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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

edited Sep 9 2011 at 17:12

asked Sep 9 2011 at 15:16

Valerio Capraro
3,585●11●25

Bill Johnson · Accepted Answer · 2011-09-09 15:40:51Z UTC

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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.

answered Sep 9 2011 at 15:40

Bill Johnson
14.3k●1●33●60

Thank you! I was just thinking to an example of that sort, while getting back home! – Valerio Capraro Sep 9 2011 at 16:12

Alessandro Sisto · Answer 2 · 2011-09-10 13:57:40Z UTC

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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.

edited Sep 10 2011 at 13:57

answered Sep 9 2011 at 15:39

Alessandro Sisto
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	thank you - also for the specification about the density. – Valerio Capraro Sep 9 2011 at 16:14
	@Alessandro re ps: What the OP wrote is equivalent even if your wording is the usual way of writing the condition. – Bill Johnson Sep 9 2011 at 20:45
	@Bill: I think it has been edited. In any case, I removed the "ps" part, thanks for pointing this out. – Alessandro Sisto Sep 10 2011 at 0:00
	I am sorry, what is quasi-homogeneity? – Valerio Capraro Sep 10 2011 at 2:19
	As I mentioned, it means that there are many isometries, meaning that there exists some $c$ such that for each $x,y$ there exists an isometry mapping $y$ to a point at distance at most $c$ from $x$ . Notice that being homogeneous is the same thing as being quasi-homogeneous with constant $c=0$ . – Alessandro Sisto Sep 10 2011 at 14:04

R W · Answer 3 · 2011-09-09 18:35:36Z UTC

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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).

answered Sep 9 2011 at 18:35

R W
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	I am sorry, what do you mean by "finite exponential growth"? – Valerio Capraro Sep 9 2011 at 18:50
	In the discrete case the exponential growth rate is defined as $\limsup \frac1n \log\|B_n(x)\|$, where $B_n(x)$ is the ball of radius $n$ centered at a point $x$. In the continuous case instead of cardinality one takes the minimal number of balls of a fixed radius necessary to cover $B_n$. – R W Sep 9 2011 at 19:03
	OK, at least in the discrete case it's the classical notion. I got a bit scared because of the "balls of fixed radius". – Valerio Capraro Sep 9 2011 at 19:43

Local finiteness and coarse bounded geometry

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