How can I prove that $d(A,B)=d(A,\bar{B})=d(\bar{A},\bar{B})$ How can I prove that $d(A,B)=d(A,\bar{B})=d(\bar{A},\bar{B})$
$\bar{A}$ is the set of all accumulation points $A$
If $d(A,B)=\inf\{ d(a,b)$ with $a\in A ,  b\in B\}=\alpha$
$d(A,\bar{B})=\inf\{ d(a,b)$ with $a\in A ,  b\in \bar{B}\}=\beta$
$d(\bar{A},\bar{B})=\inf\{ d(a,b)$ with $a\in \bar{A} ,  b\in \bar{B}\}=\gamma$
I proved that $\alpha \geq \beta \geq \gamma$
 A: For any $\epsilon >0$ there exist $x\in \bar A, y\in \bar B$ with $d(x,y)<\gamma+\epsilon $ and there exist  $x'\in X,  y'\in Y$ with $d(x',x)<\epsilon$ and $d(y,y')<\epsilon.$ Then $$\alpha \le d(x',y')\le d(x',x)+d(x,y)+d(y,y')<\gamma+3\epsilon.$$ Hence $$\forall \epsilon >0\;(\alpha<\gamma+3\epsilon)$$ which is not possible unless $$\alpha\le \gamma.$$ Since you already have $\gamma \le \beta \le \alpha$, therefore $$\alpha\le \gamma\le \beta \le \alpha\;.$$
A: For every $c>0$, there exists $x \in A, y\in \bar B$ such that $d(x,y)\leq d(A,\bar B)+c/2$. There exists $y'\in B$ such that $d(y',y)<c/2$, $d(A,B)\leq d(x,y')\leq d(x,y)+d(y,y')\leq d(A,\bar B)+c/2+c/2$. This implies that $d(A,B)\leq d(A,\bar B)$ and $d(A,B)=d(A,\bar B)$ from the equality that you have proven.
$d(A,B)=d(A,\bar B)$ implies that $d(B,\bar A)=d(B,A)$ and $d(\bar A,B)=d(\bar A,\bar B)$, the result follows from the fact that $d(A,B)=d(B,A)$.
A: Clearly, $\beta\leq\alpha$ since $B\subset\bar{B}$.
Suppose, in order to obtain a contradiction, that $\beta<\alpha$.
Then, we can find $a$ in $A$ and $b$ in $\bar{B}$ such that $d(a,b)\leq\gamma$ where $\gamma\equiv(\beta+\alpha)/2<\alpha$.
We can also find a sequence of points $(b_{n})_{n}$ in $B$ such that $b_{n}\rightarrow b$.
By continuity, $d(a,b_{n})\rightarrow d(a, b)$, which in turn implies that we can find some $n$ sufficiently large with $d(a,b_{n})<\alpha$, contradicting the minimality of $\alpha$.
We have established $\alpha = \beta$.
To get $\beta = \gamma$, we can apply the same argument, reversing the roles of $A$ and $B$.
