Infimum of distance between two sets I have come across one little snippet in my Real Analysis book and I can not get it through. It generally says:
$E\subset [0,1] \subset G$ and $E$ is closed $G$ is open. Define $\delta=\inf\{|x-y|,x\in E \text{ and }y\in G^c \}$, then $\delta>0$.
I know $\delta\ge 0$, but I do not know how to prove $\delta$ could not be $0$.
Thanks.  

As $G$ is open and $[0,1] \subset G$, there exists a connected component  of $G$, say $G_1$, containing the unit interval. Then we have $G_1=(a,b)$ with $a<0 <1 <b$ (as connected open subsets of the real line are open intervals). You should be able to conclude from here.

Note (after an edit): my original answer ("As $G$ is open and $E=\overline{E} \subset G$, around every point of $E$  there's a ball of positive radius all contained in $G$.") is incorrect as the radii can have infimum $0$, as pointed out by the OP in the comments.
