Lemma 5.3.4. Let $(X, \fU)$ be a uniform space and $\bracsn{U_n}_{0}^{\infty} \subset \fU$ such that:
$U_{0}= X \times X$.
For each $n \in \natz$, $U_{n}$ is symmetric.
For each $n \in \natz$, $U_{n + 1}\circ U_{n+1}\subset U_{n}$.
then there exists a pseudometric $d: X \times X \to [0, 1]$ such that
for each $n \in \natp$.
Proof. Let
and $d: X \times X \to [0, 1]$ by
then
For any $x \in X$, $x \in \bigcap_{n \in \natp}U_{n}$. Thus $\rho(x, x) = 0$ and $d(x, x) \le \rho(x, x) = 0$.
Let $x, y \in X$. By assumption (b), $\rho(x, y) = \rho(y, x)$. Thus $d(x, y) = d(y, x)$ as well.
Let $x, y, z \in X$, then for any $\seqf{x_j}$ and $\seqf[m]{y_j}$ with $x_{0} = x$, $x_{n} = y = y_{0}$, and $y_{m} = z$,
\[d(x, z) \le \sum_{j = 1}^{n} \rho(x_{j - 1}, x_{j}) + \sum_{j = 1}^{m} \rho(y_{j - 1}, y_{j})\]As this holds for all such $\seqf{x_j}$ and $\seqf[m]{y_j}$, $d(x, z) \le d(x, y) + d(y, z)$.
so $d$ is a pseudometric.
For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1}\subset E(d, 2^{-n})$.
Let $x, y \in X$ with $d(x, y) < 2^{-n}$. If $\rho(x, y) < 2^{-n}$, then the claim holds directly. Assume inductively that for any $x, y \in X$ with $d(x, y) < 2^{-n}$, if there exists $\seqf[m]{x_j}\subset X$ such that $x_{0} = x$, $x_{m} = y$ and $\sum_{j = 1}^{m} \rho(x_{j - 1}, x_{j}) < 2^{-n}$, then $(x, y) \in U_{n - 1}$.
Let $x, y \in X$ and $\seqf[m+1]{x_j}\subset X$ such that $x_{0} = x$, $x_{m+1}= y$, and $\sum_{j = 1}^{m+1}\rho(x_{j - 1}, x_{j}) < 2^{-n}$. Let $1 \le k < m+1$ such that $\sum_{j = 1}^{k}\rho(x_{j - 1}, x_{j}) < 2^{-n-1}$ and $\sum_{j = 1}^{k+1}\rho(x_{j-1}, x_{j}) \ge 2^{-n-1}$, then $\sum_{j = k + 1}^{m+1}\rho(x_{j-1}, x_{j}) < 2^{-n-1}$ as well. By the inductive hypothesis, $(x, x_{k}), (x_{k+1}, y) \in U_{n+1}\subset U_{n}$. Given that $\rho(x_{k}, x_{k+1}) < 2^{-n}$, $(x_{k}, x_{k+1}) \in U_{n+1}$ too. Thus