Show that $a = \limsup_{n\to\infty} a_n$.

Question

Suppose $a \in \mathbb{R}$ is such that:

1. given any $ε>0$ there exists $n_0 \in \mathbb{N}$ such that $a_n \le a+\varepsilon$ for all $n \ge n_0$

2. there is $k\ge n_0$ for which $a−\varepsilon<a_k$.

Then, prove that $a=\limsup_{n\to\infty}a_n$

It looks like a theorem and is obviously true, but I have no idea how to prove it and where to start. Any help would be appreciated.

Answer 1

Hint

Let $y_n=\sup_{k\geq n}a_k$. So you have to show that $$a=\lim_{n\to\infty }y_n.$$

Notice that $(y_n)$ is decreasing.