I had to do an exercise with this function:
$$ d_M:\Bbb C \rightarrow \Bbb R, \quad z \rightarrow \inf\{ |z-w|; w \in M| $$ with $\emptyset \neq M\subset \Bbb C$. First I proved that this function is Lipshitz. Now I'm in trouble proving that this relation $$d_M(z)=0 \Leftarrow \Rightarrow z \in M$$ holds exactly if $M=\overline {M}$ (the closure).
I already have an intuitive idea, but I don't know how to prove the relation. If we consider the right direction. If the function outputs $0$ it means $z \in M$, that makes sense, because for every $z$ the infimum will give us $0$ only if the $z$ itself is in $M$ and equal to $w$.
The other way. If $z \in M$ than $d_M(z) =0$ because for every $z$ in $M$ there will be one equal to $w$ (considering $w\in M$).
How would you prove it mathematically?
There would be a last question about drawing the graphs of $d_M$ and $d_{\Bbb C \backslash M}$ in case $M=\{ z \in \Bbb C; |z|\lt1 \}$. In this last question I can't even imagine how they would be.
Assume $M=\overline{M}$
If $d(x,M)=0$ then $x\in \overline{M}$. So $x \in M$.
Now if $x\in M$. $d(x,M)=inf\{ |x-w|; w \in M|\} \leq |x-x|=0$ so d(x,M)=0