How to prove that lim sup $a_{n} \leq b$

Question

Assume that $(a_{n})$ is a bounded sequence, prove that lim sup $a_{n} \leq b$ iff, for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ so that $n \geq N$ implies $a_{n} \leq b + \epsilon$

First of all, I'm a little unclear on what $b$ is, so I just assumed that $b \in \mathbb{R}$.

Proof: ($\Rightarrow$)

Assume that lim sup $a_{n} = a$, then by definition, for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $n \geq N$ implies that $a_{n} < a + \epsilon$ And given that $a \leq b$, it holds that $a_{n} < b + \epsilon$. Is this correct?

($\Leftarrow$) If for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ so that $n \geq N$ implies $a_{n} \leq b + \epsilon$, then...I'm not sure as how to continue from here.

Answer 1

Continuing your argument on ($\Leftarrow$):

..$\limsup a_n\leq b+\epsilon$. This is true for each positive $\epsilon$ so automatically implies that $\limsup a_n\leq b$