# Thread: Cauchy sequences proof (from Intro to Real Analysis class)

1. ## Cauchy sequences proof (from Intro to Real Analysis class)

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.

2. Because $a_n$ is a Cauchy sequence we have
$\begin{gathered} \left( {\exists N_a } \right)\left[ {m,n \geqslant N \Rightarrow \left| {a_m - a_n } \right| < 1} \right] \hfill \\ \left| {a_n } \right| < 1 + \left| {a_{N_a } } \right| \hfill \\
\end{gathered}$
.
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 \\
\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}
$
.

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

3. 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!

4. 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.

5. Originally Posted by Plato
??
No, you did the sum.
I tried to help you with the product.
Oh, okay. Got it. Thanks again.