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Closed set equivalence theorem

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi guys, this problem gave me some trouble before, but I'd like to know if I have it worked out now.....

    "If S = S[tex]\cup[/tex]BdyS, then S is closed (S[tex]_{compliment}[/tex] is open)



    2. Relevant equations
    S is equal to it's closure.


    3. The attempt at a solution
    1. Pick a point p in S[tex]^{compliment}[/tex].
    2. For all points q[tex]^{n}[/tex] in S, let [tex]\delta[/tex] = min{|p-q[tex]^{n}[/tex]|}
    3. Dfine B(p,[tex]\delta[/tex])
    4. For any point in S[tex]^{compliment}[/tex], we can produce [tex]\delta[/tex]>0 such that B(p,[tex]\delta[/tex])[tex]\subset[/tex]S[tex]^{compliment}[/tex]. Therefore, all points in S[tex]^{compliment}[/tex] are interior points; therefore, S[tex]_{compliment}[/tex] is open, and S is closed.
     
  2. jcsd
  3. Dec 10, 2009 #2
    I am having a hard time figuring out what you mean by "S[tex]\cup[/tex]BdyS"
     
  4. Dec 10, 2009 #3
    Oh! I'm sorry if I was unclear about something.....

    "S[tex]\cup[/tex]BdyS" refers to the union of the set S with its boundary, and is called Closure(S)

    It can also be referred to as the union of the interior of S with the boundary of S.

    (Someone tell me if I made an error!)
     
  5. Dec 11, 2009 #4

    HallsofIvy

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    How do you know there exist such a minimum? Not every set has a minimum. Every set of real numbers, bounded below, has an infimum (greatest lower bound). But then how do you know it is not 0?

     
  6. Dec 11, 2009 #5
    Argh, that makes it even worse for me because now I really have no idea how to finish this one.

    What I have so far is:
    1. Pick a point z in Compliment(S)
    2.Then, z is not in S, and is not in Closure(S) by hypothesis
     
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