# Show that $\Bbb{R^n}$ with the $\ell^2$-norm is complete

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.

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

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.