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Topology Proof: AcBcX, B closed -> A'cB'

  1. Jan 26, 2012 #1
    Topology Proof: AcBcX, B closed --> A'cB'

    1. The problem statement, all variables and given/known data

    Prove:
    AcBcX, B closed --> A'cB'

    and where the prime denotes the set of limit points in that set
    X\B is the set difference


    2. Relevant equations

    Theorem:
    B is closed <--> For all b in X\B, there exists a neighborhood U of b with UcX\B



    3. The attempt at a solution

    Okay, so I am having no problem showing that the set of limit points of B is contained in B, if B is closed. I feel that there must be a way to extend this fact to show that A'cB', but I am not sure exactly how to do this.

    From the theorem:
    Since B is closed, there exists a neighborhood U of b with UcX\B, for all b in X\B

    then, the intersection of B and U must be empty, AND the intersection of B and (U\{b}) must also be empty.

    From the definition of limit point, if b is contained in X\B, then b cannot be contained in B'

    or, for all x in B', x also is contained in B.

    This is where I have gotten stuck.

    Could I possibly show that B closed implies A closed, then use the same logic to show that A'cAcB ?

    Thanks
     
  2. jcsd
  3. Jan 26, 2012 #2
    Re: Topology Proof: AcBcX, B closed --> A'cB'

    Okay, so I was able to prove that if AcBcX, then A'cB'.

    Then, A'cB'cB
     
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