I have tried to read many proofs of this but I'm not sure I get it, so please bare with me.
Show that $\lim_{n \rightarrow \infty} \sup (a_n+b_n) \leq \lim_{n \rightarrow \infty} \sup (a_n)+lim_{n \rightarrow \infty} \sup (b_n).$
I care about the case when $(a_n + b_n)$ is bounded.
Take an arbitrary subsequence $({a_n}_k + {b_n}_k)$, it must converge to a limit $l$ since $(a_n + b_n)$ is bounded.
Take the subsequences $({{a_n}_k}_j)$ and $({{b_n}_k}_j)$, since $({a_n}_k + {b_n}_k)$ converges to $l$, $({{a_n}_k}_j)$ + $({{b_n}_k}_j)$ must also converge to $l$.
So we have $$\lim_{n \rightarrow \infty}({a_n}_k + {b_n}_k)=lim_{n \rightarrow \infty}({{a_n}_k}_j)+\lim_{n \rightarrow \infty}({{b_n}_k}_j) \leq \lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n)$$ which is true by the definition of supremum. (Is there some intuition here?)
Well, any subsequence $(a_n+b_n)$ that converges is bounded above by $\lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n) $, so certainly $\lim_{n \rightarrow \infty} \sup (a_n+b_n)$ is bounded above, or in other words $$\lim_{n \rightarrow \infty} \sup (a_n+b_n) \leq \lim_{n \rightarrow \infty} \sup (a_n)+\lim_{n \rightarrow \infty} \sup (b_n).$$
Questions:
Is this a correct proof?
The inequality seems obvious by itself, because the suprema of two sequences will always be an upper bound of those two sequences added together. The only other thing that can happen is that those suprema are an upper bound for $(a_n+b_n)$, but not the least upper bound - if this is correct, is this a case of 'so easy it's hard' or am I missing an essential piece of the puzzle? None of the proofs I've seen online have made any sense to me, so I'm guessing it's the latter.
The simplest way that I have found to think about the sub-additivity of $\limsup$, is to use the definition $$ \limsup_{n\to\infty}a_n=\lim_{n\to\infty}\sup_{k\ge n}a_k\tag{1} $$ The limit on the right exists since $\sup\limits_{k\ge n}a_k$ is a decreasing function of $n$.
Note that $$ \sup_{k\ge n}\,(a_k+b_k)=\sup_{\substack{j\ge n\\k\ge n\\j=k}}\,(a_j+b_k)\tag{2} $$ and $$ \sup_{k\ge n}a_k+\sup_{k\ge n}b_k=\sup_{\substack{j\ge n\\k\ge n}}\,(a_j+b_k)\tag{3} $$ Since the $\sup$ on the right hand side of $(2)$ is over a smaller set, it will be smaller than that on the right hand side of $(3)$. That is, $$ \sup_{k\ge n}\,(a_k+b_k)\le\sup_{k\ge n}a_k+\sup_{k\ge n}b_k\tag{4} $$ Taking the limit of both sides of $(4)$ yields $$ \limsup_{n\to\infty}\,(a_n+b_n)\le\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n\tag{5} $$