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Let be $M$ a compact metric space, and let $\{x_n\}$ be a dense subsequence in $M$.

We say that a set $\Lambda=\{y_1,\ldots,y_n\}$ is $\epsilon$-dense when every ball of radius $\epsilon$ contains a point of $\Lambda$.

I want to prove that for every $\epsilon$ there exists $N\in\mathbb{N}$ such that $\{x_1,\ldots,x_N\}$ is $\epsilon$- dense.

I'm trying to do this by contradiction. I'm trying to argue that it does not exist then $\{x_n\}$ is not dense. But I'm having trouble with it.

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  • $\begingroup$ What are you trying to prove? That for every $\epsilon$ there exists $N$ such that $\{x_1,\ldots,x_N\}$ is $\epsilon$ dense? $\endgroup$ – Arturo Magidin Apr 29 '12 at 22:59
  • $\begingroup$ for each $\epsilon$ fixed there exists $N$ such that $\{x_1,\ldots,x_N\}$ is $\epsilon$ dense $\endgroup$ – user27456 Apr 29 '12 at 23:01
  • $\begingroup$ So then that's not something "you can see", it's what you are trying to prove. $\endgroup$ – Arturo Magidin Apr 29 '12 at 23:03
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You can start with a finite cover $\cal B$ of $M$ of balls of radius $\epsilon/2$. Compactness of $M$ ensures that you can do this. Note that any open ball $O$ of radius $\epsilon$ would then contain a member of $\cal B$; in particular, $O$ would contain the member of $\cal B$ containing the center of $O$. So all you need to do is select an $x_i$ in each member of $\cal B$. The denseness of $\{x_i\}$ ensures that you can do this. Finally, take $N$ to be the maximum index selected and set $\Lambda=\{x_1,\ldots, x_N\}$.

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I think you can manage it directly:

Let $\epsilon>0$. Show $\bigcup_{i\in \mathbb{N}}B(x_i,\epsilon)=X$ by density.

Extract a finite subcover, then find an $N\in\mathbb{N}$ greater than all of the indices of the centers of the finite subcover. Clearly $\bigcup_{i=1}^N B(x_i,\epsilon)=X$.

For any $x\in X$, $x\in B(x_i,\epsilon)$ for some $i\in 1\dots N$, and so $x_i\in B(x,\epsilon)$.

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    $\begingroup$ Close, but not quite. You need smaller balls. $\endgroup$ – Robert Israel Apr 29 '12 at 23:04
  • $\begingroup$ How do you ensure that if we take a ball of radius $\epsilon$ outside of the coverage that you extracted this ball will contain some of the $\{x_1,\ldots,x_N\}$? $\endgroup$ – user27456 Apr 29 '12 at 23:05
  • $\begingroup$ Oh, thanks Robert :) Uh oh... is this a Lebesgue number situation? $\endgroup$ – rschwieb Apr 29 '12 at 23:07
  • $\begingroup$ @RobertIsrael I think it's OK now, I don't think my radii are too small. $\endgroup$ – rschwieb Apr 30 '12 at 0:36
  • $\begingroup$ I think this is not correct. If you take a ball $B(y,\epsilon)$ How do you ensure that $B(y,\epsilon)$ contains some $x_i$? I think you have to make balls with a smaller radius. $\endgroup$ – user27456 Apr 30 '12 at 1:35

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