Proof. Let $\seqi{d}$ be the family of all uniformly continuous pseudometrics on $X$, and $\mathfrak{V}$ be the pseudometric uniformity induced by $\seqi{d}$. By (U) of Definition 5.3.3, $\fU \supset \mathfrak{V}$.

On the other hand, let $U_{1} \in \mathfrak{U}$. By (U3), there exists $\seq{U_n}\subset \mathfrak{U}$ such that $U_{n + 1}\circ U_{n + 1}\subset U_{n}$ for all $n \in \natp$. Let $U_{0} = X \times X$, then $\bracsn{U_n}_{0}^{\infty} \subset \fU$ satisfies the hypothesis of Lemma 5.3.4. Thus there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for all $n \in \natp$,

By Definition 5.3.3, $d$ is a uniformly continuous pseudometric on $X$. Since $E(d, 1/4) \subset U_{1}$, $U_{1} \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.$\square$