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Definition Suppose $X$ is a metric space with a metric $d$ , and suppose $S$ is a subset of $X$ . Let $\varepsilon$ be a positive real number. A subset $N\subset S$ is an $\varepsilon$ -net for $S$ if, for all $x\in S$ , there is an $y\in N$ , such that $d(x,y)<\varepsilon$ .
For any $\varepsilon>0$ and $S\subset X$ , the set $S$ is trivially an $\varepsilon$ -net for itself.
Theorem Suppose $X$ is a metric space with a metric $d$ , and suppose $S$ is a subset of $X$ . Let $\varepsilon$ be a positive real number. Then $N$ is an $\varepsilon$ -net for $S$ , if and only if $$\{ B_\varepsilon(y) \mid y\in N \}$$ is a cover for $S$ . (Here $B_\varepsilon(x)$ is the open ball with center
$x$ and radius $\varepsilon$ .)
Proof. Suppose $N$ is an $\varepsilon$ -net for $S$ . If $x\in S$ , there is an $y\in N$ such that $x\in B_\varepsilon(y)$ . Thus, $x$ is covered by some set in $\{ B_\varepsilon(x) \mid x\in N \}$ . Conversely, suppose $\{ B_\varepsilon(y) \mid y\in N \}$ is a cover for $S$ , and suppose $x\in S$ . By assumption, there is an $y\in N$ , such that $x\in B_\varepsilon(y)$ . Hence $d(x,y)<\varepsilon$ with $y\in N$ . 
Example In $X=\mathbb{R}^2$ with the usual Cartesian metric, the set $$ N = \{(a,b) \mid a,b\in \mathbb{Z} \}$$ is an $\varepsilon$ -net for $X$ assuming that $\varepsilon> \sqrt{2}/2$ . 
The above definition and example can be found in [1], page 64-65.
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- G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.
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