Let $(X,d)$ be a metric space and $S \subset X$. Show that $d_S(x):=\text{inf}\{d(x,s): s \in S\}=0 \Leftrightarrow x \in \overline S .$
Notes: $\overline S$ is the closure of S. Maybe you can use that a closed set is also closed for sequences in the set? I think the difficult part is when S is open, otherwise its trivial as the closure would be equal to S.
Best Answer
If $x\notin \overline{S}$, then there exists $r>0$ such that $B_r(x)\cap S=\phi$. It follows that $d(x,s)\ge r>0$ for all $s\in S$ and hence $d_S(x)>0$. And these steps can be reversed, i.e. the steps above are actually "if and only if". So you get the proof.