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Bounded sets, Limits superior and convergence

  1. Feb 21, 2013 #1
    (Hey guys and gals!)
    1. The problem statement, all variables and given/known data
    Given a bounded set x_n and for any y_n the following condition holds:
    [tex] \limsup_{n \rightarrow ∞}(x_n+y_n) = \limsup(x_n)+\limsup(y_n) [/tex]
    Show that x_n converges.

    2. Relevant equations

    Definition of limsup(x_n) = L:
    [tex] \forall \epsilon > 0 \mid \exists N \mid \forall n > N \mid x_n < L +\epsilon [/tex]

    3. The attempt at a solution
    I started by defining the limsups as follows:
    Let limsup(y_n) = a; limsup(x_n) = b;
    According to the given conditions limsup(x_n+y_n) = a + b(see above).
    Therefore:
    [tex]
    \forall \epsilon > 0 \mid \exists N_1 \mid \forall n > N_1 \mid x_n+y_n < (a+b) + \epsilon \\
    \mbox{Taking the same Epsilon:}\\
    \exists N_2 \mid \forall n > N_2 \mid y_n < a + \epsilon \\
    n>N \mid N = \max(N_1, N_2)
    [/tex]
    Now I arrive at the following congruent inequalities(to be subtracted by their transitive property):
    [tex]
    x_n+y_n < (a+b) + \epsilon \\
    y_n < a + \epsilon \Rightarrow \\
    x_n < b
    [/tex]
    Now, I claim that since I am given that x_n is bounded, by arriving at that final inequality, I've effectively discovered its upper bound .
    Therefore:
    [tex]
    x_n<b
    [/tex]
    But, since b is the limit superior, it is also the smallest possible, real upper bound.
    And yet, here I'm stuck.
    On the one hand, the above relation means that -b is the largest possible lower bound , or the limit inferior.
    But that means that limsup(x_n)≠liminf(x_n) which would imply that x_n does not converge.
    Where should I turn next?
    Is there perhaps a better way to look at this?
    Very thankful for your attention,
    Daniel
     
  2. jcsd
  3. Feb 21, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    I would consider a specific y_n. With the right choice, it is nearly trivial.

    Your definition for limsup looks strange - every L' > L would satisfy it as well. And (-1)^n would not have the correct 1 as limsup.
     
  4. Feb 21, 2013 #3

    jbunniii

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    No, this isn't right at all. It's possible to have a sequence ##(x_n)## with a lim sup ##b## such that ##x_n > b## for all ##n##. For example, if ##x_n = 1/n##, then ##\lim \sup x_n = 0##, but ##x_n > 0## for all ##n##.
     
  5. Feb 21, 2013 #4
    Firstly thanks for your replies.
    If you can follow my reasoning(which I know can be sometimes tortuous), I deduced that x_n < b only after taking both x_n+y_n<(a+b)+ε & the fact that y_n < a + eps.
    jbunn, I can clearly see your example, you're absolutely right; the problem is that I obtained this inequality solely through the definition of the limsup and the given statement re x_n being bound.
    I am clearly at loss, now, as to how to proceed. Any pointer in this regard would be extremely helpful.
    **
    Mfb, that sounds promising. Are you sure I'm allowed to take a discrete y_n(i.e. not arbitrary)?
    Obviously it occurred to me(please don't take me as quite so daft ;)) that if y_n = -x_n, then the given relation reduces to:
    limsup(x_n+y_n) = limsup(x_n+(-x_n)) = 0 = limsup(x_n)-liminf(x_n) -->limsup(x_n)=liminf(x_n) <-> x_n converges.
    But what if y_n isn't such? The proof should touch, I believe, on the general case.
    In any event, I am rightly stumped now. Do you have any suggestions?
    I would greatly appreciate any help,
    And I am grateful for your time, and useful advice,
    Daniel
     
  6. Feb 21, 2013 #5

    jbunniii

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    Gold Member

    Well, it does say that the condition holds for any (mathematician-speak for every) ##y_n##, so I'm sure mfb is correct: you can choose any specific ##y_n## you like to get the job done.

    Do you see where your proof implicitly uses the fact that ##(x_n)## is bounded?
     
  7. Feb 22, 2013 #6
    Jbun, thanks again.
    I used the condition to state that after I arrived at x_n < b, for n > N, it is the limsup that forms the eventual, smallest possible, upper bound, in other words, the
    limsup(x_n) = inf(sup(x_N, x_(N+1), ..., x_n)).
    Therefore, I had thought(as you say, mistakenly) that since x_n < b, for those n > N, x_n can construed bound by b(in that region, at least).
    If that's the case, and since I can eliminate a finite number of outlying elements of x_n(preceeding N), all of x_n, after that N, lies squarely(or equals either) between -b, and b.
    Then I considered that by writing it thus(subtracting b, taking the absolute value of both sides):
    |x_n-b| => 0, I can firmly state that for any(!) ε>0, this: |x_n-b|<ε holds, since the former conditions is true iff 0 >= x_n-b >= 0 ->|x_n-b|=0, and |x_n-b| < ε is the implicit definition of a limit of a sequence.
    The "N" in this case, would be the matching "N" to the above relation...
    Is this getting me anywhere?
    As you say, the better way is to simply take y_n=-x_n. I tried keep it as generalised as possible, without discretisation.
    Of course, if you consider it valid, then it's surely quite trivial.
    Still, what do you think?
    Is there a better method for the general case?(if at all?)
    Thanks again!
    Much beholden,
    Daniel
     
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