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Convergence implies maximum/minimum/both

  1. Nov 4, 2012 #1
    Hello,

    Could anyone please assist in proving that given a sequence converges it has a maximum/minimum/both?
    I have hitherto written that granted it converges it must be bounded and have a supremum and an infimum. Now, how may I proceed to prove that the latter are indeed within (the neighbourhood) of the limit itself?
     
  2. jcsd
  3. Nov 4, 2012 #2

    Zondrina

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    So if your sequence has a maximum, it is bounded above. If your sequence has a min, it is bounded below.

    What do you know about a bounded sequence of real numbers?
     
  4. Nov 4, 2012 #3
    What do you mean? I understand that any bounded sequence of real numbers must have a maximum and/or minimum, but how do I formally demonstrate that? Namely, how do I show that its supremum and infimum are within L, its limit?
     
  5. Nov 4, 2012 #4

    SammyS

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    Look at some examples of convergent sequences.

    Does each have both a maximum and minimum?

    Do all have only a maximum?

    Do all have only a minimum?

    Do they have to have at least one or the other?

    Do any have both?
     
  6. Nov 4, 2012 #5
    I can think of examples for sequences having either one of the two or both, but that still does not help me in formally proving the statement.
     
  7. Nov 4, 2012 #6

    SammyS

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    Do they have to have at least one or the other (minimum, maximum)?
     
  8. Nov 4, 2012 #7
    Yes, they do.
     
  9. Nov 4, 2012 #8

    SammyS

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    How about proving the contra-positive, that a sequence which has neither a maximum nor a minimum does not converge?
     
  10. Nov 5, 2012 #9
    I am still not sure how to formulate that mathematically. May you please advise?
     
  11. Nov 5, 2012 #10

    HallsofIvy

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    if sequence an converges to A, then there exist N such that if n> N, |an- A|< 1 which is the same as saying A-1< an< A+1. Do you see how that gives upper and lower bounds for all an with n> N? And an for [itex]n\le N[/itex] is a finite set.
     
  12. Nov 5, 2012 #11
    Naturally, but how do I use that to demonstrate that the supremum and infimum are within the neighbourhood of L.
     
  13. Nov 5, 2012 #12
    This is very frustrating. I am still unable to prove that the supremum and infimum are within L. May someone please assist?
     
  14. Nov 5, 2012 #13

    SammyS

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    What do you mean by the supremum and infimum are within L ?
     
  15. Nov 5, 2012 #14
    Hence, within L+epsilon and L-epsilon.
     
  16. Nov 5, 2012 #15
    Where sequence converges to limit L.
     
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