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:

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?

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?

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!

whythey are not quasi-isometric... – user6503 Oct 27 '11 at 10:05