Show that in ($\mathbb{R},d_2$), $\mathbb{Q}^{o}=\emptyset$ but $C((\mathbb{Q})^{o})=\mathbb{R}$.
(C(A) meaning the closure of A, not sure how to do a macron in LaTeX).
My attempt: Since for any subset $S$ of a metric space $X$, $S^o=S/\partial S$, I want to show $\partial\mathbb{Q}=\mathbb{R}$ and thus have that $\mathbb{Q}^{o}=\mathbb{Q}/\mathbb{R}=\emptyset$.
$\partial\mathbb{Q}=${all $x\in\mathbb{R}|dist(x,\mathbb{Q}=\inf_{y\in\mathbb{Q}}d(x,y)$}. But $\mathbb{Q}$ is dense in $\mathbb{R}$ so $\forall x\in\mathbb{R}$, $d(x,\mathbb{Q})=0=\inf_{y\in\mathbb{Q}}d(x,y)\implies\partial\mathbb{Q}=\mathbb{R}$.
To show that $(C(\mathbb{Q})^{o})=\mathbb{R}$, I use the fact that for any set $S\subseteq X$ dense in $X$, $C(S)=X$. Thus we have $\mathbb{R}^o=\mathbb{R}$ since $X^o=X$ in $(X,d)$.