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I cant visualize/understand this convergence proof

  1. Dec 12, 2008 #1
    for an alternative definition

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

    lim sup (x_n) = sup {x: infinitely many x_n are > x }.

    Now if both are equal to p, then consider a neighbourhood (p-e, p+e) of p (for some e>0).
    As y:= p+ e/2 > p we cannot have that infinitely many x_n are > y (otherwise lim sup (x_n) >= y > p)
    So at most finitely many x_n are > y, and almost all x_n are <= y < p+e.
    Similarly with y' = y - e/2 and liminf: almost all x_n are >= y' > p-e.
    So there is some N such that n >= N implies that x_n is in (p-e, p+e).

    As e>0 was arbitrary x_n -- > x.

    The reverse is similar, suppose x_n --> x.
    If p > x, then there is some n such that x_n < p for all n >= N.
    So p is NOT in the set B := {x: infinitely many x_n > x } so
    B subset {t: t <= x} and so lim sup x_n = sup B <= x.
    If p < x, then for all but finitely x_n, x_n >= p.
    So p is not in A:= {t: infinitely many x_n < t} and so
    A subset {t: t >= x } and so lim inf x_n = inf A >= x.
    x <= liminf x_n <= limsup x_n <= x implies equality.
     
  2. jcsd
  3. Dec 12, 2008 #2

    HallsofIvy

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    What is it you do not understand?
     
  4. Dec 13, 2008 #3
    this is how i see it:
    lim inf Xn=p
    lim Sup Xn=p
    if we look at the neighborhood of (p-e, p+e) of p (for some e>0)

    i cant understand this line
    "As y:= p+ e/2 > p we cannot have that infinitely many x_n are > y (otherwise lim sup (x_n) >= y > p)"

    whats y?
    why are they doing p+ e/2 > p

    "we cannot have that infinitely many x_n are > y"

    whats y?
     
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