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Metric Space and Subsets

  1. Jun 30, 2009 #1
    Let (X,d) be a metric space A and B nonempty subsets of X and A is open. Show:
    A[tex]\cap[/tex]B = [tex]\oslash[/tex] Iff A[tex]\cap[/tex]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)[tex]\subseteq[/tex]A[tex]\cap[/tex]B(closure) but i cannot come conlusion. Can you help me please ? :(
     
    Last edited: Jun 30, 2009
  2. jcsd
  3. Jun 30, 2009 #2

    Office_Shredder

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    Umm.... I think you're missing part of the iff statement
     
  4. Jun 30, 2009 #3
    yeah sorry i forget to write empty
     
  5. Jun 30, 2009 #4
    Are you able to show that A [tex] \cap[/tex] 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.
     
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