Exercise 20, § 9

A topological space X is said to be completely Hausdorff (and the topology of X is said to be completely Hausdorff) if each pair of distinct points of X has disjoint closed neighbourhoods.

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

b) Every minimal completely Hausdorff space X (Exercise 19) is compact. (By reductio ad absurdum: consider an ultrafilter $\mathfrak{U}$ on X and the filter base formed by the open sets which belong to $\mathfrak{U}$).

c) Let $\theta$ be an irrational number > 0. For every point $(x, y) \in \mathbf{Q}^2$ and every $\alpha > 0$ let $B_\alpha(x, y)$ be the set consisting of $(x, y)$ and the points $(z, 0) \in \mathbf{Q}^2$ such that $|z - (x + \theta y)| < \alpha$ or $|z - (x - \theta y)| < \alpha$. Let $\mathfrak{B}(x, y)$ be the set of $B_\alpha(x, y)$ where $\alpha$ runs through the set of real numbers > 0. Show that there is a Hausdorff topology $\mathfrak{T}$ on $\mathbf{Q}^2$ such that, for each $(x, y)$, $\mathfrak{B}(x, y)$ is a fundamental system of neighbourhoods of $(x, y)$ for $\mathfrak{T}$. Show that, given any two points $a, b$ of $\mathbf{Q}^2$, every closed neighbourhood of $a$ meets every closed neighbourhood of $b$ in the topology $\mathfrak{T}$