Cauchy sequences proof (from Intro to Real Analysis class)

**ilikedmath** · November 15th 2008, 07:29 AM

Problem:
If { $a_n$ } and { $b_n$ } are Cauchy sequences in $\mathbb{R}$ , prove that { $a_n + b_n$ } and { $a_nb_n$ } are also Cauchy sequences.
You cannot use the Theorem 2.6.4 which states, "Every Cauchy sequence of [tex]\mathbb{R}/MATH] converges.

My work:
We can't use Theorem 2.6.4, but Theorem 2.6.2 says, "Every convergent sequence of $\mathbb{R}$ is a Cauchy sequence." So in order for me to prove { $a_n + b_n$ } and { $a_nb_n$ } are also Cauchy sequences, I need to prove that { $a_n + b_n$ } and { $a_nb_n$ } converge.

I found this:

But our professor told us to prove this claim directly by using $\epsilon$ , and think that proof above uses the theorem 2.6.4 which we are not allowed to use.

My work so far:
Assume { $a_n$ } and { $b_n$ } are Cauchy sequences in $\mathbb{R}$ . Thus $\forall$ $\epsilon$ > 0, $\exists$ $n_o$ $\in$ $\mathbb{N}$ such that $|a_n - a_m|$ < $\epsilon$ $\forall$ $n,m \geq$ $n_o$ . And also, $\forall$ $\epsilon$ > 0, $\exists$ $k_o$ $\in$ $\mathbb{N}$ such that $|b_v - b_w|$ < $\epsilon$ $\forall$ $v,w \geq$ $k_o$ .

Let $\epsilon$ > 0. Since $|a_n - a_m|$ < $\frac{\epsilon}{2}$ , also $|b_v - b_w|$ < $\frac{\epsilon}{2}$ . Thus we have $|a_n - a_m + b_v - b_w|$ $\leq$ $|a_n - a_m|$ + $|b_v - b_w|$ < $\frac{\epsilon}{2}$ + $\frac{\epsilon}{2}$ = $\epsilon$ .

This is where I got stuck/confused: How to prove { $a_nb_n$ } converges using epsilon because when I multiply $|a_n - a_m|$ $|b_v - b_w|$ , I get lost because I'm seeing too many different subscripts

. How do I go about finishing the proof using epsilons to proof convergence so I can say the sequences are Cauchy sequences?

Any help, suggestions, corrections, and tips are greatly appreciated.
Thank you for your time.

**Plato** · November 15th 2008, 09:19 AM

Because $a_n$ is a Cauchy sequence we have
$\begin{gathered} \left( {\exists N_a } \right)\left[ {m,n \geqslant N \Rightarrow \left| {a_m - a_n } \right| < 1} \right] \hfill \\ \left| {a_n } \right| < 1 + \left| {a_{N_a } } \right| \hfill \\ <br /> \end{gathered}$ .
In other words, Cauchy sequences are bounded.
So your sequences are bounded. $\left( {\forall n} \right)\left[ {\left| {a_n } \right| \leqslant A\,\& \left| {b_n } \right| \leqslant B\,} \right]$ .

Now,
$\begin{gathered} \varepsilon > 0\,\& \,C = A + B \hfill \\<br /> \left( {\exists N_a } \right)\left[ {m,n \geqslant N_a \Rightarrow \left| {a_m - a_n } \right| < \frac{\varepsilon }{{3C}}} \right] \hfill \\ \left( {\exists N_b } \right)\left[ {m,n \geqslant N_b \Rightarrow \left| {b_m - b_n } \right| < \frac{\varepsilon }{{3C}}} \right] \hfill \\ N = N_a + N_b \hfill \\ \end{gathered} <br />$ .

$\begin{gathered} n,m \geqslant N \hfill \\ \left| {a_n b_n - a_m b_m } \right| \leqslant \left| {a_n b_n - a_m b_n } \right| + \left| {a_m b_n - a_m b_m } \right| < \left| {a_m - a_n } \right|\left| {b_n } \right| + \left| {b_m - b_n } \right|\left| {a_m } \right| < \varepsilon \hfill \\ <br /> \end{gathered}$

**ilikedmath** · November 16th 2008, 01:32 PM

So what I did was not a legitimate approach to proving the problem?
I didn't look at the second part of Theorem 2.6.2 which states "Every Cauchy sequence is bounded." And that's what you used.

Thanks!

**Plato** · November 16th 2008, 01:38 PM

Originally Posted by ilikedmath

So what I did was not a legitimate approach to proving the problem?

??
No, you did the sum.
I tried to help you with the product.

**ilikedmath** · November 16th 2008, 01:40 PM

Originally Posted by Plato

??
No, you did the sum.
I tried to help you with the product.

Oh,

okay. Got it. Thanks again.

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