Prove that $\lim\limits_{n->\infty}\sup a_n\le q\iff\forall\epsilon>0\: \exists N\in\mathbb{N}:n\ge N\:\forall n\in\mathbb{N}\implies a_n< q+\epsilon$ Assume that $\{a_n\}$ is a bounded sequence and $q \in\mathbb{R}$. Show that $\lim\limits_{n->\infty} \sup a_n \le q \iff \forall \epsilon > 0$ $\exists N\in\mathbb{N}: n\ge N\:\forall n\in\mathbb{N} \implies a_n < q + \epsilon.$ 
Intuitively it all makes sense. However, I have a problem proving it formally.
 A: *

*To see that
$$
 \overline{\lim}_{n\rightarrow\infty} a_n \le q
 $$
implies
$$
 \forall \epsilon > 0\:\exists N\in\{1, 2, \dots\}\:\forall n\in\{N, N+1,\dots\},\  a_n < q + \epsilon,
 $$
assume that $\overline{\lim}_{n\rightarrow\infty} a_n \le q$, and let $\epsilon \in (0,\infty)$. Choose some $N \in \{1, 2, \dots\}$ such that, for all $n \in \{N, N+1, \dots\}$, $\sup \{a_n, a_{n+1}, \dots\} < q+\epsilon$. Then, for all $n \in \{N, N+1, \dots\}$, $a_n < q+\epsilon$.

*To see that
$$
 \forall \epsilon > 0\:\exists N\in\{1, 2, \dots\}\:\forall n\in\{N, N+1,\dots\},\  a_n < q + \epsilon
 $$
implies
$$
 \overline{\lim}_{n\rightarrow\infty} a_n \le q,
 $$
assume that
$$
 \forall \epsilon > 0\:\exists N\in\{1, 2, \dots\}\:\forall n\in\{N, N+1,\dots\},\  a_n < q + \epsilon,
 $$
and let $\epsilon \in (0,\infty)$. Choose some $N \in \{1, 2, \dots\}$ such that for all $n \in \{N, N+1, \dots\}$, $a_n < q + \epsilon$. Then for all $n \in \{N, N+1, \dots\}$, $\sup \{a_n, a_{n+1}, \dots\} \leq q+\epsilon$. Then $\overline{\lim}_{n\rightarrow\infty} a_n \le q + \epsilon$. Since $\epsilon$ was chosen arbitrarily in $(0,\infty)$, $\overline{\lim}_{n\rightarrow\infty} a_n \le q$.
