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Homework Help: Prove that if x > limsup s_n, then x is not the limit of any subsequence

  1. Aug 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Directly from the definition, for a sequence [tex](s_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}[/tex] prove that if [tex]x > \limsup s_n [/tex], then [tex]x[/tex] is not the limit of any subsequence of [tex](s_n)[/tex]. (i.e. Do not use the fact that [tex]\limsup s_n[/tex] is the supremum of the set of subsequential limits.)

    2. Relevant equations
    I have been told by my instructor that my proof will fail due to problems with inequalities --- but I fail to see where it would fail; i.e. are there any errors where [tex]>[/tex] should be [tex]\ge[/tex] or vice-versa?

    3. The attempt at a solution
    Please see the attachment.

    Thanks all!

    Attached Files:

  2. jcsd
  3. Aug 12, 2009 #2
    This should fall into place by using both facts. Try using a contradiction. I have a rough proof below:


    Suppose x > lim sup (S_n) and some subsequence (S_K(n)) has x as a limit, where K(n) is a strictly increasing function of natural numbers. If x is the limit of (S_K(n)), then for every epsilon there exists a M such that for all m > M, |S_K(m) - x| < epsilon. Since (S_K(n)) contains only terms in (S_n) and all terms of (S_n) are smaller than x, then every term of (S_K(n)) is smaller than x. Hence for any m > M, |S_K(m) - x| > 0. Let T = lim inf (|S_K(m) - x|) for all m > M. If T = 0, then x = lim sup (S_K(n)) which is less than or equal to (S_n), a contradiction. Hence T is non-zero. Since T is non-zero and positive, take epsilon to be smaller than T, say epsilon = T / 2. |S_K(m) - x| >= T > epsilon for all m > M. Hence x is not the limit of (S_K(m)), a contradiction.
  4. Aug 12, 2009 #3

    Thanks for your proof! May I ask some follow up questions --- just for the sake of knowing the motivations, etc.

    You wrote let [tex]T = \liminf (|S_{K(m)} - x|)[/tex] for all [tex]m > M[/tex] and showed that we must have [tex]T > 0[/tex] and then set [tex]\varepsilon = T / 2[/tex]. I have two questions:

    (1) What lead your motivation for constructing [tex]T = \liminf (|S_{K(m)} - x|)[/tex]? I just want to know what "sparked" you to think of using the limit inferior for this proof, as I'd never thought of introducing this when I wrote my attempted proof (see attached).

    (2) May I confirm my understanding of your last second to last sentence? You wrote that if [tex]T > 0[/tex], then we have [tex]|S_{K(m)} - x| \geq T[/tex].

    That is, if in general we have a sequence, say, [tex](a_n)[/tex] in [tex]\mathbb{R}[/tex], and if [tex]\lim a_n = a[/tex], where [tex] a > 0 [/tex], then we can conclude that there exists some [tex]N[/tex] such that for all [tex]n > N, a_n > 0[/tex]; that is, it is not necessarily true that we have [tex]a_n > a[/tex] for all [tex]n > N[/tex]. But in this case, since [tex]T[/tex] is also the limit inferior, we must have that [tex]|S_{K(m)} - x| \geq T > 0[/tex] by properties of the infimum. Is this understanding correct?

  5. Aug 12, 2009 #4
    I'm not sure if the proof is 100% correct. I mean, I did word it poorly and things do need to be fixed up. but I believe the general idea is correct.

    1. It just seemed to fit in order to find a value of epsilon: Take the smallest element of the set and take a smaller element to be your epsilon. Then no matter what you wouldn't be able to produce a M such that the limit inequality is less than the chosen number, which ends up contradicting the fact that every epsilon had such an M. (Although in this case, the number M isn't really as important as a well-chosen epsilon.)

    Keep in mind that the goal is to show that there is no convergent subsequence whose limit is x. I had to use the negation of the definition in order to prove this to be the case, since we have little else to work with.

    2. I believe your assertion is correct. It is certainly the case that A_n doesn't always has to be greater than a. Think of something such as sin(n) * exp(-n) + 1. This constantly oscillates but it still converges to 1 as n becomes arbitrarily large, because at some point it has to be within a small enough neighborhood of 1 for sufficiently large n. If it doesn't, then intuitively convergence cannot happen. (Remember, only the ultimate behavior of a sequence matters when dealing with convergence.)

    P.S. I am unable to view your attachment since it is pending approval.
  6. Aug 12, 2009 #5
    Thanks for the reply!

    I'm new to this forum and I was not aware that attachments need to be "approved" --- for future postings, I'll just directly type my message rather than attaching it.

    Thank you again for your proof and clarifications!
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