Cauchy Sequences Triangle Inequality.

[Enjoicube]

Homework Statement

assuming an and bn are cauchy, use a triangle inequality argument to show that cn=
| an-bn| is cauchy

Homework Equations

an is cauchy iff for all e>0, there is some natural N, m,n>=N-->|an-am|<e

The Attempt at a Solution

I am currently trying to work backwards on this one, since we know
for some m,n>=N where N is a natural
2e>|-an+am|+|bn-bm|
Thus, through the triangle inequality this is >=|am-an+bn-bm|=|(bn-an)-(bm-am)|>=|bn-an|-|bm-am|.

Am I going completely wrong somewhere, because I am getting very stuck here. To anyone who answers, please only hints, not full answers (I am sure this is customary). Again, many thanks.

 

[quantumdude]

Two quick comments:

I would have said that $|a_n-a_m|< \epsilon / 2$ and $|b_n-b_m|< \epsilon /2$ just to get $|(a_n-b_n)-(a_m-b_m)|<\epsilon$.

Here's a nitpicky thing that your professor might correct (mine would have). Once $\epsilon >0$ is given, there's no reason to think that the same $N$ will work for both sequences. So I would have said that there exists $N_1$ for one sequence and $N_2$ for the other, such that blah blah blah... Then I would have let $N=max\{N_1,N_2\}$.

 

[Enjoicube]

Yeah, my instructor does require this also. I can't believe the anguish this is causing me, It must be a very simple problem, and no one asked about it in class. Right now, I am supposed to be reading a history book, but this problem is back there bugging me.

 

[Enjoicube]

Not to imply that anyone read this incorrectly, but the problem is cn=abs(an-bn) not just an-bn. Just entered my head that this might not have been emphasized.