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To show it I started off from the following:

Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}^n$ so

$$ \forall \,\,\,\,\epsilon_1 >0\,\,\,\,\exists \,\,\,N_1 \in \mathbb{N}\,\,\,\ \text{such that} \,\,\,\, n,m \geq N_1 \,\,\,\, \rightarrow d(x_n,x_m) <\epsilon_1 $$ Also Let $(y_n)$ be a Cauchy sequence in $\mathbb{R}^n$ so $$ \forall \,\,\,\,\epsilon_2>0 \,\,\,\,\exists \,\,\,N_2 \in \mathbb{N}\,\,\,\ \text{such that} \,\,\,\, p,q \geq N_2 \,\,\,\, \rightarrow d(y_p,y_q) <\epsilon_2 $$

Let $N=\max(N_1,N_2),r=\max(n,p),s=\max(m,q)$ Then $$ \forall \,\,\,\,\epsilon_1>0 \,\,\,\text{and}\,\,\,\,\epsilon_2>0 \,\,\,\,\exists \,\,\,N \in \mathbb{N}\,\,\,\ \text{such that} \,\,\,\, r,s \geq N \,\,\,\, $$

$$ \rightarrow d(x_r,x_s) < \epsilon_1 \,\,\,\text{and} \,\,\,\,d(y_r,y_s)< \epsilon_2 $$

I do not know how to show $(x_n+y_n)$ be Cauchy?

If we sum them we have

$$ \rightarrow d(x_r,x_s) +d(y_r,y_s) < \epsilon_1 + \epsilon_2 $$

But I need

$$ d(x_r+y_r,x_s+y_s) < \epsilon_1 + \epsilon_2=\epsilon $$

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  • $\begingroup$ You're pretty close. Hint: What $\epsilon_1$ and $\epsilon_2$ can you choose for each Cauchy sequence so that $\epsilon_1 + \epsilon_2 = \epsilon$? Also, you can just state that r,s $\geq$ N $\endgroup$ – Joel Pereira Oct 17 '18 at 16:51
  • $\begingroup$ Minimum of both? $\endgroup$ – Saeed Oct 17 '18 at 16:53
  • $\begingroup$ Well you are given $\epsilon$ for the sequence $a_n+b_n$, so now you can choose different $\epsilon_1,\epsilon_2$ to add up to that value. Since each sequence is Cauchy, you will get a N$_1$, N$_2$ as bounds and you proceed as you did. $\endgroup$ – Joel Pereira Oct 17 '18 at 16:58
  • $\begingroup$ Can you explain how I can get to the last line, because this is my problem? $\endgroup$ – Saeed Oct 17 '18 at 17:12
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Given $\epsilon$, there exists N$_1$ such that r,s $\ge N_1$ such that $\mid x_r-x_s\mid<\frac{\epsilon}{2}.$ Similarly, there exists N$_2$ such that r,s $\ge N_2$ such that $\mid y_r-y_s\mid<\frac{\epsilon}{2}.$

Now let N = max{N$_1$,N$_2$}. So for r,s $\ge$ N, we have $$\mid (x_r+y_r)-(x_s+y_s)| = | (x_r-x_s)+(y_r-y_s )\mid <\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$

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  • $\begingroup$ we are not working in $\mathbb{R}$ we are in $\mathbb{R}^n$!!! $\endgroup$ – Saeed Oct 17 '18 at 18:07
  • $\begingroup$ Essentially, it's the same. Just understand the $|\cdot|$ as a norm in $\mathbb R^n$. $\endgroup$ – Gonzalo Benavides Oct 17 '18 at 18:18
  • $\begingroup$ @Saeed If you can show where my proof is wrong, please tell me. $\endgroup$ – Joel Pereira Oct 17 '18 at 19:52
  • $\begingroup$ $x_r$ and $y_r$ live in $\mathbb{R}^n$ and we cannot use absolute value as the metric. $\endgroup$ – Saeed Oct 17 '18 at 23:27

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