Q:

About distance of a point from a set

Consider a subset D of normed space X and y is a  vector which is not in D. So we find distance of y from set D say this distance is '£' . Now i read that if set D is closed then we take guarantee that '£'  is always positive means  £>0.  So i think if D is not closed( Open) then is  there any possibility that £ =0  as y is  not in D and if yes then how. I thing a lot about it. But  not  found any solution 

A:

If $D$ is open then the distance between any point of the boundary of $D$ and $D$ is zero, with the boundary defined as 
$$\partial D:=\overline D\setminus D^\circ$$
where if $D$ is open then $D=D^\circ$, with $D^\circ$ being the interior of $D$ and $\overline D$ the closure of $D$. Then
$$d(x,D)=0,\quad x\in\partial D$$
and obviously $x\notin D$ by the above definition. Example: let the open set of the real line (with the standard topology) $(0,1)$, then
$$d(0,(0,1))=d(1,(0,1))=0$$
and $0\notin (0,1)$, and $1\notin(0,1)$.

