# Show that in $\mathbb{R}^n$, sum of two Cauchy sequences is Cauchy.

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

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

• 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 – Joel Pereira Oct 17 '18 at 16:51
• Minimum of both? – Saeed Oct 17 '18 at 16:53
• 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. – Joel Pereira Oct 17 '18 at 16:58
• Can you explain how I can get to the last line, because this is my problem? – Saeed Oct 17 '18 at 17:12

## 1 Answer

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 $$\mid (x_r+y_r)-(x_s+y_s)| = | (x_r-x_s)+(y_r-y_s )\mid <\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$

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