Do Cauchy Sequences Imply Convergent Differences?

[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!

 

[Office_Shredder]

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?

 

[fresh_42]

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

 

[pasmith]

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 $$\begin{align*} |y_n| &= |y_n - x_n + x_n| \\ &\leq |y_n - x_n| + |x_n|. \end{align*}$$

 

[Office_Shredder]

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.

 

[PeroK]

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.

 

[PeroK]

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?

Even with strict inequality we have: $x_n = 1 - \frac 1 n$ and $y_n = 1 + \frac 1 n$.