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A Supremum Problem

  1. Sep 23, 2008 #1
    The problem statement, all variables and given/known data
    Let [itex]a_1,a_2,\ldots[/itex] and [itex]b_1,b_2,\ldots[/itex] be bounded sequences of real numbers. Define the sets X, Y and Z as follows:

    [tex]\begin{align*}
    X &=\{x \in \mathbb{R} : a_n > x \text{ for infinitely many } n \} \\
    Y &=\{y \in \mathbb{R} : b_n > y \text{ for infinitely many } n \} \\
    Z &=\{z \in \mathbb{R} : a_n + b_n > z \text{ for infinitely many } n\}
    \end{align*}[/tex]

    Show that [itex]\sup Z \le \sup X + \sup Y[/itex] and that equality holds if one of the sequences converges.

    The attempt at a solution
    Let X+Y = {x + y : x in X and y in Y}. I know that sup (X+Y) = sup X + sup Y. If I can show that Z is a subset of X+Y, then sup Z ≤ sup (X+Y) and the inequality follows.

    To this end, let z belong to Z. Then [itex]b_n > z - a_n[/itex] for infinitely many n. Since these [itex]a_n[/itex] are bounded, they have a least upper bound x. Hence [itex]b_n > z - a_n \ge z - x[/itex]. Let y = z-x and notice that y belongs to Y. Unfortunately, x does not belong to X and there seems to be no fix to this problem.

    What else can I do?
     
  2. jcsd
  3. Sep 23, 2008 #2
    By the way, sup X, sup Y, and sup Z are the definitions of [itex]\displaystyle \limsup_{n \to \infty}a_n[/itex], [itex]\displaystyle \limsup_{n \to \infty}b_n[/itex], and [itex]\displaystyle \limsup_{n \to \infty} (a_n+b_n)[/itex], respectively.
     
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