# In $\mathbb{R}^n$ show that every convergent sequence is a Cauchy

Show that every convergent sequence in $$\mathbb{R}^n$$ is a Cauchy sequence.

We want to show the following $$\forall\varepsilon>0,\exists J> 0,s.t.\forall i\in\mathbb{N}(i\ge J\rightarrow|a_i−L|<ε)$$

$$\rightarrow \forall\varepsilon>0,\exists N\in\mathbb{N},s.t.\forall j,k\in\mathbb{N},(j,k\ge N\rightarrow|a_j-a_k|<\varepsilon)$$

Something to notice is

$$|a_i−L|$$ and $$|a_j-a_k|$$ denotes the Euclidean norm in $$\mathbb{R}^n$$, rather than the absolute value, actually I think absolute value is a special case in $$\mathbb{R}^1$$ of Euclidean norm. similarly $$L\in \mathbb{R}^n$$.

Also we have

$$|a_i−L|<ε\Leftrightarrow a_i\in B(L;\varepsilon)$$

$$|a_j−a_k|<ε\Leftrightarrow a_j\in B(a_k;\varepsilon)$$

But, seems not very useful for the proof$$\dots$$

And i'm thinking that can I just take the proof of $$\mathbb{R}^1$$, and change something to make it works for $$\mathbb{R}^n$$

My attempts

Proof.

Let $$\{a_n\}_{n=0}^∞$$ be a sequence in $$\mathbb{R}^n$$

Assume $$\{a_n\}_{n=0}^∞$$ converges to $$L$$ where $$L\in \mathbb{R}^n$$

Show $$\{a_n\}_{n=0}^∞$$ must be Cauchy

Let $$j,k\in\mathbb{N}$$

By assumption, and since $$2\varepsilon>0$$ $$L-2\varepsilon

And

$$L-\varepsilon

Subtract second from first $$-\varepsilon

Implies

$$|a_j-a_k|<\epsilon\tag*{\square}$$

This is quick proof and almost the same as case in $$\mathbb{R}^1$$, is this vaild $$?$$

Any suggestion would be appreciated.

If not, how do I prove this hold in $$\mathbb{R}^n?$$

• What do you mean by the sequence of real numbers $\{a_n\}$ converges to $L\in \mathbb R^n$? Oct 11, 2019 at 1:25
• @Keshav Thanks to point out, It's a typo Oct 11, 2019 at 1:27

That's not a valid proof because those inequalities hold for real numbers but not for elements of $$\mathbb R^n$$.

You can do it as follows: given $$\varepsilon>0$$, take $$N\in\mathbb N$$ such that $$n\geqslant N\implies\lVert a_n-L\rVert<\frac\varepsilon2$$. Then$$m,m\geqslant N\implies\lVert a_m-a_n\rVert\leqslant\lVert a_m-L\rVert+\lVert L-a_n\rVert<\varepsilon.$$

• I see, basicly it's using triangle inequality for the norm in $\mathbb{R}^n$ Oct 11, 2019 at 1:34
• Yes. And this works on any metric space. Oct 11, 2019 at 1:40