Question

$(X,d)$ metrik uzay ve $A\subseteq X$ olmak üzere $d\left(\overline{A}\right)=d(A)$ olduğunu gösteriniz.

Not: $d(A): A$ kümesinin çapı

Answer 1

$A\subseteq X\Rightarrow A\subseteq\overline{A}\Rightarrow d(A)\leq d\left(\overline{A}\right)\ldots (1)$

$x,y\in\overline{A}$  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.

$(1),(2)\Rightarrow d\left(\overline{A}\right)=d(A).$