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

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SUMMARY

The discussion centers on proving that for a bounded sequence \( (x_n) \), the limit superior \( \lim \sup (x_n) \) equals the maximum of the limit superiors of its even and odd indexed subsequences, \( \lim \sup (y_n) \) and \( \lim \sup (z_n) \). The user proposes letting \( M = \lim \sup (x_n) \), \( M_1 = \lim \sup (y_n) \), and \( M_2 = \lim \sup (z_n) \), establishing that \( M \geq \max(M_1, M_2) \). The goal is to demonstrate that \( M \) is indeed equal to \( \max(M_1, M_2) \).

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Homework Statement


Let (x_{n}) be a bounded sequence. For each n \in \mathbb{N}, let y_{n}=x_{2n} and z_{n}=x_{2n-1}. Prove that
\lim \sup {x_n} = \max (\lim \sup {y_n},\lim \sup {z_n})


Homework Equations





The Attempt at a Solution



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

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