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Limit point of Sn := {1-1/n} is 1

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data


    Show that the limit point of Sn:={1-1/n} is 1.

    2. Relevant equations

    We are prohibited from using epsilon and delta

    3. The attempt at a solution

    Let Sn:= {1-1/n} and U be any open interval from (a,b) where a<1<b. Observe that Sn is always [itex]\leq[/itex] 1. Since a<1 is linearly ordered, there is a positive number d between a and 1 such that a<d<1 and 1/d > 1. Then since d < 1,
    d-1 < 1-1
    d-1 < 0
    (d-1)/d < 0 --> (d-1)/d = 1 - [itex]\frac{1}{d}[/itex] and ...

    i am stuck, it appears that i shown my Sn is always less then zero making my limit point zero not one. Any help would be appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 22, 2013 #2

    pasmith

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    This appears to be an invitation to prove the result from basic properties of convergent sequences and already known limits.
     
  4. Sep 22, 2013 #3
    Here are the definitions that we must use. We are prohibited also from using monotonic and bounded.


    definition:

    The statement that the point sequence p1, p2, . . . converges to the point x means that if S is an open interval containing x then there is a positive integer N such that if n is a positive integer and n ≥ N then pn ∈ S.

    and

    The statement that the sequence p1, p2, p3, . . . converges means that there is a point x such that p1, p2, p3, . . . converges
     
  5. Sep 22, 2013 #4

    Ray Vickson

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    So, basically, you are using ε and N (not δ, of course, since n is not staying finite).
     
  6. Sep 22, 2013 #5

    pasmith

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    Are you allowed to assume that for every [itex]a > 0[/itex] there exists a positive integer [itex]n[/itex] such that [itex]n^{-1} < a[/itex], or do you have to prove that as well?
     
  7. Sep 22, 2013 #6
    yes

    yes. This was one of our axioms
     
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