Exercise 18, § 4

A family $(A_\alpha)$ of subsets of a topological space $X$ is said to be discrete if, for each $x \in X$, there is a neighbourhood of $x$ which meets at most one set $A_\alpha$ of the family. A Hausdorff space $X$ is said to be collectively normal if, for each discrete family $(A_\alpha)$ of closed subsets of $X$, there is a family $(U_\alpha)$ of mutually disjoint open subsets of $X$ such that $A_\alpha \subset U_\alpha$ for each index $\alpha$. Every collectively normal space is normal.

a) Every open covering of a completely regular space $X$ is divisible (Exercise 16) if and only if the set of neighbourhoods of the diagonal $\Delta$ in $X \times X$ is the filter of entourages of the universal uniformity of $X$ ($§ 1$, Exercise 5). Show that if this condition is satisfied, then $X$ is collectively normal [if $(A_\alpha)$ is a discrete family of closed subsets of $X$, consider the open covering $(V_\alpha)$, where $V_\alpha = X - \bigcup_{\beta \neq \alpha} A_\beta$ for each $\alpha$].

b) Let $Y = [a, b[$ be the locally compact space defined in Chapter I, $§ 9$, Exercise 12; let $Y_0$ be the set $Y$ endowed with the discrete topology, and let $Z$ be the set $Y_0 \cup \{ b \}$, where the points of $Y_0$ are open sets, and the sets $]x, b[$, where $x$ runs through $Y_0$, from a fundamental system of neighbourhoods of $b$. Show that the space $X = Y \times Z$ is collectively normal (recall that no infinite family of subsets of $Y$ can be discrete, and use Exercise 4). Let $\mathfrak{R}$ be the open covering of $X$ consisting of $Y \times Y_0$ and the products $[a, x] \times ]x, b]$ (where $x < b$); show that $\mathfrak{R}$ is not divisible and that consequently the set of neighbourhoods of the diagonal $\Delta$ in $X \times X$ is not the filter of entourages of a uniformity on $X$ [use Exercise 12 a) of Chapter I, $§ 9$].