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.

  • $\begingroup$ 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

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.

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

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