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Homework Help: Quick check on simple supremum/infimum problems

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Please just tell if I'm right or wrong on these. I think they should be fairly easy, but I don't want to say too much and end up looking like an idiot in case there is a mistake =p

    We are supposed to say if it's bounded above and/or below. Also to indicate supremum and infimum if they exist

    2. Relevant equations
    (a) [tex]{n \in Z: \cos n!}[/tex]

    Not sure how to do this, but intersection runs from 1 to infinity
    (b) [tex]\bigcap (\frac{1}{n},1+\frac{1}{n})[/tex]

    3. The attempt at a solution
    On (a) I'm assuming that n can only be positive integers (not sure what the factorial of a negative integer means). Then it forms a subset of Reals (-1, 1) (1 and -1 are not included since factorials are not 0 or irrational) with some gaps.
    Thus it's bounded above and below and supremum = 1 and infimum = -1

    On (b), the intersection turns out to be just {1}. This is bounded above and below and infimum = supremum = 1.
  2. jcsd
  3. Jan 17, 2008 #2
    I think you entered in problem (b) wrong. If it is as you say, then the first interval is (1,2), which does not include 1. Did you mean (1-1/n, 1+1/n)?

    As for (a), if you say that 1 and -1 are the sup and inf, then you need to show that cos n! gets arbitrarily close to 1 and -1.
  4. Jan 17, 2008 #3
    Oops, correct you are.

    #1b is wrong, the set is empty hehehe. Thus no supremum and no infimum.

    And how would I show (a)? The definition of the limit? Then I would probably need to use trig identities correct?
  5. Jan 17, 2008 #4
    Indeed, as masnevets points out, in (b) the intersection is empty. In that case (I think) you would have that all real numbers serve as both upper and lower bounds, so that there would be no sup and no inf. Check it again.

    To enlarge upon masnevets other point - for (a), if you suspect that the sup = 1 and the inf = -1 you would need to verify that |n! -Pi*k| can be made as small as desired for both an odd and an even value of k. I don't know if that is possible.
    Last edited: Jan 17, 2008
  6. Jan 17, 2008 #5

    The empty is thus bounded above and below correct?

    Also, that's what I was sorta thinking for (a) and I also dunno if it's possible. More formally, I believe I need to show that the max and min of the set of limits of all convergent subsequences are respectively 1 and -1. I'm not sure if I can, because as you have said it comes down to showing that |n! -Pi*k| can be made as small as desired, which seems intuitively true, but intutition is not a proof.
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