Exercise 24, § 4

One says that a family $(A_\alpha)$ of subsets of a topological space X is discrete if, for every $x \in X$, there exists a neighbourhood of x meeting at most one set $A_\alpha$. One says that a separated space X is collectively normal if, for every discrete family $(A_\alpha)$ of closed subsets of X, there exists a family $(U_\alpha)$ of open sets pairwise without common point such that $A_\alpha \subset U_\alpha$ for every $\alpha$. Every collectively normal space is normal. a) In order that every open covering of a completely regular space X be divisible (IX, p. 105, exerc. 20), it is necessary and sufficient that the set of neighbourhoods of the diagonal $\Delta$ of $X \times X$ be the filter of entourages of the universal uniform structure of X (IX, p. 86, exerc. 10). Now prove that if this is so, X is collectively normal (if $(A_\alpha)$ is a discrete family of closed subsets of X, consider the open covering $(V_\alpha)$, where

b) Let $Y = \{a, b\}$ be the locally compact space defined in exerc. 12 of I, p. 106, $Y_0$ the set Y endowed with the discrete topology, Z the set $Y_0 \cup \{b\}$, where the points of $Y_0$ are open sets, and the sets $]x, b]$, where x runs through $Y_0$, form a fundamental system of neighbourhoods of b. Now prove that the space $X = Y \times Z$ is collectively normal (observe that in Y no infinite family of closed subsets of Y can be discrete, and use IX, p. 102, exerc. 8). Let $\mathscr{R}$ be the open covering of X formed by $Y \times Y_0$ and the products $[a, x) \times ]x, b]$ (for $x < b$); now prove that $\mathscr{R}$ is not divisible, and consequently that the set of neighbourhoods of $\Delta$ in $X \times X$ is not the filter of entourages of a uniform structure on $X$ (use I, p. 106, exerc. 12).