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Every interior point of 'the closure of S' is in Int S?

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Let S be a set in R^n, is it true that every interior point of 'the closure of S' is in Int S? Justify.

    2. Relevant theorem

    S^int = {x belongs to S: B(r,x) belongs to S for some r>0}
    The closure of S is the union of S and all its bdary points.



    3. The attempt at a solution

    My answer is yes, but I am not sure how to give a proof, anyone can give a counterexample?
     
  2. jcsd
  3. Oct 3, 2008 #2
    It can happen that [tex]\textrm{int}(S)=\emptyset[/tex] while on the other hand [tex]\textrm{int}(\overline{S})\neq\emptyset[/tex].
     
  4. Oct 3, 2008 #3

    hi jostpuur, thx for your reply, but when the first case happens, how can the second one happens like that?
    I dont quite get it, can you explain it a little bit more?
     
  5. Oct 3, 2008 #4
    If S doesn't have interior, it doesn't necessarily mean that S is somehow "thin" (like n-k dimensional manifold in n dimensional space, with k > 0), but S can also be "filling" all n dimensions in the space, but on the other hand being "infinitely full of holes" so that interior doesn't exist.
     
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