TITLE: If $d(x,y)$ is a metric on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology.
QUESTION [7 upvotes]: If $d(x,y)$  and $d'(x,y)$ are a metrics on $X$, then $d'(x,y)=\frac{d(x,y)}{1 + d(x,y)}$ and $d(x,y)$ generates the same topology.
My attempt:-
 Let $\mathscr T$ be the topology generated by $d$ and $\mathscr T'$ be the topology generated by $d'$. Let $U\in \mathscr T\implies U=\bigcup_{\alpha\in \Lambda}B_d(x_\alpha, r_\alpha).$ So, $x\in B_d(x_\alpha, r_\alpha)$. We want to prove there is a $B_{d'}(x,\delta)\subset B_d(x_\alpha, r_\alpha).$ If $r=\min\{.9,r_{\alpha}\}. So,B_{d}(x_\alpha,r)\subset B_d(x_\alpha, r_\alpha). $ I could prove that $B_{d'}(x_\alpha,\frac{r}{1-r})\subset B_{d}(x_\alpha, r)$. Let $y\in  B_d(x_\alpha, r_\alpha)$, let $\epsilon=\min\{r_\alpha-d(x_\alpha,y),.9\}.$ Then, there is an $B_{d'}(y,\frac{\epsilon}{1-\epsilon})\subset B_{d}(y, \epsilon)\subset B_{d}(x_\alpha, r_\alpha).$ Hence, I can write $U$ as the union of open balls of $d'$. Hence $U\in \mathscr T'$. If my proof is correct, I can write the converse similarly. By using $B_d(x,\frac{r}{r+1})\subset B_{d'}(x,r)$. Please help me to rectify my errors.

REPLY [2 votes]: i) $d(x_0,y)<\varepsilon$ so that $$ d'(x_0,y)\leq
d(x_0,y)<\varepsilon $$
Hence $B^d(x_0,\varepsilon)\subset B^{d'}(x_0,\varepsilon)$
ii) If $d'(x_0,y)< \varepsilon < 1/2$, then $$
d(x_0,y)< \varepsilon +\varepsilon d(x_0,y) \Rightarrow d(x_0,y)<
\frac{\varepsilon }{1-\varepsilon } < 2\varepsilon $$
$B^{d'}(x_0,\varepsilon)\subset B^{d}(x_0,2\varepsilon)$