1
$\begingroup$

I want to show that $\Bbb{R^n}$ where $\Vert x \Vert_{R^n}=\left(\sum_{i=1}^{n}\left| x_i \right| ^2\right)^{1/2}$ is complete

Here is what I've done.

Let $\{x^{(s)}\}\subseteq (R^n,\Vert \cdot \Vert_{R^n})$ be a Cauchy sequence and $\epsilon>0.$ Then, $\exists\, N\in \Bbb{N}$ s.t. $\forall r\geq s\geq N,$ \begin{align}\Vert x^{(r)}-x^{(s)} \Vert_{R^n}=\left(\sum_{i=1}^{n}\left| x_{i}^{(r)}-x_{i}^{(s)} \right|^2 \right)<\epsilon^{2}.\end{align} Hence, we have that $x_{i}^{(r)}\to x_{i}^{*}\in \Bbb{R},\;\text{as}\;r\to\infty$, since $\Bbb{R}$ is complete.

Fix $n,r\in \Bbb{N}$, then allow $t\to\infty.$ We have \begin{align}\left(\sum_{i=1}^{n}\left| x_{i}^{(r)}-x_{i}^{*} \right|^2 \right)<\epsilon^{2},\;\;\forall \;r\geq N, n\in \Bbb{N}.\end{align} For $r=N,$ \begin{align}\left(\sum_{i=1}^{n}\left| x_{i}^{(N)}-x_{i}^{*} \right| ^2\right)<\epsilon^{2},\;\;\forall \; n\in \Bbb{N}.\end{align} Hence, $x^{N}-x^{*}\in (R^n,\Vert \cdot \Vert_{R^n})$ and since $(R^n,\Vert \cdot \Vert_{R^n})$ is a linear vector space, then $x^{*}=x^{N}-(x^{N}-x^{*})\in (R^n,\Vert \cdot \Vert_{R^n}),$ and we are done!

Kindly check if I'm correct. Corrections and alternative proofs are welcome.

$\endgroup$
  • $\begingroup$ The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing. $\endgroup$ – user587192 Dec 13 '18 at 15:25
  • $\begingroup$ @user587192: Ok, I will work on that! $\endgroup$ – Omojola Micheal Dec 13 '18 at 15:27
  • $\begingroup$ @user587192: I've done something on that! I hope it's okay now! $\endgroup$ – Omojola Micheal Dec 13 '18 at 15:31
1
$\begingroup$

Looks good to me.

By the way, since each $x^*_i\in\Bbb R$ we already have $x^*=(x^*_1,\dots,x^*_n)\in\Bbb R^n$ by definition. You don't need that fact that $\Bbb R^n$ is a linear space to conclude that.

$\endgroup$
  • $\begingroup$ Oh, thanks for that! $\endgroup$ – Omojola Micheal Dec 13 '18 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.