why $\sup$ come before $(a_n + b_n) ?$ Taken from Rudin real and complex analysis  book     chapter $1$ excercise
let $\{a_n\}$ and $\{b_n\}$ be  sequence  in $[-\infty,\infty]$.Prove the following assertion
$b).\lim_{n \to \infty} \sup(a_n +b_n) \le  \lim_{\to \infty} \sup a_n + \lim_{n\to \infty} \sup b_n$
My attempt : $\sup a_n$ is upper bound  of  $a_n$  and $\sup b_n$ is a upper bound of  $b_n$.
Now $ a_n \le \sup  a_n \tag1$
and  $b_n\le \sup b_n\tag2 $
From $(1)$ and $(2)$  we can say that
$a_n + b_n \le \sup a_n  + \sup b_n$
Put  limit both side
$\lim_{n\to \infty} a_n + b_n \le \lim_{n \to \infty}  \sup a_n  + \lim_{n \to \infty} \sup b_n$
My doubt : why  $\sup$  come  before  $(a_n + b_n) ?$
 A: If you take the supremum of a sequence you get a fixed value, so it doesn't make much sense to take the limit after it. I think you're confusing $\limsup$ and $\sup$.
They're not the same thing, since $\sup\limits_{n\in\Bbb N}\{a_n\}$ is the biggest value you can approach with the sequence (could be $+\infty$ if not bounded), while $\limsup\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k\}$, which means that you consider $a_1,a_2,a_3,\ldots$, you take the biggest value approachable and you add it to your collection; but then you consider $a_2,a_3,\ldots$, you take the biggest value approachable and you add it to your collection; now consider $a_3,\ldots$ and etc. You are collecting the biggest value approachable of each $\{a_k\}_{k\ge n}$ for each $n\in\Bbb N$. Then you take the limit, which you can do because the way you collection is constructed is as a non-increasing sequence (you have to allow $+\infty$ as a value). Doing this you're making sure that you're getting either $+\infty$ if not bounded, either a value that your sequence doesn't stop taking or approaching to.
This is the key difference. For example: If you consider the sequence $0,0,0,0,0,1,0,0,0,0,0,\ldots$ then the supremum is $1$, but that value just appears once, so it would be nice if there was something that makes sure we get a value that keeps appearing or a value that our sequence keeps approaching to. That's a job for $\limsup$, since eventually it gets rid of that single instance of $1$ and give us $0$, which makes sense. Another example could be the sequence $1+\frac{1}{n}$: If you consider $\sup$ you get the biggest term, which is $2$, the first one. Considering $\limsup$ you get the actual limit if the sequence, which is $1$. You can prove that if the limit of a sequence exists, then $\limsup$ and $\liminf$ give the same value as the limit.
Now, for the result you're asking about:
$a_k\le\sup\limits_{k\ge n}\{a_k\}\;\forall k\ge n$ and $b_k\le\sup\limits_{k\ge n}\{b_k\}\;\forall k\ge n$ implies $a_k+b_k\le\sup\limits_{k\ge n}\{a_k\}+\sup\limits_{k\ge n}\{b_k\}\;\forall k\ge n$, which implies $\sup\limits_{k\ge n}\{a_k+b_k\}\le\sup\limits_{k\ge n}\{a_k\}+\sup\limits_{k\ge n}\{b_k\}\;\forall k\ge n$.
Taking limit on both sides you get $\limsup\limits_{n\to\infty}(a_n+b_n)=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k+b_k\}\le\lim\limits_{n\to\infty}(\sup\limits_{k\ge n}\{a_k\}+\sup\limits_{k\ge n}\{b_k\})\le\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k\}+\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{b_k\}=\limsup\limits_{n\to\infty}a_n+\limsup\limits_{n\to\infty}b_n$
A: At this state::
$$a_{n}+b_{n}\le sup\left(a_{n}\right)+sup\left(b_{n}\right)$$
You could say that
$$c_n = (a_n+b_n)$$
is bounded from above by $sup\left(a_{n}\right)+sup\left(b_{n}\right)$
So its sup must be less or equal than this upper bound:
$$sup\left(a_{n}+b_{n}\right)\le sup\left(a_{n}\right)+sup\left(b_{n}\right)$$
