Let $ (a_n)_{n\in \mathbb{N}} $ a bounded sequence and $ a\in \mathbb{R} $. Then $$ \lim\sup\limits_{n\to\infty}a_n=a $$ if and only if the following applies for all $ \varepsilon>0 $:
i) $ a_n<a+\varepsilon $ for all but finitely $ n\in \mathbb{N} $
ii) $ a_n>a-\varepsilon $ for infinitely $ n\in \mathbb{N} $.
Here is a proof which I cannot comprehend and I don't have an own idea how I could proof the upper statement:
The proof
Let $ g_k:=\sup_{n\geq k} a_n $ for all $ k\in \mathbb{N} $.
Show ,,=>''. Let $ \varepsilon>0 $ arbitrary. Because of $ a=\lim\sup\limits_{n\to\infty}a_n=\lim\limits_{k\to\infty}g_k $ there exists $ N_{\varepsilon}\in \mathbb{N} $ with $$ a-\varepsilon\leq g_k\leq a+\varepsilon $$ for all $ k\geq N_{\varepsilon} $. Because of $ a_n\leq g_{N_{\varepsilon}} $ for all $ n\geq N_{\varepsilon} $ we get i). $ (g_k) $ is monotonous decreasing.
So far so good. Until here I undrstand everything. But for ii) I get stuck:
1.)
Suppose ii) is wrong. Then $ a_n>a-\varepsilon $ is true for finitely $ n\in \mathbb{N} $ and so there exists an explicit $ N\in \mathbb{N} $ with $ a_n\leq a-\varepsilon $ for all $ n\geq N $.
Why??!!! This negation looks so weird to me especially the part $ a_n\leq a-\varepsilon $ for all $ n\geq N $. I don't see any reason why there exists an explicit $ N\in \mathbb{N} $ with $ a_n\leq a-\varepsilon $ for all $ n\geq N $.
Infinitely $ n\in \mathbb{N} $ means for me that there is a set $ M\subseteq \mathbb{N} $ with $ |M|=\infty $ such that $ a_{n_M}>a-\varepsilon $ for all $ n_M\in M $. For example $ M $ could look like this $ M=\mathbb{N}\setminus{\{1,5,42,99\}} $. Now if I take the same $ N $ from above: What does guarantee $ N > 1,5,42,99 $ such that $ a_n\leq a-\varepsilon $ for all $ n\geq N $?
2.)
This implies $ g_k\leq a-\varepsilon $ for all $ k\geq N $ because $ a-\varepsilon $ is an upper bound and $ g_k $ a supremum. But then we get $$ a=\lim_{k\to \infty} g_k\leq a-\varepsilon $$ which is a contradiction because of $ \varepsilon>0 $ was given arbitrary.
If accept the part 1.) then I clearly understand 2.) and I understand the argumentation in part 2.).
3.)
Show ,,<=''. Let $ \varepsilon>0 $ arbitrary. Then especially with ii) it follows $ g_k>a-\varepsilon $ for all $ k\in \mathbb{N}. $
Again: The same problem for me like in 1.). How do we know that?
4.)
This shows $ \lim\limits_{k\to \infty}g_k\geq a $.
Because of i) for all $ \varepsilon>0 $ there exists $ N_{\varepsilon}\in \mathbb{N} $ such that for all $ n\geq N_{\varepsilon} $ it follows $ a_n<a+\varepsilon $ which implies $ g_k\leq a+\varepsilon $ for all $ k\geq N_{\varepsilon} $. With that get $ \lim\limits_{k\to \infty} g_k\leq a+\varepsilon $ and inclusion $ \lim\limits_{k\to \infty} g_k=a $.
Part 4.) is clear for me.