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 $\displaystyle lim_{n\rightarrow \infty}\{a_n-b_n\}=a-b$ and that $\displaystyle lim_{n\rightarrow \infty}\{a_n-b_n\}\leq0$ since $\displaystyle a_n\leq b_n$, but I'm stuck on proving it. Any help appreciated. Thanks! (Happy)

Here is the problem:

Suppose that $\displaystyle \{a_n\}$ and $\displaystyle \{b_n\}$ are both Cauchy sequences and that $\displaystyle a_n\leq b_n$ for each n. Prove that $\displaystyle lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n$

Re: Analysis Proof Involving Cauchy Sequences

Quote:

Originally Posted by

**UNLVRich** Here is the problem:

Suppose that $\displaystyle \{a_n\}$ and $\displaystyle \{b_n\}$ are both Cauchy sequences and that $\displaystyle a_n\leq b_n$ for each n. Prove that $\displaystyle lim_{n\rightarrow\infty}a_n\leq lim_{n\rightarrow\infty}b_n$

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

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

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

If you can then let $\displaystyle c_n=b_n-a_n~.$

Re: Analysis Proof Involving Cauchy Sequences

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Originally Posted by

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

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

Re: Analysis Proof Involving Cauchy Sequences

$\displaystyle c_n$ is a snake that crawls arbitrarily close to $\displaystyle c$. On the other hand, if $\displaystyle c < 0$, then there is a finite distance between $\displaystyle c$ and 0. Eventually the distance between $\displaystyle c_n$ and $\displaystyle c$ becomes smaller than this distance. What happens then?

Re: Analysis Proof Involving Cauchy Sequences

Quote:

Originally Posted by

**UNLVRich** I've boiled it down to being able to show $\displaystyle \forall n[c_n\ge 0]~\&~\left( {{c_n}} \right) \to c\text{ then }c\ge 0$, 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 $\displaystyle (c_n)\to c$ then almost all of the terms are 'near' $\displaystyle c$.

As was pointed out in reply #4, if $\displaystyle c<0$ that would impossible if $\displaystyle \forall n[c_n\ge 0].$

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 $\displaystyle \epsilon={-c\over2}$. Then there is N such that $\displaystyle c_N-c\leq|c_N-c|<\epsilon$. But this says $\displaystyle c_N<c+\epsilon={c\over2}<0$, a contradiction.

Re: Analysis Proof Involving Cauchy Sequences

Thanks to johng, Plato and emakarov. I will meditate further on this. :)