Is $\mathbb{R}$ complete under the $\frac{t}{2^k(1+t)}$ metric?

Question

I want to show $\mathbb{R}$ is complete under the metric $\rho_k(x,y) = \frac{|x-y|}{2^k(1+|x-y|)} $ where $k \in \mathbb{N}$ is fixed (presumably this is true). My idea is to start with a Cauchy sequence $\{x_n\}$ in $(\mathbb{R}, \rho)$ and show that some homeomorphism $\phi$ transforms $\{x_n\}$ into a sequence $\{y_n\}$ that's Cauchy in $(\mathbb{R}, |\cdot|)$, which is complete so $y_n \to y \in \mathbb{R} $ under $|\cdot|$. Then I use my homeomorphism to import $y$ back into $(\mathbb{R}, \rho)$, where I use the fact that $\phi^{-1}$ is continuous to conclude $x_n \to \phi(y) $ in $(\mathbb{R}, \rho)$.

I'm stuck at the first step: constructing this homeomorphism. I know that $\phi$ must be related to the metric $\rho$ but I am not seeing the connection. Also, it would be nice if someone can tell me if this is even true so I don't waste my time proving some false statement (although I am 99% sure this is true).

Answer 1

I don't think $(\mathbb{R}, \rho)$ is a homeomorphic to $(\mathbb{R}, |\cdot|)$. If this were true then the two metrics are equivalent, i.e. there exists $C_1, C_2>0$ such that \begin{align} C_1 |x-y| \leq \rho(x, y) \leq C_2|x-y|. \end{align} It's clear $\rho(x,y) \leq |x-y|$. However, there doesn't exists $C_1$ such that $C_1|x-y| \leq \rho(x, y)$ because $0 \leq \rho(x,y)\leq 1$ and $|x-y|$ could be arbitrarily big.

Answer 2

The $2^k$ factor cannot matter as that just changes the measuring scale. You just need to apply it to $\delta$ in any $\delta - \epsilon$ proof, so let us consider $\rho(x,y) = \frac{|x-y|}{(1+|x-y|)}$ Now for $|x-y|\lt 1$ we have $\frac 12|x-y| \lt \rho(x,y) \lt |x-y|$ For completeness we only need this for small distances.