The question is
Let $(a_k)_k$ $_\in$ $_\mathbb N$ be a bounded sequence of real numbers.
Let b be a real number such that b > lim sup$_k$$_\to$$_\infty$ $a_k$
Prove that there exists N $\in$ $\mathbb N$ such that b > $a_k$ for all k > N.
Here is my attempt:
Denote $d_n$ = sup{$a_k$ | k $\geq$ n}
b > lim sup$_k$$_\to$$_\infty$ $a_k$ = $\lim_{n\to\infty}$ dn
$\exists$ N $\in \mathbb N$ such that $\forall n > N$, $d_n$ < b
$\exists$ $\epsilon>0$ such that b-$\epsilon$ > $\lim_{n\to\infty} dn$ = d [denote d = $\lim_{n\to\infty} dn$]
$\exists$ N $\in \mathbb N$ such that $\forall n \geq N$, |$d_n - d$| < $\epsilon$
$\Rightarrow$ $a_n \leq d_n < \epsilon + d = b$
Substitute n=k, we now have $\Rightarrow a_k < b$
Does this make sense?