metric space

flower3 · #1 $Old$ May 24th, 2009, 02:18 AM

let $(X,d)$ be a metric space and $A\neq \phi$ a subset of X then $x \in \overline{A}$ $iff$
$d(x,A)=0$

Gamma · #2 $Old$ May 24th, 2009, 12:02 PM

Quote:

Originally Posted by flower3 $View Post$

let $(X,d)$ be a metric space and $A\neq \phi$ a subset of X then $x \in \overline{A}$ $iff$
$d(x,A)=0$

$(\Rightarrow )$
Suppose $x \in \overline{A}$ but $d(x,A)=\delta >0$ . Then consider $B(x,\frac{\delta}{2})$ this is an open neighborhood containing x, but does not intersect A, thus $x \not \in \overline{A}$ , contradiction. d(x,A)=0 as desired.

$(\Leftarrow)$ Let d(x,A)=0. Suppose $x \not \in \overline{A}$ then there exists an open set U containing x, but does not intersect A. By definition of open set, this means there is a basis element (a ball) containing x and contained in U. That is $x \in B(x,\delta) \subset U \subset A^c$ . But this means $d(x,A)>\delta >0$ contradiction. thus $x \in \overline{A}$ as desired.

QED