Question about the proof of the fact that IR is not quasi-isomtric to IR^2 - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-23T00:15:30Z http://mathoverflow.net/feeds/question/79250 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79250/question-about-the-proof-of-the-fact-that-ir-is-not-quasi-isomtric-to-ir2 Question about the proof of the fact that IR is not quasi-isomtric to IR^2 Eric 2011-10-27T09:44:45Z 2011-10-27T09:56:39Z <p>Hello. Yesterday we proofed, that $\mathbb{R}$ is not quasi-isometric to $\mathbb{R}^2$ (both endowed with the standard Euclidean metric).</p> <p>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 $\mathbb{Z}^2$.</p> <p>This is clear.</p> <p>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</p> <p>$f(X)\cap N_r(f(0))\subset f(N_{\lambda(r+C)}(0))$</p> <p>As $f$ is further C-quasi-surjective it follows that:</p> <p>$N_{r-C}(f(0))\subset N_C(f(X)\cap N_r(f(0))\subset N_C(f(N_{\lambda(r+C)}(0))$</p> <p>This is also clear.</p> <p>Step 3.: Now $\mid N_{r-C}(f(0))\mid$ grows quadratically in r while</p> <p>$\mid N_C(f(N_{\lambda(r+C)}(0))\mid\le\mid N_C(f(0))\mid\cdot\mid f(N_{\lambda(r+C)}(0))\mid\le\mid N_C(f(0))\mid\cdot\mid N_{\lambda(r+C}(0)\mid$ </p> <p>grows ato most linearly in $r$. Thus for large$r$ we have </p> <p>$\mid N_{r-C}(f(0))\mid >\mid N_C(f(N_{\lambda(r+C)}(0))\mid$</p> <p>Now my questions are:</p> <ol> <li><p>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?</p></li> <li><p>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?</p></li> <li><p>And why does the second inequalitiy of Step 3. holds: I mean why does $\mid f(N_{\lambda(r+C)}(0))\mid\le\mid N_{\lambda(r+C)}(0)\mid$???</p></li> </ol> <p>Thaks for help!</p> http://mathoverflow.net/questions/79250/question-about-the-proof-of-the-fact-that-ir-is-not-quasi-isomtric-to-ir2/79251#79251 Answer by Julian Kuelshammer for Question about the proof of the fact that IR is not quasi-isomtric to IR^2 Julian Kuelshammer 2011-10-27T09:55:24Z 2011-10-27T09:55:24Z <ol> <li>It is the circle, or the elements of $\mathbb{Z}^2$ in that circle.</li> <li>It is the union of the circles around the points of $f(N_{\lambda(r+C)}(0))$.</li> <li>This is because $f$ is a map and therefore there are less or equal points $f(x)$ than $x$.</li> </ol>