Do Cauchy Sequences Imply Convergent Differences?

Bptrhp
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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!
 
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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}##?
 
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Office_Shredder said:
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 said:
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 <br /> \begin{align*}<br /> |y_n| &amp;= |y_n - x_n + x_n| \\ &amp;\leq |y_n - x_n| + |x_n|. \end{align*}
 
pasmith said:
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 <br /> \begin{align*}<br /> |y_n| &amp;= |y_n - x_n + x_n| \\ &amp;\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.
 
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pasmith said:
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.
 
Office_Shredder said:
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##.
 
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