limsup and liminf and it's relation to number of the indices What I mean is that, I'm trying to understand the theorem :
let $x_n$ be a bounded real sequence and set $B=\limsup{x_n}$. for any $e>0$, 
1) $x_n≤B+e$ holds for all but a finite number of $n$, 
2) $x_n≥B-e$ holds for infinitely many $n$
My approach to understand this :
1) We sure know that for an open interval centered at $B$ with a radius of $e$, for any $e>0$, infinitely many $x_n$ exist. We also know that $\limsup{x_n}$ doesn't mean the maximum value of $x_n$. So we know that there are points of $x_n$ (could be) on both left side / right side of the interval $(B-e,B+e)$ If $x_n≤B+e$ We can't say that all $x_n$ satisfies this. on the interval $(a,B+e)$ where $a$ is the greatest lower bound for $x_n$ there are definitely all but a finite number of $x_n$.
2) First of all I'm not sure what is meant by infinitely many n, does it mean that we should contain all n? Or does it mean that some infinite subset is enough, I couldn't get what is meant by "infinitely many $n$" if someone could explain that to me it would be great.
Need your comments on both of my approach, please. thanks.
 A: Your first remark (1) actually proves statement (2)! You've argued correctly that there are points of $\{x_n\}$ (indeed, infinitely many points) somewhere in the interval $(B-e,B+e)$; in particular, those infinitely many points are all greater than $B-e$. Note that this proof works for any limit point $B$ of $\{x_n\}$, not just the lim sup.
To prove statement (1), we need to incorporate the fact that $B$ is the greatest limit point of $\{x_n\}$. So suppose (1) is false, so that $x_n>B+e$ holds for infinitely many $n$. Those infinitely many points still form a bounded set, and so they have a limit point, which is then a limit point of the original sequence. However, since all of those points exceed $B+e$, the limit point we just found is $\ge B+e$—contradicting the fact that $B$ is the greatest limit point.
A: Let $s_n = \sup_{k \ge n} x_n$. Note that $s_n$ is non increasing and 
bounded below, hence has limit, and we have $B= \lim_{n \to \infty} s_n$.
Pick some $\epsilon>0$.
Then there must be some $N$ such that for $n \ge N$, we have $s_n \le B + \epsilon$. By definition of $s_n$, we have
$x_n \le B+ \epsilon$ for all $n \ge N$.
Hence (1) holds for all $n$, except possibly $1,...,N-1$.
In a similar manner, there is some $N$ such that for $n \ge N$, we have $s_n \ge B - {1 \over 2}\epsilon$ (the ${1 \over 2}$ is there for a reason).
By definition of $s_n$, there is some $k_n \ge n$ such that
$x_{k_n} \ge s_n - {1 \over 2}\epsilon$, and so
$x_{k_n} \ge B -\epsilon$.
Then it follows that the set $I = \{ k_N, k_{N+1},... \}$ is not finite, otherwise
we could pick $n' = \max(k_N, k_{N+1},...)+1$ and find a $k_{n'} \ge n'$,
which would be a contradiction. Hence the set $I$ is infinite. In particular,
there are an infinite number of $m \in I$ such that $x_{m} \ge B -\epsilon$.
