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.


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, $|x_m-x_n|<\varepsilon$.

  • $\begingroup$ I am sorry, but supermum is defined later in my book, and the fact that reals satisfy Axiom of Completeness is also not stated before this question, really sorry that I did not mention in my original question. I mentioned in question that Monotone Convergence Theorem also comes later. Because of no tools, like sup or MCT, I have to come here. Will you please give a proof that is 'primitive'? $\endgroup$ – Silent Aug 27 '17 at 12:00
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    $\begingroup$ @Silent Done. But that next time you post a question here, please state all your hypothesis from the start. $\endgroup$ – José Carlos Santos Aug 27 '17 at 12:06
  • $\begingroup$ I will. You cannot imagine how sorry I am, I just forgot to see that sup was defined later! Thank you very much for all your pains. $\endgroup$ – Silent Aug 27 '17 at 12:08
  • $\begingroup$ @Silent If I was helpful, perhaps that you could mark my answer as the accepted one. $\endgroup$ – José Carlos Santos Aug 27 '17 at 12:10
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    $\begingroup$ @Silent Please read again my answer. In the first paragraph, I proved that $(\forall n\in\mathbb{N}):x_n\leqslant y_n$. $\endgroup$ – José Carlos Santos Aug 27 '17 at 15:07

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 $y_n-y_m$.


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