Local finiteness and coarse bounded geometry - MathOverflow

Local finiteness and coarse bounded geometry

2011-09-09T15:16:18Z

I've just started learning these things and so probably my questions will be very easy. Please forgive me.

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

1) there exists $c>0$ such that the set of points $x\in X$ such that $d(x,\Gamma)\leq c$ is dense in $X$.

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 $x$.

Question 1: what is an example of metric space without coarse bounded geometry?

Well, infinite dimensional Banach spaces. But I would like something more handable.

Question 2: Is it true that locally finiteness implies coarse bounded geometry?

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.

Thanks in advance,

Valerio

Answer by Alessandro Sisto for Local finiteness and coarse bounded geometry

2011-09-09T15:39:52Z

The answer to question 2 is negative, but if you require quasi-homogeneity (i.e. you have a group of isometries with a $c-$ dense orbit form some $c$ ) then it becomes affirmative. You typically have this.

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.

Answer by Bill Johnson for Local finiteness and coarse bounded geometry

2011-09-09T15:40:51Z

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 $n\not= m$.

This gives a simple example for Q1 as welll.

Answer by R W for Local finiteness and coarse bounded geometry

2011-09-09T18:35:36Z

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).