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)