# Question about the proof of the fact that IR is not quasi-isomtric to IR^2 [closed]

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!

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## closed as off topic by Loop Space, HJRW, Mark Sapir, Igor Rivin, Bill JohnsonOct 27 '11 at 16:03

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By the way, this would be better suited for math.stackexchange.com. – Julian Kuelshammer Oct 27 '11 at 9:59
To prove that $\mathbb{Z}$ is not quasi-isometric to $\mathbb{Z}^2$, you could thinks about the number of ends of your Cayley graph. Ends are a quasi-isometric invariant, and quite clearly $\mathbb{Z}$ is 2-ended while $\mathbb{Z}^2$ is 1-ended. Of course, this doesn't answer your specific questions...but it does tell you why they are not quasi-isometric... – ADL Oct 27 '11 at 10:05
Who are 'we'? It is often said that MO is not for homework. I submit that MO is also not a substitute for asking your professor to clarify some points in an argument. Voting to close – HJRW Oct 27 '11 at 11:49
Alan - to be fair, growth is a better invariant in this case, as it also distinguishes distinct $\mathbb{Z}^m$'s. – HJRW Oct 27 '11 at 11:56
I agree with @HW and vote to close. – Igor Rivin Oct 27 '11 at 13:41

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 $x$.
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Thanks. At first I thought the brackets had something to do with the distance between two points in this circle. But it's only the cardinality of the points in this circle, which lies in $\mathbb{Z}^2$. But now it is clear. – Eric Oct 27 '11 at 15:25