Sufficient conditions for a = lim inf xn

  1. This is part of a theorem which is left unproved in "Elementary Classical Analysis" by Marsden and Hoffman.

    Let xn be a sequence in R which is bounded below. Let a be in R.

    Suppose:

    (i) For all e > 0 there is an N such that a - e < xn for all n >= N.

    (ii) For all e > 0 and all M, there is an n > M with xn < a + e.

    Show that a = lim inf xn.

    (Definition: When xn is bounded below, lim inf xn is the infimum of the set of all cluster points of xn. If xn has no cluster points, then lim inf xn = + infinity. If xn is not bounded below, then lim inf xn = - infinity.)

    I was able to use (i) and (ii) to show that a is the limit of a subsequence of xn, hence a is a cluster point. So to show that a = lim inf xn, it is sufficient to show that a is a lower bound for the set of cluster points. This is what I can't do. Any suggestions?
     
  2. jcsd
  3. LCKurtz

    LCKurtz 8,390
    Homework Helper
    Gold Member

    Well, suppose b < a is a cluster point. Doesn't that give you a problem with (i).?
     
  4. So I set e = a - b in (i) and get b < xn for all n>=N. I don't see the problem with this.

    I think this implies all cluster points x satisfy b <= x, which means b <= a, but now I'm back where I started.
     
  5. LCKurtz

    LCKurtz 8,390
    Homework Helper
    Gold Member

    Try e = (a - b)/2. Can't you show all but finitely many of the xn are bounded away from b?
     
  6. I think I get it now.

    Setting e = (a -b)/2 in (i) gives (a + b)/2 < xn for all n >= N

    ==> (a - b)/2 < xn - b for all n >= N

    ==> no subsequence of xn can converge to b, since (a - b)/2 > 0

    ==> b is not a cluster point.


    thanks!
     
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