Proof of $\limsup s_n = \lim _{N \rightarrow \infty} \sup \{s_n: n >N \}$. What does $n>{}$natural number mean? Is the solution wrong?

Question

Is the answer key wrong?
Problem $$\text{Let $(s_n)$ be a bounded sequence. Prove that $\limsup s_n = \lim _{N \rightarrow \infty} \sup \{s_n: n >N \}$. }$$

Answer key: Since $\{s_n \}$ is bounded, $m= \limsup s_n \in \mathbb{R}$. We need to show (1) $m \le \limsup \{ s_n: n>N \}$, and (2) $\limsup \{ s_n: n>N \} \le m $.

Proof of (2):
Given $\epsilon>0$, there exists $K \in \mathbb{N}$ such that $s_n<m+ \epsilon$ for all $n>K$ we have $\sup \{ s_n: n> \mathbb{N} \} \le m + \epsilon$. Since this holds for all $\epsilon > 0$, $\sup\{s_n: n>N\} \le m$ whenever $N>K$.

My question: Can someone explain the proof in plain words? In particular,

we have $\sup \{ s_n: n> \mathbb{N} \} \le m + \epsilon$.

what does it mean to have n larger than natural numbers?

Original answer below:

Answer 1

$\sup\{s_n\,:\, n>\Bbb N\}$ is a misprint. It should be $\sup\{s_n\,:\, n>N\}$.