Definition. We call $\mathbf x$ a boundary point of $S$ if every $\epsilon$-disk $D(\mathbf x, \epsilon)$ contains point of $S$ and $S'$(the complement of $S$). The set of all boundary points of $S$ is called the boundary of $S$ (notation: $\partial(S)$ ). The closure of a set is the set $\bar S=S\cup \partial(S)$.
Prove that $\partial(S)=\partial(\bar S)$.
I can show $\bar S$ and $\partial(S)$ are closed set.
Not true. Consider $S=\Bbb Q$ in the usual topology of $\Bbb R$.