Supremum of sum of two sequences: $\sup (x_n+y_n) \le \sup x_n + \sup y_n$ 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.
 A: 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 $\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n$.
A: You may also try:
(1)If A is a sub set in B, then sup(A) <= sup(B). (2)For A, B are subsets in R, sup(A + B) = Sup(A) + Sup(B). $A + B = \{c = a + b\mid a \in A, b \in B\}$. 
Then notice that $\{an + bn\}\subseteq \{an\} + \{bn\}.$
(2)The second inequality follows naturally:
$Am$ = { $an | n >= m$ }, $Bm$ = { $bn | b >= m$}, $Cm$ = { $cn = an + bn | n >= m$}.
Then, by the first inequality, Sup(Cm) <= Sup(Am) + Sup(Bm), for each m.
And, the sequence inequality holds as m -> $+00$.
