Supremum of sum of two sequences: $\sup (x_n+y_n) \le \sup x_n + \sup y_n$

Question

Prove that $\sup\{x_n+y_n\}\leq \sup\{x_n\}+\sup\{y_n\}$, if both sups are finite. Furthermore, prove that $\limsup\{x_n+y_n\}\leq \limsup\{x_n\}+\limsup\{y_n\}$ if both limsups are finite.

Answer 1

For the first part of your question, let $\alpha := \sup_n x_n$ and $\beta := \sup_n y_n$. Then simply note that

$$ x_n + y_n \leq \alpha + \beta $$

for all $n \in \Bbb{N}$. As the supremum of a set is the least upper bound, we conclude

$$ \sup_n (x_n + y_n) \leq \alpha + \beta = \sup_n x_n + \sup_m y_m. $$

Note that it can happen for the inequality to be strict. Consider for example $x_n = (-1)^n$ and $y_n = (-1)^{n+1}$. Then $\sup_n x_n = 1 = \sup_n y_n$, but $x_n + y_n = 0$ for all $n$, hence $$\sup_n (x_n+y_n) = 0 < 2 = \sup_n x_n + \sup_m y_m.$$

For the second part, as Matrin Sleziak wrote, see Prove $\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n$.