Proposition 5.3.11. Let $X$ be a set and $\seq{d_n}$ be pseudometrics on $X$, then there exists a pseudometric $d: X \times X \to [0, \infty)$ equivalent to $\seq{d_n}$.
Proof. Using Lemma 5.3.10, assume without loss of generality that for each $n \in \natp$, $d_{n}$ takes values in $[0, 1]$. Let
\[d(x, y) = \sum_{n \in \natp}\frac{d_{n}(x, y)}{2^{n}}\]
then $d$ is a well-defined a pseudometric.
Let $r > 0$, then there exists $n \in \natp$ such that $2^{-n}< r$. Take $s = r - 2^{-n}$, then $\bigcap_{k = 1}^{n}E(d_{k}, s) \subset E(d, r)$. On the other hand, for any $n \in \natp$ and $r > 0$, $E(d, r/2^{n}) \subset \bigcap_{k = 1}^{n} E(d_{k}, r)$. Therefore $\seq{d_n}$ and $d$ are equivalent.$\square$