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Proving on nested intervals

  1. Aug 4, 2005 #1
    If I_n = (0, 1/n) where n is any natural no. is a sequence of nested intervals, then the intersection of all the I_n is empty.

    I was able to get the proof similar as that of above problem but the I_n is closed, that is, I_n = [0, 1/n]... and further, the intersection of all the I_n is {0}.

    What I am trying to do is to prove by contradiction because it is the easiest way to prove it. How do I construct this?
  2. jcsd
  3. Aug 4, 2005 #2


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    let x be in the intersection
    x<1/n all natural numbers n -> x=<0
    x=<0 & x>0 -><-
  4. Aug 4, 2005 #3

    I apologize for asking this... I will "repeat" what you mentioned because I am a bit "slow"...

    If I were to prove it by contradiction, will I assume that x is the intersection...?
    Then, from the nested interval property (is it?), 0 < x < 1/n for all natural no. n.

    From here on...you mentioned that "-> x=<0
    x=<0 & x>0 -> <-".... do you mind if you can explain to me?
  5. Aug 4, 2005 #4

    matt grime

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    by the definitions of the real numbers there is no real number that is strictly positive and less than 1/n for all n, that's all.
  6. Aug 4, 2005 #5


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    More "primitively", if x is any positive real number then 1/x exists and is a positive real number. By the "Archimedean property", there exist an integer n> 1/x so 1/n< x.
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