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

drawar

## 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
$\lim \sup {x_n} = \max (\lim \sup {y_n},\lim \sup {z_n})$

## 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!

$\sup {x_n} = \max (\sup {y_n},\sup {z_n})$