Let $x_n$ and $y_n$ be bounded sequences such that $x_n \leq y_n$ for all $n$. Show $\limsup(x_n)\leq \limsup(y_n)$
$x_n \leq y_n$ for all $n$, so $\sup(x_n) \leq \sup(y_n)$ for all $n$.
As this holds for all $n$, $\limsup(x_n)\leq \limsup(y_n)$
This seems too simple to me. I tried to do this with the $\epsilon$ definition of limits, but I struggle with this strategy. Any clarification or verification is appreciated.
You have to use the definition of the limit superior: $$ \limsup_n x_n=\lim_{n\to\infty}\bigl(\sup_{k\ge n} x_k\bigr)=\inf_{\vphantom{\ge^l} n}\bigl(\sup_{k\ge n} x_k\bigr). $$ Now set $a_n=\sup_{k\ge n} x_k$, $b_n=\sup_{k\ge n} y_k$. Since $x_k\le y_k$ for all $k$, we deduce that $$a_n=\sup_{k\ge n} x_k\le b_n=\sup_{k\ge n} y_k, \enspace\text{whence }\enspace \lim_{n\to\infty}a_n\le\lim_{n\to\infty}b_n.$$