Proposition 1

Let $X$ be a set endowed with a uniform structure $U$, and for each $x \in X$ let $\mathfrak{B}(x)$ be the set of subsets $V(x)$ of $X (*)$, where $V$ runs through the set of entourages of $U$. Then there is a unique topology on $X$ such that, for each $x \in X$, $\mathfrak{B}(x)$ is the neighbourhood filter of $x$ in this topology.

We have to show that the $\mathfrak{B}(x)$ satisfy conditions (V_I), (V_{II}), (V_{III}) and (V_{IV}) of Chapter I, § 1, no. 2. That this is so for the first three of

() We recall (Set Theory, R, § 3, no. 7*) that if $V$ is any subset of $X \times X$ and $x$ is any element of $X$, then $V(x)$ denotes the set of all $y \in X$ such that $(x, y) \in V$.

these conditions follows immediately from the fact that the entourages of $\mathfrak{U}$ satisfy $(\mathbf{F}_I),\ (\mathbf{F}_{II})$ and $(\mathbf{U}_I)$. As to $(\mathbf{V}_{IV})$, let $V$ be an entourage of $\mathfrak{U}$, $W$ an entourage of $\mathfrak{U}$ such that $\dot{W} \subset V$; then if $(x, y) \in W$ and $(y, z) \in W$ we have $(x, z) \in V$, so that $W(y) \subset V(x)$ for all $y \in W(x)$, and therefore $V(x) \in \mathfrak{B}(y)$ for all $y \in W(x)$. This completes the proof.