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Uniqueness of Limits of Sequences

  1. Feb 13, 2012 #1
    Here is the proof provided in my textbook that I don't really understand.

    Suppose that x' and x'' are both limits of (xn). For each ε > 0 there must exist K' such that | xn - x' | < ε/2 for all n ≥ K', and there exists K'' such that | xn - x'' | < ε/2 for all n ≥ K''. We let K be the larger of K' and K''. Then for n ≥ K we apply the triangle inequality to get:

    | x' - x'' | = |x' - xn + xn - x'' | ≤ | x' - xn | + | xn - x'' | < ε/2 + ε/2 = ε​

    Since ε > 0 is an arbitrary positive number we conclude that x' - x'' = 0.

    ***I understand how they got to the conclusion |x' - x''| < ε. What I don't understand is how they can conclude from that that x' -x'' = 0.

    Any help is much appreciated. This isn't hw by the way. I'm just trying to better my understanding.
  2. jcsd
  3. Feb 13, 2012 #2


    Staff: Mentor

    If |x' - x''| < ε, for any small and positive number ε, then x' and x'' are the same number. IOW, x' = x''.
  4. Feb 14, 2012 #3
    In general, [itex]0\leq a<\epsilon[/itex] is satisfied for all [itex]\epsilon>0[/itex] implies
    that [itex]a=0[/itex]. Since otherwise, if [itex]a>0[/itex] then taking [itex]\epsilon=a[/itex] makes contradiction.
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