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Interior of the boundary

  1. Jul 30, 2010 #1

    c16

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    1. The problem statement, all variables and given/known data
    I need to prove that the int(U union Bdy(U))=Int(U) when U is open.


    2. Relevant equations
    Bdy(U)=closure(U) intersect closure(X-U)
    a point is in the interior if there is an open neighborhood of the point that is contained in the set.


    3. The attempt at a solution
    obviously, if x is in U and U is open, there is a neighborhood of x in U by the fact that U is open. If x is in Bdy(U), then I want to prove that every open neighborhood of x is not in S=U union Bdy(U). Now, by the definition of boundary, every open neighborhood of x intersects X-U, so obviously there are points outside of U. What I need to prove is that there is a point y in every open neighborhood of x that is not in U AND is not in the closure of U, meaning that it is not in the boundary.
     
  2. jcsd
  3. Jul 30, 2010 #2

    Dick

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    Homework Helper

    Take U to be the union of the open intervals (0,1) and (1,2). What you are trying to prove is false.
     
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