Equivalent metrics on $X = (-\pi/2, \pi/2)$

Question

$\rho(x,y) = |\tan(x) - \tan(y)| $ and $d(x,y) = |x-y|$ are equivalent metrics on $X = (-\pi/2, \pi/2)$

I observed that $|x-y| \leq |\tan(x) - \tan(y)|$ but cannot get an inequality the other way round

Answer 1

Maybe you're struggling because the result is false, as stated.

There exists no constant $\kappa$ such that $|\tan x-\tan y|\le\kappa|x-y|$ for all $x,y$ in $(-\pi/2,\pi/2)$. The RHS of that inequality is bounded by $\pi\kappa$ yet the LHS can be made arbitrarily large: consider $x:=\frac{\pi}{2}-2^{-n}$ and $y:=2^{-n}-\frac{\pi}{2}$ for a very large integer $n$.

More concretely, set $y=0$ and choose $x:=\arctan(2\pi\kappa)$. Thus, these metrics are inequivalent.

In comments, you asked if the spaces have the same topology. They do. Any open subset of $(X,d)$ is open in $(X,\rho)$ from the inequality you state, so it suffices to show open subsets of $(X,\rho)$ are also open subsets of $(X,d)$. For that, it suffices to show that for any $x\in X$ and $\epsilon>0$ there is $\delta>0$ such that $|x-y|<\delta$ implies $|\tan x-\tan y|<\epsilon$ for $y\in X$… but we know that’s true because $\tan$ is continuous.