Exercises
[0B6] (Solved on 2022-11-24) Let \(a_ n\) be a real-valued sequence, for \(nβ I\) a set of indexes; let \(r{\gt}0,tββ,π{\lt}0\); show that
\[ \sup _{nβ I}(a_ n+t)=t+\sup _{nβ I}a_ n~ ~ ,~ ~ \sup _{nβ I}(r a_ n)=r \sup _{nβ I}a_ n~ ~ ,~ ~ \sup _{nβ I}(π a_ n)= π \inf _{nβ I}a_ n~ ~ . \]1