How can you prove that $x_n$ is a constant sequence if and only if $x_n$ is both monotone increasing and monotone decreasing? I know this statement makes sense if the sequence were simply both increasing and decreasing without being monotone, that is, $x_{n+1} \geq x_n$ and $x_{n+1} \leq x_n.$ 
Actually, I am now confused how we know  $x_{n+1} \geq x_n$ and $x_{n+1} \leq x_n$ implies that $x_{n+1} = x_n$. 
 A: Hint: Logically, $x_{n+1}\geq x_n$ means $x_{n+1}>x_n$ or $x_{n+1}=x_n$. 
A: You can write $x_{n+1}\geq x_n$ and $x_{n+1}\leq x_n$ like so:
$x_n\leq x_{n+1}\leq x_n$ so $x_n=x_{n+1}$.
A: I believe it's called an antisymmetric relation - if for two two real numbers x and y, both inequalities x≤y and y≤x hold then x=y. 
In your case, if xn is both monotone increasing and decreasing, from their definitions, xn ≤ x(n+1) and x(n+1) ≤ x both must hold - hence xn = x(n+1). There's no other way the two inequalities would hold, hence xn must be constant. 
A: You question can be essentially answered by the following fact.

Suppose $a,b\in\mathbb{R}$. Then the following are equivalent:

*

*(1) $a\leq b$ and $a\geq b$;

*(2) $a=b$.


To see how (1) implies (2), suppose (2) is not true. Then one must have $a<b$ or $a>b$. What contradiction can you get?
To see how (2) implies (1), all you need to do is recalling that $a\leq b$ means $a<b$ or $a=b$.
A: You can argue like this:
If there is an $n$
such that
$x_n \ne x_{n+1}$,
then either,
by tricotomy,
$x_n > x_{n+1}$
or
$x_n < x_{n+1}$.
But
$x_n > x_{n+1}$
contradicts the assumption that
$x_n \le x_{n+1}$.
Similarly,
$x_n < x_{n+1}$
contradicts the assumption that
$x_n \ge x_{n+1}$.
Therefore,
$x_n = x_{n+1}$
for all $n$.
