I'm trying to teach myself some basic topology by self-studying from Intro to topology by Mendelson. I'm stuck on one of the exercises and can't figure out how to proceed with the proof.
The question is as below:
Let $a_1,a_2,...$ be a bounded sequence of real numbers. Since each of the sets $A_n =\{ a_n,a_{n+1},...\}$ is bounded, we may set $v_n = g.l.b. A_n$, $u_n = l.u.b. A_n$. Observe that $v_n \leq u_n$; $v_1,v_2,...$ is monotone non-decreasing and bounded above; and $u_1,u_2,...$ is monotone non-increasing and bounded below. Let $V=\lim_{n}\ v_n$ and $U=\lim_{n}\ u_n$.
Prove that there are subsequences of $a_1, a_2,....$ which converge to $U$ and $V$ respectively (thus a bounded sequence of real numbers has a convergent subsequence). Prove that $a_1,a_2,...$ converges if and only if $U=V$.
Any hints/tips would be much appreciated.
Thanks
Note that
$$U = \lim_{n \rightarrow \infty}u_n = \lim_{n \rightarrow \infty}\sup_{k\geq n}a_k,\\V = \lim_{n \rightarrow \infty}v_n = \lim_{n \rightarrow \infty}\inf_{k\geq n}a_k.$$
These are definitions of $\limsup a_n$ and $\liminf a_n.$
Since $u_n$ is non-increasing,
$$\lim_{n \rightarrow \infty}\sup_{k\geq n}a_k= \inf_{n}\sup_{k\geq n}a_k$$
For any $\epsilon >0$ there exists $N \in \mathbf{N}$ such that for $n \geq N$
$$a_n < U + \epsilon.$$
Since $U-\epsilon < \sup_{k\geq m}a_k$, for each $m \geq N$, there exists $k_m \geq m$ such that
$$U - \epsilon < a_{k_m}.$$
Hence, $|a_{k_m} - U| < \epsilon$ when $m \geq N$ and the subsequence $(a_{k_m}$) converges to $U$.
Make a similar argument for $V$.
If $U = V=L$ then there exist $N_U,N_V \in \mathbf{N}$ such that if $n \geq \max(N_U,N_V)$
$$L-\epsilon= V-\epsilon < a_n < U + \epsilon= L + \epsilon,$$
and $a_n$ converges to $L$.