Topology, Showing that two metric spaces are topologically equivalent Can someone verify if this is true?
$X=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
 . $d$
  is the standard metric on $\mathbb{R}$
  and $d'(x,y)=d\left(\tan(x),\tan(y)\right)$
 . We want to show that $\left(X,d\right)$
  and $\left(X,d'\right)$
  are topologically equivalent. I will just show the case if U
  is an open set in $\left(X,d\right)$
  then U
  is also open in $\left(X,d'\right)$
 . Let $x_{0}\in U$
 . There exists $0<r\in\mathbb{R}$
  such that $B_{r}(x_{0})=\left\{ y\,|\,d(x_{0},y)<r\right\} \subseteq U$
 . Since $\tan(x)$
  is a monotonic function and continuous, there exists $0<r'\in\mathbb{R}$
  such that the the greatest $\delta>0$
  that satisfies $d(x_{0},y)<\delta\Rightarrow d(\tan(x_{0}),\tan(y))<r'$
  , is smaller than $r$
 . Since the function is monotonic $r'$
  exists. And therefore $B_{r'}^{'}(x_{0})=\left\{ y\,|\,d^{'}(x_{0},y)=d(\tan(x_{0}),\tan(y))<r'\right\} \subseteq B_{r}(x_{0})\subseteq U$
 . Therefore $U$
  is open in $\left(X,d'\right)$
 .
The part that I am not sure of, justifying the existence of r'
 . Is there a way of choosing r'
  more specifically ? Thanks!

I will just comment on this paragraph here:

Since $\tan(x)$ is a monotonic function and continuous, there exists
  $0<r′\in\mathbb{R}$ such that the the greatest $\delta>0$ that
  satisfies $d(x_0,y)<\delta\Rightarrow d(\tan(x_0),\tan(y))<r'$ , is
  smaller than $r$ . Since the function is monotonic $r'$ exists.

I don't know what you need $\tan(x)$ to be monotonic for, and I also doubt that monotonicity makes the argument about "the greatest $\delta>0$..." work. Let me know, if I just fail to understand your idea.
The way I think about it is:
I want to prove existence of an $r'$, such that
$$B'_{r'}(x_0)\subseteq B_r(x_0).$$
This means that whenever $d(\tan(x_0),\tan(y))<r'$, I must have $d(x_0,y)<r$. But this property is continuity of $\tan^{-1}(x)$, the inverse of tangent. Letting $v_0=\tan(x_0)$, and $w=\tan(y)$, I know by continuity of $\tan^{-1}(x)$ that there exists a $\delta>0$, such that
$$d(v_0,w)<\delta\Rightarrow d(\tan^{-1}(v_0),\tan^{-1}(w))<r,$$
which is the same as
$$d(\tan(x_0),\tan(y))<\delta\Rightarrow d(x_0,y)<r.$$
In the above, I can thus choose $r'=\delta$.
