$A\subseteq X\Rightarrow A\subseteq\overline{A}\Rightarrow d(A)\leq d\left(\overline{A}\right)\ldots (1)$
$x,y\in\overline{A}$ ve $\epsilon>0$ olsun.
$\left.\begin{array}{rr} x,y\in\overline{A} \\ \\ \epsilon>0 \end{array} \right\}\Rightarrow \begin{array}{c} \\ \\ \left. \begin{array}{rr} (\exists x_0,y_0\in A)\left(d(x,x_0)<\frac{\epsilon}{2}\right)\left(d(y,y_0)<\frac{\epsilon}{2}\right) \\ \\ d(x,y)\leq d(x,x_0)+d(x_0,y_0)+d(y_0,y)\end{array} \right\} \Rightarrow \end{array}$
$\Rightarrow d(x,y)\leq d(x,x_0)+d(x_0,y_0)+d(y_0,y)<\frac{\epsilon}{2}+d(x_0,y_0)+\frac{\epsilon}{2}=\epsilon +d(x_0,y_0)\leq \epsilon +\sup_{x_0,y_0\in A}d(x_0,y_0)$
$\Rightarrow \underset{d\left(\overline{A}\right)}{\underbrace{\sup_{x,y\in \overline{A}}d(x,y)}}\leq \epsilon +\underset{d(A)}{\underbrace{\sup_{x_0,y_0\in A}d(x_0,y_0)}}$
$\Rightarrow d\left(\overline{A}\right)\leq \epsilon +d(A)$
olur. $\epsilon>0$ keyfi olduğundan $$d\left(\overline{A}\right)\leq d(A)\ldots (2)$$ bulunur.
$(1),(2)\Rightarrow d\left(\overline{A}\right)=d(A).$