TITLE: Prove that subsequence converges to limsup
QUESTION [12 upvotes]: Given a sequence of real numbers, $\{ x_n \}_{n=1}^{\infty}$, let $\alpha =$ limsup$x_n$ and $\beta = $ liminf$x_n$. 
Prove that there exists a subsequence $\{ x_{n_k}\}$ that converges to $\alpha$ as $k \rightarrow \infty$. 
Not sure how to start this without since I'm not given that the subsequence is bounded..

REPLY [2 votes]: $A:=\lim_{n \rightarrow \infty} \sup a_n=\lim_{n \rightarrow \infty} A_n$, where $A_n=\sup ${$a_k|k \ge n$};
Let  $L$ be the set of all limits of convergent subsequences.of $(a_n)$.
1) We show that  $\forall \epsilon >0$ there exists an $a_n$ s.t. $|a_n-A| <\epsilon$.
Since $\lim_{n \rightarrow \infty} A_n =A$, there is a $n_0$ s.t. for $m \ge n_0$
$|A_m -A|<\epsilon$.
2)Since
$A_{n_0}= \sup${$a_k| k\ge n_0$}, there is 
a $k  \ge n_0$ s.t. 
$|a_k-A_{n_0}| <\epsilon$.
Set $m=k$ in 1), and we get
$|a_k-A| \le |a_k-A_{n_0}| +|A_{n_0}-A| <2\epsilon$.
2) Given that in any $\epsilon$ neighbourhood of $A$ there is an $a_n$, we can construct a subsequence $a_{n_k}$ that converges to $A$, i.e. $A \in L$. (JohnD's answer)