Let's consider only bounded sequences.
Let ($x_n$) be a bounded sequence then I am using the following definition for limsup of the sequence ($x_n$):
limsup($x_n$)= inf $V$=inf{$v: x_n >v$ for at most a finite natural number $n$}=$x^*$, say. Consider the following:
$(a) \Rightarrow (b)$: Since $x^*$ is inf $V$ hence for any $\epsilon \gt 0 , \exists v\in V:$ $x^* \le v\lt x^*+\epsilon \implies x^*\in V \implies $ there can be atmost a finite number of $n\in N$ for which $x^*+\epsilon\lt x_n$. $\forall \epsilon \gt 0 $ $ x^*-\epsilon \notin V$. Hence $x_n \gt x^* -\epsilon$ for infinitely many $n$.
Now my claim is that $(b)$ implies that $x_n$ is convergent, below is the proof:
From $(b)$ above, let $ n=K $ (this K either exists or doesn't exist because of "atmost" condition) be the number for which $x^*+\epsilon\lt x_n$.
(A) $\forall n\ge K$, we have $x_n \lt x^* +\epsilon$
(B) Now we know that $x_n \gt x^* -\epsilon$ for infinitely many $n \implies \exists M \in N : \forall n \ge M$, we have $x_n \gt x^* -\epsilon$
From (A) and (B), for $n \ge L$=sup{$K,M$}, we have
$x^* -\epsilon \lt x_n \le x^* +\epsilon \Rightarrow |x_n-x^*|\lt \epsilon \implies$ $lim (x_n)=x^*$
I know that if $X_n=$ sup{$x_k: k\ge n$}, then $lim (X_n)=x^*=$ inf{$X_m, m\in N$}. That's why the proof above doesn't seem right to me. I am struggling to know what is wrong in the above proof. Please help. Thanks in advance.
The key is that the 2 statements $x^*+\epsilon<x_n$ and $x^*-\epsilon<x_n$ do not cover all the possibilities for the $x_n$. In particular, you don't know anything about the case where $x^*-\epsilon >x_n$. This could be finite or infinite. You can look at the example of $x_n=\sin{n}$ for an example where both regions contain an infinite number of $x_n$.