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we were learning about epsilon-dense sets: If $S$ is a subset of a metric space, then $T$ in $S$ is $\epsilon$-dense in $S$ for given $\epsilon$ if for any $s$ in $S$, there is $t$ in $T$ s.t. the distance between $s$ and $t$ is $<$ $\epsilon$. I was wondering if this means that every set $X$ is actually $\epsilon$-dense in $X$ for every $\epsilon>0$ by definition?

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    $\begingroup$ Yes, any set is dense in itself (choose $s=t$). $\endgroup$ – Ian Coley Dec 8 '13 at 19:11
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    $\begingroup$ Such sets are also called $\epsilon$-nets. $\endgroup$ – Brian M. Scott Dec 8 '13 at 19:29
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Yes.

To ask if a given set is $\epsilon$-dense in itself is to ask whether, given an element $x\in X$ you can find another element $y\in X$ such that $d\left(x,y\right)<\epsilon$.

Well, you can just choose, for each $x$ the element $x$ itself, because it is an axiom of metric space theory that $d\left(x,x\right)=0$, and we know that $0<\epsilon$.

Thus, for every $\epsilon>0$, any set is $\epsilon$-dense in itself.

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