# Analysis Proof Involving Cauchy Sequences

• Jun 24th 2013, 03:56 PM
UNLVRich
Analysis Proof Involving Cauchy Sequences
So I teach high school math, but I love to study math in my spare time. I'm charging my way through an analysis book, trying to solve all of the problems. Been looking at this one for a couple of days now and I need to get un-stuck. It's clear to me that $lim_{n\rightarrow \infty}\{a_n-b_n\}=a-b$ and that $lim_{n\rightarrow \infty}\{a_n-b_n\}\leq0$ since $a_n\leq b_n$, but I'm stuck on proving it. Any help appreciated. Thanks! (Happy)

Here is the problem:
Suppose that $\{a_n\}$ and $\{b_n\}$ are both Cauchy sequences and that $a_n\leq b_n$ for each n. Prove that $lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n$
• Jun 24th 2013, 04:46 PM
Plato
Re: Analysis Proof Involving Cauchy Sequences
Quote:

Originally Posted by UNLVRich
Here is the problem:
Suppose that $\{a_n\}$ and $\{b_n\}$ are both Cauchy sequences and that $a_n\leq b_n$ for each n. Prove that $lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n$

Because $\mathbb{R}$ is complete we know that $\left( {{a_n}} \right) \to a\;\& \;\left( {{b_n}} \right) \to b~.$

All you need to show is that $a\le b$.

Have you (can you) shown that if $\forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0~?$

If you can then let $c_n=b_n-a_n~.$
• Jun 24th 2013, 08:32 PM
UNLVRich
Re: Analysis Proof Involving Cauchy Sequences
Quote:

Originally Posted by Plato
Have you (can you) shown that if $\forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0~?$

I've boiled it down to being able to show $\forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0$, but there is something that I'm missing. I'm not sure how to show that.
• Jun 25th 2013, 04:15 AM
emakarov
Re: Analysis Proof Involving Cauchy Sequences
$c_n$ is a snake that crawls arbitrarily close to $c$. On the other hand, if $c < 0$, then there is a finite distance between $c$ and 0. Eventually the distance between $c_n$ and $c$ becomes smaller than this distance. What happens then?
• Jun 25th 2013, 05:02 AM
Plato
Re: Analysis Proof Involving Cauchy Sequences
Quote:

Originally Posted by UNLVRich
I've boiled it down to being able to show $\forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0$, but there is something that I'm missing. I'm not sure how to show that.

I take you at your word that you are self studying basic analysis. Here is very useful concept.
First, "almost all" means all but a finite collection.

So if $(c_n)\to c$ then almost all of the terms are 'near' $c$.
As was pointed out in reply #4, if $c<0$ that would impossible if $\forall n[c_n\ge 0].$
• Jun 25th 2013, 07:34 AM
johng
Re: Analysis Proof Involving Cauchy Sequences
Hi,
Maybe this will help. Try proof by contradiction. Namely, assume c < 0. Then in the definition of limit take $\epsilon={-c\over2}$. Then there is N such that $c_N-c\leq|c_N-c|<\epsilon$. But this says $c_N, a contradiction.
• Jun 25th 2013, 10:28 AM
UNLVRich
Re: Analysis Proof Involving Cauchy Sequences
Thanks to johng, Plato and emakarov. I will meditate further on this. :)