Exercise 5, § 1

The topology induced by $\mathscr{U}$ is finer than $\overline{\mathscr{T}}$ if and only if, for each $x_0 \in X$ and each neighbourhood $V$ of $x_0$ in the topology $\overline{\mathscr{T}}$ there is an index $i$ and a real number $s > 0$ such that $f_{i, x_0, x'} > s$ for all $x \in \mathbf{C}V$.

4) Let $X$ be a non-Hausdorff uniformizable space and let $R$ be the relation "$y \in \{ x \}$" between two points $x, y$ of $X$.

a) Show that $R$ is an equivalence relation on $X$, that every continuous mapping of $X$ into a Hausdorff space is compatible with the relation $R$ (Set Theory, R, § 5, no. 7), and that the quotient space $X/R$ is completely regular.

b) If $\mathscr{U}$ is any uniformity compatible with the topology of $X$, then the Hausdorff uniformity associated with $\mathscr{U}$ (Chapter II, § 3, no. 9) is defined on the quotient space $X/R$ and is compatible with the topology of $X/R$. The topological space $X/R$ is called the completely regular space associated with the uniformizable space $X$.

5) Let $X$ be a uniformizable space. Show that the family of all pseudometrics on $X$ which are continuous on $X \times X$ defines a uniformity on $X$ compatible with the topology of $X$. This uniformity $\mathscr{U}_0$ is called the universal uniformity on $X$; it is the finest of all uniformities compatible with the topology of $X$. If $Y$ is any uniform space and if $f : X \to Y$ is a continuous mapping, then $f$ is uniformly continuous with respect to the universal uniformity on $X$; this is not the case for any other uniformity on $X$ compatible with its topology. If there exists a uniformity $\mathscr{U}$ on $X$, compatible with its topology and such that $X$, endowed with this uniformity, is a complete space, then $X$ is also a complete space with respect to any uniformity $\mathscr{U}'$ which is finer than $\mathscr{U}$ and coarser than $\mathscr{U}_0$. (Observe that the Cauchy filters are the same for $\mathscr{U}'$ as for $\mathscr{U}_0$.)