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Prove that lim sup(x_n) = max(lim sup(y_n), lim sup(z_n))

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex](x_{n})[/itex] be a bounded sequence. For each [itex]n \in \mathbb{N}[/itex], let [itex]y_{n}=x_{2n}[/itex] and [itex]z_{n}=x_{2n-1}[/itex]. Prove that
    [itex]\lim \sup {x_n} = \max (\lim \sup {y_n},\lim \sup {z_n})[/itex]


    2. Relevant equations



    3. The attempt at a solution

    Don't know if I'm at the right path but I've tried letting [itex]M= \lim \sup {x_n}[/itex], [itex]M_{1}= \lim \sup {y_n}[/itex], and [itex]M_{2} = \lim \sup {z_n}[/itex] and see that [itex]M \geq \max (M_{1}, M_{2})[/itex]. How do I proceed from here to prove that [itex]M = \max (M_{1}, M_{2})[/itex]? Thank you!
     
  2. jcsd
  3. Apr 20, 2013 #2
    Does this observation help?
    $\sup {x_n} = \max (\sup {y_n},\sup {z_n})$
     
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