How to prove that these are Cauchy sequences? 
Let $x_1,x_2,x_3,\dots$ is a non-decreasing and $y_1,y_2,y_3,\dots$ is a non-increasing sequence, and they are real sequences. If $|x_n-y_n|\le\frac{|x_1-y_1|}{2^n}$ for each $n\in \Bbb N$, then show that $\{x_n\}$ and $\{y_n\}$ are equivalent Cauchy sequences.

I can show that these sequences are equivalent. But, I could not show that they are Cauchy. I tried using Monotone Convergence Theorem, but then noticed that it is proven later! 
I can see how each consecutive term of sequence, e.g., $\{x_n\}$ come closer, but can't show that every term beyong a point can come as close we want. 
My book defines greatest lower bound after this.
 A: You need to have $(\forall n\in\mathbb{N}):x_n\leqslant y_n$. In fact, if $x_n>y_n$ for some $n$, then, since $(x_n)_{n\in\mathbb N}$ is non-decreasing and $(y_n)_{n\in\mathbb N}$ is non-increasing, then $m\geqslant n\Longrightarrow|x_m-y_m|\geqslant|x_n-y_n|$, which is impossible, since $\lim_{n\to\infty}|x_n-y_n|=0$.
So, the sequence $(x_n)_{n\in\mathbb N}$ is non-decreasing and it is bounded above by $y_1$. Therefore it converges and this implies that it is a Cauchy sequence. A similar argument proves that the sequence $(y_n)_{n\in\mathbb N}$ is a Cauchy sequence.
A direct proof of the fact that $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence can be obtained as follows: since $(x_n)_{n\in\mathbb N}$ is bounded above, you can consider the number$$s=\sup\{x_n\,|\,n\in\mathbb N\}$$Take $\varepsilon>0$. Pick $p\in\mathbb N$ such than $s-x_p<\varepsilon$. If $m,n\geqslant p$, then $x_m,x_n\in[x_p,s]$ and therefore $|x_m-x_n|<\varepsilon$.

Here is another proof. Take $\varepsilon>0$. Pick $p\in\mathbb N$ such that $\frac{|x_1-y_1|}{2^p}<\varepsilon$. Then, if $m,n\geqslant p$, both $x_m$ and $x_n$ belong to $[x_p,y_p]$. Therefore, $|x_m-x_n|<\varepsilon$.
A: Assume that $m>n$. Then $$x_n\leq x_m\leq y_m\leq y_n\ ,\tag{1}$$ because $\delta:=x_m-y_m>0$ would imply $x_k-y_k\geq \delta$ for all $k\geq m$, violating the assumption that  $\lim_{k\to\infty}|x_k-y_k|=0$.
Now let an $\epsilon>0$ be given. There is an $n_0$ such that ${\displaystyle{|x_1-y_1|\over 2^{n_0}}<\epsilon}$. If $m>n\geq n_0$ it then follows from $(1)$ that
$$0\leq x_m-x_n\leq y_n-x_n\leq{|x_1-y_1|\over 2^n}<\epsilon\ ,$$
and similarly for $y_n-y_m$.
