Showing that $T(x)\geq{}x$ for a distance-preserving transformation $T$

Question

For a metric $d$ defined as $d(x,y)=|y-x|$ on $\mathbb{R}_{>0}$, and a distance-preserving transformation mapping $(\mathbb{R}_{>0},d)$ to itself, we are asked to show that $T(x)\geq{}x$ for all $x\in\mathbb{R}_{>0}$. And to show that equality only holds when $T$ is an isometry.

I've just started a course in Euclidean geometry and I've been stuck on this question for a couple hours, I know that for a distance-preserving transformation it holds that $d(x,y)=d(T(x),T(y))$ thus $|y-x|=|T(y)-T(x)|$ but I don't see how then the inequality arises, and why is it an equality when $T$ is bijective. Appreciate any help,

Edit: Due to an inconsistency remark, am making clear that $\mathbb{R}_{>0}$ here is the set of all positive integers, "distance-preserving" as $d(x,y)=d(T(x),T(y))$ for a transformation $T$ and an isometry as a transformation that is both bijective and distance-preserving.

Answer 1

Let $T\colon\mathbb{R}_{>0}\rightarrow\mathbb{R}_{>0}$ be such that $|T(x)-T(y)|=|x-y|$ for all $x,y\in\mathbb{R}_{>0}$. Work out the following facts:

1. $T$ is injective.
2. $T$ is continuous and, hence, monotonic.
3. $T$ is differentiable with $T^{\prime}=\pm1$.
4. There is a $b\in\mathbb{R}_{\ge0}$ such that $T(x)=x+b$ for all $x\in\mathbb{R}_{>0}$.

The claim then follows. These steps all follow from the definitions involved and properties of $T$. If you have trouble with any of the steps, feel free to ask.