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Constructing an interval by uniting smaller intervals

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data
    We have [itex]C_n = [1-\frac{1}{n},2-\frac{1}{2n}] [/itex] and [itex]C = C_1 \cup C_2 \cup C_3 \cup ...[/itex] and are asked to describe the interval C and then prove that it is actually what we say it is.


    2. Relevant equations



    3. The attempt at a solution

    I am guessing that [itex] C = [0,2) [/itex] and to prove this, I need to show that the infimum and minimum is 0, supremum is 2, maximum does not exist.

    Am I right in describing this interval? And what other features of this interval would I need to prove in order to prove my description is valid?
     
  2. jcsd
  3. May 10, 2013 #2

    Bacle2

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    Why do you think 2 is not included in the interval? What is the limit of 2-(1/2n)? If 2 is not included,

    in C, can 2- 1/2n converge to 2?
     
  4. May 10, 2013 #3
    I don't really understand what you mean. 2 cannot be in C because 2-1/2n -> 2 as n -> ∞, so C_n = (1,2) as n -> ∞, wouldn't it?
     
  5. May 10, 2013 #4

    SammyS

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    I doubt that Bacle2 is saying that 2 is in interval C. He's just asking how to prove that you can have both
    2 is not in C.

    and

    ##\displaystyle \lim_{n\to\infty}(2-1/(2n))=2 ## ?​
     
  6. May 10, 2013 #5
    Well, now I have no idea.

    How else could I show that 2 is not in C? Unless 2 IS in C...
     
  7. May 10, 2013 #6
    If I make [itex]A[/itex] the set of all maxima of all [itex] C_n = \{ 2-\frac{1}{2n}\}[/itex] for all natural numbers n,

    Then the maximum of this set [itex]A[/itex] does not exist since it is infinite. However the least upper bound would be the limit = 2. Am I correct in thinking this?

    Then sup(C) = sup(A) = 2. Is this reasoning correct?
     
  8. May 10, 2013 #7

    Bacle2

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    Sorry if my statement was not clear. What I meant was that if 2 were included in the union, then it would be in some of the intervals, by def. of union. Then, re this last, the issue of convergence ,and (2-1/2n) being strictly increasing, come into play.
     
    Last edited: May 10, 2013
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