Suppose that for all $n$ natural numbers, $d\_{n}$ is a pseudometric on set $X
$. Define $d=\sum\_{n=1}^{\infty }a\_{n}\frac{d\_{n}}{1+d\_{n}}$, where $\left(
a\_{n}\right) $ is a sequence of positive numbers such that $\sum\_{n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}a\_{n}<\infty $.


It is claimed that the uniformity $\mathcal{U}\_{d}$ generated by
pseudometric $d$ is the same the uniformity $\mathcal{U}\_{P}$ generated by
pseudometrics $P=\left\{ d\_{n}:n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
\right\} $, that is $\mathcal{U}\_{d}=\mathcal{U}\_{P}$.


At page 237 in Topology Book by W. W. Fairchild, C. I. Tulcea,


<https://archive.org/details/topology0000fair/page/236/mode/2up>



A question about uniformities generated by pseudometrics
One direction:
let $\varepsilon>0$ and taken $N$ such that $\sum\_{n>N}a\_n<\frac12\varepsilon$. Take $\delta>0$ such that $\delta\cdot\sum\_{n\le N}a\_n<\frac12\varepsilon$. Now if $d\_n(x,y)<\delta$ for all $n\le N$ then
$$
d(x,y)< \delta\cdot\sum\_{n\le N}a\_n+\sum\_{n>N}a\_n<\varepsilon
$$
(because $d\_n(x,y)/(1+d\_n(x,y))\le d\_n(x,y)$ always).


For the second direction: you have $a\_n\cdot d\_n(x,y)/(1+d\_n(x,y))\le d(x,y)$. Then from $d(x,y)\le\frac12a\_n$ you get $d\_n(x,y)/(1+d\_n(x,y))\le\frac12$ and then $d\_n(x,y)\le1$. And from this you get $\frac12d\_n(x,y)\le d\_n(x,y)/(1+d\_n(x,y))$. In summary: if $d(x,y)\le\frac12a\_n$ then $\frac{a\_n}2d\_n(x,y)\le d(x,y)$, so you have an explicit formula for $\delta$ in terms of $\varepsilon$ to show uniform continuity of the identity.


