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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

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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.

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  • $\begingroup$ For the second direction, I understand that Given $\varepsilon >0$ ve $n\in \mathbb{N}$ $\endgroup$
    – Mehmet Onat
    Commented Mar 1, 2023 at 20:47
  • $\begingroup$ Let's choose $\delta \leq \min \left\{ \frac{a_{n}}{2},\frac{a_{n}\varepsilon }{2}\right\} $. $\endgroup$
    – Mehmet Onat
    Commented Mar 1, 2023 at 20:47
  • $\begingroup$ We have $\frac{a_{n}d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }\leq d\left( x,y\right) $. Suppose that $d\left( x,y\right) <\delta $. Then $\frac{a_{n}d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }<\delta \leq \frac{a_{n}}{2}\Rightarrow \frac{d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }<\frac{1}{2}$. $\endgroup$
    – Mehmet Onat
    Commented Mar 1, 2023 at 20:48
  • $\begingroup$ $\Rightarrow d_{n}\left( x,y\right) <1\Rightarrow 1+d_{n}\left( x,y\right) <2 $. From this, we get $\frac{d_{n}\left( x,y\right) }{2}<\frac{d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }$. Hence $\Rightarrow \frac{ a_{n}d_{n}\left( x,y\right) }{2}<\frac{a_{n}d_{n}\left( x,y\right) }{ 1+d_{n}\left( x,y\right) }\leq d\left( x,y\right) $ $\endgroup$
    – Mehmet Onat
    Commented Mar 1, 2023 at 20:48
  • $\begingroup$ $\Rightarrow d_{n}\left( x,y\right) <\frac{2}{a_{n}}d\left( x,y\right) < \frac{2}{a_{n}}\delta \leq \frac{2}{a_{n}}\cdot \frac{a_{n}\varepsilon }{2} =\varepsilon\Rightarrow d_{n}\left( x,y\right) <\varepsilon $. $\endgroup$
    – Mehmet Onat
    Commented Mar 1, 2023 at 20:49

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