
lim sup
Let $\displaystyle (x_n)$ be a bounded sequence. For each $\displaystyle n \in \mathbb{N}$ , let $\displaystyle y_n=x_{2n}$ and $\displaystyle z_n=x_{2n1}$. Prove that
$\displaystyle \limsup{x_n}=\max({\limsup{y_n},\limsup{z_n}})$.
I tried using the identity $\displaystyle \max(x,y)=\frac{1}{2}{(x+yxy)}$, but it can't seem to work... can anyone help me out?

Re: lim sup
$\displaystyle \limsup_n x_n=\lim_n\sup_{k\geq n}x_k=\lim_n\max (\sup_{2k\geq n}x_{2k},\sup_{2k+1\geq n}x_{2k+1})$. Put the $\displaystyle \lim$ into the $\displaystyle \max$ to get the result.