Exercise 28, § 4

a) Let $Y$ be a compact space, $X$ a subspace of $Y$, $U$ an open neighbourhood of the diagonal $\Delta$ in $X \times X$, $V$ an open subset of $X \times Y$ such that $V \cap (X \times X) = U$. Now prove that if, in $X \times Y$, the closed sets $\Delta$ and $C V$ are normally separated (IX, p. 87, exer. 15) then $U$ is an entourage of the universal structure of $X$ (use IX, p. 84, exer 4).

b) Let $X$ be a completely regular space. Now prove that the following conditions are equivalent:

α) $X$ is paracompact.

β) The product of $X$ and every compact space is normal.

γ) If a filter $\mathscr{F}$ on $X$ is such that, for every continuous mapping $f$ of $X$ into a metrizable space $M$, $f(\mathscr{F})$ has an adherent point in $M$, then $\mathscr{F}$ has an adherent point in $X$.

(Using $a$), now prove first that condition $\beta$ implies that the neighbourhoods of the diagonal $\Delta$ in $X \times X$ form a fundamental system of entourages of the universal uniform structure $\mathscr{U}_omega$. To prove that $\beta$ implies $\gamma$, reason by contradiction: consider a compact space $Z$ containing $X$ (for example $\beta X$) and the set $A$ of points adherent to $\mathscr{F}$ in $Z$, and note that there exists a neighbourhood $U$ of $\Delta$ in $Z \times X$ such that $\overline{U}$ does not meet $A \times X$. Let $d$ be a continuous distance on $X$ such that the set of $(x, y) \in X \times X$ such that $d(x, y) < 1$ is contained in $U$; let $X_d$ be the metric space associated with $d$ (IX, p. 11), $f : X \to X_d$ the canonical mapping. By applying the assumption $\gamma$ to $f(\mathscr{F})$, now prove that there would exist on $X$ a filter $\mathfrak{R}$ finer than $\mathscr{F}$, having an adherent point in $Z$ not contained in $A$. To see that $\gamma$ implies $\alpha$), reason by contradiction, supposing that there is a covering $\mathscr{R}$ of $X$ without a finer locally finite open covering; the set of complements of the finite unions of sets of $\mathscr{R}$ is then a filter basis $\mathscr{B}$ without adherent point; there exists consequently a continuous mapping $f$ of $X$ into a metrizable space $M$ such that $f(B)$ has no adherent point. The exteriors of the sets of $f(\mathscr{B})$ then form an open covering of $M$; using the fact that $M$ is paracompact, form a locally finite open covering of $X$ finer than $\mathscr{R}$, contradicting the assumption.)