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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!

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Enclose formulas between dollar signs. –  DonAntonio Dec 28 '13 at 19:12

3 Answers 3

up vote 1 down vote accepted

It looks good, for the most part, but two of your inequalities not what they should be. In particular, we want $|c_n-c_m|<\epsilon,$ and a correct application of triangle inequality shows us that $$|c_n-c_m|=|c_n-c+c-c_m|\le|c_n-c|+|c-c_m|<2\left(\frac\epsilon2\right)=\epsilon,$$ as desired. A nonstrict inequality is also fine, but be sure it's going the right direction.

Added: Also, you should have $$\lim_{n\to\infty}|a_n-b_n|=|a-b|.$$

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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.

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1  
+1 Good point, although an arbitrary metric space won't have subtraction, so something more specific like "normed vector space" might be better. –  Erick Wong Dec 29 '13 at 0:28
    
This is part of the second chapter of an introductory analysis book, the one on Series and convergence. There aren't even complex numbers anywhere in the book ;) amazon.com/… –  Shug Dec 30 '13 at 0:53
    
@user902834 There's a whole chapter on the the topology of $\mathbb{R}$ and a section on metric spaces in the last chapter... There should be some note on what a complete metric space is. –  hallaplay835 Dec 30 '13 at 14:13

In the case that the metric space isn't complete you can use the following inequality:
given a metric $d$ then:
$|d(x,w) - d(y,z)| \leq d(x,y) + d(w,z)$ (it is easily proven with the triangle inequality).
and then:
$|d(x_n ,y_n) - d(x_m ,y_m)| \leq d(x_n ,x_m) + d(y_n,y_m)$
and finally you choose the appropriate $N$ such that $d(x_n ,x_m)< \frac \epsilon 2 \space \And\space d(y_n ,y_m)< \frac \epsilon 2 $

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