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