Assume $\{a_n\}\;$ and $\;\{b_n\}\;$ are Cauchy sequences. Use a triangle Inequality Argument to prove $\{c_n=|a_n-b_n|\}\;$ is Cauchy.
So in the answer key, the author proved it by taking $|c_n-c_m|$ and substituting in $(a_n-b_n)$ and $(a_m-b_m)$ respectively. I had tried this but had an algebra fail with distributing the negative sign and so it wouldn't come out for me. However, I came up with this proof and would like to know if it's logically coherent.
We must show that for any $\epsilon>0$, there exists a natural number $N$ such that when $m,n \geq N$ then $|c_n-c_m|\geq \epsilon$.
Let $\epsilon>0$. By the Algebraic Limit Theorem, $\lim |a_n-b_n|=a-b=c$, where $a=\lim a_n$ and $b=\lim b_n$.
Since $(c_n)$ is convergent to $c$, by definition for any $\epsilon>0$ there exists an $N$ such that when $n \geq N$, $|a_n-c|< \epsilon/2$.
Let $n,m > N$. Then $|c_n-c_m|=|c_n-c+c-c_m| \geq |c_n-c|+|c-c_m| = 2(\epsilon/2) = \epsilon$.
I realize that I kind of skirted around the main point of the question which was meant to use the fact that the sequences are Cauchy rather than the algebraic limit theorem and assume that they are convergent (even though they obviously are), but I got excited because I found my own proof and so I wanted someone to critique it for me. Thanks a lot!
Best Answer
As far as I can see, your proof is correct. However, maybe the author was trying to prove this for sequences in any metric space. If $\{a_n\}, \{b_n\} \subset \mathbb{R}, \mathbb{C}$, or any other complete metric space, then your proof will be valid since all Cauchy sequences in those metric spaces are convergent (by definition). However, if the metric space you are working with is not complete, then you cannot assume that some arbitrary sequences $\{a_n\}, \{b_n\}$ are convergent, because they may not be.