Let $(X,d)$ be a metric space, then $d' = \frac{d}{1+d}$ and $d$ are equivalent; furthermore let $B_\theta(x, \epsilon) = \{y \in X\ |\ \theta(x,y) < \epsilon\}$ for some metric $\theta$ over $X$. I wanna ask you guys if my proof is correct.
Edit 1:
The question posted here ask if, given a metric $d$, $d' = \frac{d}{1+d}$ is also a metric; my question is different as by hypothesis I already know that $d$ and $d'$ are metrics, so given that I'm asking if my below proof, showing that $d$ and $d'$ are equivalent metrics, is correct.
Proof:
Since $d' = \frac{d}{1+d} \le d, \ \forall x,y \in X$, then $d < \epsilon \implies d' < \epsilon$. So $y \in B_d(x, \epsilon) \implies y \in B_{d'}(x, \epsilon)$.
Conversely, let $\delta = \frac{\epsilon}{1+\epsilon}$. Then $d' < \delta \implies d = \frac{d'}{1-d'} < \frac{\delta}{1-\delta} = \epsilon$. So $y \in B_{d'}(x,\delta) \implies y \in B_d(x,\epsilon)$.
Edit 2:
As pointed out by Berci, this is the conclusion of the proof.
Finally, since the set of all open balls in a metric space form a basis for the space and $\forall x \in X, \forall \epsilon > 0$ we have that $B_d(x,\epsilon) \subseteq B_{d'}(x, \epsilon)$ and $B_{d'}(x, \delta) \subseteq B_{d}(x, \epsilon)$, $d$ and $d'$ induce the same topology on $X$.
