Have I correctly proven $\limsup_{n \to \infty} x_n+\liminf_{n \to \infty} y_n \le \limsup_{n \to \infty} (x_n+y_n)$?

Question

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 $\lim_{n \to \infty} (\sup\{x_k:k\ge n\}+ \inf\{y_k:k\ge n\}) \le$ $\lim_{n \to \infty} \sup\{x_k +y_k:k\ge n\}=\limsup_{n \to \infty} (x_n+y_n)$.

Answer 1

Everything looks good - good job! Since you asked whether you could simplify it further I tried to write out your proof with as few lines as possible and I included it below. Though, if your Prof is insisting every line be absolutely clear - then your proof might be a little better to submit.

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LHS $=\limsup _{n \rightarrow \infty} x_n+\liminf _{n \rightarrow \infty} y_n$

$\quad \quad =\lim_{n \to \infty} (-\inf\{-x_k:k\ge n\})+ \lim_{n \to \infty}(\inf\{y_k:k\ge n\})$, from reflection and def'n

$\quad \quad =\lim_{n \to \infty} (-\inf\{-x_k:k\ge n\}+ \inf\{y_k:k\ge n\})$

$\quad \quad =\lim_{n \to \infty} (\sup\{x_k:k\ge n\}+ \inf\{y_k:k\ge n\})$

$\quad \quad \leq \lim_{n \to \infty} (\sup_k\{x_k + y_\alpha :k\ge n, \forall\alpha\geq n\})$

$\quad \quad \leq \lim_{n \to \infty} (\sup\{x_k + y_k \})$

$\quad \quad = \limsup_{n \to \infty} (x_n+y_n)$

Answer 2

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