Showing that $\mathbb{R}\times [0,1]$ is quasi-isometric to $\mathbb{R}$

Question

I have a question that asks me to show that $\mathbb{R}\times [0,1]$ is quasi-isometric to $\mathbb{R}$

I have having trouble showing what I have is a quasi isometry. My map is simply:

$f:\mathbb{R}\times [0,1]\rightarrow \mathbb{R}$ such that $f(a,b)=a$

This is clearly quasi surjective but in order to show that it is a quasi-isometry i need to consider $d(f((x,y)),d(f(x',y'))$ but I am struggling to come up with bounds on this?

Here I define a quasi-isometry to be a map $f:X\rightarrow Z$ such that we have the following two proerties:

Property 1- Quasi-Embedding

$\exists \lambda, C$ with:

$\frac{1}{\lambda}d(x,y)-C\leq d(f(x),f(y))\leq \lambda(x,y)+C$ for all $x,y\in X$

Property 2- Quasi-surjective

There exists $K\in \mathbb{Z}$ such that for all $z\in Z$ exists $x\in X$ such that $d(f(x),z)\leq K$

Thanks for any help

Answer 1

The simplest way is probably to set $\lambda = 1, C = 2$. Then use the following:

1) $d\left((a,b),(a,0)\right)$ is at most $1$

2) $f$ restricted to $\mathbb{R} \times \{0\}$ is an isometry

3) The triangle inequality

This should get you the Quasi-Embedding property you need.