Prove that subsequence converges to limsup 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..
 A: I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it. 
Let us define the terms first. 
$$\lim\sup s_n :=\lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$$
Let $A_N := \{s_n:n>N\},~~ v_N := \sup A_N.$
Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that 
$A_i \subseteq A_j$ when $i>j$. 
$\therefore \sup A_i \leq \sup A_j \Rightarrow v_N$ is a non-increasing (monotonous) sequence.
Thus, $\lim \sup s_n$ is the infimum of the set of $\{v_N\}_{N=1}^\infty.$ 
$$\inf v_N = \lim\sup s_n \tag{d1}$$
$$\therefore~~ \exists \ v_N: v_N < \lim \sup s_n + \epsilon \ \forall \epsilon > 0\tag 1$$
Using the definition of 
$v_N$,
$$\exists s_n: s_n > v_N - \alpha ~~ \forall \alpha > 0\tag 2$$
Using (d1) and (1),
$$\lim \sup s_n \leq v_N < \lim \sup s_n + \epsilon ~~ \forall \epsilon>0 \tag a$$
Using (2) and previous definition of $v_N$,
$$v_N \geq s_n > v_N - \alpha~~ \forall \ \alpha > 0 \tag b$$
Using (a) and (b),
$ \limsup s_n - \alpha \leq v_N - \alpha < s_n \leq v_N < \lim \sup s_n + \epsilon$ where $\alpha,\epsilon$ are arbitrary and can be taken as small as needed. This shows that $s_n$ exist such that they tend to $\lim_{N\rightarrow \infty}v_N$, i.e. $\lim \sup s_n$.
A: $A:=\lim_{n \rightarrow \infty} \sup a_n=\lim_{n \rightarrow \infty} A_n$, 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)
A: Since $\alpha=\limsup x_n$, by the definition of $\limsup$, there is some $x_{n_1}$ with $|x_{n_1}-\alpha|<{1\over 2}$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_{n_2}$ with $|x_{n_2}-\alpha|<{1\over 2^2}$. Continuing, for each $k\in\mathbb{N}$, there is some $x_{n_k}$ with $|x_{n_k}-\alpha|<{1\over 2^k}$.
Then $\{x_{n_k}\}\subset \{x_n\}$ and $x_{n_k}\to \alpha$ as $k\to\infty$.
