove that there exist in the minimal separated space X thus defined closed subspaces which are not minimal.