0
$\begingroup$

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 $2\varepsilon>0$ $$L-2\varepsilon<a_j<L+2\varepsilon$$

And

$$ L-\varepsilon<a_k<L+\varepsilon$$

Subtract second from first $$-\varepsilon<a_j-a_k<\varepsilon$$

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 $\mathbb{R}^n?$

Thanks for your help.

$\endgroup$
2
  • 2
    $\begingroup$ What do you mean by the sequence of real numbers $\{a_n\}$ converges to $L\in \mathbb R^n$? $\endgroup$
    – Keshav
    Oct 11, 2019 at 1:25
  • $\begingroup$ @Keshav Thanks to point out, It's a typo $\endgroup$
    – Ethan
    Oct 11, 2019 at 1:27

1 Answer 1

2
$\begingroup$

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$$m,m\geqslant N\implies\lVert a_m-a_n\rVert\leqslant\lVert a_m-L\rVert+\lVert L-a_n\rVert<\varepsilon.$$

$\endgroup$
2
  • $\begingroup$ I see, basicly it's using triangle inequality for the norm in $\mathbb{R}^n$ $\endgroup$
    – Ethan
    Oct 11, 2019 at 1:34
  • $\begingroup$ Yes. And this works on any metric space. $\endgroup$ Oct 11, 2019 at 1:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .