Prove that $\lim_{n \to \infty} \sup (x_n) =\min\{A \in \Bbb{R} : \forall\epsilon>0 \{n \in \mathbb{N}: x_n \geq A+ \epsilon \} $ set is finite} Any tips solving this?
I think I need to firstly show that $\lim \sup x_n$ is element of the given min{...} set and secondly there are no smaller elements in the given set. 
Should I open the set with minimum definition? Or something like this:
$x_n > a+ \epsilon$ for finite number of sequence $x_n$ elements $\forall \epsilon$...
Thanks!
Edit: sequence $x_n$ is bounded. 
 A: 
Definition $1$ : $\limsup x_n$ is the supremum of the set of all (extended real) limit points of $x_n$ (and $-\infty$ if the set is empty). If real, then it is also a limit point, by the fact that the set of limit points of any set is closed.

If you do not know definition $1$ then proceed to definition $2$. You must be knowing at least one of these, and it is not difficult to show they are equivalent.
The boundedness of $x_n$ ensures that the limit supremum is a real number, hence we need not worry about infinities.
With this in mind, let us call the set $\{a \in\mathbb R : \forall \epsilon > 0 \mbox{,the set } \{n \in \mathbb N : x_n \geq a + \epsilon\} \mbox { is finite} \} = S$. We want to show that $\inf S \geq \limsup x_n$, and that $\limsup x_n \in S$. This will show that $\limsup x_n$ is the minimum of $S$, as desired.

Suppose $t < \limsup x_n$. Let $\epsilon = \frac{\limsup x_n - t}2$ . Then, take the subsequence $x_{n_k}$ which converges to $\limsup x_n$ : note that there exists $N$ such that $K > N $ implies  $x_{n_K} > t + \epsilon$ eventually,  since $t + \epsilon < \limsup x_n$, and $x_{n_k}$ is converging to the right hand side. This forces that $\{n \in \mathbb N : x_n \geq t+\epsilon\}$ is infinite, since $n_N, n_{N+1}$ etc. belong to this set.
Consequently, we have shown that $t \notin S$. In otherwise, for every $s \in S$, we must have $s \geq \limsup x_n$. This also gives $\inf S \geq \limsup x_n$.

We will now show that $\limsup x_n \in S$. Suppose not. Then, there exists $\epsilon > 0$  such that the set $\{n \in \mathbb N : x_n \geq \limsup x_n  + \epsilon\}$ is infinite. Now, take the elements of this set and order them like $m_1,m_2,...$. So we have $x_{m_i} \geq x_n + \epsilon$ for all $i$. The $x_{m_i}$ is a subsequence of $x_n$, hence bounded, but then any limit point of $x_{m_i}$ is a limit point of $x_n$, which must be strictly larger than $\limsup x_n$, since every element of $x_{m_i}$ is larger than $\limsup x_n + \epsilon$. This contradicts the definition of $\limsup x_n$, showing that $\limsup x_n \in S$.
Consequently, we have shown that $\limsup x_n \in S$ and every other element of $S$ is larger than $\limsup x_n$. It follows that $\limsup x_n = \min S$ as desired.


Definition $2$ : $\limsup x_n = \lim_{n \to \infty} (\sup_{k \geq n} x_k)$.

Notation is the same as for definition $1$. We will use the same proof plan as for definition $1$ as well. Note that the sequence $\sup_{k \geq n} x_k$ is  a decreasing sequence, since the set over which supremum is being taken is getting smaller with increasing $n$.

Let $t \in S$. Then, fix $\epsilon > 0$. By definition of being in $S$, we see that $\{n \in \mathbb N : x_n \geq t + \epsilon\}$ is finite. Therefore, $t + \epsilon\geq \sup_{n \geq N+1} x_n$ for $N$ being the maximum of this set. But since $\sup_{k \geq N} x_k$ is decreasing, the right hand side is automatically greater than $\limsup x_n$, which is the limit of the decreasing sequence it is part of. So $t + \epsilon \geq \limsup x_n$, for all $\epsilon$, giving $t \geq \limsup x_n$.

Now, to show that $\limsup x_n \in S$, fix $\epsilon > 0$. Suppose $\{n : x_n \geq \limsup x_n + \epsilon\}$ is infinite. Order the elements of this set as $m_1,m_2,...$, then note that $\sup_{k \geq m_i} x_{k}\geq x_{m_i} \geq \limsup x_n + \epsilon$ for all $i$.  As a subsequence of the sequence $\sup_{k \geq n} x_k$ , this sequence must also converge to $\limsup x_n$, but it cannot be always greater than $\limsup x_n + \epsilon$, so there lies the contradiction.
Hence, we complete our proof in the same way.
