I want to prove: $\limsup_{n \to \infty} x_n+\liminf_{n \to \infty} y_n \le \limsup_{n \to \infty} (x_n+y_n)$. Is the proof I have written below correct? Have I overcomplicated the proof, overlooking a simpler method? Note that it is important, at my level of ability, to include and justify every step.
First note that $\limsup_{n \to \infty} x_n=-\liminf_{n \to \infty} (-x_n)$ follows directly from the reflection principle of $\sup$/$\inf$ and the definiton of $\limsup$ and $\liminf$.
$\limsup_{n \to \infty} x_n+\liminf_{n \to \infty} y_n=$
$-\liminf_{n \to \infty} (-x_n)+\liminf_{n \to \infty} y_n=$
$-(\liminf_{n \to \infty} (-x_n)-\liminf_{n \to \infty} y_n)=$
$-(\lim_{n \to \infty} (\inf\{-x_k:k\ge n\}- \inf\{y_k:k\ge n\}))=$
$\lim_{n \to \infty} (-\inf\{-x_k:k\ge n\}+ \inf\{y_k:k\ge n\})=$
$\lim_{n \to \infty} (\sup\{x_k:k\ge n\}+ \inf\{y_k:k\ge n\})$
but, $\sup\{x_k:k\ge n\}+ \inf\{y_k:k\ge n\} \le \sup\{x_k:k\ge n\}+y_\alpha$ for all $\alpha \ge n$
and because $\sup\{x_k:k\ge n\}+y_\alpha=\sup\{x_k +y_\alpha:k\ge n\}$ for all $\alpha \ge n$
we must have: $\sup\{x_k:k\ge n\}+ \inf\{y_k:k\ge n\} \le \sup\{x_k +y_k:k\ge n\}$ for all $n$
therefore $\lim_{n \to \infty} (\sup\{x_k:k\ge n\}+ \inf\{y_k:k\ge n\}) \le$
$\lim_{n \to \infty} \sup\{x_k +y_k:k\ge n\}=\limsup_{n \to \infty} (x_n+y_n)$.
Best Answer
I use the properties $$\sup(A) = -\inf(-A)$$ and $$\sup_{n\in I}(a_n+b_n) \leq \sup_{n\in I} a_n + \sup_{n\in I} b_n \text{ for } I\subseteq\mathbb{N}$$
You can then show for $I\subseteq\mathbb{N}$ $$ \begin{align*} & \sup_{n\in I}(x_n) \\ & = \sup_{n\in I}(x_n+y_n - y_n) \\ & \leq \sup_{n\in I}(x_n+y_n) + \sup_{n\in I}(-y_n) \\ & = \sup_{n\in I}(x_n+y_n) - \inf_{n\in I}(y_n) \end{align*} $$
You can then use this to show the desired result by applying this to the sup/inf sequences of the limsup/liminf