# Proof for Cauchy sequences

Bptrhp
Homework Statement:
Let ##(x_n)## and ##(y_n)## be Cauchy sequences in ##\mathbb{R}## such as ##x_n-y_n\rightarrow 0##. Prove that if exists ##K>0## such as ##|x_n|\leq K,\forall \,n\in\mathbb{N}##, then there exists ##n_0\in\mathbb{N}## such as ##|y_n|\leq K, \forall \,n>n_0## .
Relevant Equations:
##|x_n|\leq K,\forall \,n\in\mathbb{N}##
I've started by writing down the definitions, so we have

$$x_n-y_n\rightarrow 0\, \Rightarrow \, \forall w>0, \exists \, n_w\in\mathbb{N}:n>n_{w}\,\Rightarrow\,|x_n-y_n|<w$$
$$(x_n)\, \text{is Cauchy} \, \Rightarrow \,\forall w>0, \exists \, n_0\in\mathbb{N}:m,n>n_{0}\,\Rightarrow\,|x_m-x_n|<w$$
$$(y_n) \,\text{is Cauchy} \, \Rightarrow \,\forall w>0, \exists \, n_0\in\mathbb{N}:m,n>n_{0}\,\Rightarrow\,|y_m-y_n|<w$$

I tried using properties of the absolute value and the only vaguely useful result I got is ##|x_n-t_n|\leq C+|t_n|##. I can't see how to use this to prove the desired result.
Any hints? I appreciate any help!

Staff Emeritus
Gold Member
What about ##x_n=K## for all n, ##y_n = K+1/n##? It seems to prove the statement false. Is the inequality supposed to be strict?

Mentor
2022 Award
What if ##x_n=1=K## and ##y_n=1+\dfrac{1}{n}##?

Office_Shredder
Homework Helper
2022 Award
What about ##x_n=K## for all n, ##y_n = K+1/n##? It seems to prove the statement false. Is the inequality supposed to be strict?

What if ##x_n=1=K## and ##y_n=1+\dfrac{1}{n}##?

Your example is not a counter-example: The quantifier in the premise is existential, so if $K = 1$ doesn't work you should take a larger value of $K$. In this case $K \geq 2$ and $n_0 = 1$ works.

@Bptrhp: You have a bound on $|x_n|$ and you need to find a bound on $|y_n|$. So use the triangle inequality in the form \begin{align*} |y_n| &= |y_n - x_n + x_n| \\ &\leq |y_n - x_n| + |x_n|. \end{align*}

Staff Emeritus
Gold Member
Your example is not a counter-example: The quantifier in the premise is existential, so if $K = 1$ doesn't work you should take a larger value of $K$. In this case $K \geq 2$ and $n_0 = 1$ works.

@Bptrhp: You have a bound on $|x_n|$ and you need to find a bound on $|y_n|$. So use the triangle inequality in the form \begin{align*} |y_n| &= |y_n - x_n + x_n| \\ &\leq |y_n - x_n| + |x_n|. \end{align*}
You don't get to pick K. It says if there exists ##K## such that ##K\geq |x_n|## then stuff about it is true. We gave an example of such a K, so the stuff about it should be true.

fresh_42
Homework Helper
Gold Member
2022 Award
Your example is not a counter-example: The quantifier in the premise is existential, so if $K = 1$ doesn't work you should take a larger value of $K$. In this case $K \geq 2$ and $n_0 = 1$ works.
I don't see that at all.