Is the Sum of Two Cauchy Sequences Also Cauchy?

[cragar]

Homework Statement

Assume $x_n$ and $y_n$ are Cauchy sequences.
Give a direct argument that $x_n+y_n$ is Cauchy.
That does not use the Cauchy criterion or the algebraic limit theorem.
A sequence is Cauchy if for every $\epsilon>0$ there exists an
$N\in \mathbb{N}$ such that whenever $m,n\geq N$
it follows that $|a_n-a_m|< \epsilon$

The Attempt at a Solution

Lets call $x_n+y_n=c_n$
now we want to show that $|c_m-c_n|< \epsilon$
Let's assume for the sake of contradiction that
$c_m-c_n> \epsilon$
so we would have
$|x_m+y_m-x_n-y_n|> \epsilon$
$x_m> \epsilon+y_n-y_m$
since $y_n>y_m$
and we know that $x_m< \epsilon$
so this is a contradiction and the original statement must be true.

 

[Office_Shredder]

Homework Statement

Assume $x_n$ and $y_n$ are Cauchy sequences.
Give a direct argument that $x_n+y_n$ is Cauchy.
That does not use the Cauchy criterion or the algebraic limit theorem.
A sequence is Cauchy if for every $\epsilon>0$ there exists an
$N\in \mathbb{N}$ such that whenever $m,n\geq N$
it follows that $|a_n-a_m|< \epsilon$

The Attempt at a Solution

Lets call $x_n+y_n=c_n$
now we want to show that $|c_m-c_n|< \epsilon$

It's always good to be careful with wording. We want to show that if n and m are big enough, that this inequality holds.

since $y_n>y_m$

This is probably not true, especially since you haven't even said what n and m are besides arbitrary numbers!

and we know that $x_m< \epsilon$

This is also probably not true since there's no reason to think the limit is zero

To get the contradiction you're going to want to use the triangle inequality on |(x_n-x_m)+(y_n-y_m)|

 

[cragar]

ok thanks for your response.
So I take $|(x_n-x_m)+(y_n-y_m)| \leq |x_n-x_m|+|y_n-y_m|$
let's assume that $|x_n-x_m|+|y_n-y_m| > \epsilon$
I am going to rewrite it as $A+B> \epsilon$
so now we have $A> \epsilon -B$
Can I just say this since we know that $A< \epsilon$ and $B< \epsilon$ since $\epsilon$ can be any number bigger than zero, then both of these values should be less than $\frac{\epsilon}{2}$
therefore $A+B< \epsilon$
I have a feeling my last step is not ok

 

[Office_Shredder]

Can I just say this since we know that $A< \epsilon$ and $B< \epsilon$ since $\epsilon$ can be any number bigger than zero, then both of these values should be less than $\frac{\epsilon}{2}$
therefore $A+B< \epsilon$

This is the crux of the argument. It's not the whole proof of course - A and B aren't always that small. Feel free to post a full proof if you want it checked over for errors

 

[cragar]

Ok , Am I thinking about this in the right way.

 

[Office_Shredder]

Well, I can't read your mind but the part I quoted is the basis of the a correct argument for the proof. You just have to add the fact that these are sequences - A and B aren't always that small, but as long as n and m are big enough they are (by definition of the x's and the y's forming Cauchy sequences)