Local finiteness and coarse bounded geometry - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:22:53Z http://mathoverflow.net/feeds/question/75009 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75009/local-finiteness-and-coarse-bounded-geometry Local finiteness and coarse bounded geometry Valerio Capraro 2011-09-09T15:16:18Z 2011-09-10T13:57:40Z <p>I've just started learning these things and so probably my questions will be very easy. Please forgive me.</p> <p>A metric space $(X,d)$ is called locally finite if every bounded set is finite. A metric space is said to have coarse bounded geometry if there is $\Gamma\subseteq X$ such that</p> <p>1) there exists $c>0$ such that the set of points $x\in X$ such that $d(x,\Gamma)\leq c$ is dense in $X$. </p> <p>2) For all $r>0$, there exists $K_r$ such that, for all $x\in X$, $|\Gamma\cap B_r(x)|\leq K_r$, where $B_r(x)$ stands for the ball of radius $r$ about $x$.</p> <blockquote> <p><strong>Question 1:</strong> what is an example of metric space without coarse bounded geometry? </p> </blockquote> <p>Well, infinite dimensional Banach spaces. But I would like something <em>more handable</em>.</p> <blockquote> <p><strong>Question 2:</strong> Is it true that locally finiteness implies coarse bounded geometry?</p> </blockquote> <p>Maybe I have misunderstood, but in a published paper I have found a sentence that looks implicitly assume that the answer is positive. It might be trivial, but I am not quite convinced. </p> <p>Thanks in advance,</p> <p>Valerio</p> http://mathoverflow.net/questions/75009/local-finiteness-and-coarse-bounded-geometry/75015#75015 Answer by Alessandro Sisto for Local finiteness and coarse bounded geometry Alessandro Sisto 2011-09-09T15:39:52Z 2011-09-10T13:57:40Z <p>The answer to question 2 is negative, but if you require quasi-homogeneity (i.e. you have a group of isometries with a <code>$c-$</code>dense orbit form some <code>$c$</code>) then it becomes affirmative. You typically have this.</p> <p>Also, to construct examples as in question 1 you can consider non-quasi-homogeneous spaces. Hope this helps, I can be more explicit on this point if you need clarifications.</p> http://mathoverflow.net/questions/75009/local-finiteness-and-coarse-bounded-geometry/75016#75016 Answer by Bill Johnson for Local finiteness and coarse bounded geometry Bill Johnson 2011-09-09T15:40:51Z 2011-09-09T15:40:51Z <p>Q2--no. Let $A_n$ have cardinality $n+1$ for $n=0,1,...$. Specify all distances between distinct points in the same $A_n$ to be one, and the distance between a point in $A_n$ to a point in $A_m$ to be $n+m$ when $n\not= m$. </p> <p>This gives a simple example for Q1 as welll.</p> http://mathoverflow.net/questions/75009/local-finiteness-and-coarse-bounded-geometry/75025#75025 Answer by R W for Local finiteness and coarse bounded geometry R W 2011-09-09T18:35:36Z 2011-09-09T18:35:36Z <p>One more comment (which also implies answers to your questions): bounded geometry implies that the space has a finite exponential growth rate (defined, say, with respect to covers by balls of a fixed radius).</p>