TITLE: Prove $\operatorname{dist}(\overline{A},\overline{B}) = \operatorname{dist}(A, B)$
QUESTION [5 upvotes]: This is the last question on the exercise sheet and I am having real trouble formalizing my intuitions.
It should be obvious.  Since the closure of a set is the set of all points in the universe with distance zero to the set, then there should be no difference between finding the distance between the set and the distance between their closures (since zero is the additive identity). I feel like I could take advantage of the triangle inequality to formalize the picture in my head, but I can't seem to grasp it concretely.
Working Definitions:
$$ d(x, A) = \inf_{a \in A}\{d(x,a)\} $$
$$ \overline{A} = \{x \in X : d(x, A) = 0\} $$
$$ \operatorname{dist}(A, B) = \inf_{b \in B}\{d(b, A)\} $$

REPLY [3 votes]: Hints:
"$\leq$": Since $A\subseteq \overline{A}$ and $B\subseteq \overline{B}$ then $d(\overline{A},\overline{B})\leq d(a,b)$ for all $a\in A$, $b\in B$. Take infimum.
"$\geq$": Prove by triangle-inequality that for any $x,y\in X$ we have
\begin{equation*}
d(A,B)\leq d(A,x)+d(x,y)+d(y,B),
\end{equation*}
and conclude that for any $a\in\overline{A}$ and $b\in\overline{B}$ we have $d(A,B)\leq d(a,b)$. Take infimum.