# Metric Space and Subsets

1. Jun 30, 2009

### seyma

Let (X,d) be a metric space A and B nonempty subsets of X and A is open. Show:
A$$\cap$$B = $$\oslash$$ Iff A$$\cap$$B(closure)= empty
Only B closure
it is easy to show rigth to left but how can i use A's open property I try to solve with contradiction s.t. there exist r>0 Br(p)$$\subseteq$$A$$\cap$$B(closure) but i cannot come conlusion. Can you help me please ? :(

Last edited: Jun 30, 2009
2. Jun 30, 2009

### Office_Shredder

Staff Emeritus
Umm.... I think you're missing part of the iff statement

3. Jun 30, 2009

### seyma

yeah sorry i forget to write empty

4. Jun 30, 2009

### VeeEight

Are you able to show that A $$\cap$$ B' is empty (where B' is the set of limit points for B)? What is the definition of A being open (use the neighborhood definition)? What is the requirement that a point be an element of B' (use the deleted neighborhood definition)?

Using these definitions, you can work a proof by contradiction.