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Homework Help: What is the meaning of this definition

  1. Dec 12, 2008 #1
    lim inf (x_n) = inf {x: infinitely many x_n are < x }

    i cant understand what they are saying here?

    i was told to see lim inf as the infimum of all the limits of all the subsequences

    they say something else
    ??
     
  2. jcsd
  3. Dec 12, 2008 #2

    HallsofIvy

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    Look at all x with the property that there are infinitely many numbers in the sequence {x_n} less than x. Take the "infimum" (greatest lower bound).

    For example, suppose the sequence is x_n= 1/n:
    1, 1/2, 1/3, 1/4, 1/5, ...
    If x is 0 or any negative number, all the numbers in the sequence are greater than x- an infinite number. If x is positive, 1/x is a positive number so (Archimedian property) there exist some integer N such that N> 1/x which means that 1/N< x. From that, in n> N, 1/n< 1/N< x so there are only a finite set of numbers in 1/n less than n. The set of all numbers, x, such that infinitely many members of the sequence are less than x is precisely the non-positive numbers, [itex](-\infty, 0][/itex]. It's inf is 0, the limit of the sequence.

    Or take a_n= -1/n for n odd, n/(n+1) for n even: -1, 2/3, -1/3, 4/5, -1/5, 6/7, -1/7... With an infinite number of negative numbers in there, it is easy to see that there exist an infinite number of terms of that sequence less than any non-negative number. But if x< 0, then -x is positive, so there exist N> -1/x or x> -1/N. There exist only a finite number of terms of the sequence less than x. Now the set of all x such that infinitely many x_n< x is the set of all negative numbers: [itex]-\infty, 0)[/itex]. That set does not include 0 but includes number arbitrarily close to 0 so its infimum is still 0. Of course, the two "subsequential limits" are the limits of the subsequences {-1/n for n even} and {n/(n+1) for n odd} which are 0 and 1. The smaller of those, 0, is the lim inf.
     
  4. Dec 13, 2008 #3
    you are saying:

    lim inf (x_n) = inf {x: infinitely many x_n are < x }

    means that there are infinite number of members which are smaller then X

    this is a definition of upper bound
    so this X is a Upper bound
     
  5. Dec 13, 2008 #4
    No, an upper bound is greater than or equal to all [tex]x_n[/tex], not just infinitely many of them.
     
  6. Dec 13, 2008 #5

    HallsofIvy

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    No I didn't say that and No that is NOT the definition of upper bound!
    The definition of "X is an upper bound of S" is that ALL members of S are less than X, not just an infinite number of them.
     
  7. Dec 13, 2008 #6
    infinitely is not all
    ??
     
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