Let \(\{a_k\}_{k=1}^\infty\) be a bounded sequence of real numbers and set \[ E:=\left\{s\in\mathbb R:\{k\in\mathbb N:a_k\ge s\}\text{ is finite}\right\}. \] Prove that \[ \limsup_{k\to\infty}a_k=\inf E. \]
Solution
Let \[ L:=\limsup_{k\to\infty}a_k =\lim_{N\to\infty}\sup_{k\ge N}a_k. \] Since \((a_k)\) is bounded above, \(E\ne\varnothing\): every number larger than \(\sup_k a_k\) lies in \(E\).
1. \(L\le \inf E\).
Fix \(s\in E\). By definition of \(E\), there is \(N\) such that \[ a_k<s\qquad(k\ge N). \] Therefore \[ \sup_{k\ge N}a_k\le s. \] Since the tail suprema decrease to \(L\), \[ L\le \sup_{k\ge N}a_k\le s. \] Thus \(L\) is a lower bound for \(E\), and hence \(L\le\inf E\).
2. \(\inf E\le L\).
Let \(\varepsilon>0\). Since \[ \sup_{k\ge N}a_k\downarrow L, \] there is \(N\) such that \[ \sup_{k\ge N}a_k<L+\varepsilon. \] Hence \[ a_k<L+\varepsilon\qquad(k\ge N), \] so only the finitely many indices \(k<N\) can satisfy \(a_k\ge L+\varepsilon\). Therefore \[ L+\varepsilon\in E. \] It follows that \[ \inf E\le L+\varepsilon. \] Letting \(\varepsilon\downarrow0\) gives \(\inf E\le L\).
Combining the two inequalities, \[ \boxed{\limsup_{k\to\infty}a_k=\inf E}. \]