\begin{array}{l}{\text { Theorem } 2-17: \text { Let }\left\{a_{n}\right\} \text { be a bounded sequence of real numbers. Then: }} \\ {\begin{array}{lll}{\text { (a) } \limsup {a_{n}}= L \text { if and only if, for any } \varepsilon>0, \text { there are infinitely many terms of }} \\ {\left\{a_{n}\right\} \text { in }(L-\varepsilon, L+\varepsilon) \text { but only finitely many terms of }\left\{a_{n}\right\} \text { with } a_{n}>L+\varepsilon .}\end{array}}\end{array}
Now the author proves directly that $\limsup\{a_n+b_n\} \leq \limsup\{a_n\}+\limsup\{b_n\}$.
\begin{array}{l}{\text { By Theorem } 2-17, \text { there are at most finitely many terms of }\left\{a_{n}\right\} \text { that exceed }} \\ {K+\varepsilon / 2 \text { . Call those that do exceed } K+\varepsilon / 2 \text { the } n_{1}, \ldots, n_{n} \text { terms. At most finitely }} \\ {\text { many terms of }\left\{b_{n}\right\} \text { exceed } L+\varepsilon / 2 . \text { Call those that do exceed } L+\varepsilon / 2 \text { the } m_{1}, \ldots, m_{s}} \\ {\text { terms. Now if } n \neq n_{1}, \ldots, n_{r}, m_{1}, \ldots, m_{s} \text { , then }} \\ {\qquad a_{n}+b_{n}<\left(K+\frac{\varepsilon}{2}\right)+\left(L+\frac{\varepsilon}{2}\right)=K+L+\varepsilon . }\end{array}
I was wondering how this might be proven indirectly, say, by contradiction. Could someone start me out on the proof? It is not as obvious how to proceed using an indirect argument
Suppose that exists $c>0$ such that $$\limsup_n a_n+b_n=\limsup_n a_n+\limsup_n b_n+c$$ call $A:= \limsup_n a_n,\ \ B:=\limsup_n b_n$, then by the theorem exists $n_0$ such that $$\forall n>n_0:\ b_n\le B$$ so $$\forall n>n_0:\ a_n+b_n\le B+a_n$$ then, $$c=\limsup_n a_n+b_n-B-A=\limsup_{n>n_0} a_n+b_n-A-B\le \limsup_n a_n+B-B-A=$$ $$=\limsup_n a_n-A=0$$ contradiction