Distance between two sets equals distance between the closure of the sets

[CfM]

My problem is as follows: If we define d(A,B) = inf{ d(x,y) : x in A and y in B }, show that d(clos(A),clos(B)) = d(A,B), where clos(A) is the closure of A

My attempt at a solution was this: Since A is a subset of the closure of A, then d(A,B) must be less than or equal to the distance between the closure of A and the closure of B, but the other inequality seems to elude me. I thought about splitting it into cases, since if x belongs to clos(A) then x is in A or x is a limit point of A (y in clos(B) implies y is in B or y is a limit point of B), but got nowhere. I really have no idea how to begin this other part. Any help would be greatly appreciated

 

[lippyka]

. A:Let $d(A,B) = \inf\{d(x,y) : x\in A, y\in B\}$. We will show that $d(A,B)$ is equal to $d(\overline{A},\overline{B})$Proof: Let $\epsilon>0$. By definition of infimum, there exists $x'\in A$ and $y'\in B$ such that $d(x',y')<d(A,B)+\epsilon$. Since $x'\in A$, either $x'\in \overline{A}$ or $x'$ is a limit point of $\overline{A}$. Similarly, since $y'\in B$, either $y'\in \overline{B}$ or $y'$ is a limit point of $\overline{B}$.In either case, there exists $x\in \overline{A}$ and $y\in \overline{B}$ such that $d(x,y)<d(A,B)+\epsilon$. Therefore $d(A,B)\leq d(\overline{A},\overline{B})$.To show the other direction, let $\epsilon>0$. By definition of infimum, there exists $x'\in \overline{A}$ and $y'\in \overline{B}$ such that $d(x',y')<d(\overline{A},\overline{B})+\epsilon$. Since $x'\in \overline{A}$, either $x'\in A$ or $x'$ is a limit point of $A$. Similarly, since $y'\in \overline{B}$, either $y'\in B$ or $y'$ is a limit point of $B$.In either case, there exists $x\in A$ and $y\in B$ such that $d(x,y)<d(\overline{A},\overline{B})+\epsilon$. Therefore $d(\overline{A},\overline{B})\leq d(A,