Hello. 
Yesterday we proofed, that $\mathbb{R}$ is not quasi-isometric to $\mathbb{R}^2$ (both endowed with the standard Euclidean metric).


Step 1.: $\mathbb{R}$ is q.i. $\mathbb{Z}$ and $\mathbb{R}^2$ is q.i. to $\mathbb{Z}^2$, so we only need to show $\mathbb{Z}$ is q.i. to $\mathbb{Z}^2$.


This is clear.


Step 2.: Indeed suppose that $f:\mathbb{Z}\mapsto \mathbb{Z}^2$ is a $(\lambda,C)$-quasi-isometry for some $\lambda\ge1$ and $C\ge0$. As $f$ is $(\lambda,C)$-quasi-isometric embedding it follwows that $\frac{1}{\lambda}d\_X(x,y)-C\le d\_Y(f(x),f(y))$ for all $x,y\in X$. This implies that for any $x,y\in X$ we have $d\_X(x,y)\le\lambda(d\_Y(f(x),f(y))+C)$.
Chosing $x=0$ the implies that


$f(X)\cap N\_r(f(0))\subset f(N\_{\lambda(r+C)}(0))$


As $f$ is further C-quasi-surjective it follows that:


$N\_{r-C}(f(0))\subset N\_C(f(X)\cap N\_r(f(0))\subset N\_C(f(N\_{\lambda(r+C)}(0))$


This is also clear.


Step 3.: Now $\mid N\_{r-C}(f(0))\mid$ grows quadratically in r while


$\mid N\_C(f(N\_{\lambda(r+C)}(0))\mid\le\mid N\_C(f(0))\mid\cdot\mid f(N\_{\lambda(r+C)}(0))\mid\le\mid N\_C(f(0))\mid\cdot\mid N\_{\lambda(r+C}(0)\mid$ 


grows ato most linearly in $r$. Thus for large$r$ we have 


$\mid N\_{r-C}(f(0))\mid >\mid N\_C(f(N\_{\lambda(r+C)}(0))\mid$


Now my questions are:


1. What is $\mid N\_{r-C}(f(0))\mid$? Ist it the area of an circle of radius $r-C$ and mddle point $f(0)$ or what does this absolute value brackets mean in this thense?
2. If it is. Is $\mid N\_C(f(N\_{\lambda(r+C)}(0))\mid$ also a cirlce? I don't think so, because $f(N\_{\lambda(r+C)}(0))$ does not have to be connected. Right? But anyway we are looking at the area of this thing under these "absolut value brackets", right?
3. And why does the second inequalitiy of Step 3. holds: I mean why does $\mid f(N\_{\lambda(r+C)}(0))\mid\le\mid N\_{\lambda(r+C)}(0)\mid$???


Thaks for help!



Question about the proof of the fact that IR is not quasi-isomtric to IR^2
1. It is the circle, or the elements of $\mathbb{Z}^2$ in that circle.
2. It is the union of the circles around the points of $f(N\_{\lambda(r+C)}(0))$.
3. This is because $f$ is a map and therefore there are less or equal points $f(x)$ than $x$.


