Suppose {$x_n$} is a bounded sequence. Defind $a_n$ and $b_n$ as in Definition 2.3.1. Show that {$a_n$} and {$b_n$} are bounded. Suppose {$x_n$} is a bounded sequence. Defind $a_n$ and $b_n$ as in Definition 2.3.1. Show that {$a_n$} and {$b_n$} are bounded. 

no idea to solve this question.. 
 A: Since $x_n$ is a bounded (by this, we mean above and below) sequence, the sequence $\{x_k : k \geq n\}$ is subsequence of $x_n$, hence is also bounded. Therefore, if we define $a_n = \sup \{x_k : k \geq n\}$, then this sequence exists and is bounded. Furthermore, if $n \geq m$, then we see that $a_n \leq a_m$, as the size over the set in which we are taking supremum has become smaller.
Now, $a_n$ is a sequence, which is bounded and decreasing. Hence, it follows that $a_n$ is a convergent sequence (tell me if you don't know how  this follows, but first try an $\epsilon-\delta$ argument yourself).
Similarly, the sequence $b_n$ will also be convergent, because it will be bounded and increasing (because we are taking infimum,  not supremum here). 
Hence, not only are $a_n,b_n$ bounded, but the associated limits also exist (and are called limit superior and limit inferior respectively). 
EDIT:
In response to comment, we first need a candidate for the limit. Since $a_n$ is decreasing and bounded, we take the candidate for the limit as $L = \inf a_n$.
Now, pick $\epsilon > 0$. We know that by definition of infimum, $L+\epsilon$ is not a lower bound, so there exists $N$ such that $a_N < L+\epsilon$. However, because $a_n$ is decreasing, $m > N \implies a_m \leq a_N < L + \epsilon$. By definition of limit convergence, we get the limit as $L$.
You can do a similar thing for $b_n$, but the candidate limit will be $\sup b_n$, since the $b_n$ are increasing.
