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Thread: derived sets

  1. #1
    Aug 2008

    derived sets

    prove that the derived set of T = { 1/m + 1/n | m,n are natural numbers} is the set U = {1/n | n is a natural number}. first show that U is a subset of the derived set of T. To prove that T has no other limit points, show that there are only finitely many points of T in (1/N, 1/(N-1)) that are not of the form 1/N + 1/n.

    i am having trouble with the second part of this proof. the derived set of a set S as i understand is the set of limit points of the set S. a limit point x is a point such that any open interval that contains that point x also contains infinitely many points of the set S.

    i don't understand how proving the second statement proves that T has no other limit points. so from what i understand, we're trying to see if there is a limit point between 1/N and 1/(N-1). lets say we choose N = 2 so we have the interval (1/3, 1/2). so lets pick a point that cannot be expressed as 1 / some natural number. but it seems to me that no matter how small an interval you put around it, you will always find infinitely many numbers in that interval of that form 1/m + 1/n for some m and n by just choosing m and n big enough.

    so how is it that the only limit points of the set T are only ones of the form 1/n for all natural numbers n?
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  2. #2
    Junior Member
    Feb 2010

    Re: derived sets

    Firstly I think that T^{'}=\{\frac{1}{n}\}\cup\{0\}
    Why aren't there any other limit points? Between 1/n and 1/(n-1) there are infinitely many points of T and they are of the form 1/n + 1/m with m>n(n-1) and for any real number r between 1/n and 1/(n-1) there is a minimun distance to a point of T since r falls between two numbers of of form 1/n + 1/m
    Last edited by facenian; Jul 1st 2011 at 02:54 AM. Reason: gramma fixing
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