Why the limit $\rho_2(i,-i)=0$?

Question

Why the limit $\rho_2(i,-i)=0$?

Answer 1

For any $y\in S^1$ \ $\{(0,1)\}$ there is a unique $T(y)\in (-3\pi/2, \pi/2)$ such that $y=(\cos T(y),\sin T(y)).$

And $T: S^1$ \ $\{(0,1)\}\to (-3\pi /2, \pi /2)$ is a homeomorphism. So the composite function $(Th):\Bbb R \to (-3\pi/2,\pi /2)$ is a homeomorphism.

Consider any homeomorphism $H: \Bbb R \to (a,b)$ to a bounded real interval $(a,b).$ Now $H$ must be monotonic because it is continuous and injective. We must have either

(I). $ \lim_{i\to \infty} H(i)=a$ and $\lim_{i\to -\infty} H(i)=b,$... or

(II). $ \lim_{i\to \infty}H(i)=b$ and $\lim_{i\to -\infty}H(i)=a.$

If neither (I) nor (II) held then the monotonicity of $H$ would imply that $H$ is not surjective onto $(a,b)$.

Apply this with $H=Tf$ and $(a,b)=(-3\pi /2,\pi /2).$

Answer 2

Note that since $h:\mathbb{R}\to S^1\backslash\{(0,1)\}$ is a homeomorphism then

$$\lim_{x\to\infty} h(x)=(0,1)=\lim_{x\to\infty} h(-x)$$

And since $\rho$ is continuous with respect to the Euclidean (natural) metric then the thesis easily follows by simple evaluation:

$$\lim\limits_{i\to\infty} \rho_2(-i, i)=\lim\rho\big(h(-i), h(i)\big)=$$ $$=\rho\bigg(\lim \big(h(-i), h(i)\big)\bigg)=$$ $$=\rho\big(\lim h(-i), \lim h(i)\big)=$$ $$=\rho\big((0,1), (0,1)\big)=0.$$