[Math] $\epsilon$- dense subsets

general-topologymetric-spacesreal-analysis

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.

Best Answer

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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