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A subset A of a compact metric M is called a $\epsilon$ net if it satisfies the following conditions

(1)$\epsilon$ dense: the neighborhood of A is the entire M

(2)$\epsilon$ separate: $\forall x, y \in A$, $d(x,y)>\epsilon$

It is a well known fact that for any $\epsilon$, there is a fintie $\epsilon$ net.

And I wonder whether there is an uniform bound for cardinalities of all the $\epsilon$-nets of a given compact metric space(fixed $\epsilon$). May be exist a comapcat metric space,just constructing one, who have a series of $\epsilon$ net and the cardinality of these series of $\epsilon$ net are unbounded.

I think the question is negative and should involve the Hausdorff measure, dimenson and volume, but now I am confused. It will be so nice for some people to give me a answer.

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What do you mean by "the number of these series of ϵ net are unbounded"? –  Trevor Wilson Sep 10 '12 at 4:01
It means that for example , it have a series of finite $/epsilon$ nets ,$A_i$' and the number of the element in $A_i$ are 5 ,8 ,100' ...., 10000,..... (not bounded). –  Zhongmin Jin Sep 10 '12 at 4:41
In its current form the question is not very clearly stated. When you ask for a uniform bound -- the word is "uniform", not "unique" -- are you allowing $\epsilon$ to vary? Are you asking about the cardinality of an $\epsilon$-net, or the number of possible $\epsilon$-nets for a given $\epsilon$? –  Yemon Choi Sep 10 '12 at 7:34
Sorry for my poor explanation. The word "uniform" is best. The $\epsilon$ is fixed and given. And I also mean the cardinality of an $\epsilon$-net. –  Zhongmin Jin Sep 10 '12 at 7:54
I have rewote the statement. Thank you for Choi's recommendation. –  Zhongmin Jin Sep 10 '12 at 8:01

2 Answers 2

up vote 6 down vote accepted

Are you asking whether there is always an upper bound on the cardinality of an $\epsilon$-separated set of points in a compact metric space $X$? If so, the answer is yes.

Find a finite $\epsilon/2$-net $N$. Let $S$ be an $\epsilon$-separated set of points. Then every point of $S$ is in $B_{\epsilon/2}(x)$ for some $x\in N$, and no two points of $S$ lie in the same $B_{\epsilon/2}(x)$, so $|S|\leq |N|$.

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I think you are right. It is the right answer for me, thanks. Nice trick. I have too limited reputation to vote you answer. Sorry. –  Zhongmin Jin Sep 10 '12 at 8:41

I interprete the universality as follows: For given $\epsilon$, is there a natural number $N=N_\epsilon$ such that for all compact metric spaces $K$ of diameter $\le 1$ there exists an $\epsilon$-net of cardinality $\le N$. The bound on the diameter is necessary, for otherwise compact intervals in $\mathbb R$ would give counterexamples.

The answer is no: Let $I=[0,1]$ be the unit interval. On the set $I^n$ instal the metric attached to the norm $$ ||a||=\max_j|a_j|. $$ Then $I^n$ has diameter $1$. For $\epsilon=1/4$ an $\epsilon$-ball in $I^n$ has at most euclidean volume $1/2^n$, therefore you need at least $2^n$ such balls to cover $I^n$ which has euclidean volume 1.

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You statement is right, however, I think there are some different between you statement and my question. In my question, the metric space and the $\epsilon$ is fixed. The only variable is the $\epsilon$-net. Sorry for my misleading question. I think i should rewote it again. –  Zhongmin Jin Sep 10 '12 at 8:28

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