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Math Help - Question on monotone convergence theorem problem

  1. #1
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    Question on monotone convergence theorem problem

    Hey guys,

    Can you check whether my proof of the problem?

    Show that a monotone sequence has at most one cluster point.

    pf)

    i) {a_n} is monotone increasing and bounded above.

    By monotone convergence theorem, {a_n} -> x

    If {a_n} is monotone decreasing and bounded below

    b_n = -a_n and apply the same as above.

    a_n -> x

    Since a_n converges to x, a_n only has x as its cluster point.

    ii) {a_n} is monotone increasing and not bounded.

    There is N in N such that |a_N|  \geq M, \forall M>0

    Assume a_n -> x.

    For any epsilon > 0 There is a_N such that |a_N - x | < 1, when n>N.

    |a_N| - |x|  \leq |a_N - x| < 1

    M = Max { |a_1|, |a_2|, ... |a_N|, |a_n| +1}

    |a_n|  \leq M

    Since a_n is unbounded, contradiction.

    Since a_n is unbounded and diverges, a_n has no cluster point.

    If a_n is monotone decreasing and unbounded, let b_n = - a_n and apply the same procedure.
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  2. #2
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    Re: Question on monotone convergence theorem problem

    Quote Originally Posted by rokman54 View Post
    Can you check whether my proof of the problem?

    Show that a monotone sequence has at most one cluster point.

    Frankly, I follow little of what you posted. Why are you doing so may different things.
    Look carefully at what you are asked to show.
    Suppose that (a_n) is a a monotone sequence.
    Without loss of generality (wlog) we can assume the sequence is non-decreasing.

    For proof by contradiction, suppose the sequence has two cluster points, p~\&~q.
    Again wlog we can assume p<q. Let c=\frac{q-p}{2}>0.

    The open intervals (p-c,p+c)\cap(q-c,q+c)=\emptyset.

    By definition of cluster point, each of those intervals must contain infinitely many points of (a_n).
    Is the possible for a non-decreasing sequence?
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