The metrics $d_1, d_2$ induce the same topology on $X$ Let $T:\mathbb{R}\to S^{1}\setminus\{(0, 1)\}$ be the map given by 
\begin{equation}
T(t)=\left(\frac{2t}{1+t^2}, \frac{t^2-1}{1+t^2}\right)
\end{equation}
and let $X=T(\mathbb{Z})$. Let $d_1$ be the discrete metric on $X$, and $d_2$ be the metric on $X$ induced by Riemannian metric on $S^1$.
 I recently received a review report on my paper and Referee gave the above example and claimed that
$d_1$ and $d_2$ induce the same topology on $X$.
I don't understand it.
Please help me 
Thanks
 A: Yes, they induce the same topology, which is the topology for which every set is an open set. Note that the topology induced by the Riemannian metric on $S^1\setminus\{(0,0)\}$ is just the usual topology. For this topology, $T(\mathbb{Z})$ is indeed a discrete set.
A: In the discrete topology (the one induced by the discrete metric), every singleton $\{x\}$ is open (which implies that every set is open).  Now you have to see that every singleton is open as well for the topology induced by $d_2$. You can't get finer than the discrete topology, so you do not have to check anything the other way round.
For $z \in \mathbb Z$, you can see that by picking a $\varepsilon$ small enough, you can ensure that $B^{S^1}_\varepsilon\left(T(z)\right)$ contains only $\left(\frac{2z}{1+z^2}, \frac{z^2-1}{1+z^2}\right)$ as the image of an integer by $T$ on $S^1$ (note how $\frac{z^2-1}{1+z^2}$ is monotonic with $|z|$).  Of course, the open ball on $S^1$ is open in $S^1$.  Hence the singleton $\left\{T(z)\right\} = T(\mathbb Z) \cap B^{S^1}_\varepsilon(T(z))$ is also open.
